# Can A Cold Object Warm A Hot Object?

Guest Post by Willis Eschenbach

Short answer? Of course not, that would violate the Second Law of Thermodynamics … BUT it can leave the hot object warmer than it would be if the cold object weren’t there. Let me explain why this is so.

Let me start by introducing the ideas of individual flows and net flows. Suppose I owe you twenty-five dollars. I run into you, but all I have is a hundred dollar bill. You say no problem, you have seventy-five in cash. I give you the hundred, you give me the seventy-five, and the debt is paid.

Now, there are two equally valid ways to describe that transaction. One way looks at both of the individual flows, and the other way just looks at the net flow. Here they are:

Figure 1. Net flows and individual flows. The individual flows are from me to you, \$100, and from you to me, \$75. The net flow is from me to you, \$25.

What does this have to do with cold and warm objects? It points out a very important distinction, that of the difference between individual flows of energy and the net flow of energy, and it relates to the definition of heat.

Looking at Figure 1, instead of exchanging dollars, think of it as two bodies exchanging energy by means of radiation. This is what happens in the world around us all the time. Every solid object gives off its own individual flow of thermal radiation, just as in the upper half of Figure 1. We constantly radiate energy that is then being absorbed by everything around us, and in turn, we constantly absorb energy that is being radiated by the individual objects around us.

“Heat”, on the other hand, is not those individual flows of energy. Heat is the net flow of energy, as represented in the bottom half of Figure 1. Specifically, a heat flux is the net flow of energy that occurs spontaneously as a result of temperature differences.

Now, the Second Law of Thermodynamics is only about net flows. It states that the net flow of thermal energy which we call “heat” goes from hot to cold each and every time without exception. However, the Second Law says nothing about the individual flows of energy, only the net flow. Heat can’t flow from cold to hot, but radiated energy absolutely can.

When an object emits radiation, that radiation goes on until it hits something that absorbs it, whereupon it is converted to thermal energy. The individual temperatures of the emitting and absorbing objects are not significant because these are individual energy flows, and not the net energy flow called “heat”. So there is no violation of the Second Law.

Here’s the thing that keeps it all in balance. If I can see you, you can see me, so there are no one-way energy flows.

Which means that if I am absorbing radiation from you, then you are absorbing radiation from me. If you are warmer than me, then the net flow of energy will always be from you to me. But that says nothing about the individual flows of energy. Those individual flows only have to do with the temperature of the object that is radiating.

So how do we calculate this net energy flow that we call “heat”? Simple. Gains minus losses. Energy is conserved, which means we can add and subtract flows of energy in exactly the same way that we can add and subtract flows of dollars. So to figure out the net flow of energy, it’s the same as in Figure 1. It’s the larger flow minus the smaller flow.

With all of that as prologue, let me return to the question that involves thermal radiation. Can a cold object leave a warm object warmer than it would be without the cold object?

While the answer is generally no, it can do so in the special case when the cold object is hiding an even colder object from view.

For example, if a person walks between you and a small campfire, they hide the fire from you. As soon as the fire is hidden, you can feel the immediate loss of the radiated energy. At that moment, you are no longer absorbing the radiated energy of the fire. Instead, you are absorbing the radiated energy of the person between you and the fire.

And the same thing can happen with a cold object. If there is a block of wood between you and a block of ice, if you remove the wood, you’ll get colder because you will be absorbing less radiation from the ice than you were from the wood. You no longer have the wood to shield you from the ice.

Why is all of this important? Let me offer up another graphic, which shows a simple global energy budget.

Figure 2. Greatly simplified global energy budget, patterned after the Kiehl/Trenberth budget. Unlike the Kiehl/Trenberth budget, this one is balanced, with the same amount of energy entering and leaving the surface and each of the atmospheric layers. Note that the arrows show ENERGY flows and not HEAT flows.

These ideas of individual flows, net flows, and being shielded from radiation are important because people keep repeating over and over that a cold atmosphere cannot warm the earth … and they are right. The temperature and the radiation are related to each other by the Stefan-Boltzmann equation. When we apply the S-B equation to the 321 W/m2 of downwelling “back radiation” shown in the graphic above, it tells us that the effective radiating level is somewhere around freezing, much colder than the surface.

BUT a cold atmosphere can leave the earth warmer than it would be without the atmosphere because it is hiding something even colder from view, the cosmic microwave background radiation that is only a paltry 3 W/m2 …

And as a result, with the cold atmosphere shielding us from the nearly infinite heat sink of outer space, the earth ends up much warmer than it would be without the cold atmosphere.

To summarize …

• Heat cannot flow from cold to hot, but radiated energy sure can.

• A cold atmosphere radiates about 300-plus W/m2 of downwelling radiation measured at the surface. This 300-plus W/m2 of radiated energy leaves the surface warmer than it would be if we were exposed to the 3 W/m2 of outer space.

My best regards to all,

w.

My Usual Request: When you comment, please QUOTE THE EXACT WORDS THAT YOU ARE DISCUSSING, so that we can all understand the nature of your objections.

My Second Request: Please keep it civil. Speculation about the other person’s motives and cranial horsepower are greatly discouraged.

Further Reading: My post entitled “The Steel Greenhouse” looks at how the poorly-named “greenhouse effect” work, based on the principles discussed above.

Math Notes: There’s an excellent online calculator for net energy flow between two radiating bodies here. It also has the general equation used by the calculator, viz:

with the following variables:

and Q-dot (left-hand side of the equation) being the net flow.

Now, when the first object is totally enclosed by the second object, then area A2 is set to a very large number (I used a million) and the view factor F12 is set to 1. This is the condition of the earth completely surrounded by the atmosphere. For the general case, I’ve set area A1 to 1 square metre. Finally, I’ve made the usual simplifying assumption that thermal IR emissivity is 1.0 for the surface and the atmosphere. The emissivity values are greater than 0.9 in both cases, so the error is small. With those usual assumptions, the equation above simplifies as follows, courtesy of Mathematica:

But sigma T ^ 4 is simply the Stefan-Boltzmann radiation for the given temperature.  That is why, in the energy budget above, we can simply add and subtract the energy flows to produce the budget and check to see if it is balanced.

## 1,939 thoughts on “Can A Cold Object Warm A Hot Object?”

1. Joel O'Bryan says:

“If I can see you, you can see me…”

I think that assumes we are both “looking” at the same EM wavelength (or overlapping EM bands).

• Joel O'Bryan says:

In talking science and making definitive statements, it is always is good practice to acknowledge what assumptions are being used.

• AndyG55 says:

If the front plate is at equilibrium with its input and surroundings, the first effect of adding a cooler plate (green) very close behind the first (blue) will be to cool the blue plate while warming the green plate. Both plates will aim to reach the same temperature. Then they will both warm back up to the original equilibrium temperature, assuming the input stays the same.

At no point with the temperature of the original plate get higher than its original temperature.

REALITY , not some bungled explanation from a rabbit. !

• Tim Folkerts says:

“the first effect of adding a cooler plate (green) very close behind the first (blue) will be to cool the blue plate “

You seem to be confused about radiation vs conduction. If the two plates were physically placed into contact (basically making a single, thicker plate), then your description is pretty accurate. But that is not the scenario under consideration. Willis’ post clearly and thoroughly explains what happens with radiation.

• AndyG55 says:

You seem to be confused about energy transfer.

Where did I say they were touching??

If its 2 miles away it will have no affect whatsoever.

Energy transfer rate is related to the temperature difference and distance between objects.

• Tim Folkerts says:

Andy says: “If the front plate is at equilibrium … “
First, to be precise we need to say the front plate is in a “steady-state” condition, where Qin (the total heat in) equals Qout (the total heat out). In this case, the temperature of the warmer (blue) plate will be constant. (For a true ‘equilibrium’ situation, every part would be the same temperature and there would be no heat flows).

“adding a cooler plate (green) very close behind the first (blue) will be to cool the blue plate while warming the green plate. ”

Why? How does the energy loss from the warm plate INCREASE when the cool (green) plate is added? Certainly not by conduction, since you said they are not touching. Certainly not by radiation either. Radiation loss from the blue plate DECREASES since the blue plate is radiating to a warmer region than it had been (the cool green plate vs the much colder background).

• Tim Folkerts says:

Tony, It continues to be mildly bemusing that you treat a discussion at a third-rate blog that disagrees with every thermodynamics text ever written to be the only authoritative source.

• Tony says:

It’s a thought experiment from one blog discussed on another. Don’t get carried away. Other perspectives are not forbidden here, thankfully.

• Tony says:

And people at this blog usually see through such tactics as poisoning the well, so you might want to give that sort of thing a miss.

• Suppose you have two bodies (A and B), and the warmer one (B) has a parabolic mirror around it directing incoming radiation from A onto B. In this case, is it possible for cooler A to warm B (B not counting the mass of the mirror)? (As with many things, I have no clue, but it seemed like it might be an interesting question.)

• Bob Shapiro November 28, 2017 at 11:46 am

Suppose you have two bodies (A and B), and the warmer one (B) has a parabolic mirror around it directing incoming radiation from A onto B. In this case, is it possible for cooler A to warm B (B not counting the mass of the mirror)? (As with many things, I have no clue, but it seemed like it might be an interesting question.)

Sorry, but if the mirror focuses the rays of A onto B, it will focus the rays of B onto A, so no change in temperature.

TANSTAAFL …

w.

• Leo Morgan says:

Your comment is both true and surprising.
There’s very few observations that satisfy both of those criteria.

I sat down to type an explanation as to why you were wrong; reconsidered, then re-reconsidered.

The word ‘visible’ is mildly ambiguous in this context. We sometimes think of a brighter object as being more visible than a darker object. One side might well be outputting more photons at a particular wavelength than the other, ie be brighter, and thus in a sense be more ‘visible’, while at the same time at a different wavelength the converse is true. However, that’s not the sense of visibility under discussion. The alternate nuance of the word, the one under discussion, is what my old instructors referred to as inter-visibility, or ‘line of sight’.

I have heard it alleged that the earth puts out more energy at radio wavelengths than the sun does.For the sake of discussion, let’s assume that claim is true.That would be an example in which two objects are brighter than each other at different wavelengths, and therefore in one sense ‘more visible’ than the other at that wavelength. However, at each wavelength, if one object can be ‘seen’ by another at that wavelength, then the converse is true, and it can be seen from the first at that wavelength.

Whatever the wavelength, whatever the obstructions, the same line of sight applies to both sides. (A statement which I think is a fair paraphrase of Willis’s point.)

As an aside, I’m grateful to Willis for addressing this issue. Some of our fellow sceptics do misunderstand energy flows. Their ignorance is an embarrassment to all sceptics. Even worse, it’s often ascribed to all of us as a group.

• menicholas says:

If I understand correctly, what you are saying is 9in effect)…if I am made of air but have eyes that can somehow see visible light nonetheless…I can see you, but you cannot see me.
Is that it?

• Jaakko Kateenkorva says:

As an aside, I’m grateful to Willis for addressing this issue. Some of our fellow sceptics do misunderstand energy flows. Their ignorance is an embarrassment to all sceptics. Even worse, it’s often ascribed to all of us as a group.

Relax. Catastrophic Anthropogenic Global Warming err Climate Change err Climate Disruption err Climate-Related Shock err Carbon err “Cannot settle the name” Apocalypse a.k.a CACA is 97% political.

As long as freedom of conscience and thought remain in the universal declaration of human rights, there is no need to be embarrassed about non-compliant thoughts of individuals. The concept of “consensus science” is an oxymoron anyway.

If we focus on the remaining 3%, any concept starting with Average Global exits internationally established metrology standards de facto and, thus, also the modern scientific methods. Irrespective of subsequent clarifications, such as, “energy flows”, “energy budget”, “air temperature”, “air composition”, “sea level”, “skeptic psychoanalysis”, “social cost of carbon” etc. They may be helpful for general understanding similarly to philosophy, as long as they are acknowledged as such.

• Willis, what the article ignores is that there must also be a GHG effect due to conduction if there is one due to radiation. And the GHG effect due to conduction is much larger than then one due to radiation, because conduction affects all gasses, not just GHG.

The key to understanding is to realize that conduction involves the transfer of virtual photons, which in all respect except 1 are identical to real photons. Virtual photons cannot act at a distance.

Thus a molecule of N2 accepts a virtual photon from the surface which can then be conducted upwards or back-conducted to the surface warming the surface. The one difference between N2 and CO2 is that CO2 can radaite to space while N2 cannot. Thus the CO2 GHG due to radiation provides a net cooling as compared to N2 GHG effect due to conduction.

2. co2islife says:

“Can A Cold Object Warm A Hot Object?”

The answer is yes. Electromagnetic radiation flys through the 0°K outer space, once that radiation hits a molecule that absorbs that wavelength, it is thermalized. A match in a freezer is very cold, but striking it on sandpaper will ignite it to a very hot temperature. When energy is changed in form, yes, cold objects can warm a hot object. The GHG theory takes EM IR radiation and thermalizes it, changing a cold EM wave, into a hot vibrating molecule. That is why the CO2 IR signature is identified way up in the atmosphere where it is about -80°C.

• David Ramsay Steele says:

Of course this is correct, but watch out for one problem in communication. Some parts of the world know only “safety matches”, so they have no experience of a match which will ignite by striking on sandpaper. I was once talking to a Swedish person and recounting a trick we used to play at school in England. We would dig a hole in the end of a stick of chalk, put a broken off head of a match in that hole and cover it up with chalk dust, then wait for the teacher to write on the blackboard and have the end of the chalk explode into flames. The Swede couldn’t understand how this could work, then I realized the Swede had never encountered a non-safety match and I explained that at that time in England non-safety matches were the norm.

• HotScot says:

David Ramsay Steele

As were blackboards, now a non PC term. But whiteboard is OK, it’s non discriminatory, apparently.

• co2islife says:

LOL, great story

• lemiere jacques says:

well you can’t say that….. thermalization implies a lot of molecules…it is a collective concept… but well you re right..
can a cold object warm a hotter objet ..by heat flux..no.
the heat goes from cold to hot..

• co2islife says:

Yes, that is true, but my comment was about the GHG effect. People often use the cold can’t warm a hot object, which is true for heat flux, but not the conversion of energy from one form to another. That is the relevant issue when discussing the GHG effect. The GHG effect is converting cold EM radiation to warm kinetic energy.

• Crispin in Waterloo says:

co2islife

Agreed, it can ‘warm’ a ‘hot object’. The problem with Willis’ piece is that it has many problems with the definitions. “Flow” has a normal meaning different from the way the term is used above. For someone coming to the issue afresh, a set of definitions is needed and they have to conform to certain norms. The reason is that as soon as one takes an argument outside the text, the terms have to match. Radiated energy does not “flow”. That’s the point of giving it another term with its own definition. One can speak of radiative balance and net gain, but not flow. There is no flow going on.

There are multiple issues with the piece: EM travelling through space ‘hitting’ or ‘encountering’ an object does not mean it will be ‘absorbed’. It might be reflected or refracted. It is not always true that adding energy to a molecule raise its temperature, it may in some cases and not in others. In some cases the net effect is it lowers the temperature by inducing emission of a photon of greater energy than the one striking it. That is how extraordinarily low temperatures are achieved – by hitting the atoms with additional energy.

And so on and on. The presentation is generally true: that the surface cooling can be reduced by GHG’s through the mechanism of preventing radiation from the surface proceeding directly to space. This will prevent some of the cooling of the surface. It does not guarantee (at all) that it will prevent cooling of the atmosphere. The atmosphere (heated by the surface or by radiation) cools by radiation from GHG’s and without them, it could not.

There is in all these arguments about back-radiation, confusion abounding. The root problem is the BIG FAT ERROR which is the assumption that the appropriate comparison is a planetary body with no atmosphere at all receiving insolation from a star, and a planet with an atmosphere with GHG’s in it. That is a completely inappropriate comparison when one is speaking of the influence of the relative concentration of GHG’s in an atmosphere. It should be obvious that the appropriate comparison is between an atmosphere without GHG’s and one with them.

If one were to suggest that an atmosphere with no GHG’s at all was going to be cold, I would reply that it a stunning and obvious error. The surface would be warmed freely by incoming radiation and the surface would warm the atmosphere, which could not cool by radiation, only convection of heat to the surface when the surface was colder than the atmosphere (at night). The atmosphere would just get hotter and hotter until it warmed the radiating night side surface enough to dispose of a quantum of heat sufficient to balance that received by the atmosphere on the sun-side. A no-GHG atmosphere would be hot as Hades because it cannot cool by radiation. Adding GHG’s to it does shield the surface from directly cooling to space, but it also dramatically cools the atmosphere which can now radiate freely in all directions. This would in turn greatly reduce the surface heating at night from the hot atmosphere. Remember that the temperature of our atmosphere at a high elevation is very high because those widely dispersed molecules cannot cool by radiation or convection.

In short, the cartoon with the GHG’s and back radiation is fundamentally flawed because it is based on the assumption that without GHG’s the Earth and its atmosphere would be much colder. How many times have you seen the comparison and ’33 degrees C of warming’ by GHG’s? That is stuff and nonsense. It is always a comparison of a body without any atmosphere and one with both an atmosphere+GHG’s in it. Never is it the appropriate comparison in which one can evaluate the effect of GHG’s themselves.

Don’t confuse the sun-side surface temperature with the average surface temperature, and the sun-side atmosphere’s temperature with its average, nor with the surface temperature on the day or night side.

A planet with no atmosphere at all (like the moon) would have a surface temperature on average 33C colder than our surface.
A planet like Earth with an atmosphere with no GHG’s at all would have an atmosphere that is much warmer than ours is presently.
The claim for 33C of warming by GHG’s is false.
A planet like Earth with an atmosphere and some GHG’s would have an atmosphere that is as warm as it is now.
The temperature of a no-GHG atmosphere will be strongly influenced by the period of rotation because that affects the temperature of the surface that heats it.

• Miro says:

thank you. I thought I was the only one :)

• Aphan says:

I’m so glad it’s not just me! The terms make no sense….”a cold object hiding an even colder object from view?” Flow? Net flow of heat or net flow of energy? Can low energy electromagnetic waves being emitted by a cold object cause an increase in temperature of an object that can only be heated by high energy electromagnetic waves?

The temperature of an object is determined by how much energy it absorbs vs how much it emits. Absorbs the same amount it emits? Temp is stable. Absorbs more than it emits? Temperature increases. Absorbs less than it emits-temperature drops. It doesn’t matter if the absorbed energy comes from a colder object, a hotter one, or a hundred objects of differing temperatures…it’s still about the ratio of energy absorbed to energy emitted.

Our atmosphere might be a “cold object” between Earth and the “even colder object” of space, but that colder object of space ALSO holds the “hotter object of the Sun”. The moon has almost no atmosphere and on the daylight side it’s temperature reaches 127 C and on the dark side it’s temp plunges to -173 C. Our “cold” atmosphere buffers us from both the intense cold of space AND the intense heat of the Sun. Using Willis’s terms, it “hides something colder from view AND something hotter from view”. But all of the heat warming the surface AND the atmosphere comes from the Sun. If it wasn’t there, not only would the Earth get colder than it currently does, but it would also get warmer than it currently does.

The 2nd law of thermodynamics is about the SPONTANEOUS flow of heat between objects always from a warmer to a colder. You CAN change the flow of heat but only with an additional/outside source providing additional work…but that’s not what the 2nd LOT refers to.

Uggggg…

• DMacKenzie says:

Crispin, generally look forward to your comments from Waterloo, Ulan Batar, etc., but in this one, you are in error when you say a non GHG atmosphere cannot cool by radiation. A pure diatomic oxygen and nitrogen atmosphere is certainly not a GHG, but would be completely transparent to infrared radiation. Therefore the planetary surface directly “sees” outer space at 3 degrees Kelvin and radiates energy to it quite nicely.

• Crispin in Waterloo says:

DMacKenzie

Because there is no chance of us finding a true non-GHG atmosphere (in my view anyway) please treat the discussion as being about an ideal atmosphere with and without GHG’s. I agree that O2 has some ability to radiate, but it is inconsequential (kills some lovely logical tricks!) but the importance of the lesson is, I hope, evident.

The discussion everywhere I look is about the moon (no atmosphere) and the Earth with an atmosphere and GHG’s, with the difference in temperature attributed to the GHG’s alone. That is a fundamental error far worse than comparing apples and oranges.

Is it obvious that the entire edifice of CAGW is based on a comparison so logically baseless? If no one can point out another source for my analysis, I think I will claim priority. I am going to bring up the idea of modeling it next week at the Chinese Academy of Sciences and see if they will devote some time to a simple calculation of the equilibrium temperature of an Earth-like planet with no GHG’s, say, an argon atmosphere of 1 bar and a sandy desert surface with an emissivity of 0.93. They have a monster of a computer just down the road.

• DMacKenzie says:

Crispin, I think you don’t need CAS. Manabe and Strickler covered the basics in their classic 1964 paper referenced by Spencer (search drroyspencer Why 33 deg), also Lindzen (search Some Coolness Concerning Global Warming). Your stated result is correct, but I find Spencer and Lindzen explain it in terms more relatable to the average audience.

• DMacKenzie says:

Sorry Crispin, that read quite a little ruder than I intended…

• Crispin in Waterloo says:

DMacKenzie

I checked the paper by Spencer and he does not get the point at all. He repeats the error (as I understand it). The error is he says a planet with no atmosphere (case 1) and a GHG atmosphere (Case 3) are 33 degrees different but does not consider the atmospheric temperature of Case 2: a planet with a full atmosphere without any GHG’s. If he wants to calculate the effect of adding CO2, then he has to compare it with having none (Case 2), not having “no atmosphere at all” (Case 1). Case 1 v.s. Case 3 is a silly comparison.

I tried to come up with a suitable analogy today but it is difficult because the claim and the example (repeated by Spencer) is so inappropriate.

Considered this:

I want to determine the effect of the mass of yeast added to a loaf of bread baking in a bread pan, and its influence on the surface temperature of the pan measured at the centre of one side.

I propose to alter the amount of yeast added to the dough, which will alter the porosity of the bread, which will affect the heat conducting rate through the pan into the dough, which will affect the pans surface temperature. I want to see the effect of doubling the mass of yeast on the pan temperature, (which will be influenced by the porosity of the bread).

According to the CAGW method, to get a baseline I measure the temperature of an empty pan and compare it with a pan containing 1 kg of dough and 1 g of yeast, and then claim that the temperature difference is caused by the yeast! This is a reasonable analogy of their mental experiment underlying the claim for the GH effect.

They are not conducting an experiment I described at all which was to assess the effect of increasing the mass of yeast (CO2). They are testing two completely different things and claiming that the net effect is caused by one tiny portion of the dough.

People have swallowed this misconception hook, line and sinker. I could show you a thousand copies of the error. The greenhouse cartoon embeds their conceptual error. The correct comparison is an atmosphere with a pressure of 1 bar, and no GHG’s and one with GHG’s. First, get the equilibrated temperature without, then add CO2 and recalculate.

The assumption inherent in the error is that without GHG’s the atmosphere would be as cold as the average surface of a planet with no atmosphere at all. Nonsense! Ever heard of heat gain by radiation and heat disposal by conduction to a gas? There is an example in every house in Canada with a forced air furnace. All the heat from burning fuel is passed through a hot metal plate to a gas on the other side. A surface of a planet with an atmosphere without GHG’s would be heated even better than with them. That how surface would heat the atmosphere, which has no way to dump the heat save back to the surface when the planet turns away from the sun. Adding CO2 might change the temperature, up or down, we don’t know, but my silly calculator says it will be hotter just above the ground to have no GHG’s at all.

1360 W/m^2 daytime at the equator. Emissivity of sand, 0.93. What is the temperature reached when it is in radiative balance? It would heat up to about 170 C. At present sand does not get much above 70 C in a desert with 50 C air. The temperature of the atmosphere next to the ground would be in the region of 140 C. I have no idea what the nighttime air temperature minimum would be, but it would be a heck of a lot higher than 15 C because gases do not cool effectively downwards to a surface by conduction (REF: Bejan, A, 2005 “Convection heat Transfer”).

Because that atmosphere could not cool by radiation (by definition) it would simply stay hot and get even hotter the next day, if it managed to cool a bit at night, until it was eventually in equilibrium. Adding CO2 to such an atmosphere turns the atmosphere itself into a radiating body. It introduces back radiation but it introduces a shade at the same frequency. The net effect of turning the atmosphere into a gigantic radiator would be to cool it, which would draw down the surface temperature. In radiative balance, the surface would no longer be the only emitter so it would be cooler, by more than 100 C.

So where is this putative 33 degrees of heating? There is no 33 degrees for GHG’s. Without an GHG’s it would be a heck of a lot hotter because …..physics! Ask Bejan. The bottom line is that water provides evaporative cooling and is a powerful GHG. CO2 has to be a very minor player because there is so little of it (and all the usual asides). 20 ppm CO2 does not provide “6 degrees of warming” as the chart often shows.

Summary:
Case 1 No atmosphere
Case 2 Atmosphere
Case 3 Atmosphere with GHG’s

Which has the warmest average surface temp; which has the highest atmospheric temperature 1.5 m above the ground?

Case 2.

• Crispin in Waterloo:

Summary:
Case 1 No atmosphere
Case 2 Atmosphere
Case 3 Atmosphere with GHG’s

Which has the warmest average surface temp; which has the highest atmospheric temperature 1.5 m above the ground?

Case 2.

I’m sorry, but that is demonstrably and provably not true, as I showed in A Matter Of Some Gravity. Here’s the short version.

You say that a planet with an atmosphere which neither radiates nor emits thermal IR will be warmer than the S-B temperature with no atmosphere. IF that were the case, then the surface, which is the only part of the system capable of radiating heat, must be radiating more energy than it is receiving. Of course this violates the laws of thermodynamics …

Read the post for the long version.

w.

• wildeco2014 says:

The planet viewed from space will be at S-B with a non radiative atmosphere but the surface beneath the atmosphere will be above S-B in order to supply the necessary kinetic energy to fuel the inevitable convective overturning.

• Crispin in Waterloo says:

co2islife

I am posting this separately – sorry for duplication of my long single comment appears later.

Agreed, it can ‘warm’ a ‘hot object’. The problem with Willis’ piece is that it has many problems with the definitions. “Flow” has a normal meaning different from the way the term is used above. For someone coming to the issue afresh, a set of definitions is needed and they have to conform to certain norms. The reason is that as soon as one takes an argument outside the text, the terms have to match. Radiated energy does not “flow”. That’s the point of giving it another term with its own definition. One can speak of radiative balance and net gain, but not flow. There is no flow going on.

There are multiple issues with the piece: EM travelling through space ‘hitting’ or ‘encountering’ an object does not mean it will be ‘absorbed’. It might be reflected or refracted. It is not always true that adding energy to a molecule raise its temperature, it may in some cases and not in others. In some cases the net effect is it lowers the temperature by inducing emission of a photon of greater energy than the one striking it. That is how extraordinarily low temperatures are achieved – by hitting the atoms with additional energy.

And so on and on. The presentation is generally true: that the surface cooling can be reduced by GHG’s through the mechanism of preventing radiation from the surface proceeding directly to space. This will prevent some of the cooling of the surface. It does not guarantee (at all) that it will prevent cooling of the atmosphere. The atmosphere (heated by the surface or by radiation) cools by radiation from GHG’s and without them, it could not.

• I think the temperature of “space” at any point is reasonably considered the temperature corresponding to the the total radiant energy flux impinging on it , or equivalently , dividing by a lightsecond , the energy density at the point . That gives the temperature of a gray , ie : flat spectrum , object at that point , eg : the ~ 278.6 +- 2.3 from peri- to ap- helion on our orbit .

That the “cold can’t heat warm” issue is still in need of discussion given the simple classic energy flow law shows its stagnation . One of the first steps forward would be to recognize that those emissivities are not really scalars ; they are averages over spectra . Simply getting agreement on the experimentally testable computations at http://cosy.com/Science/warm.htm#EqTempEq seems nigh on impossible — and it is only with that full spectral computation that the radiative equilibrium of a radiantly heated object , eg : a planet , can be calculated to the 4 decimal place variation this entire fiasco is about .

Then it is found , unless someone comes up with some new physical laws , the is no way a spectral phenomenon can “trap” heat in excess of the equilibrium .

• Now, I’ve been looking for an answer to this question for a long time and no-one has been able to provide one, which leads me to believe it is false: How is it possible that 3 of 4 CO2 molecules out of 10000 can warm the rest by 1 degree (for simplicity’s sake) without each of those CO2 molecules at least absorbing and transferring 2500 deg C or more of heat in the process?? If there is anyone that claims this is indeed happening, then where can we test that by measuring it? We have all the molecules available after all.

• Crispin in Waterloo says:

co2islife

You really have to include water vapour. Activating 20,400 ppm of GHG’s (water and CO2) can easily move energy around. Reflecting clouds can match that, by the way.

• Gary Pearse. says:

CO2 islife:’insignificant’ 0.04% is a very poor way to couch an argument about this molecule. The reason I say this should be evident in your moniker! It is, as far as we, and the entire biosphere, atmosphere (O2) and hydrosphere (O2 and CO3) is concerned, the heaviest lifter in the atmosphere.!

3. Tom Halla says:

Nice succinct argument.

4. co2islife says:

If I take Cold Sodium and mix it with cold water, I get a very hot reaction.

• Joel O'Bryan says:

That was a hydrogen explosion you heard.

2Na(s) + 2H2O(l) –> 2NaOH (l) + H2(g)

• Joel O'Bryan says:

The “smoke” is steam, it is the water vapor that formed when the H2 combusted with atmospheric O2 to make hot H2O.

• Urederra says:

nope

• co2islife says:

Yep, my point what that energy being changed in form can in fact take a cold object and warm it. The GHG effect isn’t about heat flux, it is about converting cold EM radiation to warm kinetic energy, and that does in fact happen.

• That is just a chemical reaction liberating stored chemical energy. Nothing to do with this topic.

• Catcracking says:

of course, agree.

• co2islife says:

It has everything to do with this topic. The comments of cold warming hot have to be put in the context of the GHG effect, where in fact cold EM radiation is thermalized into kinetic energy and in fact does warm the atmosphere. Willis is confusing heat flux with energy being changed in form. This is a common false argument made by people trying to argue against the AGW theory. They simply don’t fully understand the concept.

• Crispin in Waterloo says:

J Richard

Just to quibble, it is not stored energy – not at all. It is just sodium. It is not like a canister of compressed air. When sodium makes contact with water, it makes new chemistry, releasing energy, but that energy was not stored in either the sodium nor the water.

OTOH, perhaps we can view all matter as stored energy in that matter exists and could be released upon annihilation, right? I wonder if we could calculate the total energy in the universe including all the sensible matter, all the shell of Dark Matter that surrounds it, and all the much larger shell of the Even Darker Matter that surrounds that.

If energy can be transferred between these types of matter, then it is on-topic for a discussion on GHG radiative heat transfer/obstruction.

• Crispin,

It is stored energy, because making pure sodium needs a lot of energy to separate it from any other ion (OH-, CL-,…). Part of that energy is given back when Na reacts with water… Comparable with loading a battery, where you store electrical energy into chemical energy and back…

• Gary Pearse. says:

CO2, you are using the cold object’s chemical energy to accomplish this. If you destroyed the sodium completely, the E=mc^2 would warm it up even more. If a cold asteroid enters the cold atmosphere, it, too, will warm both up- friction and kinetic energy when it hits the ground. If you turn on a refrigerator, heat from the icebox will make a warm room warmer, but you have to do work to accomplish this. Hammer cold steel with a cold hammer….. I think Willis was referring to a more passive relationship between objects!

• co2islife says:

Yes he is, but all these comments have to be put in the context of the GHG effect. The GHG effect does in fact take cold EM radiation and convert it to warm kinetic energy. The GHG is about energy changing in form, not heat flux, and that doesn’t apply to the GHG effect.

5. I get so tired of hearing this stupid argument from the anti-ghg crowd. Thanks for quantifying it.

• Yeah, I’ve been going through this on another thread, and I thought it deserved its own clear explanation … I suspect that there will always be people saying it is impossible.

Best to you and yours, my friend, and thanks for the endless work you’ve done to keep this site fresh and vital,

w.

• Take an indoor pool. Heat it to 30C. Fill the room with 100% CO2. Will the pool’s water temp go above 30C? Nope. Impossible. CO2’s IR wavelength is very cold.

• J. Richard Wakefield November 24, 2017 at 7:59 pm

Take an indoor pool. Heat it to 30C. Fill the room with 100% CO2. Will the pool’s water temp go above 30C? Nope. Impossible. CO2’s IR wavelength is very cold.

Huh? Nobody said the pool’s water temp would go up in that situation, that’s a straw man.

w.

• No, it’s not. The IR from the pool would get absorbed by the CO2 and be re-emitted back to the pool, and raise it’s temp. That is the EXACT argument AGW people claim happens. It’s in that flat earth graphic you posted (which is completely false of how the planet’s energy system works.)

• J. Richard Wakefield November 24, 2017 at 8:09 pm Edit

No, it’s not. The IR from the pool would get absorbed by the CO2 and be re-emitted back to the pool, and raise it’s temp. That is the EXACT argument AGW people claim happens. It’s in that flat earth graphic you posted (which is completely false of how the planet’s energy system works.)

Please re-read the section about how a cold object can leave a warm object warmer ONLY if the cold object is hiding something that is even colder.

And this is the case for the planet, where the atmosphere is hiding the 3 W/m2 heat sink of outer space.

But in your thought experiment, this is NOT the case. Without the CO2 you get radiation from the roof of the room. With CO2 you get radiation from the CO2, which is at the same temperature as the roof … so there is no change in downwelling radiation when you introduce the CO2.

Regards,

w.

• reallyskeptical says:

“The IR from the pool would get absorbed by the CO2 and be re-emitted back to the pool”

No. To the pool, yes, but to the ceiling and the walls as well. The pool would still cool, but at a rate slower than without CO2.

Now if there was a constant source of energy, say an electric heating element or the sun, the temp of the pool would be higher with the CO2 than without.

• “And this is the case for the planet, where the atmosphere is hiding the 3 W/m2 heat sink of outer space.”

I have no idea how the atmosphere can hide anything. What does “hide” mean?

“But in your thought experiment, this is NOT the case. Without the CO2 you get radiation from the roof of the room. With CO2 you get radiation from the CO2, which is at the same temperature as the roof … so there is no change in downwelling radiation when you introduce the CO2.”

That depends on what the roof is made of. Make the roof an absorber of IR, then none of that from the roof would go back to the pool. Or is that your “hiding”?

• “No. To the pool, yes, but to the ceiling and the walls as well. The pool would still cool, but at a rate slower than without CO2.

Now if there was a constant source of energy, say an electric heating element or the sun, the temp of the pool would be higher with the CO2 than without.”

You can prove that? That would happen regardless of the gas in the room. But you are correct, insulation slows the rate of heat loss. That is NOT the same thing as making the heat source hotter.

This is one of the major flaws in that flat earth graphic. The planet is not getting a constant flow of energy from the sun on any given surface area. The planet is rotating. That acts like a thermostat. If the planet rotated more slowly, the surface would absorb more sun’s energy, and the temp would be higher, Rotate slow enough and the planet would be too hot for life during the day.

Rotate faster and the surface wouldnt be able to absorb enough energy to keep the nice tropic temps we enjoy.

This is why that flat earth graphic is a completely wrong depiction of the energy flow.

• reallyskeptical says:

“The IR from the pool would get absorbed by the CO2”

Those are your words not mine. If the energy gets held up by CO2 and then readmitted with some of the energy going back to where it came from, of course it will preserve the pools temp.

and no one is saying the temp of the pool would heat higher. That would be stupid.

And it the sun’s energy was daily not constant, that doesn’t change the argument.

• menicholas says:

“Rotate faster and the surface wouldnt be able to absorb enough energy to keep the nice tropic temps we enjoy.”
Can you prove that?
Days are shorter, but so are the nights.
The temp would be more even, most likely. But the temps are very even in the tropics anyway.

• AndyHce says:

Back radiation from the colder object is absorbed by the hotter object.
The hotter object is heated by that absorbed radiation.
Being heated, it now radiates more per unit time than it was radiating before because rate of radiation is related to temperature.
Thus this increased temperature increases its rate of cooling.
The question is, what is the net flow:
Does the increased absorbed energy result in an output increase that is equal to, less than, or greater than the back radiation absorbed?

• Paul Aubrin says:

The emissivity of the atmosphere is not 1. Actually the atmosphere is not a grey body (=surface). by the way, it has no proper surface.

• Tony says:

“Back radiation from the colder object is absorbed by the hotter object.
The hotter object is heated by that absorbed radiation“

“Can A Cold Object Warm A Hot Object?
Willis Eschenbach / 9 hours ago November 24, 2017
Guest Post by Willis Eschenbach
Short answer? Of course not, that would violate the Second Law of Thermodynamics”

• Carbon BIgfoot says:

If I recall neither Willis or Anthony have a BS ( or studied Thermodynamics ) and as a result will remain in their fantasy world with these unicorn examples of science.

• Ed Bo says:

Bigfoot:

I have studied thermodynamics (at MIT) and have both bachelors and masters degrees. I assure you they are absolutely correct on this subject.

A lot of universities use thermodynamics as a “washout” course to quickly eliminate the people who just are not capable of this type of rigorous analysis. At first I thought it was cruel. Now when I look at many of the comments here, I see why that is so necessary to keep those people away from real-world systems where they could cause serious harm.

• The issue is how the interior of a sphere can be made hotter than the radiative equilibrium temperature of the sphere by some spectral phenomenon in apparent violation of the Divergence Theorem which is another way of expressing that heat comes to equilibrium — essentially Fourier’s differential eq ( here’s a YT on it I recently watched : https://www.youtube.com/watch?v=NHucpzbD600 ) .

My brain requires simplicity . Show me the quantitative equations on a sphere . Then I can write the code and explore the parameter space .

No planets , no clouds , atmospheres . Just spherical shells of ( ae spectra ; transparency ) and power source and sink spectra . Actually it can be simplified to the 1 dimensional case something like this :

Show for what “Atmospheric Filter” and Surface ae spectrum ( b + c ) is greater than the equilibrium lumped spectrum .

So far as I know , the Schwarzschild differential ( http://www.barrettbellamyclimate.com/page47.htm ) is the definitive differential for radiant – mass heat transfer . So show under what subspace of parameter values does it “trap” a higher energy density on the side away from a source .

• Ian H says:

Thanks Willis for your very lucid explanation. There are some people who just refuse to accept this no matter how clearly it is explained. It is frustrating arguing with these people and I’ve just about given up trying to convince them. They have a false mental picture that heat flows between objects like water and are stuck on the idea that water doesn’t flow uphill.

I see they are now resorting to some sort of argument from authority by arguing you lack a PhD in science. While an argument from authority is a false argument, if they want to go down that route no problem. There are many here who do have a PhD in science, including myself, who would fully endorse your explanation. I doubt those arguing with you could come up with even one to support their position.

Ultimately a PhD is just a piece of paper. What is important is the spirit of open inquiry, curiosity and rational investigation that underlies science. You certainly have those qualities and what you write is usually both interesting and informative. I have no doubt you could have earned such a piece of paper for yourself if you had taken the time to do so; but then you would have led a much less interesting life.

• Catcracking says:

JRD,
I don’t understand your boundary conditions,
Are you talking about a closed room that cannot radiate or loose any heat to the outside? i.e. a closed system?
Was the CO 2 that was added at the same temperature of the water?
If same , in a closed system then there would be no change in the temperature of the water or the gas?
If one was at a greater or lower temperature then I expect in a closed system, both would end up at the same temperature with time?
Are the boundary conditions different than I assumed?

• menicholas says:

I agree…who said the pool would get warmer if the CO2 was also at 30 degrees?

• A C Osborn says:

Who said the pool would get warmer if the CO2 was at -180C?

• Jaakko Kateenkorva says:

I get so tired of hearing this stupid argument from the anti-ghg crowd. Thanks for quantifying it.

Although I’ve never used the argument, that’s a pity. But perhaps the heavenly quantifications of the pro-ghg crowd will compensate enough to maintain your balance positive also without them.

Now when “metrology”, measurement science, is no longer considered a typo in WUWT, the same recognition could perhaps now be given also to “measurand” i.e. a quantity intended to be measured, an object being measured, a physical quantity or property which is measured.

• richard verney says:

Please re-read the section about how a cold object can leave a warm object warmer ONLY if the cold object is hiding something that is even colder

But what is the even colder object, that the cold object is hiding?

Space as such is not necessarily cold, at any rate not as we understand temperature since there are all but no molecules, thus no kinetic energy.

• Ed Bo says:

Richard:

“Space”, while having incredibly low density, has incredibly vast extent. The low density means there is no conductive are convective heat transfer to speak of.

However, astronomers have made incredibly detailed measurements of the background radiation from space, and it exactly (to within the precision of the measurements) matches that of a blackbody, both in magnitude and spectrum, at 2.725K (+/-0.0001).

Now, you have have semantic and philosphical arguments as to whether space “has a temperature” of ~3K, but for the purposes of radiative heat transfer calculations, it is completely correct to treat it as the equivalent (at least) of an ambient at ~3K, providing 3 microwatts/m2 (0.000003 W/m2).

On the other hand, as we sit here typing our comments, we are bathed in an ambient of ~400 W/m2 from our surroundings that are near 300K. This is what is fundamentally confusing J Richard Wakefield above with his pool example — extra CO2 over the pool under the roof is not at a much different temperature from the roof.

• David Middleton says:

Anthony Watts on November 24, 2017 at 7:34 pm

I get so tired of hearing this stupid argument from the anti-ghg crowd. Thanks for quantifying it

+1,000

6. This is so simple to test, and prove you wrong. Take a blow torch, measure the temp inside the flame. Then light a candle near by. Does the torches temp go up? Nope. But try it to make sure.

The downward IR from CO2 attempting to warm the warmer surface is like trying to piss into a fast flowing river.

• daved46 says:

Two problems with your contention. First I doubt you’ve ever done the experiment. The temperature change would be small and how would you mearsure it? But the big problem is that, as in another arch example above, you’re dealing with chemical reactions and it’s going to be hard to create a good experimental set up. You’d have to put the blow torch inside something, set the torch to a constant flow of gas and equilibrate the temperatue on the outside surface, then add the candle outside and see if you have to reduce the flow of gas to keep the temperature constant. I’m sure you would, but it would a difficult experiment to perform.

• No, if Willis is correct, the burning hot gasses in the torch should be absorbing the IR from the candle flame, making those molecules in the torch flame hotter. If he is correct.

What you are proposing is different. You have the torch and the candle heating a third object. Of course that third object’s temp will be higher than just one heat source!

This is all academic. Willis should propose an experiment to prove his position. Conceptual analogies only work if the premise can be shown empirically to be correct.

• daved46 says:

“No, if Willis is correct, the burning hot gasses in the torch should be absorbing the IR from the candle flame, making those molecules in the torch flame hotter. If he is correct.”

My point was that you were presenting an experiment that is impossible to carry out, so I was suggesting one which could be carried out. Provide us with the actual details of how the experiment you proposed would operate and why you claimed “Nope” when both Willis and I would claim “Yes”?

• Crispin in Waterloo says:

J. Richard Wakefield

“No, if Willis is correct, the burning hot gasses in the torch should be absorbing the IR from the candle flame, making those molecules in the torch flame hotter.”

Good grief. The candle slightly reduces the rate at which the hot gases cool. The flame temperature is fixed by the energy released in the chemical reaction which is based on the energy needed to pull the fuel molecules apart and the energy gained by assembling them into new molecules like CO2 and H2O.

As soon as the combustion molecules form they release photons cooling the molecule and heating the gases surrounding it. Those gases in turn shed heat in all directions. At the same time they are picking up photons from all surfaces ‘in sight’, the energy of which depends on the source temperature. Suppose we have a CO flame burning to CO2. The theoretical temperature is 2200 C. Pointing two CO flames at each other will not generate a central temperature of 4400 C. Why?

Measuring a flame temperature is a good analogy for an atmosphere. Put a thermocouple into a flame. It will get very hot and emit visible photons, perhaps it will look white if it is hot enough. Whatever the temperature device says is the temperature, the flame is actually much higher because the thermocouple cools radiatively. Place a shield over the tip. This is called a “shielded thermocouple”. Place it in the flame and the temperature reading is much higher – never quite as high as the actual flame, but higher because the radiation is reflected back to the source (the metal of the thermocouple) limiting its ability to cool. The tip is like the Earth’s surface heated by the sun (flame). The reflector is like CO2 around the Earth. CO2 sends some radiation back, except that it isn’t nearly as effective as a reflective tube.

So, what’s the beef? The temperature of the thermocouple tip will never be hotter than the flame just because it is in radiative balance with the mirrored tip. The Earth’s surface (not the atmosphere) would cool faster if there were no GHG’s in the atmosphere. For an explanation of why the atmosphere would not react in the same manner, see my long post above.

• commieBob says:

Here’s an interesting link. MIT scientists have built an incandescent light bulb whose envelope passes visible light but which reflects infrared radiation back at the filament. The result is that it takes less electricity to heat the filament to the desired temperature … a lot less electricity.

There is no practical difference between the mirror and an infrared emitter. The resulting radiation hitting the filament is indistinguishable between the two.

How is this an example of a colder object causing a warm object to be warmer? The filament is emitting visible light. That requires a high temperature of around 3000 K. link The infrared reflected back at the filament would be generated by a much lower temperature of less than 800 K. That’s around the temperature where metal starts to glow red. link

The trick is not to confuse temperature (measured in degrees) with power (measured in BTU per second or calories per second). Most of the power radiated by the filament is infrared.

The cold and warm objects are not different than antennas. Antennas typically have large noise voltages and therefore radiate a significant signal at the noise frequencies. That doesn’t keep a miniscule signal from exciting the antenna and being detected. In fact it is possible that an antenna can be radiating many watts of signal while, at the same time, being used to detect microwatt signals.

Just because you can’t measure the effect of a candle on a blowtorch it doesn’t mean the effect doesn’t exist. It’s like gravity. A comet flying past the Earth will exert gravitational attraction on the planet, you just won’t be able to measure the effect. Similarly, if I pee into a fast moving river, you won’t be able to detect the greater flow. On the other hand if 100 million of me pee in the same river, the effect will be measurable.

• “Similarly, if I pee into a fast moving river, you won’t be able to detect the greater flow. On the other hand if 100 million of me pee in the same river, the effect will be measurable.”

All that is doing is increasing the cold source (you peeing) into a hot source (many people peeing).

• Fascinating. Someone asked for a practical example. The MIT light is an excellent example. The cold outer shell adds heat to the filament leaving it warmer than it would be otherwise.

w.

• commieBob says:

J. Richard Wakefield November 24, 2017 at 8:56 pm

… All that is doing is …

If each of us pees 100 cc in 20 seconds, the total volume is 10 million litres. There are 1000 litres in a cubic metre, so that’s 10,000 cubic metres total, in 20 seconds. That’s 30,000 cubic metres per minute. That’s about a seventh of the discharge of the Amazon river. link Impressive! We will be able to measure the effect.

Thanks be to God, there are not a hundred million of me. If we reduce the hundred million, at what point do we say that there is no effect? Why do you pick that particular number?

The effect will be there even if you can’t measure it.

You made the bold statement that Willis is wrong. Then you gave a couple of examples of experiments that would be hard to measure. That doesn’t actually prove that Willis is wrong, does it. Not only that but you have the burden of proving that your examples are apt. How about providing some numbers to back up your assertion.

• Ian W says:

The trick is not to confuse temperature (measured in degrees) with power (measured in BTU per second or calories per second). Most of the power radiated by the filament is infrared.

Excellent. Unfortunately, 75% or so of the Earth’s surface is water and a good proportion of the remainder is covered in plants that transpire. Infrared does not heat water it is absorbed by the first molecule and raises its energy level and latent heat eventually leading to the water molecule evaporating and taking its heat with it. So 75% or more of the surface of the Earth will be cooled by ‘downwelling’ radiation. The dry portions of the rest of the Earth may be raised in temperature but following Stefan Boltzmann will increase their radiation by the 4th power (modified by emissivity) and thus radiate any increase from downwelling infrared away rapidly. As is demonstrated by the rapid increase in radiation from the bulb filament, that, absent any input electricity would cool and go dark, it cannot continually circulate the infrared.

Both you and Willis make the same mistake of an reasoning in the abstract and avoiding the fact that a volume of water will cool from infrared due to increased evaporation and that most of the Earth’s surface is water or plants transpiring water. The subsequent convection and release of latent heat are the elephant in the room that you are carefully avoiding discussing.

• Brett Keane says:

In effect, IR enables more visible/UV, while having correspondingly reduced exittance itself from the system, the bulb. Lovely physics, but no ghe except on the horticultural sense. Had this before, haven’t we?

• Crispin in Waterloo says:

Willis Eschenbach

“Fascinating. Someone asked for a practical example. The MIT light is an excellent example. The cold outer shell adds heat to the filament leaving it warmer than it would be otherwise.”

Exactly. Now the big problem: The element is the source of the emission. On planet Earth, there are two sources after the energy is put in: first, the surface and second, the atmosphere itself. They are both sources of IR.

If there were no GHG’s present in the atmosphere, there would be no second source. The surface would be the sole source of IR emission and the hot surface would warm the atmosphere because of contact with it. The atmosphere would thereby cool the surface a bit, and during the daytime, be unable to get rid of the heat gained. The remaining IR would pass through to space. Daytime after daytime, the atmosphere would warm and not be able to cool. At night the surface would radiate freely into space and be warmed by the hot atmosphere which would cool a bit depending on its circulation and the length of the night.

It is untrue, in spite of thousands of claims to the contrary, that the atmosphere would be cold because it contained no GHG’s. In the daytime it would be as hot as the surface, certainly in the vicinity of the surface on the sunny side. Unable to cool by radiation, the temperature in the atmosphere would be the same as the surface at the bottom and then cooler with altitude according to the lapse rate.

Consider how different this is from the claims made in tens of thousands of articles on ‘the GH effect”. Without GHG’s, there is no ‘reflector’ and the atmosphere cannot cool by radiation, even though it would be constantly heated by the hot surface. This special case is more like a light bulb with xenon gas in it. In that case the filament is also “hotter” (than a vacuum bulb) because the gas is heated by contact with the filament. Therefore the filament runs hotter without a reflector. Different scenario, same enhanced visible radiation effect. The xenon can only cool by contact with the quartz bulb = inefficient.

When GHG’s are added to the atmosphere, the surface cooling by contact continues, the “reflector” moves into position, and the atmosphere itself begins emitting IR to space. What will be the net effect of this additional loss? Will there be a net decrease in the system’s temperature with the addition of IR radiative gases, or will the system’s temperature as a whole increase? Speaking of the atmosphere alone, will it cool because it gained the ability to shed heat directly to space?

You see what I am emphasizing? The correct comparison is an atmosphere with and without GHG’s, not an atmosphere with GHG’s and a planet with no atmosphere at all. The light bulbs can have an increased emission in the visible range by putting in a reflector, or by adding a non GHG gas. IMAX projectors use high pressure xenon in the light source. Halogen headlights in cars uses a mix of inert gases and a reflector inside the bulb where it can be the most effective as a temperature enhancer.

• coaldust says:

“If there were no GHG’s present in the atmosphere, there would be no second source. The surface would be the sole source of IR emission”

This is incorrect. The non-greenhouse gasses are still radiating IR. Everything radiates IR unless its temperature is absolute zero. The non-GHG doesn’t absorb IR frequencies, but that doesn’t stop it from radiating IR. The atmosphere would absorb energy where it contacts the surface (conduction) and these warmer molecules/atoms would rise (convection). So the atmosphere would still radiate to space and to the planet. Thus the planet surface would be cooler with a non-GHG atmosphere than with no atmosphere because it would lose some energy to the atmosphere through conduction, and not get it all back.

• Crispin in Waterloo,

The surface is heated by the sun, thus there is an indirect source of energy, as good as in the example of the flame. If that wasn’t the case, everything would would cool to space temperature, with GHGs only a little slower…

• Tim Folkerts says:

coaldust says: “The non-GHG doesn’t absorb IR frequencies, but that doesn’t stop it from radiating IR. “
It turns out that materials absorb IR exactly as well as they emit IR. “For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.” https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation

If — as you claim here — the non-GHG doesn’t absorb absorb IR frequencies, then it also doesn’t emit those frequencies.

“The non-greenhouse gasses are still radiating IR. Everything radiates IR unless its temperature is absolute zero.”
While this is sort of true, it is wildly misleading. The amount of radiation emitted by N2 or O2 is minuscule compared to the IR emitted by N2. An atmosphere of pure N2 would radiate orders of magnitude less IR than a similar atmosphere with a little H20 and a little CO2. This has been confirmed by innumerable experiments and is understood on a theoretical level.

• Crispin in Waterloo says:

Ferdinand, you are not adding information. The electric bulb has energy from a power station. So what? It doesn’t impact the argument. We are discussing shifting the emitted spectrum using an insulator or reflector.

• Crispin,

Sorry, my wrong… I am a little late in the discussion and overlooked the background in this case…

• RayG says:

I suspect that many WUWT readers have pissed into a fast flowing river. Probably into the ocean as well.

• Something commonly done in engineering combustion is to preheat the incoming gases. This can be done with something much cooler than the flame (often exhaust gas), but will increase the temperature of the flame.

• Preheating is not the same as making a hot source hotter. All you are doing in the preheat is making a cold object hotter before you make it into a much hotter object.

• Catcracking says:

Nick ,
Are you saying that the heat of combustion is the same but the final temperature of the gas is higher, then I agree. In most cases the fuel temperature is probably not heated, but where possible the waste heat is used to preheat the “gas or fluid” being heated, of course there are always exceptions.

• “All you are doing in the preheat is making a cold object hotter before you make it into a much hotter object.”
The object is the flame. If you could make it hotter with waste heat, that would already be the supposed paradox.

It’s the same deal. You have an object which becomes hot through some source – sun for earth, enthalpy of combustion for flame. Anything that adds heat – DWLWIR for earth surface, warmer incoming gas for combustion, makes it hotter, basicallly because it has to shed more heat against the same thermal resistance. That extra heat can easily come from a cooler object, such as a preheater.

• menicholas says:

I heard a thought experiment a while back involving a light bulb inside a mirrored box…mirrored on the inside of the box.
Would it keep getting brighter and brighter inside the box?

• daved46 says:

Until the melting point of the mirror was reached, or the filament of the bulb evaporated, etc. There are practical limits to everything. When I was in the heat treating business we’d put pads of nichrome wire with ceramic beads on the outside of the wire on the pipes we were heating and high temp insulation on the outside, but where the bead coated wire had to poke through the insulation you had to be careful and let some of the heat escape or the wire would get too hot and melt the nichrome wire destroying the pad. You have to know your limits.

• Crispin in Waterloo says:

menicholas

Good point. In fact the brightness would change if you cannot see all the radiation. The energy in is fixed, right? Suppose for a moment it was 100W and we insulate it perfectly and see what happens. The bulb would start off with a normal spectrum – mostly IR and some visible. As the temperature rose the mix of wavelengths would change, not the amount of output energy which is fixed at 100W. As the temperature rues the frequency would keep increasing in order to shed 100W starting at a higher temperature. This would carry on until something mechanically failed.

An insulated box can reach a very high temperature inside with a constant input of electrical energy. You just have to keep the heat in.

• That experiment was here on WUWT some years ago. The net result was that the filament got hotter and a hotter filament has a higher resistance. That was measured as a small drop in amperes (at a constant voltage)…
As only part was reflected and the bulb was cooled by outside air, the filament didn’t burn up, but I am pretty sure that without outside cooling that would have happened.

• LdB says:

Every microwave oven in existence defies the above :-).

Thermal emissions are not hot or cold they aren’t anything until they are absorbed, they can be straight reflected or pass straight thru with no interaction. They are called thermal because of there origin not because they are hot. Thermal emission are not hot or cold and no different to any other RF emission.

If we look at Willis rather silly example stuff above our microwave oven breaks his law in that net heat moves from cold to hot. From a the stupid classical physics rubbish you have something at room temperature or zero (depends what temperature stupidity you try to put on the RF signal) making something very very hot AKA your food.

The key point is when working with EM waves the concept of hot and cold go out the window and EM waves aren’t hot or cold.

• A C Osborn says:

You seem to be somewhat confused.
The Standard Dictionary understanding of “Thermal Emission” is that it comes from a “Hot” Object.
As you say EM is neither Hot or Cold just at one or more Frequencies and only creates heat when it excites atoms in an object it strikes.

• A C Osborn says:

But a Microwave does not have energy coming from a “Cold” Object, It has Energy coming from the Energy applied the cold object which converts it to EM.
Try putting your dinner in the Microwave without switching it on to see how warm it gets.

• LdB says:

The point you are missing is that Willis Eschenbach artcile is not even about the greenhouse effect which doesn’t remotely work like that and you can’t account for the GreenHouse Heat in that way no more than you can a microwave oven.

There are a multitude of ways to produce heat via resonance and GreenHouse Effect happens to use one of them. Greenhouse effect occurs because of resonant heating, like dielectric heating (microwave), Induction heating (Magnetic field in any conducting material) and a number of other effects they have absolutely nothing to do with heat transfer as such.

If your answer involves heat transfer it is wrong, and it is like trying to find how the heat gets into the water molecules in a microwave. The answer is dielectric effect in microwave ovens and electromagnetic resonance in GreenHouse effect it really isn’t that hard.

7. Bill Treuren says:

So is there any solar cycle induced effect that could hinder/change the apparent black sky near absolute temperature from being “seen” by earth. Thus altering the outward net flux of radiation.
We have heard that the solar cycle has a near zero impact on the energy balance of earth thus nothing to see there, we have heard about frequency changes of light through the cycle, but little definitive stuff really, or maybe just unqualified, so maybe it is a simple apparent background temperature change.

I’m not the scientist so am able to ask these questions without shame.

8. I didn’t read the article through, but you made a mistake in your opening analogy and chart.

You left out the “original ” flow of the \$25 “loan” from me. The original \$25 “flowed” from me to you, is missing in the “all flows” category and cancels out the “net flow” from you to me in the “net flows” category.

Maybe you explain the discrepancy in the article, but why use a flawed analogy at all?

• Aphan November 24, 2017 at 8:03 pm

Aphan November 24, 2017 at 8:03 pm Edit
I didn’t read the article through, but you made a mistake in your opening analogy and chart.

You left out the “original ” flow of the \$25 “loan” from me. The original \$25 “flowed” from me to you, is missing in the “all flows” category and cancels out the “net flow” from you to me in the “net flows” category.

Maybe you explain the discrepancy in the article, but why use a flawed analogy at all?

Flawed? The original reason for the \$25 debt is absolutely immaterial to the concepts explained. Maybe I assumed someone else’s debt. Maybe I bought \$25 worth of goods from the man. None of that matters. I’m merely trying to explain one single transaction, an exchange of money (or radiation).

w.

w.

• If you want to be clear, then explain one single transaction without adding “absolutely immaterial ” details about a prior transaction.

If you want to be scientific, don’t use words like “hide” and “see” to describe things that dont/can’t perform the act of hiding or actually be seen. Don’t interchange “radiation”, “heat” and “temperature” as if they are the same thing.

The atmosphere slows down the rate at which the Earth cools. But should the Sun disappear, the Earth with the same atmosphere will eventually cool until it reaches equilibrium with space. So in the end, that colder object would NOT “leave Earth warmer”.

Here’s a link to an experiment on this topic for anyone interested:

• commieBob says:

Aphan November 24, 2017 at 11:04 pm

… The atmosphere slows down the rate at which the Earth cools. But should the Sun disappear, the Earth with the same atmosphere will eventually cool until it reaches equilibrium with space. So in the end, that colder object would NOT “leave Earth warmer”.

You’re talking about the heat death of the universe. In the end there will be no cooler objects.

• commiebob,

Nope. Never mentioned the universe. I’m talking about the Sun in our universe, in our solar system, our galaxy and the fact that our atmosphere generates no “heat” of it’s own to warm something else with.

9. Brett Keane says:

In both cases we are changing the situation by adding mass. The first added mass has no surface because it is a different phase of matter., which has that characteristic. Nor can it act as a Black Body, they say. Physics is more interesting because it has many such tricks.

S-B and Planck Relation are claimed to be designed for objects in equilibrium, radiating to 0Kelvin. This has been claimed to make GH factors dodgy.

Anyone caring to comment on what effects the above have on this post, feel free.

• Brett Keane November 24, 2017 at 8:05 pm Edit

I don’t understand that.

In both cases we are changing the situation by adding mass. The first added mass has no surface because it is a different phase of matter., which has that characteristic.

True. And … so what?

Nor can it act as a Black Body, they say. Physics is more interesting because it has many such tricks.

I have no idea who “they” are, but “they” are wrong. There is no reason that a gas cannot act as a black body.

S-B and Planck Relation are claimed to be designed for objects in equilibrium, radiating to 0Kelvin. This has been claimed to make GH factors dodgy.

Citation? I fear I don’t believe that at all … for example, the S-B equation in the math notes is specifically for two objects which are NOT in equilibrium, nor are they radiating to 0 K. That’s the equation that’s used in the online calculator … which you apparently claim is wrong, wrong, wrong …

Anyone caring to comment on what effects the above have on this post, feel free.

No effects, as the statements are either untrue or immaterial.

Regards,

w.

• Paul Aubrin says:

“I have no idea who “they” are, but “they” are wrong. There is no reason that a gas cannot act as a black body.”
A gas can only radiate in its absorption bands. thus it is not a blackbody.

• A couple of points- radiation does not have a temperature. It has a quantum of energy per photon based on the frequency. There is not “cold” infrared radiation from a cold object. If it is infrared it has less energy than if it is visible.

Regarding the “energy” diagram. Why is there no energy radiated from the stratosphere to space, or from the troposphere to the stratosphere? Photons aren’t radiated uniformly in one direction from a molecule.

• commieBob says:

The Earth, with or without an atmosphere will radiate the same. Assuming that all energy comes from the sun, the planet will eventually achieve equilibrium and will radiate as much energy as it receives.

• David A says:

Yes, there are only two ways to change the T of an object in a radiative equilibrium. Either change the input, or change the residence time of the energy entering the object. The residence time of energy in the object depends on only two factors. The W/L of the input, and the materials encountered.

• Paul Aubrin says:

“The Earth, with or without an atmosphere will radiate the same. ”
In both cases the earth will radiate with a blackbody spectrum. The Earth with its atmosphere will radiate very differently.

http://climatemodels.uchicago.edu/modtran/

10. eyesonu says:

Willis,

Nice condensed and simple straight forward explanation.

• Dave Fair says:

Yes, thank you very much, Willis.

I seem to remember some time ago when someone attempted to estimate the fraction of the downwelling radiation derived solely from man’s contribution to the CO2 in the atmosphere. Anybody know?

As I remember, it was a very small fraction; lost in the noise.

11. KM says:

According to the figure, the surface pays 392 W/m^2 to the atmosphere (“surface radiation”) and gets 321 W/m^2 (“back radiation”) in change.

Net radiative energy flow is therefore supposedly 71 W/m^2 from the surface to the atmosphere.

Assuming the Surface has an average temperature of 15 °C, and solving Stefan-Boltzmann for 71 W/m^2, shows that the atmosphere must have a temperature of around 1 °C. It must also be fully opaque, so that no energy radiates directly to space.

I’m using this online S-B calculator: https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html

Problems:
1. The atmosphere is not opaque.
2. The average temperature in the troposphere (which represents 99% of the total atmospheric mass) is -15 °C.

Let the flaming begin. :)

• Nah, no flaming. You say:

Problems:
1. The atmosphere is not opaque.

Other than the “atmospheric window”, it is very near to opaque, with almost all of the upwelling radiation absorbed more than once before it makes it to space.

2. The average temperature in the troposphere (which represents 99% of the total atmospheric mass) is -15 °C.

The average temperature in the troposphere is not relevant. What is relevant is the temperature in the effective radiating layer. About 75% of the downwelling radiation at the surface comes from the first hundred metres of the atmosphere, which biases the effective radiating layer down in the warmer temperatures.

w.

• Catcracking says:

Willis, I agree with your article 100%. One thing as an engineer, some people seem to assume this is a one step process which it is not especially when no additional outside sun radiation is added when it is night, otherwise it would not cool down overnight. There is continuous radiation and back radiation in my mind and the earth keeps cooling down after the sun goes down since the CO 2 radiates a portion of energy out to space each “step”
What am I missing?

• KM says:

If you are correct, then how come on a cloudless night the temperature drops much more than on a cloudy night?

Clouds are much higher up than a few hundred meters you claim is responsible for 75% of the downwelling radiation. Also clouds are visibly opaque but the clear sky is transparent.

• KM November 24, 2017 at 9:35 pm

If you are correct, then how come on a cloudless night the temperature drops much more than on a cloudy night?

The greenhouse effect works because some of the upwelling IR is absorbed by the atmosphere. When the atmosphere subsequently radiates the heat, about half goes up, and the other half goes down to the surface. This leaves the surface warmer than it otherwise would be.

Now, obviously, the more upwelling IR is absorbed, the more downwelling IR comes back to the surface. So which one will absorb more upwelling IR?

1. Clear skies

2. Clouds

Clouds are basically black bodies for IR. If they have any thickness they don’t let any IR get through to the other side. So clouds absorb more than clear skies. As a result, they absorb more, and as a result they radiate more, both downwards and upwards. This leads to the phenomenon you describe above.

On a clear night, the IR is much freer to escape to space, so less of it comes back down to warm the ground.

w.

• KM says:

Willis, I understand the point you’re trying to make but it doesn’t seem to add up.

You first claim that the atmosphere is near to opaque (i.e. essentially acts as a black body).

Then you state that 75% of the back radiation happens within the lower hundred meters. If this were true then only a small part of the remaining 25% of the total outgoing IR would reach higher altitudes where the clouds are forming.

In your latest post you claim that the clear sky is transparent to IR, but clouds are not. You are contradicting your first statement that the atmosphere itself (in the absense of clouds) is near to opaque.

• Dave Fair says:

Along the lines of my prior comment, the downwelling radiation fraction from strictly man’s contribution to atmospheric CO2 concentrations must be minuscule.

• One has to add TIME to this discussion. There is no such thing as a one-and-done with energy. What happens over time.

Then you have different places to describe. 2 feet underground, 1 inch underground, right at the surface, the atmospheric molecules that touch the surface, 2 metres high, 50 metres, 1 km, 6 km, 10 kms, 70 and 100 kms high. And you need to describe it over time as more energy is coming in and more energy is leaving over that time.

A cold object warms a warmer object through EM radiation? How much and for how long and what happens to both objects over time.

• richard verney says:

The greenhouse effect works because some of the upwelling IR is absorbed by the atmosphere. When the atmosphere subsequently radiates the heat, about half goes up, and the other half goes down to the surface. This leaves the surface warmer than it otherwise would be.

Please explain the GHE on Mars.

The Martian atmosphere, on a numerical basis, has an order of magnitude more CO2 molecules than does the atmosphere of Earth. Even taking into account water vapour, there are more molecules of so called GHGs in the Martian atmosphere, than there are in Earth’s atmosphere.

Further not only are there numerically more molecules of so called GHGs, they are much more tightly spaced together, since Mars is a smaller sphere and the atmosphere has a smaller volume.

The upshot of this is that on Mars, there is a greater prospect that an upwelling IR photon emitted from the surface will be absorbed by a so called molecule of GHG in the Martian atmosphere, and then a greater prospect that a re-radiated photon from that so called GHG molecule will then be absorbed by another molecule of so called GHGs, than is the case on planet Earth.

Put simply, it is far more difficult for an IR photon emitted from the surface of Mars to find its way out to TOA and there be radiated to space. The journey from surface to space will be much more impeded than a like journey on planet Earth, and more photons (in relative terms) will be back radiated towards the surface.

And yet there is no measurable (radiative) GHE on Mars. Why is that?

PS. Obviously Mars is further from the sun, so receives less energy from the sun than does planet Earth. Since the Martian atmosphere is less cloudy, in relative terms more of the incoming solar irradiance finds its way to the surface, compared to the position in planet Earth.

I am not suggesting that Mars ought to have a (radiative) GHE of 33 degC as claimed for Earth. There is less solar irradiance so obviously the (radiative) effect should be less, but it considered to be so close to zero that it is considered that Mars has no GHE.

• gregfreemyer says:

Willis, Great main article and I very much appreciate you writing it. I’ve fought the same fight; very frustrating.

But your comment about the effect of clouds is misleading. You imply the atmosphere is near opaque for all infrared. That’s false.

In the above the first curve is the emission spectra from the sun and is mostly visible light. The earth surface blackbody emission is also shown. Note that a significant part of the earth surface’s blackbody radiation makes it to space during clear skies. It isn’t absorbed even once.

Clouds change that of course and you get significantly enhanced absorption of upwelling radiation, and a corresponding significantly enhanced downwelling radiation.

• David A says:

This assertion ” 75% of the back radiation happens within the lower hundred meters, is not relevant as far as I know, in that in the dense lower atmosphere all the energy, conducted, radiated and convective, acts like conducted and convective, as these conducted exchanges happen far more rapidly in the denser lower atmosphere.??

• Toneb says:

“Please explain the GHE on Mars.”

To thin an atmosphere.
It does actually have one but it amounts to just ~ 5K.
The atmospheric pressure at the surface is just 6mb
0.6% of Earth’s.
It’s atmosphere is 100x less massive.
It is much colder than Earth’s atmosphere – a min of 50K throughout it’s profile.
Therefore has a weaker LWIR radiative effect.

• tty says:

Richard Veney:

GHE is weak on Mars because there is very little of the really important GHG, i e water, in the atmosphere, which is therefore much more transparent to LWIR than on Earth:

Making the CO2 peaks a bit taller won’t have much effect compared to taking the other four gases away. And as a matter of fact the CO2 absorption peaks will actually be a bit narrower since there is lower atmosphere pressure (=less pressure broadening) and lower temperatures (less doppler broadening) on Mars.
:

• KM says:

A clear and a cloudy sky both have the same amount of CO2.

However, clouds are opaque and therefore block the line of sight. The result is that fewer CO2 molecules can be reached by the surface radiation.

So we have the following:
Clear night sky, with more CO2 in the line of sight: LOWER temperature.
Cloudy night sky, with less CO2 in the line of sight: HIGHER temperature.

Looks like anti-correlation to me.

I agree that a cold opaque object (a cloud) can “warm” a hot object (the surface) by insulating it from the colder space. CO2, however, cannot.

• richard verney says:

There are some that claim it is the GHGs in Earth’s atmosphere that keep the Earth’s surface warm. However the substantial difference between Earth and Mars is that Earth has large quantities of Nitrogen, Oxygen and some Argon in its atmosphere. If one were to remove from Earth’s atmosphere all the non GHGs, then Earth would have an atmosphere of about the same mass and density to that of the Martian atmosphere..

If one were to remove all the non GHGs from Earth’s atmosphere, Mars will have slightly more molecules of GHGs. It will have a lot more CO2, and quite a bit less water vapour. Earth will have little CO2 and quite a bit of water vapour.

You state:

The atmospheric pressure at the surface is just 6mb
0.6% of Earth’s.
It’s atmosphere is 100x less massive.

But that is relevant if one considers that it is the pressure of Earth’s atmosphere that keeps the planet surface warm, or if you consider that it is the thermal mass and thermal inertia of all the Nitrogen, Oxygen and Argon in the atmosphere is what keeps the surface warm by insulating and slowing down the heat loss from the surface.

• Richard Verney points out simple truths here that should give one pause. Always pay attention to Richard’s posts.

You want an experiment? Don’t experiment with Mother Earth they sometimes say. But we have all these planets with atmospheres near-by that have already done the experiment for the past 4.4 billion years.

The GHG religion is disproven by all these other experiments.

Something else is going on. Rack your brain for what they might be. Think about photons and gases and electron shells absorbing photons or not and the time-scales that energy flows by and the collision rates of gaseous atmospheres and the rotation rates of planets and how long they are absorbing solar photons in a daytime period and the Stephan-Boltzmann equations/implications which is the best and most descriptive theory ever invented.

• Crispin in Waterloo says:

Willis

“The greenhouse effect works because some of the upwelling IR is absorbed by the atmosphere. When the atmosphere subsequently radiates the heat, about half goes up, and the other half goes down to the surface. This leaves the surface warmer than it otherwise would be.”

“Than is would be” in what condition, exactly? Than it would be if there were no greenhouse gases in the atmosphere? The assumption that the atmosphere would not be warm if there were no GHG’s is contradicted by evidence. I will give two examples:

At the top of the atmosphere there is precious little CO2 or other GHG’s. The temperature of what few molecules there are is very high. Why? Because the molecules are heated by radiation and are not emitting in the IR so they remain hot. Really hot. They can only cool by bumping into another colder molecule which can emit in IR range, but they are few and far between. This indicates that in principle a gas can be heated and the temperature will rise if it is unable to cool radiatively. An atmosphere with no GHG’s would be hotter than one with GHG’s because it can’t cool except against the ground.

Second example: The surface would be heated to a higher temperature if there were no GHG’s in the atmosphere. Agreed? The surface would cool, in part, by heating the air in contact with it (convective cooling). That heated air would rise and the heat would spread through the system. Heating would continue every day in bright sunshine (no clouds of water vapour). Unable to cool by radiation, the atmosphere would continue to warm. Radiative equilibrium would only be reached when the surface heating by the hot night air equaled the heat gained from the daytime surface heating of the air. An atmosphere with no GHG’s would be hotter than one with GHG’s.

Adding GHG’s to an atmosphere that was part of a planetary system that was already in radiative equilibrium would cool that atmosphere, becoming cooler that it would otherwise be. This is not a direct contradiction of your last sentence which mentions the surface, but the point should be evident. Let’s say the surface may or may not be warmer with GHG’s, depending on how hot it had to be at night to get rid of the heat imparted to the atmosphere during the daytime and downwelling IR. The atmosphere will definitely be cooler with GHG’s added. The fooferaw about CAGW is about the air temperature, not the temperature of the surface.

This is related to your mention of the ‘window’ The window at ground level is small. At 100m it is larger. At 1 km it is much larger again. Adding more emitters throughout the atmosphere will not cause it to get warmer than it would be with no emitters at all because with none it can’t cool except at the bottom.

I think the only way to prove my point is to model a GHG-free atmosphere using FEA and see how hot the air becomes by the time the system reaches radiative equilibrium. People are so busy claiming that a GHG-free atmosphere would be cold, while simultaneously arguing that the surface would absorb 1 kW/m2 and be hot enough to send it into space as IR. Crikey. What temperature is that? Well, the air next that heated surface would be just as hot as that surface. “Just as hot” is not “cold”.

None of this has anything to do with what the surface temperature would be if there were no atmosphere at all. How long as the IPCC and its minions been skirting around this rather large hole in their GHG explanation?

• Kristian says:

Toneb says, November 25, 2017 at 7:34 am:

“Please explain the GHE on Mars.”

To thin an atmosphere.
It does actually have one but it amounts to just ~ 5K.

No. It doesn’t “actually have one”. If anything, it’s NEGATIVE. T_e on Mars is ~211K, while T_s is ~203K.

12. Anne Ominous says:

Just a minor correction, but I think an important one.

“Heat” is NOT the “net flow of energy”. That results in radiative heat TRANSFER, from one object to another. It’s not the same as “heat”.

I don’t like to quibble, but in this particular matter, which is easily confused, precise language is called for.

• Crispin in Waterloo says:

Agreed Anne.

Further:
Heat flows through objects. Radiative energy transfer is not heat flow. The term ‘flux’ can refer to heat flow or radiant energy.

“In physics, and in particular as measured by radiometry, radiant energy is the energy of electromagnetic and gravitational radiation.[1] As energy, its SI unit is the joule (J). The quantity of radiant energy may be calculated by integrating radiant flux (or power) with respect to time. The symbol Qe is often used throughout literature to denote radiant energy (“e” for “energetic”, to avoid confusion with photometric quantities). In branches of physics other than radiometry, electromagnetic energy is referred to using E or W. The term is used particularly when electromagnetic radiation is emitted by a source into the surrounding environment. This radiation may be visible or invisible to the human eye.[2][3]”

It contains an error, however. The definition says “…radiant flux (or power) with respect to time.” Radiant flux with respect to time is power. One can’t have “radiant power with respect to time”. That would be radiant flux with respect to time with respect to time. That’s like saying “Watts per second”.

Anyway, the common argument that cold and hot objects ‘can’t heat each other’ is rooted in the misbelief that the radiant energy “flows” which is to say, conducts from cold to hot, which is not going to happen. Terms and definitions matter.

• Tony says:

No, radiative energy transfer CAN be heat transfer, when it’s from a warmer object to a cooler object. Heat transfer generally, by definition, is a transfer of energy from a warmer object to a cooler one, by any means (conduction, convection or radiation). Energy flows both ways, true, but heat only travels in one direction – from warmer to cooler. By definition.

There is a flow of ENERGY going from cooler to warmer even with conduction; however nobody seems to concern themselves with this “back conduction”, since more energy is always going the other way, and so HEAT of course transfers that way overall (warm to cool). Funny how everyone’s so bothered with “back radiation” and not “back conduction”…but there you go.

13. “When an object emits radiation, that radiation goes on until it hits something that absorbs it, whereupon it is converted to thermal energy.”

Nope. It can be re-emitted instantaneously by the much hotter molecule. Temperature is the average motion of molecules. It’s measured by the collisions of those molecules on the thermometer. Incoming colder IR frequencies will not make the motions of the molecules increase. Hence will not make the hotter object hotter. It cant even increase the energies of the hotter object increase, because the hotter object is ejecting billions of photons more than is coming in. Hence the immediate re-emission.

• Anne Ominous says:

The NET radiative heat transfer (and I think “net” is the confusing factor here) will always be from hotter to colder. Always. Anything else violates conservation of energy.

Willis’ first diagram illustrates this. 100 one way, 75 the other, the NET transfer is 25.

Nobody is denying that it does go both ways. But it is the net that matters.

• Toneb says:

“Nobody is denying that it does go both ways.”

I’m afraid it seems that some do.

• Toneb says:

“Incoming colder IR frequencies will not make the motions of the molecules increase. Hence will not make the hotter object hotter. ”

We keep coming back to the same fundamental mistake.

No No NO the hotter object recieving the LWIR from the colder does NOT get hotter.
It just cools at a slower rate such that after a uit time it will be warmer than otherwise.

What is so difficult about the concept of cooling more slowly???

See Willis’ cartoon.
It is the NET flow that has to be considered.

I’ve used this analogy here in a recent thread…..

A water tank looses 10 gal/hr from a leak.
it is a 90 gal tank.
It would be empty in 9 hours.
But there is a feed into it of 1 gal/hour.
So 10-1 = an overall loss from the tank of 9 gal hr.
SO it takes 10 hours to empty.

At no time does the overall quantity of water in the tank go UP.
But after 9 hours it has more water in it than if there were not a 1 gal/hr feed to it.
Substitute for hotter/colder and net transfer of LWIR.

• A C Osborn says:

But you have added another water supply. Not just the original water.

• Back radiation from the colder object is absorbed by the hotter object.
The hotter object is heated by that absorbed radiation.
Being heated, it now radiates more per unit time than it was radiating before because rate of radiation is related to temperature.
Thus this increased temperature increases its rate of cooling.
The question is, what is the net flow:
Does the increased absorbed energy result in an output increase that is equal to, less than, or greater than the back radiation absorbed?

• Tony says:

“Back radiation from the colder object is absorbed by the hotter object.
The hotter object is heated by that absorbed radiation“

“Can A Cold Object Warm A Hot Object?
Willis Eschenbach / 9 hours ago November 24, 2017
Guest Post by Willis Eschenbach
Short answer? Of course not, that would violate the Second Law of Thermodynamics”

• tty says:

“But you have added another water supply. Not just the original water.”

Do you seroiusly think the result would be different if you took the 1 gal/h from the leaking water?

• Crispin in Waterloo says:

A C Osborn

How is this:

You are sitting in a boat that springs a leak; water is flowing into the boat at 1 gallon per second. It is going to sink in 15 minutes.

You bail furiously using a small bucket, returning 1/2 a gallon per second to the lake.

Does the boat take longer to sink when you bail or does it still sink in 15 minutes?

• Richard says:

When I pour my hot tea into my cup , the cup dictates how hot the tea is but does not make the tea hotter than the incoming temperature.

• A C Osborn says:

Crispen, to do this I have to Expend Extra Energy.
Where is the Energy coming from in Back Radiation, you say CO2.
I say prove it by Measurement.
I can prove the Watts coming from the Sun by measurement.
Show me how to measure 340W/m2 of DWILR.
Dr Spencer tried it but he did not and cannot have a Control Object, so when the object got colder than Ambient he said that proved DWLIR because it would have got colder, but would it have been -340W/m2 colder.
That is an Assumption not an Observation, because without a Control he doesn’t know how cold it would have got.
Why is there no Empirical Evidence, only theory.

14. John says:

Probably for the first time, I disagree with Willis. I concede the concept of ‘net flow’, but there seems to be a mixing of conduction law, where thermal energy always flows to cooler regions, and radiation law, which says photons are emitted from one place and received at another. That ‘another’ place is warmed, period. If some body absorbs photons, it becomes warmer. Then, it radiates at a higher level, the old 4th power law. The sun is warmed by the earth’s radiation. It cannot be otherwise. The concept of ‘net flow’ is immaterial with radiation, because radiation ALWAYS increases thermal energy of whatever absorbs it. And that absorber will always radiate at an even higher rate – which can increase the thermal energy of the original emitter.
Just think of shining two flashlights at each other. Precision (very) thermometers on each. Turn one off and log the measurements. Reverse them, and record the measurements. Turn them both on and record the measurements. They will both increase when both are on, but radiation is the only method of transfer.
Now, place one flashlight on a piece of ice. Repeat the measurements. The readings will essentially repeat – because radiated energy has no concept of from where it was emitted, not where it is received.

• “If some body absorbs photons, it becomes warmer.”

Nope, not when the object the IR is intercepting is hotter. “Hotter” (temperature) is a measure of molecular motion (vibration, rotation, translation). Cold IR wavelengths cannot add to the motion of the molecules. Either nothing happens, or more likely, it is reemitted instantaneously.

• J. Richard Wakefield November 24, 2017 at 8:35 pm

“If some body absorbs photons, it becomes warmer.”

Nope, not when the object the IR is intercepting is hotter.

I’m sorry, J. Richard, but in this case you are simply wrong. I challenge you to find one serious scientific reference that says that a photon from a colder object will NOT be absorbed by a second object that is hotter than the first.

I’ve backed up my ideas. I’ve given a website, a calculator, and an equation that says quite clearly that photons ARE absorbed regardless of temperature.

Now it’s your turn … but let me ask you politely to not repeat your incorrect statement until you provide something to back it up. We’re not into science by assertion here, you won’t do your cause any good by insisting you are right and endlessly repeating your claims.

w.

• “photon from a colder object will NOT be absorbed by a second object that is hotter than the first.”

Design an experiment were we can see your claim happening. Math and numbers are just what you THINK should be happening. Prove that is what is happening.

• J. Richard Wakefield November 24, 2017 at 9:00 pm

Design an experiment were we can see your claim happening. Math and numbers are just what you THINK should be happening. Prove that is what is happening.

Read about the MIT lightbulb above, which uses exactly this same principle. The cooler shell around the filament makes the filament hotter than it would be without it.

In addition, pick up any college thermo textbook, and it will say what I’m saying. Or you could just look again at the web page I linked to. Seriously, it contains a calculator, a full explanation, and a complete equation for the calculations of what you claim can’t possibly happen. Play with the calculator, and consider the answers that it gives. For our situation (one object surrounded by another) set area a1 to 1, and area a2 to a million. Then put in different temperatures for the object and the surroundings, and think about what is happening.

Finally, let me repeat what I said to you above, which you have totally ignored:

I challenge you to find one serious scientific reference that says that a photon from a colder object will NOT be absorbed by a second object that is hotter than the first.

I’ve backed up my ideas. I’ve given a website, a calculator, and an equation that says quite clearly that photons ARE absorbed regardless of temperature.

Now it’s your turn … but let me ask you politely to not repeat your incorrect statement until you provide something to back it up. We’re not into science by assertion here, you won’t do your cause any good by insisting you are right and endlessly repeating your claims.

It’s time to put up or shut up, J. Richard. You need to find some authorities that say that somehow photons carry information with them about the exact temperature of whatever emitted them, and that warmer objects can somehow sense that temperature information in the photon and reject the photons from colder places …

w.

• Windchaser says:

“Cold IR wavelengths cannot add to the motion of the molecules. Either nothing happens, or more likely, it is reemitted instantaneously.”

The warmer body doesn’t “know” about the temperature of the other body, nor does it matter. If it’s opaque in that spectrum (which rarely depends on the temperature of a material within a given phase), it will absorb the energy. And energy absorbed is energy absorbed.

Here’s a real-world example. Water, and many metals, are pretty decent absorbers in the microwave spectrum. The microwave spectrum is “colder” than the infrared; lower-energy, but that doesn’t stop them.

Will a metal or water at room temperature, emitting in infrared, absorb microwaves? You betcha. I recommend you go try it out in your microwave for yourself.

Again: if light is hitting a body, the body does not care about where the light came from or what temperature the originating body was. It only depends on the optical properties of the absorbing body, which are generally not very sensitive to that body’s temperature.

• angech says:

J. Richard Wakefield
” Either nothing happens, or more likely, it is reemitted instantaneously.”
needs a little work,
If nothing happens you are denying absorption happens.
Trying to claim it is instantaneous to satisfy your argument is sophistry.
Instant re emission is basically claiming it was never absorbed at all as well.

If you accept absorption then the object has gained energy.
The science says that there is usually a delay between absorption and emission which is measurable.
The object being more energetic is capable of producing more heat.

• “If some body absorbs photons, it becomes warmer.”

Nope, not when the object the IR is intercepting is hotter.

When I read that argument from the Sky Dragon Slayers, phrased as cooler photons reflect off of hotter objects, I gave up on them. It is the only claim they can make, as the only way not to absorb the energy of the photon is to not absorb the photon. There’s plenty of data showing blackbody objects don’t work that way. If they did, my IR thermometer couldn’t measure the temperature of cooler objects.

https://wattsupwiththat.com/2013/05/27/new-wuwt-tv-segment-slaying-the-slayers-with-watts/

https://wattsupwiththat.com/2013/05/28/slaying-the-slayers-with-watts-part-2/ (real experimental data!)

https://principia-scientific.org/why-did-anthony-watts-pull-a-bait-and-switch/

That last post is referenced byt the WUWT post, but at least some of the links go to a current news URL that no longer exists.

• tty says:

“Nope, not when the object the IR is intercepting is hotter.”

Never wondered why a radio receiver works? The antenna is almost infinitely “hotter” than the photons it “intercepts”. Remember that the cosmic background blackbody radiation at 3 K peaks at a wavelength of about 2 mm, so even UHF radio waves are much, much colder than 3 K.

• Count to 10 says:

Also, consider that IR lasers can be used to heat a target to very high temperatures.

• Paul Aubrin says:

Photons are not small bullets fired by a hot body in all directions. They are interaction. The interaction will take place only if the destination can accept exactly one quantum of energy from the emitter.

• Crispin in Waterloo says:

Count to 10

Lasers are also used to cool gases to near absolute zero. In fact that is the standard method now to get milli-Kelvins.

• J. Richard Wakefield

Nope, not when the object the IR is intercepting is hotter.

Richard, a CO2 laser may get as “warm” as about 100ºC, but the “warm” IR waves coming out of that laser can heat up steel and melt it at 1200ºC…
There are many thousands of these lasers proving daily that you are completely wrong in that point…

• Tony mcleod says:

Hundreds of examples of the Dunning-Kruger effect.
I mock you Willis and you too Anthony for your feeble attempts to herd these cats you’ve gone out of your way to attract and cultivate.

Live by the sword, die by the sword.
Bwahahahaha.

• Mike says:

I don’t know (much) what I’m talking about; but I seldom let that stay me from talking. It does, however slow me down! Usually.

It seems to me that the two-way flow of IR energy is exactly analogous to two waves propagating in opposite directions on a conductor. Neither wave, when launched, has any knowledge of the conditions at the far end of the line – whether the destination is hot or cold, near or far, matched or unmatched. The finite velocity of light makes this so. Assume two radiators some distance apart, with unequal energies being radiated, each toward the other. One is “bright”, the other “dim”.

S_B tells us that the energy flow is proportional to the differences in T^4 of the endpoints. This tells me that the bright source will radiate less energy toward the dim source than it would toward a dark body, because the dim source isn’t cold, it’s only dim. The consequence is that the bright source must radiate more in other directions – i.e, get hotter – because of the dim source. Else the bright source is not at zero net energy. A la the self-heating light bulb.

A bolometer inserted between the sources at some intermediate point will be heated, on one side, by the bright source, and on the “dark” side by the dim source. The two sides are shaded from each other assuming only bolometer opacity to the radiated wavelength(s). In the absence of the dim source, the bolometer assumes a temperature determined by the heating of the bright side, minus any cooling attributable to the now-elevated temperature of the dark side. Now let the “dark” side also be illuminated, such that there is energy impinging on the dim side. What is the net energy flow at the dim side? It’s determined by energy intercepted (from the dim source) minus energy radiated by the bolometer. The bolometer is a summing node. It assumes whatever temperature is necessary to achieve zero net energy. Now move the bolometer to epsilon above the bright object, and tell me what, besides relative energies, changes? I believe, by analogy, this makes the bright body a summing node also. Which is true of all situations of thermal equilibrium. The incident radiation on both sides of the bolometer is radially directed toward the other object. The radiation from the bolometer is scattered. Thus the point-to-point energy flows from the original objects is disrupted.

Radiated energy transfer is a two-way street, with the net flow from hot to cold; conductive transfer is always hot to cold. Does the presence of the bolometer affect the temperatures of the radiating objects? Almost certainly. The bolometer, or a conductor in the case of conductive transfer, is mass that was added to the original system, thus changing the system.

• Brett Keane says:

John
November 24, 2017 at 8:29 pm: You are measuring the temp of the measuring devices. Never that of the light. Light is EMF, not KE, and the thermometer on ice will cool in spite of your beam, which has very little power.

15. I have a wood heater. One log alone won’t burn (very well usually) but two logs burn on the sides facing each other where they aren’t losing heat to the colder walls of the heater.

• richard verney says:

That is so, but you are altering the draft, convection and flame front, so one would expect to see a different result.

I do not consider that that example proves the radiative issue under discussion.

16. donald penman says:

If the atmosphere were to expand when it is warmed and contract when it is cooled which being a gas it can do this if not prevented by anything then there is no fixed amount of radiation returned to the surface. The ideal gas laws do not apply to a planets atmosphere because there is nothing outside of the atmosphere to stop it increasing its volume.

• Convection happens. The expanding warmed air at the surface rises and cools as it expands.

• jIM a says:

Thanks, Willis.. always thought that was dumb. Math-wise it works, but sense it doesnt make. I would surely like to see someone describe how a molecule rejects photons.
JRW,, In the swimming pool example you forget one small factor. The photon never makes it past the first few molecules of water. It may heat that molecule and heated water evaporates more readily, cooling the other surface molecules.
It is the exchange that matters, and the practical effect is that water heats air, not the other way around.
Taken to the ultimate projection you would have a perpetual motion process.

And

• John says:

Actually, rising air doesn’t cool at all, unless it contacts something cooler, or radiates. It apparently cools because of lowered density (pressure)…fewer air molecules are present. Remember, the ideal gas laws work for all gases, including CO2.

• Windchaser says:

“The ideal gas laws do not apply to a planets atmosphere because there is nothing outside of the atmosphere to stop it increasing its volume.”

The Ideal Gas Law doesn’t say anything about fixed volume or not. It applies regardless.

The Ideal Gas Law only stops applying when your gas stops being thermodynamically ideal (e.g., close to condensation or disassociation).

17. donald penman says:

The way I see it is when the atmosphere warms up it then convection occurs but what if the warm air rises further because it is warmer before it falls and cools then the atmosphere is expanding even with convection

18. Extreme Hiatus says:

“And as a result, with the cold atmosphere shielding us from the nearly infinite heat sink of outer space, the earth ends up much warmer than it would be without the cold atmosphere.”

I get your point. But the terminology obscures understanding. What is “cold”? If you had written “colder” atmosphere – relative to the earth – then it would make complete sense. It is all relative.

I’m sure what you are (correctly) trying to say is that the earth is warmer than it otherwise would be because it is surrounded by a relatively colder atmosphere which ‘insulates’ it – keeping it simple here – from even colder space. Sure. But you didn’t actually say that.

• But here’s the rub. That “colder than Earth” thing-the atmosphere-ALSO keeps the Earth cooler than it would be would be by insulating Earth from some of the Sun’s radiation. There’s even a term and definition for it:

“Thermal insulation. 1 : the process of insulating against transmission of heat. 2 : material of relatively low heat conductivity used to shield a volume against loss or entrance of heat by radiation, convection, or conduction.”

Here’s a textbook explanation from “Physical Science Concepts for Middle School”-
https://www.ck12.org/book/CK-12-Physical-Science-Concepts-For-Middle-School/section/5.17/

“Materials that are good conductors of thermal energy are called thermal conductors. Metals are very good thermal conductors. Materials that are poor conductors of thermal energy are called thermal insulators. Gases such as air and materials such as plastic and wood are thermal insulators.”

Too bad Willis didn’t just say that or start there.

19. eyesonu says:

Willis,

I can see that you have a hand full on this one, but you have the persistence and patience of Job.

• Tony says:

Or, another way of looking at it: the stubbornness and inability to admit an error of Michael Mann.

20. The ‘simplified’ diagram of radiation flows through the atmosphere is just that ‘simplified’.
Because MOST heat is transported through the atmosphere in the form of WATER VAPOUR ENTHALPY.
Try looking at CIMSS tropical cyclones page and see with your own eyes the immense heat pump carrying heat in the form of EXCITED WATER VAPOUR MOLECULES….high into the atmosphere (and also to frigid winter polar regions). So just to recap these ‘radiation budgets’ are undoubtedly scientifically correct…they are also to all extents and purposes irrelevant or to put it another way…negligible.
http://tropic.ssec.wisc.edu/tropic.php

• You are correct Charles in stating that most of the heat transferred off the surface is by moist convection. Of the massive (nearly 400W/m2) surface in vacuo radiative potential only about 30W/m2 is transferred to the atmosphere. THIS IS THE ELEPHANT IN THE ROOM.
The talk of ‘all the energy radiated is absorbed’ and similar are irrelevant because the atmosphere is hardly heated at all by long wave radiation. The surface never loses much to the colder atmosphere and the atmosphere receives very little energy this way. The opacity decouples because at close proximity no significant thermal gradient exists over the diurnal cycle. Hence very little heat is transferred in high opacity bands. The atmosphere’s heat content is 90% the result of direct absorption of sunlight and latent heat transfer and exhibits a thermal profile completely independent of long wave opacity.

21. We had a demonstration of this in a physics lesson mumble years ago with two parabolic reflectors facing each other, a thermometer at one focus, nothing at the other. Allowed to stabilise. Place a hot object at the empty focus and watch the temperature at the other focus rise. Repeat, and after things stabilise place an ice cube at the empty focus and watch the temperature fall. We had discovered cold radiation!

Except we hadn’t, of course, we had revealed the fact that when the thermometer system is in equilibrium the heat flows in and out are matched,but when you disturb the environment of the thermometer a new equilibrium is reached.

The ultimate test would be placing empty space at absolute zero at the empty focus, letting it stabilise and then replace space with an ice cube. Heating from ice…

JF

• Julian Flood November 24, 2017 at 9:40 pm

The ultimate test would be placing empty space at absolute zero at the empty focus, letting it stabilise and then replace space with an ice cube. Heating from ice…

Exactly. The ice is warmer than what was there before, so the other object ends up warmer as well.

Good to hear from you, Julian, long time.

w.

• Hi de hi.

I accidentally got a job (saving my country in small ways) so I’ve been posting less.

Talking of warming…

There must be lakes/almost-closed bodies of water which exhibit oil/surfactant pollution warming — a large lake with a growing city on its banks should, if I’m correct, warm faster than a pristine one. Maybe we should carry out the experiment, imitating the city runoff with a few oil tankers. Finland has lots of lakes, and there’s that big empty place north of the USA, that would also be suitable. And what about the Great Lakes, they must be pretty polluted in places. More research money… err… more research is needed.

There’s a nice small lake next to UEA. I’ve been tempted.

JF

• Hugs says:

A very good example! Thanks for you, Willis and Anthony for hosting this.

22. Nicholas Denman says:

The argument made sense until the leap of logic right at the end. At no time does the hotter object become even hotter. It cools slower. Therefore how does this explain the earth being 33degrees HOTTER than from the level of solar radiation alone. This persons own logic reveals how impossible the greenhouse theory is.

• Ed Bo says:

Nicholas:

Let’s take Willis’ monetary analogy further. You are earning \$240 each week. For a long time, you are spending \$240 each week (let’s say 12 \$20 purchases), so your bank balance is constant from week to week.

Now, you start getting change back from each of your \$20 purchases. You are still making 12 of these purchases each week. Does your bank balance increase?

By your logic, it doesn’t. By Willis’ logic it does. (If you really believe it wouldn’t, I want to be your banker!)

Now, if you lost your job and your income, the change you get from your purchases, would just slow the rate at which you deplete your bank account. But that is not analogous to what we are discussing here.

23. angech says:

When you comment, please QUOTE THE EXACT WORDS THAT YOU ARE DISCUSSING,
“Heat”, on the other hand, is not those individual flows of energy. Heat is the net flow of energy, as represented in the bottom half of Figure 1. Specifically, a heat flux is the net flow of energy that occurs spontaneously as a result of temperature differences.-

Heat and heat flux are two different entities. Heat and temperature are related.
heat flux is the net flow of energy .
Heat is surely equivalent to the temperature of an object as in this object is hotter i.e. has a higher temperature than another.
Why not say this?
Yes heat can flow between them but heat is the temperature of an object surely??
Or surely not.

• Heat is the transient flow of energy between two bodies of different temperature.

• angech says:

So I cannot ask for a hot cup of coffee?

• angech: Do you not know the difference between ‘heat’ and ‘hot’? Let’s be clear, ‘heat’ is the transient flow of energy between two bodies of different temperature, whereas ‘hot’ is a non-scientific term used to describe the temperature of a body. Neither a ‘hot’ body nor a ‘cold’ body contain any ‘heat’ because ‘heat’ is a transient phenomenon.

• I would disagree Philip. Heat is the (total) thermal energy in an object. The Pacific Ocean has a large heat content. This does not imply any net flow of energy. Heat transfer encompasses the various mechanisms that can transfer heat from one body to another; conduction, convection, radiation, mass flow etc.

• mkelly says:

Nuwurld you are incorrect and Phillip is correct. Heat is a energy in transit. It was part of internal energy before it left and will be internal energy again when it arrives.

• “Heat transfer physics describes the kinetics of energy storage, transport, and energy transformation by principal energy carriers: phonons (lattice vibration waves), electrons, fluid particles, and photons.[1][2][3][4][5] Heat is energy stored in temperature-dependent motion of particles including electrons, atomic nuclei, individual atoms, and molecules. Heat is transferred to and from matter by the principal energy carriers. The state of energy stored within matter, or transported by the carriers, is described by a combination of classical and quantum statistical mechanics. The energy is also transformed (converted) among various carriers. The heat transfer processes (or kinetics) are governed by the rates at which various related physical phenomena occur, such as (for example) the rate of particle collisions in classical mechanics. These various states and kinetics determine the heat transfer, i.e., the net rate of energy storage or transport. Governing these process from the atomic level (atom or molecule length scale) to macroscale are the laws of thermodynamics, including conservation of energy.”

Makes sense to me. Why have “heat transfer” if heat is already “transfer”

A fire ‘gives off’ heat. It is a direct loss from its ‘heat’ content. By conservation the ‘heat lost’ by the fire can be traced to ‘heat gained’ by the surroundings. All in Joules.

What is gained by specifying that heat only applies to ‘energy that can be thermal that is in transit’? Obviously this thermal energy was lost by the emitter, Joule for Joule and might never by thermal again. Could become chemical potential of gravitational potential so no heat transfer.

• …might never be thermal again. Could become chemical potential or gravitational potential so no heat transfer.

• Crispin in Waterloo says:

Good grief you guys are frustrating to read.

Heat is form of energy. Thermalised energy.
Energy flux is a flow of energy: it could be by chemistry, gas flow, conduction, radiation or anything else you can imagine.
Energy flow is normally used to describe conduction, not radiation. Water does not flow back and forth between two connected tanks, it goes in one direction at a time. Radiated energy is not like that. To the extent it can. it emits all the time. It is not true that everything emits until there is a ground state of 3 degrees C. Some things are unable to emit IR are remain hot. (The universe is a strange place.)

Thus flux is used, not flow when discussion radiation.

Water is a thing, a form of H2O.
Water flow is a flux, not ‘water’.
So heat is not a flow or a flux.

Something can be hot (as pointed out) without heat flowing anywhere. The explanation of why the atmospheric molecules at very high elevation are so hot is they (not CO2 or H2O) gain energy from insolation and cannot radiate it so the temperature goes up and stays up! Particles in space can be very hot unless they collide with something colder. If they can emit in IR at that temp they do, whose which have the ability. Many atoms and particles do not. Their emissivity is essentially zero so they stay hot.

24. Gary M says:

I’ve done my fair share of radiography. A major cause of poor quality radiographs is “back scatter”. This is where objects behind the film absorb and re-emit ionising radiation that has already passed through the object that is being radiographed and the radiographic film. The re-emited radiation travels in all directions, including back to the film and the primary source of radiation. If precautions are not taken to filter out this radiation before it reaches the film it will lead to an overall darkening or fogging of the film and a consequent loss of image quality. This effect, although it involves much shorter wavelength electromagnetic radiation, is analogous to the so-called greenhouse effect.

25. Gabro says:

The USAF used blow torches to cool down the skin of SR-71 Blackbirds returning from high altitude, high speed (high Mach number) recce missions.

• why?

• Actually, to reduce the rate at which the skin cooled to ambient. Cool almost any very hot metal too rapidly, and you get fracturing, from microscopic ones up to spectacular explosions. (So saying that they were cooling the metal is technically accurate; its temperature was going down while they were playing the torch over it – but they were cooling it more slowly than the naked air would.)

• Curious George says:

26. angech says:

With your steel greenhouse I take the outer shell down to a 1 molecule thick layer and place it adjacent to the outer layer, no gap but as molecules do not touch there is still radiation.
The outer layer still emits to space the heat of the surface.
I do not “see” the surface as being twice as hot to send back the same amount of energy as it sends out but we know it does send back the same amount.
Hence double that amount must be coming through for it to emit that amount and send it back but the temperature we measure for that energy is purely that of the outgoing energy.
Something to consider for the shell argument?
After all there is a lot of internal radiation, can we call it down welling radiation, in a conductive blackbody.
Should it be called conduction when a lot of it is radiative, just not over a very long distance and not visible.

• Anne Ominous says:

This is simply the “net” argument all over again.

An object at a particular uniform temperature will radiate from all surfaces.

And yet: if you have (in isolation, in total dark vacuum) an inner energy source constantly radiating energy X, warming up a passive shell surrounding it, the outside of that shell will also radiate a TOTAL of X.

Always. Without fail. The universe could not exist otherwise.

• Tony says:

Exactly. And, since the shell will always have a larger surface area than the inner energy source object, the shell must always be at a lower temperature than the inner energy source object. Same total energy radiated from larger surface area = lower flux (and also, all else being equal, lower temperature). So, a 235 W/m2 inner sphere resulting in a 470 W/m2 outer shell? Not going to happen.

• Black holes are not real?

• Tony says:

Dolphins are not real?

• angech says:

Thanks Anne, but as I pointed out, there is a paradox.

• Tony says:

A paradox that is resolved once you understand that the surface of the outer shell cannot possibly come to double the temperature of the sphere in the first place, for the reasons explained. The steel greenhouse analogy is flawed in this regard from the very outset.

• Tony says:

(since the sphere is the only energy source, the only way it could warm to a temperature producing more than its maximum radiative output of 235 W/m2, is if the outer shell was warmer; yet the outer shell must always be at a lower temperature than the inner energy source object due to its greater surface area)

27. Good luck. It seems that there are a lot of people who do not understand the argument. Temperature (of a region) is proportional to the average kinetic energy of the molecules (in that region). But there are always molecules with below average kinetic energy in the same region with molecules of average and above average kinetic energy.

28. I’ve added a note to the head post with a link to my post on the Steel Greenhouse. It goes through these questions from a different perspective.

w.

• Sparky says:

You want to try nested steel greenhouses.

YES!! Thought experiments (I am presuming no one here has actually built a steel greenhouse but if you have lets see the pictures please) should always be tested via logical extension. Nested steel greenhouse is the next step. What results do we get? What conclusions can we draw?

• gbaikie says:

“Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand metres above the surface, as shown in a cutaway view in the picture above”

Let’s change it. Imagine if earth had this “thin black steel shell, located a few thousand metres above the surface,”
Now there are few ways to do this and it does not matter *much* which way it’s done.
I going to pick having magically strong thin black steel shell which withstand more than 1/2 atm
of outward pressure and shell is going to be at 1/2 atm pressure- 5.5 km elevation.
So other chopping off a few mountain tops, I can keep earth, earth. Though missing half it’s atmosphere mass, though has the same surface air pressure of 1 atm.

Of course Earth isn’t warmed much from it’s molten core, and has to get it’s warmth from the steel shell.
So how warm is steel shell.
Well if it was was an ideal thermally conductive blackbody it “should” be about 5 C and radiate
1367 / 4 = 341 75.
But ideal thermally conductive blackbody is magic- and you are using steel which only magic added is it’s strong enough to withstand a huge amount of force. Such magic might gotten from nanotechnology which if build something molecule by molecule you could theory make a material stronger- though we have get into to how thin is thin and unless thin is meter thick, mere nanotechnology might not enough magi- but details we will ignore.
But steel or copper, silver or even diamond doesn’t give enough thermally conductivity- one could invoke some kind magical plasma based system, perhaps. But let’s look at steel as it’s the namesake.
So steel with a good blackbody coating is going to have lower average temperature than the uniform temperature of an ideal thermal conductivity blackbody.
Or in sunlight with sun at zenith, an ideal blackbody will be about 5 C. With steel it’s going to be much hotter than this and radiate into space much more energy as compared to the ideal.

At Earth surface and where sunlight would be at zenith, the steel sky will be quite warm, and elsewhere steel sky would be cooler.
How big is the hot spot, how big is the hottest spot of hot spot,
Roughly hot spot is same as solar peak hours- 3 hours before and after noon.
An hour is 15 degree longitude. So 45 degree both east and west of the point of zenith.
Each degree is about 111 km. 111 times 45 is 4995 km. So roughly a radius of 5000 km.

So if sun over equator 45 degree north, south, east and west. So if in UK, you don’t see the hotspot. You have wait for summer to see it- and then you will see it for couple hours a day.
Now at the equator you see it for 6 hours of 24 hour day.
How does the hot spot effect someone standing a equator at noon. More than the entire sky is
hot. How hot? Well roughly the hot as it could be is about 120 C,
It takes some time for the outer surface at 120 C to heat the inner surface- and the thinner the steel the less time. Let’s assume it thin enough to do this fairly quickly.
Next, there is 1/2 atm of pressure on the inner side- how cold it would be would effect how much heat is transferred to the air. The air not going to heat well because it’s like hot air against the ceiling of room in a house.
For rough idea with current earth, if air temperature was 20 C, it’s 6.5 C cooler per 1000 meter.
So 5.5 times 6.5 C is 37.75.
So air temperature was 20 C at surface it’s 17.5 C at shell. Now can’t have air at say 100 C and a foot away have air at -17.5. Or 100 C will make gradient of heat, maybe 20 C per meter.
So like the steel it takes some time to form this gradient.
So skip some these details what the effect if entire sky is 120 where closest is 5.5 km from you?

Well, it would seem if you were only 3 km from shell the radiant heat would more than if 5.5 km from it, though air is 3 time 6.5 C cooler than 20 C if surface air was 20 C= .5 C
So could like being in winter and stand next to a fire.
Or if at surface or at 3 km elevation there would direction the heat is coming from, and if at 3 km more heat is coming directly over head but unlike being at surface more heat coming all side save the below you.
Anyway rough fairly warm- but unless the air is warm, not particularly uncomfortable- or living constant darkness would be more of problem.
And if air was 20 C, the radiant heat would warm surface and warmed surface would warm the air.
And it seems the highest air temperature would not be at surface but at the ceiling.

• gbaikie says:

Well I cut it short. And to remain brief, I think if lived in such steel greenhouse, you want to live on the ceiling to remain warmer.

• Steel is a thermal conductor. Air is a thermal insulator.

• gbaikie says:

Steel is a thermal conductor. Air is a thermal insulator.

Yeah, but steel is also insulator.
Or steel has conductivity of 54 W/m K
Copper is 386 W/m K
Air is 0.024 W/m K
Water is 0.58 W/m K
diamond is 1000 W/m K

I guess to quantify, look at formula:
https://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.html
And just using calculator there
Per square meter which is 1 meter thick
temperature difference of 120 C to 0 C
Copper can conduct:46320 watts
Steel: 6480 watts
Air: 2.88 watts
Water: 69.6 watts
Diamond: 120000
But air which 1 mm thick: 2880 watts
1 cm: 288 watts:
So steel is 120 C it warm air near, and develop heat gradient,
So if steel is 1 cm thick [very thin]
And top is 120 and bottom is 119 C
It gives 5400 watts
2 cm is :2700 watts
What about 120 to 119.5 how much watts go thru 1 cm:
That’s also 2700 watt.
So have 2 cm steel and sunlight of 1367 watts can heat surface to 120 C
and can heat 1/2 way down to 119.5 [or more]
And 119.5 to 119 C is again 2700 watts
Where is it 119.9 C in the 2 cm of of steel?
120 to 119.9 at 1 cm depth is 540 watts
And 1/2 cm should double, and it is: 1080
and 3.9 mm it’s 1385 watts
Somewhere around .1 K per 3.9 mm
Or 10 mm is 119.7 and 20 mm is .5 C
So 20 mm or 2 cm it’s 119.5 if surface it 120 C and radiating
close to 1367 watts, but some of that 1367 watts or
119.5 C is 392.65 K is 1348 watts. some of the 1348 watts
is lost to warming the air. And the warmer air gets, the less
heat is lost. Or if air within couple mm is close to 119.5 C
it won’t conduct heat to it- but doubt it would get to that point.
But warmer the air was say 1 meter away from surface, it seems
more likely it could happen.
Of course if there convection due gravity, hotter and lighter gases
would rise. But lesson of fire safety is if in a fire, crawl out rather than
walk out, because you can get very hot air closer to ceiling.
So if there isn’t any wind [and there could be] one can trap a small amount of
hot air near the ceiling.
This assuming, less an inch of the magical steel is strong enough.

29. Fig 2 is complete nonsense. The surface does not absorb twice as much energy from the atmosphere as it does from the sun.

“If I can see you, you can see me, so there are no one-way energy flows. Which means that if I am absorbing radiation from you, then you are absorbing radiation from me.” More nonsense. What you are seeing is light reflected from the other person. Switch off the light and you can’t see the other person, even if he is radiating infra-red.

• Toneb says:

“More nonsense. What you are seeing is light reflected from the other person. Switch off the light and you can’t see the other person, even if he is radiating infra-red.”

So you are saying that (say) a pyrometer in a room at ambient 20C cannot see an object at the bottom of a chest freezer at -20C?
Or that when pointed at a clear sky it cannot see/measure it’s temp?
Roy Spencer shows you can here …

• No I didn’t say that. I was talking about people seeing each other – ie just the visible spectrum, which is what people see as a reflection off other people, not an emission from them.

• Phoenix44 says:

That is correct. Take a totally dark room. Shine a light on you. I can see you, you cannot see me.

• Ed Bo says:

Oh, for chrissakes, Phiip!

He was just expressing colloquially the FACT that the path that electromagnetic radiation takes from Point A to Point B is the same as it takes from Point B to Point A. That (technically true wavelength by wavelength), combined with the fact that absorptivity and emissivity are the same for each wavelength, validates his argument.

30. Rick C PE says:

Willis is correct. For those who think that the temperature of a body receiving radiant energy does not affect the temperature of the hotter radiating body, here is a simple experiment.

Connect an electric resistance heating element to a constant power output power source. Attach a temperature sensor (e.g. a thermocouple) to the heating element and measure its temperature in a large open area until it comes to equilibrium. Say a 1000 Watt element heats to 500 C in this situation.

Now suspend the heating element in a small sealed ceramic box under vacuum (so the only mode of heat transfer is radiation) and monitor the temperature of the box surfaces and the heating element from the time the element is energized. You will see that the temperature of both the box and the element increase until the heat loss from the box equals the heat input to the element (i.e. 1000 watts). If the box starts at 20 C and the element at 500 C and the box inside surfaces end up at 300 C then the heating element must increase in temperature to 549.43 C in order to maintain the steady 1000 watt output. And the outside of the box must transfer 1000 watts to its surroundings although this heat flux will be by a combination of radiation, convection and conduction.

Initial condition SB Temperature term = (500+273.1)^4 – (20+273.1)^4=3.49 E11
Equilibrium SB Temperature term = (549.427+273.1)^4 – (300+273.1)^4=3.49 E11

This experiment is relatively easy to do if you have access to a laboratory vacuum furnace. The results would be only slightly different in a furnace with air at ambient pressure.

• In my opinion Willis has used the wrong relationship to answer the question he has posed. He should have used Planck’s Law, not the Stefan-Bolzman’s law.

The reason is that he is considering an object A emitting at wavelength X and
B an object emitting at wavelength Y
where Y is longer (cooler) than X

Has object B caused object A to emit at a wavelength shorter (warmer) than the wavelength A would emit at if B did not exist?
Or alternatively, if object B were to cease to exist, would object A emit at wavelength longer than X?

Willis need three assumptions, no heat transfer
by conduction
by convection or
by reflection.

The Stefan-Bolzman Law cannot answer this question because it is derived from Planck’s Law by integrating across all wavelengths.

This is explained on another skeptical blog, The Science of Doom,
here https://tinyurl.com/y8d7gor8
here https://tinyurl.com/y77qnqpl

The experiment proposed by Rick C PE describes a closed system. My understanding is that Willis is not imposing that condition.

31. Well done. I see the usual arguments above, all stemming from the completely incorrect “greenhouse” tag on the effect of atmosphere (any atmosphere). It should properly be called the “shield” effect, allowing a more rational discussion.

For instance – it is obvious that making your shield more effective – i.e., adding a gas that makes it reflect (or absorb and emit) more heat towards the surface – will make the surface warmer than it would be otherwise.

No rational “skeptic” denies that adding more shielding CO2 does increase the surface temperature. Less energy escapes directly to the far colder surroundings with that additional CO2 in the atmosphere.

The inconvenient truth, however (for alarmists) is that the net effect is very small. Two facts come into play – the effect of additional CO2 is logarithmic. Generously, humans have added approximately 150 ppm of CO2 to the atmosphere, bringing its concentration to around 400 ppm. In order to add as much CO2 shielding effect again as we already have, we would have to burn more than twice as much fuel as we have already in our entire history since the start of the industrial revolution. That actually is not an easy thing to accomplish, even if we try very hard. Which, looking at the recent history from even the rather dodgy measurement systems, we aren’t.

So the alarmists rely on the slight increase in temperature from the CO2 shield effect to vaporize a far more effective shield gas – dihydrogen monoxide (water). This is touted as the “fatal multiplier” that will roast our world.

But the alarmists just can’t win. Mother Nature (or God, or quantum physics; whatever fountain you drink from, it’s the same flavor) has arranged things so that the “fatal multiplier” is actually a “saving divisor.” The slightly higher temperature at the surface does evaporate more water into the atmosphere. Fortunately (for humanity in general, not the alarmist’s grant prospects, or long term investors in “green” companies), this temperature also creates stronger upwelling as the hot air rises, carrying the water vapor along with it. Until it reaches the upper atmosphere – where the water vapor condenses, releasing energy where there is virtually no shield between it and the far cold reaches. This natural brake on the temperature means that, at the most, the Earth can warm an average of 14 degrees (Fahrenheit – about 7 degrees Celsius). Not that we can manage to add enough CO2 to get that high – you need 7,000 ppm for that – burning forty-five times as much fuel as we have already. Their best hope is for massive vulcanism, although that just might tip us over into an ice age from aerosols before the additional CO2 can get to work.

(The honest climate researcher, of course, knows that even the above is highly simplified – but covers the most significant drivers of climate. The “fiddly bits,” such as changes in albedo, lower in the less icy northern latitudes – somewhat offset by more water surface; outgassing of CO2 from warmer water; sinking of CO2 both geologically and biologically; et cetera, et cetera, et cetera.)

• Tony says:

“No rational “skeptic” denies…”

No true Scotsman denies…

• ‘Tis a bit of a difference between denying who is the rightful King, and denying a physical phenomenon. As all too many Scots with too much Scotch in them have discovered. You might escape the usurper’s men for your intemperate words, but not the ground when you trip on your feet instead of your tongue.

• Tony says:

No, yours was a perfect example of the no true Scotsman fallacy.

• Looking at your other posts, it should not surprise that you have no better idea of the “no true Scotsman fallacy” than of physics.

Not having the patience of a Job – or even a Willis – last reply here.

• I would point you to the last few words of that definition (which is a good one, by the way). The objective facts of physics do not change – not for alarmists – not for skeptics – not for Scotsmen.

• Tony, you embarrass yourself.

More clearly-
“The No True Scotsman (NTS) fallacy is a logical fallacy that occurs when a debater defines a group such that every groupmember possess some quality. ”

Writing observer did not state or imply that all skeptics are rational, so in order to be a “true” skeptic one must have that quality. By including the qualifier “rational” before the word “skeptic”, he implies that some skeptics are NOT rational.

• Tony says:

The group was skeptics, the exclusion is to dismiss those that disagree with the GHE as irrational. Not difficult to understand.

• I Came I Saw I Left says:

Even disregarding the effect of H2O vapor upwelling, it seems that there is a leap of logic in alarmists’ reasoning. Are they ignoring the logarithmic characteristic of CO2 (which I assume is due to saturation effect?), or (if not ignoring that) are they assuming (or do they possibly have evidence for) that the increase in H2O vapor due to (CO2 caused) warming would not follow the CO2 increase linearly?

• The narrative is that more CO2 causes more H2O, which causes more H2O, which causes even more H2O.

If the pesky molecule would only stay put on or near the surface, they’d be right.

• tty says:

“The narrative is that more CO2 causes more H2O, which causes more H2O, which causes even more H2O.”

Which will cause more convection of wetter air flattening the lapse rate and causing more condensation removing more heat from the surface. It isn’t even clear that the H2O feedback is positive.

• Dave Fair says:

It is, in the minds of climate modelers, tty. Observations do confound them, though.

• co2islife says:

“No rational “skeptic” denies that adding more shielding CO2 does increase the surface temperature. Less energy escapes directly to the far colder surroundings with that additional CO2 in the atmosphere.”

I deny that, CO2 slows the cooling, the incoming visible radiation provides the warming. Never will you see the GHG affect warm the earth at night…never (ignoring conduction and convection). CO2 is a blanket, not a heating source. The incoming radiation warms the earth, CO2 slows its cooling.

• I Came I Saw I Left says:

I’m leaning your way. It slows the cooling.

• co2islife says:

Yep, unless it is from a chemical reaction where energy is changed in form, you won’t heat a body above the temperature of the radiating body.

• I Came I Saw I Left says:

For some reason I like the word retards rather than slows. Here’s a question: Does radiant heat act like electric current? IN other words, for there to be radiant energy transfer (heat) does there have to be a difference in potential of some form? Kind of like current (analogous to heat) doesn’t flow until there is a voltage differential. Might sound like a dumb question to some people, but yano I’m just a student of the masters.

• co2islife says:

Great point, the atmosphere is like an electrical resistor. It impedes the path to outer space.

• Ed Bo says:

ICISIL:

My heat transfer texts, in their chapters on radiative heat transfer, are full of “electric circuit” diagrams with resistors for what they call “network analysis” of these transfers.

Remember that a “transparent” atmosphere (no GHGs) provides no resistance to radiative energy transfer.

“Slows” is a misleading term; “retards” may a little better. I prefer “reduces”, myself.

• Excuse me? You just restated exactly what I said.

Less energy escapes directly to the far colder surroundings… Emphasis added!

No, CO2 does not heat the surface (nor does any other component of the atmosphere). But what both Willis and I are trying to get through heads is that the atmosphere affects the net heat flow, reducing the net flow of heat from the surface (whether day or night) to space.

Ignoring, of course, the effect of convection, which transports surface heat (in the form of latent heat in water vapor) to an altitude where there is far less atmosphere – thus increasing the net outflow of heat from the entire system.

• It’s called a Thermal Insulator. From a grade school website:

“Materials that are good conductors of thermal energy are called thermal conductors. Metals are very good thermal conductors. Materials that are poor conductors of thermal energy are called thermal insulators. Gases such as air and materials such as plastic and wood are thermal insulators.”

“One way to retain your own thermal energy on a cold day is to wear clothes that trap air. That’s because air, like other gases, is a poor conductor of thermal energy. The particles of gases are relatively far apart, so they don’t bump into each other or into other things as often as the more closely spaced particles of liquids or solids. Therefore, particles of gases have fewer opportunities to transfer thermal energy. Materials that are poor thermal conductors are called thermal insulators.”

https://www.ck12.org/book/CK-12-Physical-Science-Concepts/section/5.17/

• I’m sorry Writing Observer-

You state first-
“No rational “skeptic” denies that adding more shielding CO2 does increase the surface temperature.”

Then state- “No, CO2 does not heat the surface (nor does any other component of the atmosphere).”

There is a difference between something that “heats” something else-causes it’s temperature to increase-and something that slows down the rate at which something COOLS.

“But what both Willis and I are trying to get through heads is that the atmosphere affects the net heat flow, reducing the net flow of heat from the surface (whether day or night) to space.”

Affect it YES….reduces the net flow of heat from the surface to space? Nope…merely slows the rate of flow. Why? Because the atmosphere ALSO reduces the flow of heat from the SUN to the surface. It slows down the rate at which the Sun can warm the surface in a 12 hour period, and then slows down the rate at which that warmth returns to space, but it does NOT reduce the NET flow of heat from the surface or from the SUN. It does not “cool” the Sun anymore than it “warms” the Earth. It’s a SUCKY conductor of thermal energy. It’s an insulator.

If I pour hot coffee (100 F) into a plastic thermos (that contains an air pocket inside it’s walls for insulation) that is “cold” (compared to the coffee) but is at room temperature (let’s say 70 F), and then put that thermos outside in a snow bank in “colder” 10 F weather, does the “cold” previously room temperature thermos have the ability to WARM the already HOT coffee by “hiding it” from the colder snow and outdoor temps or does it simply slow down the rate at which the coffee INSIDE THE CONTAINER cools???

If I poured the original 100 F coffee onto a saucer made of metal, and placed it in the same spot outside in 10 F, the rate at which the coffee cools would be MUCH faster. Because metal is a great conductor of thermal energy. Far better than the air pocket inside the plastic thermos was. But the fact remains that the coffee began to COOL the instant the heat source (stove, coffee pot) was removed.

The net heat transfer in both scenarios is exactly the same…one just took a lot longer than the other. The net transfer of \$25 dollars to “you” from “me” in Willis’s scenario would have been the EXACTLY the same whether “me” paid “you” one penny a day for 2,500 days or handed me \$1.00 for 25 days, or wrote me a freaking check for \$25 and handed to me in 20 seconds!!!

Words matter. Scientific principles matter. Spontaneous heat flow will ALWAYS be from a warmer object to a colder object. Net heat transfer does not END between two objects of different temperatures until both are in equilibrium and reach the same temperature.

• co2islife says:

Yes, the whole question is does a change of 100 ppm CO2, 0.0001% of the atmosphere, alter the Net heat flow, that is the whole question. Using the thermos example, you fill the Thermos with Hot Coffee, and then let it cool over night, and they you add the same amount of energy over the next day to warm the Coffee again, and then let it cool. Over time, does that does adding the same amount of energy to a thermos throughout the day result in gradual warming because the energy leakage is less than the energy added each day. The marginal difference is what would result in the warming.

• Ed Bo says:

Aphan, you say: “Because the atmosphere ALSO reduces the flow of heat from the SUN to the surface.”

No! The whole point of the “greenhouse” metaphor (and it is just a metaphor) is that the atmosphere is far more transparent to solar (visible and shortwave IR) radiation than it is to terrestrial longwave IR.

• Yes!! The atmosphere DOES absorb incoming heat and radiation. Learn something:

https://earthobservatory.nasa.gov/Features/EnergyBalance/page4.php

“About 29 percent of the solar energy that arrives at the top of the atmosphere is reflected back to space by clouds, atmospheric particles, or bright ground surfaces like sea ice and snow. This energy plays no role in Earth’s climate system. About 23 percent of incoming solar energy is absorbed in the atmosphere by water vapor, dust, and ozone, and 48 percent passes through the atmosphere and is absorbed by the surface. Thus, about 71 percent of the total incoming solar energy is absorbed by the system.”

• AndyG55 says:

“and it is just a metaphor”

No, Its a FALLACY !!

32. Dennis Sandberg says:

Only statement above that has meaning in the real world:
So just to recap these ‘radiation budgets’ are undoubtedly scientifically correct…they are also to all extents and purposes irrelevant or to put it another way…negligible.
http://tropic.ssec.wisc.edu/tropic.php

33. Don K says:

I think that you’re dead right Willis. AFAICS, the only reason that the Second Law of Thermodynamics isn’t written in terms of net flow is that the second law was derived in the nineteenth century from the ideal gas equation pV=nRT using mental models where all heat flow was by convection/conduction. Back flows are (usually) still there with convection and conduction. But they are difficult to observe and measure.so the second law is stated in terms of heat flows without the term “net”. They are, in fact, net flows.

When one deals with energy transfer by radiation, one has to make a few adjustments to one’s mental models including recognizing that flows are net flows, and also including the property of emissivity at the source and destination. If one doesn’t do that, one will find that perpetual motion is not only possible, but easy.

I think many of us might prefer to live in a world where perpetual motion machines can be built and do useful work. But it’s clearly not the world we actually live in.

Suppose the surface temperature of the single powered 100 W plate is T0. If another identical black 100 W plate is brought close to and parallel to the first, this time the surface temperature will rise by a factor of the 4th root of 2 = 1.1892. This time, “back-radiation” will heat up both plates because those photons come from separate power sources. This may be seen when the two plates are brought together, forming a single plate with a 200 W internal power supply, so that 100 W will be emitted from each of the two remaining surfaces to the surroundings. This is double the 50 W from each of the two surfaces of the original 100 W plate, requiring a higher surface temperature for emission by the Stefan-Boltzmann Law.

• Patrick MJD says:

Seriously good post (From a mobile?) however, could do with some formatting, just a bit too busy for my eyes in one big chunk as it is.

35. Richard111 says:

Here is an observation that demonstrates ‘delayed’ cooling. Wait for a warm, calm, sunny day and note the temperature. As the sun sets below the horizon the temperature starts to fall. While the sky remains clear overhead the temperature continues to fall. Then a cloud bank arrives overhead. The temperature stops falling! In fact it will start to rise. I have noted a temperature increase of over 5C under these conditions. Remember, it is pitch dark, the sun is on the other side of the world. So what is causing the warming? It is not the wind which I said was calm, it is in fact back radiation from the cloud base. This radiation from the cloud, impacting the surface, is reducing the cooling rate of the surface. Subsurface heat from the previous day is still rising. As the surface temperature rises the net radiation increases, the back radiation from the cloud increases.
The heat capacity of the ground is considerable. To really experience this go live in a desert region for a while. Here sunlight, day after day, really does warm up the ground yet just before dawn the local temperature will be close to freezing. The outer surface of the rock is trying to shrink. Often a piece of rock spats off with a loud crack. Find that rock and feel the new exposed surface. It will be quite warm!
Eventually there will be a lot of sand. Dig into that sand on a cold dawn and you will fell a lot of warmth. Dig a little trench, lie down, and enjoy some wonderful star gazing. Mind the scorpions!

There are clearly 2 schools of thought.

1. Willis is right and ALL e-m radiation that “hits” an object is absorbed and thermalized. Temperature changes are thus based on NET radiation (in versus out).

2. Willis is not even wrong. ONLY e-m radiation from a hotter object can be thermalized in a cooler object.

This needs to be resolved before one moves on to any discussion about CAGW, atmospheric physics, etc. You need to get the FUNDAMENTAL workings of radiation and heat transfer correct FIRST. Otherwise you will be building everything else on a false foundation.

For those so inclined I invite you to invent an experiment to test which explanation is correct, or perhaps find the results of an experiment which has already been done. The experiment must clearly differentiate between the 2 fundamentally different explanations of course.

This is a useful excercise generally as you will often find arguments in science where there are 2 competing views. They are not always resolvable via reference to literature as each camp obviously has well developed arguments for its own case. In all cases it pays to carefully study BOTH sides even if you are pretty sure yourself what you believe. I recommend it in this case.

Good luck with it.

I would be interested to hear of any experiments regarding e-m radiation being thermalized always/sometimes.

• mkelly says:

If em was thermalized all the time the hottest spot in America would be the base of the antenna of a 50000 W radio station. But it isn’t.

• mkelly,

If you live just under that station, you can light a fluorescent tube by only adding an antenna at one side and a ground cable at the other side…

• mkelly says:

That may be true but it does not make it hot. We used to do that with radars in the Navy. The old APS20 would if given a chance heat soup but it would not hezt air.

• The Reverend,

Many thousands of CO2 lasers prove daily that point 2 is wrong. These emit IR at between 9.600 en 10.640 micrometer. That is comparable to the peak radiation of a black body at around a human’s body temperature, not really “hot”.

According to point 2, that could warm up steel to maximum the same temperature, not further up.
In the real world CO2 lasers heat up steel and melt it at 1200ºC…

Which proves that Willis is right…

“The radiation from the cooler source doesn’t have the high frequency light-wave energy components which would be required to fill up the higher-energy microstates that the warmer source already has filled up, let alone the higher frequency states beyond to result in an increase in temperature. You can see it a little bit in the Planck curve plot above, that the higher-temperature object has a microstate population that is “activated” at higher frequencies, i.e. shorter/smaller wavelengths, than the lower-temperature object has. (The warmer one also has a larger population in each frequency microstate as compared to the cooler one.)

The integration over all the active microstates of a system determines its macrostate, and the macrostate is the thing you actually measure to take a temperature. If you activate higher-frequency microstates, then you shift the macrostate to a higher temperature. But you can only activate the higher-frequency microstates with the frequencies required to activate them, which are obviously their frequencies.
And so because the radiation from a cooler source lacks the higher-frequency microstates that the warmer source of radiation already has, then the cooler source can not have the effect of raising the temperature of the warmer source. And obviously the same would be true of two systems with identical temperatures – they could have no effect on each other’s microstate population, hence can not heat each other.”

All the math is there too, and a link to the steel greenhouse debunking.

• Aphan:

you can only activate the higher-frequency microstates with the frequencies required to activate them, which are obviously their frequencies.

A CO2 laser only sends IR at a frequency of around 10 μm. Steel at 1000 K has its peak frequency at around 5 μm, thus the CO2 laser beam can only activate some 30% of the iron atoms in the steel, if that story was right.

That means that only individual atoms with a “temperature” – or energetic level – between 0 K and about 300 K (the corresponding frequency level of the beam) can be heated by the laser beam. Or an average increase to about 1050 K for steel as a whole. Not enough to melt it. In reality steel is melted by a laser beam of only 10 μm waves up to 1500 K.

Something wrong with the theory?

• A C Osborn says:

Ferdinand, What is the Power of the Laser, not the Frequency?

Ordinary Cold Water can be used to make a hole through steel, how do you think that works?
It works by having enough Pressure, is not Power the same as Pressure?

• A C Osborn,

A water “beam” can dig a hole in steel due to kinetic energy. That is mechanical energy, as good as drilling or using a hammer and chisel.

Power is not kinetic energy, it is electrical energy which in a laser is transformed into electromagnetic energy, which if absorbed by an object is transformed into vibrational energy of the atoms/molecules. Vibrational energy is temperature…

37. Mark - Helsinki says:

“BUT it can leave the hot object warmer than it would be if the cold object weren’t there. ”

No need for a large article

The cold object provides a temperature differential potential to allow heat to move from the hot object to the cold object, no colder object, no loss of heat from the warmer object to the colder object.

Done

• Mark,

Well said – at last a sign of sanity in this crazy debate. Life is so simple unless one has a compulsive desire to over-complicate it. It seems that almost everyone else here is bent on grand-standing and mindless prevarication.

Simple standard physics says…

1. The surface of an object X at a temperature Tx will assert a radiative potential Px where Px is proportional to the 4th power of Tx (S-B equation).

2. The surface of an object Y at a temperature Ty will assert a radiative potential Py where Py is proportional to the fourth power of Ty (S-B equation).

3. If the surfaces of the two bodies X and Y are facing one another, and if Tx > Ty, then the rate of transfer of radiative energy between them is simply Px – Py; and the direction of energy flow is from the warmer body X towards the cooler body Y.

In the whole of physics is there anything easier to grasp than that? It applies under all situations everywhere in the universe. In particular it applies in the case of the earth’s warmer surface facing its cooler atmosphere.

In Willis’s energy budget diagram, Px = 392W/m^2 and Py = 321W/m^2. One conceptual trick that I use is never to think of these two numbers as independent energy flows, but as calculated potential flows. After all, they clearly do not exist in isolation. On the contrary they are completely bound up together by the geometry of the two facing surfaces. If one thinks of them as independent, one can easily get into the silly position that some people have done of thinking that the 321W/m^2 of ‘back radiation’ is somehow violating the 2nd law because it is larger than the incoming Sun’s radiation of 169W/m^2 absorbed at the surface. In reality it is always only the net radiation (in this case 392 – 321 = 71W/m^2) that counts.

It is also vital to appreciate that all of the above is true irrespective of any non-radiative energy transfers that may also be occurring at the same time. In Willis’s energy budget diagram, the non-radiative energy transfers from surface to atmosphere are 22W/m^2 (sensible heat) and 76W/m^2 (latent heat). Adding these on to the net radiative energy flow of 71W/m^2 makes a total of 22+76+71 = 169W/m^2. This exactly balances the amount of Solar radiation absorbed by the surface, as must be the case for steady-state temperatures. So the non-radiative and radiative energy flows coexist happily.

Bingo!

• Dave Fair says:

David, a simplified discussion and a question:

1) Addition of greenhouse gasses (GHG) to the system would increase back radiation, thereby reducing the net 71W/m^2 surface emissions.

2) The surface must heat in order to restore the 71W/m^2 necessary to balance input/output.

3) The net contribution of man’s production of CO2 (net of sinks) would necessarily heat the surface in order to get H2O feedbacks, positive, negative or neutral.

4) There are massive and constantly changing energy flows within the earth’s climate system. Much of this is the result of chaotic weather systems.

Is the radiative contribution of man’s net CO2 lost in the noise?

• Dave Fair says:

David, thank you for simply stating what should be obvious to an educated person.

It should be noted that fluxes in Trenberth’s diagram are based on averages and are presented as in a steady state. But our water world is turbulent and chaotic, with all of Trenberth’s noted fluxes changing on all time scales, except possibly the isolation of the growth of CO2 equivalent fluxes for analytical purposes.

I’m not a researcher so I keep asking simple questions: In isolation, how much of the indicated back-radiation is the result of man’s contributions to CO2 equivalent fluxes? What is the resulting theoretical warming from such contributions, in isolation. Is this theoretical warming even discernible in our turbulent and chaotic water world?

• Again, it’s had a name and definition for decades. It’s called a Thermal Insulator. Thermal insulators are the opposite of thermal conductors. Air is a lousy conductor of thermal energy.

38. “The temperature and the radiation are related to each other by the Stefan-Boltzmann equation” Only for black bodies and by extension for the mythical gray bodies, too. For objects where absorption / emission is frequency dependent (that is, quite a LOT of them), where the radiation is not fully thermalized (which is the case for the atmosphere), it does not apply. Ex falso, quodlibet.

Pseudo sciences crap on the boundary conditions and will apply at ease for example theorems even if the assumptions they start from when proved are plain false. With the principle of explosion, such pseudo sciences can derive anything they intend to.

• wildeco2014 says:

All the ‘extra’ radiation held within an atmosphere as a consequence of the heat capacity of the atmosphere is utilised in potential energy form ( not heat) for the purpose of holding the atmospheric gases off the surface against the force of gravity.
Thus none of that ‘extra’ energy is available for further warming the surface above S-B.
Such further warming of the surface arises not from DWIR but from the return of potential energy to kinetic energy beneath descending convective columns of air.
The cold atmosphere does heat the warmer surface but due to the gas laws that warming is a result not of DWIR but of such conversion of PE to KE

Quite – there is a whole industry of it on both sides of the debate. A great shame.

• Ed Bo says:

Adrian, you say: “For objects where absorption / emission is frequency dependent (that is, quite a LOT of them), where the radiation is not fully thermalized (which is the case for the atmosphere), it [the SB equation] does not apply.”

Absolutely not true! Explanations of radiative heat transfer almost always start with blackbodies, next covering graybodies, because these simplifications make it easier for the beginning students to focus on the important concepts rather than the mathematical minutiae.

If you look at the equation Willis posted for radiative transfer between graybodies, everything inside the big left parentheses is needed for graybodies and not for blackbodies (as he shows).

For “real world” bodies, the emissivity is not constant over wavelength, as it is for idealized blackbodies (e=1.0) and graybodies (e less than 1.0). So you must evaluate each wavelength band invidually. The MODTRAN database does this to moderate resolution; the HITRAN database does this to high resolution. But the physical principle is absolutely the same in all cases.

• That’s a bunch of anti-logical arguments. You either don’t know what Stefan-Boltzmann law is or you only play ignorant. Stefan-Boltzmann is about total energy, not about ‘lines’. And it’s about full thermalization, not about non-equilibrium situations. For your info, energy going ‘in’ for a line can go out through other ‘lines’. Despite radiation not being fully thermalized, quite a bit of CO2 collisions will lead to non radiative transitions. The energy going ‘in’ can go ‘out’ after a sensible time, it does not need to go out instantly and there is no ‘radiative balance’. That renders Stefan-Boltzmann law useless. The energy can go ‘in’ as radiation in a place at one temperature and can go out in a different place with quite a different temperature. Stefan-Boltzmann law requires a single thermodynamic temperature, that is, equilibrium, which is not the case for Earth’s atmosphere.

MODTRAN/HITRAN is a red herring and for your info it has enough error in results (especially where H2O is more involved) to be able alone to give the ‘garbage in’ for models to be exponentially amplified in the exponentially grown pile of shit the models output.

• Ed Bo says:

You say that the SB equation applies “only for blackbodies and by extension for the mythical gray bodies, too.” It’s only slightly more of an extension to “objects where absorption / emission is frequency dependent”.

If you’re trying to be a pedantic nitpicker, you should be correct in your nitpicking. You should be talking about absorptivity and emissivity, which are different things from absorption and emission.

Besides, the graybody approximation is a VERY good approximation for most solid and liquid substances at earth ambient temperatures.

You imply that when there is frequency dependence of a and e, it is because “radiation is not fully thermalized” (although it might be awkward phrasing on your part). Two points. First, a/e frequency dependence is fundamentally independent of full thermalization. They are separate issues.

Second, the atmosphere throughout the full troposphere fully thermalizes absorbed radiation, counter to your assertion. A molecule excited by absorbed radiation is about a million times more likely to collide with another molecule while still excited than it is to re-emit before a collision. (Of course, this is even more true of solid and liquid substances.)

Besides, your arguements are fundamentally irrelevant to the qualitative argument presented.

• “the graybody approximation is a VERY good approximation for most solid and liquid substances at earth ambient temperatures” It is extremely bad and it’s again a red herring, the cliamastrological religion has delusions about the atmosphere, which has a spectrum which is very far from that of a black body.

“the atmosphere throughout the full troposphere fully thermalizes absorbed radiation” As I explained, it doesn’t do that as required for a black/gray body, so your anti-logic is anti-logic and nothing else.

“your arguements are fundamentally irrelevant to the qualitative argument presented” They are fundamentally relevant, because such ‘qualitative’ anti-arguments will treat wildly non-linear systems as being linear, non-equilibrium systems as being at equilibrium, white as black and false as truth.

Ex falso, quodlibet.

• Oh, and since you appear to have absolutely no idea what thermalisation means, start from here to rise yourself above the total ignorance you exhibit: https://en.wikipedia.org/wiki/Thermalisation Maybe you’ll figure out why I stressed that it’s not equilibrium.

39. Mark - Helsinki says:

Transfer of kinetic energy. Temperature is a misnomer. We must talk of it in terms kinetic energy

Temperature can be cumulative in directional kinetic energy application, increasing pressure but it requires an external input of kinetic energy to achieve this, it does not happen in passive kinetic energy exchange.

Passive kinetic energy exchange is decided by the difference in kinetic energy between the two objects,

Leave two blocks next to each other,
*one cooler one hotter,
*The kinetic energy transfer is passive, the warmer block loses kinetic energy to the cooler block (difference potential)

Smash the cooler block into the warmer block
* You are inputting kinetic energy into the exchange that is greater than the kinetic energy of passive transfer * The cooler block uses the kinetic enegy from momentum to warm the warmer block.

All about passive vs non passive transfer of kinetic energy

• Mark - Helsinki says:

The cooler block uses the kinetic enegy from momentum to warm the warmer block *on impact

• Mark - Helsinki says:

Which increases the rate of kinetic energy. KE\Time is what changes and overrides passive transfer

40. 4TimesAYear says:

Ok, this is a subject I must confess to struggling with – but I very much appreciate this:

“These ideas of individual flows, net flows, and being shielded from radiation are important because people keep repeating over and over that a cold atmosphere cannot warm the earth … and they are right. The temperature and the radiation are related to each other by the Stefan-Boltzmann equation. When we apply the S-B equation to the 321 W/m2 of downwelling “back radiation” shown in the graphic above, it tells us that the effective radiating level is somewhere around freezing, much colder than the surface.”

Terms like “backradiation” and “downwelling” sort of add to the confusion by making it sound like one should be able to feel it – when it’s just shielding us from the cold of outer space – which is what we learned in grade school.
At the same time, the atmosphere as a whole functions, not just to shield us from the cold of outer space, but as a heat dispersal system as well.

41. Mark - Helsinki says:

It is impossible physically for a cool object to warm a warmer object unless the cooler object can make use of additional energy from outside of the two object equation.

Otherwise kinetic energy will always be passive and always be dictated by difference potential.

• Exactly!! There has to be an additional source of energy to perform the work required.

42. RAH says:

I grew up in my fathers steel fabrication business. By the time I was 12 I was welding and fabricating and sometimes blacksmithing using a forge. One thing I know is that if one heats the end of a piece of a bar of carbon steel of any solid shape or compound that is a foot or a few feet long until it is cherry red and then quenches the hot end while holding the cooler end, the end that they are holding will become hotter very quickly after they have submerged the heated end in water or oil.

• Patrick MJD says:

I love the way 5% carbon steel can be made into a chisel (Brown) or a spring (Blue) simply based on heat during hardening and colour during tempering. Now, this is a very very old memory.

• RAH says:

You can straighten bowed or twisted “I” beams, even very heavy ones, using only a rosebud tip on an oxy-acetylene torch and some water soaked rags. One just has to know where to apply the heat.

• It’s because steel is an excellent conductor of thermal energy, liquids less so, air even less than that.

43. Mark - Helsinki says:

The truth is, as things stand, it is impossible for us to even remotely know what earth’s energy balance is.

Impossible for us to know where all the kinetic energy goes.

Any claims otherwise are either utterly delusional or dishonest.

44. Mark - Helsinki says:

Logical examination is dead in science, your solution must be logical first and foremost
Too many get lost in the mathematics and lose all logical structure in their solution

See: actual physical singularties, produced from thin air by some of our “greatest minds”. There is obviously no such thing in the physical world

45. Mark - Helsinki says:

Relative mass is bunk, it requires transfer of kinetic energy between objects to manifest the extra kinetic energy from momentum

So an object moving at say half the speed of light has no relative mass, the mass is exactly the same as a static object, the energy that is changed is kinetic energy and that is only manifested when there is a kinetic energy transfer.

The so called extra mass that is in relative mass, does not belong to or come from that object, it is external kinetic energy provided by another source.

• tty says:

I can assure you that even the relatively modest speed of an electron in a CRT (like in an old-fashioned TV) will make its movement through the electric field come out significantly wrong if you don’t take the relativistic mass increase of the electrons into account. I did an experiment on this when reading freshman physics, so I know it from personal experience.

• Mark - Helsinki says:

you obviously dont understand, relative mass is mass+pseudo mass ie mass and momentum

they can be calculated as one, BUT the Electron does NOT have more mass

• Mark - Helsinki says:

As I clearly explained, if there is physical interaction then this kinetic energy comes into play, and in a CRT that is the case.

If the electron does not physically interact then the case is different, maybe read first then reply :P

46. Mark - Helsinki says:

If you work this problem of energy balance in terms of kinetic energy, it becomes clear that we have no way to work out this problem and are left with mathematical fantasies and severe top level averaging that does not represent the problem, or provide a solution.

We’re dealing with ghosts here

47. arfurbryant says:

Can A Cold Object Warm a Hot Object?

Can a cold object make a warm object warmer than it would have been if the cold object wasn’t there?

Only if it insulates the warm object. If you insulate a warm object, the rate of cooling decreases.

What has this got to do with Carbon Dioxide’s contribution the ‘Greenhouse Effect’?

Hardly anything in practice. Carbon Dioxide is not an effective insulator for two reasons:
1. It is a poor insulator by itself.
2. There is not enough of it to make an appreciable difference.

To Willis’s figures:

Figure 1:
This is NOT a good analogy!
According to Willis, doubling the number of ‘Yous’ would give \$150 to the ‘Me’. Hence, by implication, ‘warming’ the ‘Me’.
Here is a counter- question: How many cold objects does it take to warm a warmer object?
Answer – it can’t be done.
In Fig 1, the ‘net’ flow (Heat) is always from ‘Me’ to ‘You’, no matter how many ‘You’s there are.
In the climate debate, doubling the ‘You’ (CO2) will have no measurable effect on the temperature of the ‘Me’.

Why?
Because the spending power of the \$75 will never overcome the spending power of the \$100. ‘Me’ is richer than ‘You’. ‘Me’ can afford to give all the ‘You’s their \$75 back and ‘Me’ would STILL be richer than all the ‘You’s. Equate richer for warmer and you have the true picture.

Why does the Sun’s radiation heat the Earth? Because the energy of the radiation from the Sum is sufficient to increase the thermal energy of the Earth’s receiving molecules.
Can atmospheric CO2 not do that? NO.

Hence Figure 2 is also wrong:
The 321 W/m2 that REACHES the surface is NOT absorbed for thermal gain. The ‘absorbed by the surface’ is irrelevant in a thermal context if the Earth is warmer than the atmosphere.
This is why you can surround a hot object by billions of cold objects and the hot object will never (EVER) get hotter.

Don’t conflate insulation with insolation.
Comparing atmospheric CO2 with a blanket, thermos, whatever, is like saying that a string vest made of 99.96% air and 0.04% cotton is an effective insulator. Even that would be wrong because the atmospheric CO2 doesn’t actually prevent heat loss, so the effectiveness of the cotton is further reduced. CO2 does not trap heat. Backradiation is not heat. Remember the ONLY source of energy in this process is the Sun.

<i.["Anthony Watts: "I get so tired of hearing this stupid argument from the anti-ghg crowd. Thanks for quantifying it."]

Mr Watts, I have a lot for respect for the work you have done in making this website. However, just because you don’t understand an argument doesn’t make that argument stupid. Wilis has not ‘quantified’ your objection.

The science is far from settled!

Kind regards to all.

Arfur

• daved46 says:

“Can a cold object make a warm object warmer than it would have been if the cold object wasn’t there?

Only if it insulates the warm object. If you insulate a warm object, the rate of cooling decreases.”

And you don’t think blocking the warmer object from the coldness of space is insulation?

BTW a real definition of “insulation” might be useful here.

• arfurbryant says:

[“And you don’t think blocking the warmer object from the coldness of space is insulation?”]

A blocking material would act as an insulator but CO2 does not block anything, therefore it cannot either warm the Earth’s surface nor effectively reduce the rate of cooling. CO2 emits radiation as effectively as it absorbs. CO2 may delay the LWIR loss to space but the time delay is tiny compared to the length of time taken to increase the amount of CO2.

As to the definition, I am (obviously) referring thermal insulation, which I would define as a material capable of reducing the rate of cooling of a warm object.

• Arfur: I agree mostly but I think that when an insulator is in equilibrium between the heat gain from the warm object and the heat loss to the cold surroundings it does not any more slow the cooling of the warm object. I guess that is the situation in the atmosphere.

• arfurbryant says:

I sort of agree but the rate of change (build up) of the insulating material is important in this context. It is not as if we have suddenly surrounded the Earth with a thermos-type shell (or a steel shell for that matter). Our situation is a very slow increase in a trace gas which has virtually no insulating properties and which exists in sparse form (each CO2 molecule is surrounded by approx 2500 non-radiative molecules).

Also, this transfer should be considered in a realistic time-frame. It takes a comparatively long time to make the measured increase in atmospheric CO2. The Mauna Loa dataset shows an increase of 90 parts per million in about 70 years. Thats just over ONE part per million each year! Even if CO2 was an effective insulator (and it’s not), how much effect could it have at that rate of increase? That is a whole different debate! :)

48. menicholas says:

Willis, good post, and thank you for attempting to clear this up.
I suspect the effort is in vain, however. Sad as that may be.
The people that refuse to let this penetrate their dense outer bony layers of cranium have heard it before…it bounces off like rain from a duck’s back.

• menicholas says:

Now, if you could extend you laser-like focus to the subject of dowsing…

• menicholas says:

Oops, sorry…I did not see your second request on my first reading.

49. Richard111 says:

Another interesting experiment. A 3mm mild steel plate about half a square metre. Lay this on a slab of polystyrene. Place these outside, a couple of feet above ground. I had a digital thermometer screwed to the plate. The plate is exposed to the sky but insulated from the ground below. Note local air temperature and temperature of plate. Needs a reasonably calm clear morning. As the sky turns blue overhead BEFORE sunrise, note the temperature of the metal plate. It will start to rise sooner than the local air temperature.
That blue sky overhead is warming stuff on the ground. Be interesting to know how blue sky warming effects the oceans, especially as that ‘blue’ colour will have good penetration into water.

• menicholas says:

It may be blue but it is not very bright.
Get a light meter, point it at the blue re-dawn sky.
Then do the same during full daylight an hour later, and again at noon.
BTW, did you do this experiment Richard?
What was the change in temp?

• Richard111 says:

I did the experiment more than once. The mild steel plate weighed 11.5 kg and warmed up about 2 degrees more than the surrounding air. After that the temperatures rose together. I assume the plate was now losing heat to the air. The point being that ‘blue’ light is diffused sunlight and has energy to do some work. If the sun was allowed to shine on the plate then the temperature rocketed.

• tty says:

It doesn’t really matter a lot if a photon goes direct or bounces off a molecule of air before being absorbed, so the result is not very surprising. Also the air isn’t really that much warmed by the sun, but rather by re-radiated and conducted heat from the ground.
Another interesting phenomenon. In still air the ground will normally be coldest slightly after sunrise. This is because it has very high emissivity in the IR spectrum, but reflects visible light fairly well, so it takes a while before the sunlight catches up with the IR emission.
Glass has very high IR emissivity, which is the reason that you may have to scrape ice off it even when the air (and ground) temperature in the morning is slightly above freezing.

• menicholas says:

You may have to put the light meter at the bottom of a long tube.

• Richard: The thermometer on the slab of polystyrene but without the steel plate. How would the temperature behave?

• Richard111 says:

Esa-Matti: Have previously photographed polystyrene at night and notice it is highly reflective of the IR emitted by the camera, thus assume this is part of the insulation property therefore the polystyrene is unlikely to warm.

• Esa-Matti Lilius says:

Richard: I was asking because I think that the slab of polystyrene insulates also the thermometer from the cooler ground. Therefore, my guess is that you would have noticed that thermometer without the steel plate would also show higher temperature than the local air temperature.

50. Facepalm facepalm facepalm. It hurts too much to read the whole of Willis’s self confusion on this subject.

Just take the energy budget diagram near the end. Look at the numbers on the left side versus the right side. Compare to the money example. Is the “net” flow from the earth to the atmosphere or from the atmosphere to earth? Did the Earth end up with more cash or less cash? Did the atmosphere give the earth change from \$100 or did it give more cash back than was given to it?

169 from the sun into the earth where the hell do the right side numbers come from if you are not advocating cold heating hot?

There is NOTHING in physics that says the energy given off from a cooler object must be absorbed by a warmer object. That energy is NEVER added to the warmer object. The energy from an ice cube is not added to your total energy if you stand next to it, pick it up or put it in a Willis pseudoscience post. Your energy output will remain the same. It won’t jump from 169 to 392 like in the embarrassing energy budget diagram.

To raise the average surface temperature to what we observe you need something else. You can raise temperature by transferring heat from hot to cold OR you can introduce Work into your system.

The greenhouse effect does neither, so it is wrong.,

• There is absolutely no difference between “NET flows from hot to cold” to “heat flows from hot to cold” the word “net” is superfluous and simply a tool of confusion. The instant that the energy from the hotter object increases in any thought experiment with a cooler one, the “net” flow is no longer from hot to cold, is it? So you are proposing a cooler object increasing the temperature of a warmer one.

• Toneb says:

“The instant that the energy from the hotter object increases in any thought experiment with a cooler one,”
Still confusion…

The energy of the hotter object is not increased!!
It is just decreasing more SLOWLY.
At no point does the energy/temp of the hotter object go UP.
Just goes down more SLOWLY.

” So you are proposing a cooler object increasing the temperature of a warmer one.”

NO NO and a zillion NOES.

“The greenhouse effect does neither, so it is wrong.,”

Then please explain why the Earth, when considering the energy it absorbs from the Sun is 33C warmer than it should be.
Why when viewed from space it does (255K).
Yet we live at 288K.

• Toneb. The Earth is not 33C warmer than it should be. The calculation to derive 33C is based on modelling a flat Earth with no day and night and with the average energy received from the sun divided by four because the cross-sectional area of the earth is one quarter of the surface area. With those ridiculous assumptions you get the false result of 33C.

• Toneb says:

Phillip:
Sorry but it is. Satellites measure Earth’s temp at 255K

• Tony says:

Phillip Bratby: “Toneb. The Earth is not 33C warmer than it should be”

Toneb: “Sorry but it is. Satellites measure Earth’s temp at 255K”

Tony: Exactly. The Earth is not 33 C warmer than it should be. Thanks Toneb.

• “The energy from the hotter object is not increased. It is just decreasing more slowly”.

NO!!!!!

Look at the NUMBERS on the energy budget diagram. The sun (on that diagram) warms the surface of the earth to a value of 169W/m2. The atmosphere INCREASES that amount to 392W/m2. It has not cooled. It has WARMED.

The values on the left are what the sun provides. The values on the right are trying to explain (through bad science) what is observed.

The mechanism described CANNOT BE! Cold doesn’t INCREASE the temperature of something hotter! You need WORK to do that.

Gravity provides that work, not recycling of energy against the well established direction of heat flow.

• You cannot raise the temperature of an object by “cooling it more slowly”. You can cool to a different end temperature, but that is not the same as cooling to a higher temperature than what you started with. The compete jumbled up word salad that the greenhouse effect acolytes use, is what causes the idiocracy. You can’t heat the earth up to a higher temperature than that which the sun is capable of heating it. Not without additional work.

• Toneb says:

“Tony: Exactly. The Earth is not 33 C warmer than it should be. Thanks Toneb.”

Earth absorbs solar energy such that it radiates at 255K
It does, as seen from space.
On the surface it is 288K (as an average).
33K warmer.
The GHE.

• Tony says:

Let’s try putting my reply in the right place.

Toneb, what on Earth makes you think the calculated 255 K temperature should apply to the Earth’s surface, and not the Earth as a whole (which you agree is observed to be 255 K)?

• Toneb says:

“Toneb, what on Earth makes you think the calculated 255 K temperature should apply to the Earth’s surface, and not the Earth as a whole (which you agree is observed to be 255 K)?”

It does apply to the Earth as a whole (as satellites see the “whole Earth” while they measure an irradiance of 225K.
But actually the max radiation is being emitted at ~8km – the level the GHE has raise the effective radiance level to (on average -18C there).

• Toneb says:

“You cannot raise the temperature of an object by “cooling it more slowly”. You can cool to a different end temperature, but that is not the same as cooling to a higher temperature than what you started with. ”

You are, I think, fixated on the 288K being hotter than the 255K as though the GHE has heated the Earth (provided extra energy). It is an insulation effect, a slowing of emitted LWIR while Solar SW is still incoming (nano/micro seconds worth though it might be). The excess being mostly being stored and later (in the case of oceans – hence the thermal inertia) heating the atmosphere.

All the while the Earth is radiating. Cooling in the absence of the Sun. The GHE doesn’t (instantaneously) heat anything.

Which is hotter the surface or the atmosphere above (most generally)?

There is (nearly – some exceptions in WAA, warm air advection) always more energy being radiated from the Earth’s surface than any GHE can overcome.

It is not “new” energy. It is that that was first emitted from the surface (mainly).
The 33K higher temp has not been “switched on” – it is the result of the GHE slowed energy being stored in the oceans.
It cannot be “switched on” in the sense of switching on 33K.

• Toneb says:

An analogy:

Get into bed under a very high tog duvet.
You are going to feel warmer than a thinner one, yes?
Quite possibly you will get hot and start to sweat, yes?

Is the high tog duvet heating you?
Is it electrically powered?

You are still radiating heat (incoming solar SW) yet the duvet slows your cooling (GHE). You get hotter as there is an imbalance.

Additionally will the temp on the top of the duvet be hotter or colder?
GHE theory observed as stratospheric cooling.

• To make the blanket analogy more apt you need to raise or lower the height and therefore temperature of the blanket as it moves up or down along the lapse rate slope.
Thus the presence of radiation to space from the blanket would weaken the vigour of convection beneath it and lower the blanket to a lower warmer height whereupon the blanket would radiate faster to space INSTEAD of warming the surface beneath.
The same applies to the so called steel shell concept.
The behaviour of gases as opposed to solids is quite different to the behaviour of solids such as steel or blankets.
You have to take account of convection and the lapse rate slope which are determined primarily by uneven conduction from the surface to ALL atmospheric gases.

• gbaikie says:

— Toneb
November 25, 2017 at 2:15 am

Phillip:
Sorry but it is. Satellites measure Earth’s temp at 255K–

Excluding interior of earth, the warmest place on Earth is the surface.
The surface is where the sunlight is absorbed.
The satellites are obviously measuring the surface incorrectly if
they indicate the surface [which is average temperature of about
15 C [288 K] is 255 K.
Of course earth radiates 240 watts per square, and if earth was
a blackbody that would be 255 K at it’s surface where it’s absorbs
the sun’s energy. Earth of course reflects about 30% of sunlight-
anything reflecting 30% of sunlight is not a blackbody.
Blackbodies also radiate the most amount energy at a given temperature.
The earth surface has water evaporating and heat being loss due convection heat transfer
to the atmophere. Or if blackbody were losing heat by evaporation and convection heat
loss, it couldn’t radiate the most amount energy of it’s temperature. Or because
it doesn’t radiate the most at given temperature- it’s not a blackbody.
Or it would have to be in a vacuum for blackbody to be a blackbody.

• You’re playing (er, comparing) Trenberth’s flat earth average cartoon with actual measured global satellite measurements? And just “how” did your source convert spherical measurements to a flat earth average square meter value for absorbed energy, reflected energy, and measured energy at the satellite altitude?

I speculate that the only reason the word “net” got included in this was because they needed it to catch all the little balls (aka photons).

• Tony says:

So Toneb, your answer is that basically no matter what, the GHE is responsible. A temperature of 255 K (effective temperature) is calculated, you at first assume that should apply to the surface, therefore the 33 K difference is due to the GHE. When challenged on that, you switch to saying it’s because the GHE has raised the effective radiating level. No matter what, this non-existent difference has to be explained by a non-existent phenomenon. The conclusion is arrived at first – there is a GHE. Then the circumstances can be adapted to fit this conclusion. So there is no need to accept the premises unless you already accept the conclusion.

• Toneb says:

“So Toneb, your answer is that basically no matter what, the GHE is responsible. ”

For the +33K?
Yes.
Fundamentally because of the non-condensing GHG’s.
Without them than WV would comdense out as snow.
It is what the science tells us.
Do you have any that shows it isn’t?

https://pubs.giss.nasa.gov/docs/2010/2010_Lacis_la09300d.pdf

https://www.giss.nasa.gov/research/briefs/schmidt_05/

“The size of the greenhouse effect is often estimated as being the difference between the actual global surface temperature and the temperature the planet would be without any atmospheric absorption, but with exactly the same planetary albedo, around 33°C. This is more of a “thought experiment” than an observable state, but it is a useful baseline. Another way of quantifying the effect is to look at the difference between the infrared radiation emitted at the surface of the Earth, and the amount that is emitted to space at the top of the atmosphere. In the absence of the greenhouse effect, this would be zero (in other words, no difference). In actuality the surface emits about 150 Watts per square meter (W/m2) more than goes out to space.”

• Toneb says:

“The satellites are obviously measuring the surface incorrectly if
they indicate the surface [which is average temperature of about
15 C [288 K] is 255 K.”

The classic “with one bound he was free” answer.
Sorry but that level of, err, gainsaying, is beyond response – but ……
So they are measuring the temp of Venus “incorrectly” as well?.

Well yes of course!

How can a satellite measure the temp of Earth as being higher than the energy it receives from the Sun can provide?

What goes in comes out …..
However the GHE CAN keep it in via an ‘insulation’ effect.
Hidden from the view of the satellite.
What the planet cannot do is emit MORE than it receives.
Which is what you are saying it does.

Venus has an effective temp of 220K
Yet its GHE adds 510K at the surface to take it to 730K.
It is measured from space at 220 as that is what the SUN warms it to (in>>out).
The extra 510 has built up over millenia because of it’s CO2 atmosphere as it tries to get out via T^4.

This again reveals a fundamental misunderstanding of the nature of the GHE.

https://atmos.washington.edu/2002Q4/211/notes_greenhouse.html

• Tony says:

I’m just pointing out that your logic is circular. Not making any point about why the atmosphere is warmer at the bottom than the top. You originally started out by saying:

“Then please explain why the Earth, when considering the energy it absorbs from the Sun is 33C warmer than it should be.
Why when viewed from space it does (255K).
Yet we live at 288K“

and I’m telling you that a theory that the surface is warmer than the effective temperature because of a GHE is not evidenced by the observation that the surface is warmer than the effective temperature. All that is demonstrated is that the surface is warmer than the effective temperature. But you wrote your comment as if this very observation alone was sufficient to provide evidence that the GHE was the cause.

• Toneb,
The blanket analogy:

That blanket must be more of an afgan, crocheted with open holes in it, rather than a “high tog”. Only about 4% of that blanket can absorb your body heat and “return” it to you. You must also rotate beneath the blanket steadily, and place a blowing air fan under it with you.

Now how warm do you get ? Warm enough to sweat? Does the temperature of your skin even increase?

• Seems strange no one has mentioned the effect of gravity on the atmosphere and the resulting adiabatic lapse rate.

• Toneb says:

“There is NOTHING in physics that says the energy given off from a cooler object must be absorbed by a warmer object. That energy is NEVER added to the warmer object.”

And here we have the “sentient photon” myth.

Tell me, how does a photon emitted from a colder object know that it cannot be absorbed by the hotter object?

“The greenhouse effect does neither, so it is wrong.”

The D-K syndrome rages in some.
You are.

• Toneb says:

Both object absorb/emit LWIR in the same narrow frequency band.
They are not different in that sense.
A thermalisation via atomic vibrational modes occurs.
Else Pyrometers would not work….

“Infrared radiation can have a wavelength of a fraction of a micron up to several hundred microns. Infrared thermometers measure infrared with a wavelength of between 4 and 14 microns. As it is the surface of an object that emits infrared, an infrared thermometer will not measure its internal (core) temperature.”

• Tony says:

You’ve written some words. I’m not sure to what end, or who you were responding to, but OK.

• How does the energy from a block of ice “know” not to make you warmer?

It doesn’t know squat. It just obeys the laws of physics, as does everything in the universe. You can argue over the why of it as much as you like and come up with whatever theory best fits. As long as you make your theory fit the observation and not the other way around.

So cold doesn’t transfer heat to hot. That is the UNIVERSAL observation. Whatever the theories about what photons are do or “know” needs to fit inside this basic fact.

• menicholas says:

Atoms and molecules can only absorb photons with certain discreet energy levels.
The incoming photon must be able to boost an atom or molecule into a higher energy state, otherwise it will just keep going.
Photons are not little billiard balls.

• menicholas says:

According to the theory of relativity, as an object approaches the speed of light, time slows down for that object.
At the speed of light, time would stop.
Photons are going the speed of light.
Does time stop for them?
Are they emitted at the same instant, from the perspective of the photon, as they reach their destination?
Could this explain some of the mysterious behavior we know they exhibit?
If they cannot be absorbed, could it be they are never emitted?

• menicholas says:

Also, photons are not simple particles…they behave like waves as well.
We know they have a wavelength…but do they also have breadth?
Are they getting wider in the direction of travel, perhaps?
We know photons can become entangled, exhibiting action at a distance.
And other stuff no one really understands.
because we do not have the proper way of thinking about them.
Or there would be no mysteries.
So trying to understand them in some intuitive mechanical way is never going to accurately describe their behavior properly.

• tty says:

“It doesn’t need to “know”. Photons from warmer objects have a different frequency and wavelength to photons from cooler objects. Nothing magical about it.”

The average frequency differs (for blackbody radiation) but to determine the temperature of the emitting object from a single photon does require magic according to present-day physics.

I have a photon with a wavelength of, say, 1.697 micron, will you please tell me the temperature of the emitter?

• gbaikie says:

Toneb
November 25, 2017 at 1:28 am

“There is NOTHING in physics that says the energy given off from a cooler object must be absorbed by a warmer object. That energy is NEVER added to the warmer object.”

And here we have the “sentient photon” myth.

Tell me, how does a photon emitted from a colder object know that it cannot be absorbed by the hotter object?

Anything radiating heat is cooling, cold can’t cool to something hot.

• Tony says:

“The average frequency differs (for blackbody radiation) but to determine the temperature of the emitting object from a single photon does require magic according to present-day physics“

Thankfully there is no requirement for the photon to do so.

• AndyG55 says:

“The D-K syndrome rages in some.”

And YOU are a prime example. Your baseless ego boosts you FAR beyond what you are ever capable of.

• menicholas:

Atoms and molecules can only absorb photons with certain discreet energy levels.

That is true for GHGs, not for black bodies: a black body absorbs every photon of every frequency and sends out photons over a wide Gaussian range, with its peak frequency depending of only its temperature. See Wien’s displacement law:
https://en.wikipedia.org/wiki/Wien's_displacement_law

• Tony says:

Toneb, what on Earth makes you think the calculated 255 K temperature should apply to the Earth’s surface, and not the Earth as a whole (which you agree is observed to be 255 K)?

• Hugs says:

So much anger, so much strawman, and so many errors. So little time in life. I think my children win again.

• Tony says:

You’ll get past that anger. It will be OK.

• Brett Keane says:

Yes, you have to wonder what some people are on, to confabulate otherwise.

51. You have all forgotten Maxwell’s Demon, that could easily direct the more excited molecules continually to the warmer object for us. I nominate we call it the Al Gore’s Demon, and there must be myriads of them around doing AGW’s ( Al Gore’s Warming ) work. https://en.wikipedia.org/wiki/Maxwell%27s_demon

52. Bryan says:

Willis says
“Short answer? Of course not, that would violate the Second Law of Thermodynamics … BUT it can leave the hot object warmer than it would be if the cold object weren’t there. Let me explain why this is so.”

If we are talking about TWO objects thermally isolated from the universe the answer is emphatically no.

What you have done is introduced an extra object

“it can do so in the special case when the cold object is hiding an even colder object from view.”
So now we have THREE objects.
Thermal energy in two form of EM radiation will be emitted and absorbed between three objects now instead of two.
The hottest object will also emit higher frequencies than the other two objects so a question of radiation quality is also important.
It is also helpful to consider the matter in terms of insulation
If extra insulation(2nd object) is introduced the hottest object(1st object) will lose heat slower than before and so its own temperature will not drop quite so fast.
So if the 1st object is a person they will feel warmer for longer.
You are right to emphasise that heat flow will only occur spontaneously from a higher to a lower temperature.
If someone claims that the cold object heats the warmer object you can be sure they are not physicists.

53. Ed Zuiderwijk says:

Actually, it would violate the First Law of thermodynamics.

54. Sandy In Limousin says:

Part of the problem for a non-technical person is using temperature scales based on the freezing point of water. Sub-zero (or sub 32′) temperatures are assumed to have no “heat” which is not the case. It is sometimes quite difficult to explain that the Arctic in winter is radiating energy in the form of heat into space. The “temperature of space” is 3’K and the Arctic 250’K does not compute for many people.

I’ve probably displayed my lack of understanding here too!

• menicholas says:

Heck, most people have no understanding that coolness is not a thing.
Only lack of hotness.

• Oh I don’t know. I might think your coolness is added upon by your hotness…and not a lack thereof.
Sorry…I needed the humor. :)

• menicholas says:

+++

• menicholas says:

Well, at least someone recognizes a joke wearing a science fact disguise.

55. Radiative gases within an atmosphere distort the dry adiabatic lapse rate slope to the warm side which reduces the vigour of upward convection.
The radiative gases therefore settle at a lower warmer height than would otherwise have been the case and radiation to space is greater from that lower warmer height than it otherwise would have been.
Therefore, conductive adjustments occur so as to adjust the amount of radiation to space from radiative gases which causes a neutralising of any surface warming potential.
Willis and those who agree with him would be right if radiation were the only means of energy transmission but it is not. Conduction and convection affect transmission of energy and also effect transformation of energy between KE (heat) and PE (not heat) with the net outcome being long term maintenance of hydrostatic equilibrium within planetary atmospheres.
The mass of an atmosphere (whether radiative or not) conducting and convecting provides insulation which allows the surface to rise above S-B.
The kinetic energy required to achieve such surface warming is from the reconversion of PE to KE in descending columns. That process is missing from the Trenberth diagram used by Willis which is why it has to be compensated for in the diagram by a proposed surface warming effect from DWIR.
That diagram shows sensible heat and latent heat via thermals cooling the surface in ascent but shows nothing for the warming effect in descent.
There is a simple well known observation that helps to choose between the radiative and non radiative surface warming proposals.
Observation shows that for planets with atmospheres in hydrostatic equilibrium the temperature within the atmosphere at the same pressure are similar regardless of vast composition differences adjusted only for distance from the sun.
The radiative theory cannot explain that whereas the conductive/convective theory does.

• Philip Mulholland says:

Stephen
Latent heat transport can only ever occur as part of a process of mass movement. I totally agree with you, it is not a radiative process. What I had never noticed in the Trenberth diagram, until you pointed it out just now, is that there is no downward return vector to balance the upward mass transport of air that occurred during precipitation. To balance the movement cycle downward motion must be present and clearly include both the cold rainfall as well as the falling dry air heated by adiabatic compression. And no, these two thermal processes (cold falling rain and warmed falling air) do not cancel out, there is a net heating of the atmosphere by latent heat of condensation as the phenomenon for the Chinook wind proves.

• tty says:

“What I had never noticed in the Trenberth diagram, until you pointed it out just now, is that there is no downward return vector to balance the upward mass transport of air that occurred during precipitation.”

It is always that way, and there are two explanations:

1. It is very difficult to measure.

2. If it was included it would make it obvious that the LWIR part of the process (including “back radiation”) is a quite small part of the whole heat exchange process in the atmosphere. Notice that the “convection arrows” are always shown as quite puny and discontinuous)

• Not puny.
All the PE required to keep the entire mass of the atmosphere off the surface against gravity is involved.
That PE reservoir is the source of additional kinetic energy at the surface providing the observed surface temperature enhancement above S-B.

• Philip,

You say: What I had never noticed in the Trenberth diagram, until you pointed it out just now, is that there is no downward return vector to balance the upward mass transport of air that occurred during precipitation.

That is because you were to aware that the two non-radiative energy transfers from the surface, for sensible heat (22W/m^2) and for latent heat (76W/m^2), are the net flows of energy due to these two phenomena. This is explained in the original paper that accompanied the original Trenberth diagram (I don’t know where Willis got his diagram from but the figures are in more or less the same ballpark as Trenberth’s and are definitely net figures.)

Even Willis accepts that the net effect of convective overturning is zero so you cannot assert that the Trenberth diagram represents a net outturn.
Trenberth just erroneously includes the upward leg alone (leaving out the downward leg) and compensates for that omission by incorrectly asserting a surface warming effect from DWIR.
I have told you why and how convection changes to adjust the degree of radiative loss from GHGs to space so as to avoid destroying atmospheric hydrostatic equilibrium by warming the surface.

• tty says:

“Even Willis accepts that the net effect of convective overturning is zero so you cannot assert that the Trenberth diagram represents a net outturn.”

If Willis accepts that he is dead wrong. The reason is that convection carries molecules up so high that a large part of the latent heat freed by condensation/freezing radiates out into space and leaves the system. The same is true for part of the “dry” radiative cooling of the convected air. The entire mass, air, rain and snow ultimately comes back down, but it has always lost some energy on the way. The proportion lost is not large, but it is considerably larger than the energy lost by LWIR radiation.
The Earths atmosphere is a heat engine, but like all heat engines it is not perfect, it constantly loses heat to the surrounding environment.

• tty

You are mixing up the radiative energy exchange with the non radiative energy exchange.
My point relates only to the adiabatic portion of the energy flow which is non radiative and fully reversible so myself and Willis are correct that convective overturning constitutes a net zero energy flow.
Where I differ from Willis is in pointing out that it was non zero during the first convective overturning cycle when the atmosphere was formed.
It is that non zero component that enables the entire mass of an atmosphere to participate in the warming of the surface above S-B.
Even if GHGs do work as proposed the mere fact that the entire mass of an atmosphere is involved renders GHGs insignificant but as I pointed out above the GHGs distort the lapse rate slope so as to radiate more effectively to space from a lower warmer level instead of warming the surface.

• tty says:

“You are mixing up the radiative energy exchange with the non radiative energy exchange.”

Not me, nature.

• They are jumbled up in Nature but they are still discrete processes and operate as a counterpoint to each other in the way that I described.
Otherwise no long term hydrostatic equilibrium such that planets with any radiative component in their atmospheres (whether gaseous or particulate) could not retain those atmospheres.

• Cassio says:

Stephen Wilde November 25, 2017 at 2:39 am:

Conduction and convection affect transmission of energy and also effect transformation of energy between KE (heat) and PE (not heat) with the net outcome being…

This may be irrelevant to your argument, Stephen, but I feel I should point out for the sake of clarity that your distinction between “heat” and “not heat” appears false to me.

“Heat” is not simply KE. In physics and thermodynamics, the heat content (a.k.a. enthalpy) of a body is defined as the sum of the KE and the PE contained by the particles that make up the body.

However, the body’s absolute temperature is simply a function of the average KE of its constituent particles.

Hence, the molecules of a block of ice at 0°C possess the same average kinetic energy as those in the equivalent amount of water at 0°C, but they contain far less heat and thus, far less potential energy too.

There is a simple well known observation that helps to choose between the radiative and non radiative surface warming proposals.
Observation shows that for planets with atmospheres in hydrostatic equilibrium the temperature within the atmosphere at the same pressure are similar regardless of vast composition differences adjusted only for distance from the sun.
The radiative theory cannot explain that whereas the conductive/convective theory does.

I don’t think the phenomenon you describe is as well-observed as you may think it has been. The data currently available to us about the temperature/pressure/chemical composition profiles of other planets’ atmospheres are still very scanty and uncertain. I think we would need to be much more certain of the reality of this alleged phenomenon than we are before we could use it as a test-criterion for the comparison of radiative versus conductive/convective theories. We might reach that degree of scientific enlightenment in a few centuries, perhaps, if we get lucky, but we’re nowhere near it now.

• KE alone is heat. KE plus PE is energy so your point is wrong.

Many solar system bodies have had their surface temperatures measured and they are all very close to that predicted from the gas laws alone so that point is wrong too.

• Cassio says:

Stephen Wilde November 25, 2017 at 10:46 am:

KE alone is heat. KE plus PE is energy so your point is wrong.

My point is not wrong. Internal KE alone is also energy and that determines the body’s absolute temperature. PE is also energy and must be included in any calculation of the body’s total heat content.

Look, the fundamental definition of the “heat content” (a.k.a. “enthalpy” – symbol H) of a body is that of the “internal energy” (symbol – U) of its constituent particles plus the product of its pressure (symbol – p) and its volume (symbol – V). We may write this as:

H = U + pV

In this formula, the term U (internal energy) represents both the average kinetic energy (KE – determined by internal speeds of motion) and the average potential energy (PE – determined by the strengths of internal electromagnetic fields) of the constituent particles, so it is unavoidable that there is some PE in the heat content.

Then we have the pV expression, which represents the work that the body’s environment has had to do on it to bring it to the state that it’s in. That also adds directly to the PE of the body’s constituent particles (and indirectly to their average KE by increasing their average PE).

So the heat content of a body does contain both KE and PE components. QED.

Many solar system bodies have had their surface temperatures measured and they are all very close to that predicted from the gas laws alone so that point is wrong too.

Really? How many solar system bodies would that be exactly, Stephen? How accurate are the measurements of their global mean temperature and pressure profiles? And how many of those have greenhouse atmospheres? How many do not?

And finally, where are you getting your planetary data from? I hope it’s not some national, or multinational government agency like NASA or ESA, because their data is notoriously unreliable and untrustworthy for serious scientific purposes. It’s completely uncheckable, you see.

56. Mack says:

You’ve introduced one of Trenberth’s looney Earth’s Energy Budget Diagrams, Willis. According to Trenberth you are only getting an average of 169 w/sq.m of solar radiation impinging on the Earth’s surface, (you know, in certain places, melting the tar on roads in summer) ,but now you’ve also got 321 w/sq.m of this “backradiation” belting down from the atmosphere 24/7.
Is your hometown Loonyville, Willis? The people of Loonyville leave their bacon and eggs out on the porch overnight to have them cooked by the morning from the 321w/sq.m atmospheric backradiation.

• Alan D McIntire says:

I originally started to develope this issue when I read some posting purporting to prove the Stefan-Boltzmann law “wrong” based on lunar temperatures. You might find this link, regarding Newton’s law of cooling, of interest.

The law gives this equation:

T(t) = Ta + (T0 -Ta)*1/(e^kt)

Where T(t) gives Temperature, T, as a function of time, t,
Ta is ambient background temperature, and T0 is the starting temperature of the body warming up or cooling off.

mass atmosphere = 5* 10^18 kg=5*10^21gm
temp atmosphere 255K (effective radiating temp to space- underestimates heat content)
specific heat 1.01 joules/gm C
5* 10^21*1.01*255= 1.288 * 10^24 joules

radius earth = 6400km= 6.4*10^6 meters.
area earth = 4 pi r^2 =514,718,540,364,021.76
240 watts/sq meter = 240 joules/sec per square meter
60 sec/min*60 min/hr*24hr/day=86,400 secs per day

5.147* 10^14 sq meters*240 joules/sec/sq meter *8.64*10^4 secs/day= 1.067*10^22 joules per day radiated away
1.067*10^22/1.288*10^24 = 0.83%

So the daily loss of heat of the atmosphere is less than 1% per day. That makes sense when you realized that although
temperatures may swing by 20 degrees K or more during the 24 hour day/night cycle, meteorologists are still able to make fairly accurate estimates of daily highs and lows for about a week- because of that temperature stability.

The above is to show that it is reasonable to assume a constant long wave
flux from the atmosphere over the course of a day.

We get an AVERAGE of 342 watts per day from the sun over earth’s surface. That works out to an average of 648 watts during the day and zero during the night.
You may have noticed that average temperatures do not increase to anywhere NEAR (648/390.7)^0.25 *288 K = 327 K during the day, and do not cool to anywhere NEAR 0 K during the night. That’s partly because MOST of the sun’s radiation gets absorbed by the atmosphere, which helps to moderate day/night temperatures. That 168 watts directly from the sun, as opposed from the sun to the atmosphere to earth’s surface, is entirely plausible.

• tty says:

“That’s partly because MOST of the sun’s radiation gets absorbed by the atmosphere”

But it is MOSTLY because of the oceans. Compare the diurnal temperature variation on Earth, Mars and the Moon. Which is the odd man out?

Or even compare the diurnal temperature swings in e. g. Central Asia and Polynesia.

• Alan D McIntire says:

True, thanks to the high specific heat of water, oceans act as a moderating influence, but my argument was that Trenberth’s hypothesis that most of the sun’s direct energy goes directly into the atmosphere still stands; else, how do you account for the ability of meteorologists to fairly accurately predict the weather for roughly a week in advance?

• Mack,

You are perpetrating the common howler of failing to offset the 321W/m^2 back radiation figure against the 392W/m^2 forward radiation figure to yield a net upward flow (i.e. from surface to atmosphere) of 71W/m^2. Those two numbers cannot be cherry-picked independently.

Adding that true upward 71W/m^2 flow to the 22W/m^2 upward sensible heat flow and the 76W/m^2 upward latent heat flow gives a total upward energy flow of 169W/m^2 which EXACTLY matches the 169W/m^2 of Solar energy flowing down into the surface. As it must, for steady state temperatures to prevail!

For more details, please refer to my post of November 25, 2017 at 9:35 am.

• Dave Fair says:

It would be interesting if someone could resurrect the figure of how much the net 71W/m^2 is reduced by the molecules of CO2 released by man.

• Mack says:

Well David,
You are going to have to explain this “common howler” to Dr Roy Spencer, because he is in doubt as to whether or not those energy flows even exist. He’s marked them with a big “X” here….
http://www.drroyspencer.com/2015/06/what-causes-the-greenhouse-effect/
Quoting Roy…”In the classical KT global energy budget diagram, the energy flows I have marked with an “X” would not exist without GHGs.”
I would like to say to the contrary, that, if those energy flows didn’t exist, then GHGs would not exist.
Roy goes on to say…”Now, recall I said temperature is a function of rates of energy gain and energy loss. Thus those energy flow arrows marked with an “X” in the above diagram represent huge flows of energy which can affect temperature, IF THEY REALLY EXIST”
So do they exist or do they not exist, David?
Also, I find it amazing the ever changing and extreme range variety of all these watts per sq.meters being bandied about by everybody…no actual measurements, just numbers seemingly out of thin air.
My guess is that the energy numbers marked with an “X” were pulled from the kiwi troughers ass.

• Mack,

On November 25, 2017 at 10:08 pm you say: Well David,
You are going to have to explain this “common howler” to Dr Roy Spencer, because he is in doubt as to whether or not those energy flows even exist.”

I have no intention of doing anything of the sort. I put you right on an elementary conceptual error (which all too many people make) but apparently that has squashed your pride. The downward radiation must be subtracted from the upward radiation to get the net figure representing the ACTUAL flow of radiative energy, which is always from hotter to cooler body (surface to atmosphere). I even referred you to my earlier post of November 25, 2017 at 9:35 am for the rationale, which is based on standard textbook physics. And all you can do is to ignore the math and ignore the physics, rather than benefitting from the explanation.

However despite your obvious interest in grandstanding (“I would like to say to the contrary, that, if those energy flows didn’t exist, then GHGs would not exist.” etc. etc.), I will nevertheless indulge you by stating the following:

The 321W/m^2 downward radiation and the opposed 392W/m^2 upward radiation cannot be treated separately because they cannot be independently employed to do separate work. They are simultaneous and opposite – so only the net residual radiation, which is 61W/m^2 upwards, is capable of doing work. This is logically obvious when you think about it. Otherwise there would be an absurd imbalance at the earth-atmosphere interface. Which there isn’t. Otherwise the earth would have melted. Which it hasn’t.

Or, of course, you might be right and the diagram is wrong, in which case you have just discovered the most embarrassing scientific howler of all time.

Please just get over the fact that you (along with many other people, if that makes you feel better) have made an elementary mistake. If you can’t face up to that, you will for ever wander around in the wilderness of climate bloggery, blaming everyone else around you (me, Roy Spencer, Uncle Tom Cobley and all…) as you blunder on and on.

Or, for a change, you could think a bit and learn something new.

• Toneb says:

“You’ve introduced one of Trenberth’s looney Earth’s Energy Budget Diagrams, Willis. According to Trenberth you are only getting an average of 169 w/sq.m of solar radiation impinging on the Earth’s surface, (you know, in certain places, melting the tar on roads in summer) ,but now you’ve also got 321 w/sq.m of this “backradiation” belting down from the atmosphere 24/7.”

It’s two-way.
Chicken and egg.

With there is 321 from GHE there is also that added to 169 solar to emit 490 from the surface >>>> which is then partially backradiated at 321 W/m2 and so on.

From Trenberth’s Energy budget diagram
Energy out:

22 + 76 + 392 = 490

Energy in:

169 + 321 = 490

The 321 is back-radiated having first being thermalised by the 169 SW absorbed AND the 392 added from the LWIR GHE.

• Hi Toneb,

You cherry-picked your numbers and got the wrong answer. Don’t be embarrassed, it’s a common mistake. I have laid out the correct math, which balances perfectly with the incoming solar absorbed at the surface, a few items up-thread in my first reply to Mack (David Cosserat November 25, 2017 at 10:08 am).

Cheers,
David

• Hi Toneb,

My profoundest apologies.

You are on the side of the angels. I should have read your previous post much more carefully. Of course you were quoting Mack’s misconceptions, not your own, and putting him right…

Best,
David C

57. Yes it can, with help. A reverse refrigerator. Air conditioning in a cold climate. All you need is a heat pump. What we will rely on for elctrical heating during the next ice age, hopefully all nuclear powered by then. Heat is removed from the colder body, making it colder, that heat is transferred to the hot body by the pump, making it warmer. First done in Glsagow taking heat out of the water supply, I don’t see that mentioned any more, just from ground or air, as in “sourced”. Performance factors are up to at round 4 as regards heat shifted to power in, Rather important given the cost of electrical energy for heating versus gas, about 4 times in the UK. Partly because gas heating is >90% efficient now but power stations energy conversion to electricity is a lot less, 60% at best for closed cycle gas turbine, 40% for the open cycle. Must dash….

• brianrlcatt

Once you introduce a powered heat pump you are relying on convection and the gas laws rather than radiation so you simply confirm my point.
With a man made heat pump a separate source of electrical power is required.
For an atmosphere above an irradiated surface the necessary power is provided by gas parcels of different energy contents and densities working with and against the gravitational field as they each seek a height commensurate with their respective weights and energies.
The process nets out to zero energy consumption overall but for Earth that still requires an ‘extra’ kinetic energy worth 33C (or whatever) at the surface to drive the global pattern of convective overturning. Much more KE at the surface of Venus is required to drive convective overturning in such a heavy atmosphere.
And so it is with all planets with atmospheres.
Willis discounts convective overturning as a reason for the surface temperature enhancement on the basis that being a zero sum process it cannot add extra energy.
In fact it is only a zero sum process once hydrostatic equilibrium is achieved. The process of creating a convective overturning cycle is not a zero sum process. Rather it draws surface KE into atmospheric PE in preference to that surface KE being radiated to space and thereafter recycles that block of energy up and down indefinitely. It is an insulating process because conduction and convection transfer energy slower than radiation.
The downward leg of convection then heats the surface in addition to continuing insolation.
Even a wholly non radiative atmosphere will develop convective overturning due to uneven surface heating leading to density differentials across the surface so those who say that such an atmosphere would become isothermal are wrong.

• Bryan says:

Yes, headline should include ‘spontaneously’ but most people assumed that anyway.

58. A body in vacuum emits from the surface at some temperature. Add a cold fluid which gets heated by the body. Now, the same energy that was emitted by only the twodimensional surface, is shared with the volume of fluid. So, now we have both a surface emitting and a threedimensional cold fluid also emitting. But the energy supply from the source is constant, so emission from both fluid and solid body is sharing the constant limited heat flow from the source. In no way, this is a situation where the solid gets warmer.

The heat source has constant limited power(sun). Temperature is a measure of average kinetic energy per molecule. How can addition of cold fluid mass, which absorbs energy from the heat flow in a larger volume of molecules, result in higher average kinetic energy per molecule, than a smaller volume with less molecules at higher temperature?

It is not right. Just like adding water to a pot on a boiler plate makes the pot colder, adding a cold fluid to a warm solid body in space, makes the body colder.

The atmosphere is a heat sink, added to the ultimate heat sink of space.

Without the shell of the atmospheric cold fluid, the surface T would be:

(TSI/2pi*r^2)/4/3pi*r^3=510W/m^2=308K

With the shell of cold fluid:

(TSI/2pi*r^2)/(4/3pi*r^3)^2=383W/m^2=287K

Adding a heat absorbing cold fluid to a hot body cools it, only.

Prevost made the conclusion that the emission of a body depends on the internal state only. The atmosphere is definately not part of the solid surface internal state.

• Uncle Gus says:

No.

(I don’t think I can constructively add anything to that. Just No.)

• Aww shit! Good argument. I loose.

Keep denying the calculations, having faith in blankets and unicorns in greenhouses is your melody.

• So Prevost was wrong then?
The emission of a body depends on the external state?

Come on, deny more of proven and applied physics. It´s fun to read.

59. Thomas Graney says:

In the history of this blog, there has never been a simpler, clearer, or more scientifically correct post; yet, we have all this irrelevant nonsense. Get a grip people.

• Uncle Gus says:

Bless you, my son!

• I agree. Everyone knows that dry ice, water and cold fluids in general, makes its heat source create more energy from nothing.

• Ed Bo says:

lifeisthermal:

Let’s say you’ve been outside on a cold winter’s day (255K, -18C, 0F) without heavy enough clothing. You have gotten hypothermia and your body temperature has dropped to 35K.

I offer to let you come into my house at 293K (20C, 68F), saying it will let you recover your body temperature up to 37C. By your logic, you would turn me down, because this is still colder than your body temperature, and it couldn’t make your internal “heat source create more energy from nothing.”

Seriously?

• You describe a situation where a change in surrounding temperature results in less absorption of heat from my body. If I move to a surrounding environment where there is higher temperature, the surroundings will absorb less heat from my body. I will transfer less heat to the surroundings.

In what way does this support a claim where a cold atmosphere and increasing amounts of heat absorbing dry ice, causes increased temperature?

The principle of the gh-blanket is that increased heat absorption from the solid earth causes increasing output of power from earth surface. This is the exact opposite of your example.

The solid earth, is supplied with energy from solar heating. Without an atmosphere, it only has to heat itself. Adding a cold fluid which is heated exclusively by the solid body, not the sun, means that earth has to heat the atmosphere, in addition to the solid, simultaneously.

Earth provides the energy, from solar heating, for both surface emission and atmospheric emission, at the same time.

This is an inevitable consequence of continous heat emission from both bodies, cold fluid and warm solid. The sun heats the solid earth, which emits and transfers heat to the atmosphere. 4*244+383=1361W/m^2=TSI.

• Ed Bo says:

lifeisthermal:

You agree that moving from an ambient of 255K to 293K will result in lower thermal losses from your body, permitting the thermalization of your metabolic power to increase your body temperature. This is true even though the ambient temperature of 293K is less than your (reduced) body temperature of 308K.

By direct analogy, “moving” the earth’s surface from an ambient of 3K (which it would have with a transparent atmosphere) to one of, say, 255K, will permit the earth surface’s thermalization of solar radiation to increase its temperature. This is true even though the ambient temperature of 255K is less than the surface temperature.

The two examples are directly comparable, not opposite!

A transparent atmosphere cannot transfer energy to space, because it has no emissions. It does not require any continuous power input from the surface to maintain its temperature.

60. indio007 says:

Can a current of photons travel in the opposite direction to another current of photons of greater intensity? I do not think so and infrared thermometers too.

• Uncle Gus says:

Yes they can.

What, do you think they *collide* with each other?

• Uncle Gus says:

And infrared thermometers too what?…

• I remember helping out a presenter at public slide show. He had two projectors, and the audience was sort of in the way given how he left projector aimed at the left screen and right to right. I switched the two projectors so the beams crisscrossed which avoided hitting the audience. Worked fine, of course.

While there are photon-photon interactions, it takes an “astronomically” high flux to do anything measurable. I think it has been observed near supernovae.

As for infrared thermometers, some cold clear night this winter take yours outside and aim it directly upwards. It’s not seeing a black body, so the reading won’t reflect the true sky temperature, but whatever it sees will be a steady flux. However, it’s also seeing IR flux from the body of the thermometer and from the polyethylene lens, it uses an internal thermometer to compensate for that “noise.” The lens will cool quickly in the cold air and its flux will go down. The thermometer will reflect that by showing the sky temperature going down.

You may need an IR thermometer that will read low enough, I have a Kintrex IRT0421. Some people have found they can map the sky temperature to the water vapor content of the air column. Good tool.

• An IR-thermometer works by measuring a gradient across a thermopile or thermocouple. It only measures the gradient inside itself. Then it extends that gradient into the surroundings with the s-b equation, by calculating the rate of transfer to different parts of the surroundings. It can only calculate the rate of transfer, from T1-T2. So, in the case of colder surroundings, it only measures transfer FROM the device. Because T2 is smaller, and the value is negative. Which will show as darker on the screen.
The manual will tell you this: that in the case of measuring cold atmosphere, the transfer to the device, the incoming flux, is negative. Which means: there is no incoming flux. The manual also says that the negative influx of heat, the non-existent incoming flux, is used to calculate a fictive incoming flux based on what temperature the receiver of heat transfer has, calculated by outgoing transfer of heat.
What this means is: the non-existing incoming flux, is imagined from the outgoing transfer and the device shows it as positive, even though it is not there. Because it uses the s-b equation and it can only calculate net-transfer. “Net” is the only transfer there is, because “net” is heat. And transfer of energy between bodies of different temperatures is heat. And there is no other transfer which can be calculated without breaking the 2nd law of T.

Another interesting fact: a thermocouple/IR-thermometer only have a range of some 25m into the atmosphere. So it doesn´t give much information about heat transfer in the earth system.
Optical measurement on the other hand, shows spectral distribution. But the problem is: they only show how heat flow decreases from the co2, by a very distinct “bite” which decreases intensity of emission in those wavelengths.

• The manual will tell you this: that in the case of measuring cold atmosphere, the transfer to the device, the incoming flux, is negative.

I’m sure the manual told me no such thing. If it did, my eyes would have bugged out, my pulse quickened, and I’d be on an epic rant.

Which means: there is no incoming flux.

Then my thermometer would display only one value from source colder than it.

Here’s the manual, https://www.homedepot.com/catalog/pdfImages/ef/effd4a5b-8452-4095-98ce-39265a0adde2.pdf . It’s disappointingly brief, and never mentions flux.

61. Philip Mulholland says:

Willis,
I probably flatter myself if I consider that I am the gadfly the prompted this post, however as I always want to learn you will have to indulge me while I go through the following observations and resulting questions.
From first principles Science is the process of rejecting false ideas. So I will start by making statements which I hope we can easily identify as being false.
1. All forms of matter absorb (that is extinguish) all forms of light (electromagnetic radiation).
Trivially easy to prove false, because this statement denies the existence of transparent materials and also denies the existence of short frequency gamma rays and long frequency radio waves that can pass unchanged through material bodies of a specific composition. (Notice that already the caveats are creeping in), so matter in its various forms acts as a selective filter of electromagnetic radiation.
2. Absorption is the only process by which light interacts with matter. Again trivially easy to prove false because this statement denies the existence of reflection and refraction and not to mention the already discarded contention of no transmission.
3. Colour is a thermal process. Hum, now I am already getting into trouble because do I mean the emission of a specific wavelength from a hot object or the reflection of a non-absorbed wavelength from the surface of a cold body? Just for good measure what is meant here by hot and cold? Are we talking about the average kinetic property of a bulk material object or the specific translational momentum of an individual molecule, which supposedly has no temperature because temperature is a statistical measure of bulk particle motion, vibration, flexure, translation, rotation and not to even consider electron shell interactions of various kinds? (And no I don’t like the statement that an individual molecule does not have temperature because we measure the energy of translation of individual thermal particles using electron Volts).
Now for some questions:-
1. So what is colour? (Sorry dear Cos, I will spell that word the ol’ way). Let us start with the colour of an object perceived using reflected light. What is the process by which selective reflection occurs? Is it a process of absorption and immediate readmission or a boundary property of a material in which a process (perhaps similar to liquid surface iridescence?) occurs in a solid state material? Oh and just for good measure don’t forget that that liquids can also have colour, such as red-brown bromine or blue liquid oxygen, so is it that colour with reflected light is an electron shell process?
2. And what is black? We perceive carbon as being black and therefore having no colour, but this is wrong, carbon does have a colour, it is coloured infra-red, it is simply that we cannot see these reflected wavelengths and so describe carbon material as having no colour.
3. And then we come to the subject of absorption spectra. Helium was first discovered in the coronal gases of the sun because the astronomer Sir Norman Lockyer observed a unique yellow absorption band, evidence of an unknown element that selectively removed this specific wavelength from sunlight thereby dimming that wavelength when viewed from Earth. Is this not a similar dimming effect to that we observe with infra-red absorption spectra?
4. And what about the coloured light emitted by the discharge in a sodium light for example? Is this light the same process of light as I see reflected off the yellow walls of my bathroom? Well no. One colour comes from emission from a hot source and the other from selective reflection from a cold surface, so there are clearly different processes at work here that seem to me to be discounted at infra-red wavelengths.

So the key question for me is this; -How do we know that the selective infra-red frequencies that are blocked by the atmosphere are absorbed and therefore thermalise the absorbing gas and none of these infra-red wavelengths are reflected, in a perhaps recoilless process (or perhaps not i.e. radiation pressure?) that if so happened, might presumably “colour” the gas but have no thermal effect?

• A C Osborn says:

You Point 1, that is where the MIT Light bulb is wrongly classified, the IR screen does not absorb and then re-radiate the photon it reflects it. In other words they have found an Infra Red Mirror that allows white light to pass through, which is very clever physics indeed.

• Philip Mulholland says:

Thank you AC Osborn.
It has now occurred to me that because frequency is a conserved property, while wavelength varies with the refractive index of the transmission medium, I should not use wavelength to describe EM radiation.

• Uncle Gus says:

Your style is very familiar. After a bit, I realised who it reminded me of – AE Van Vogt, a science fiction writer famous for making up his science off the top of his head, from ideas he got in his dreams.

To answer your question – yes, it might be reflected instead of re-emitted (although I don’t really think this is a controversy; there has been quite a bit of research done on the subject). But that makes no difference. We’re talking about back-radiation; radiation sent *back* towards the Earth from the atmosphere. Reflected, re-emitted, it still gets sent back.

And don’t tell me it’s the wrong wavelength or “colour” or whatever. It’s infra-red, and practically everything absorbs it to some extent, even objects that are slivered or painted white.

• A C Osborn says:

Well if “practically everything absorbs it” I wonder how it gets all the way through the atmosphere to impact the earth’s Surface from 100K away?
So you do not believe that “Wavelength” makes any difference?
OK.

62. John M. Ware says:

I agree with Willis’s article, but I have an English quibble. The expression “than me” should be “than I.” “Than” is a subordinating conjunction, followed by a clause with its own unstated subject and verb. “He is taller than me” should really read “He is taller than I am.” The verb “am” is understood. “Than” is not a preposition, as “near” is; “near I” is obviously incorrect, and “near me” is correct, because “me” is the object of the preposition and has no following understood verb. “Than” can, indeed, be followed by “me” because “me” can be used as an object: “He likes Jenny better than [he likes] me.” On the other hand, “I” can be the correct pronoun if the sense is “He likes Jenny better than I [like her].”

• Kip Hansen says:

John ==> eGads! that’s what my high school English teacher said after my speech in front of the whole student body accepting the office of Student Body Vice-President! She said it, of course, in front of the whole assembly — couldn’t help herself.

(I have, somewhat reluctantly, finally forgiven her after fifty some-years)

• Hugs says:

I’m faster than he. Oh no, I never could utter that. Germans might use als er, but Swedes could use än honom, ‘than him’. I’m not sure why I use than at least sometimes as a certain kind of preposition. In my own language, you can say something like I am her faster, or I am faster than (s)he. The logic probably is the same, it’s just that who said and why ‘than’ could not be a preposition requiring me/him/her.

Usually speakers are right, even if they are wrong. Than is by its etymology a conjunction, relative to when and then, so that would mean than I is historically correct. But what’s the ‘än’ though?

63. N. Jensen says:

Color me confused.

There seems to be a general agreement on that there are 4 CO2 molecules in the atmosphere for every 10.000 other molecules.

Still, these 4 molecules are supposedly able to warm 10.000 of a different kind ?

By what mecanism, exactly ?

Sounds like a Perpetuum Mobile, to me.

• They generally don’t, they generally reemit an IR photon of the same energy. However, if they collide with any other air molecule first, the vibrations between the CO2 atoms (that holds the energy from the photon) will be partially transferred as kinetic energy due to pushing both molecules away, and that means the temperature goes up. Over the next small fraction of a second subsequent collisions will heat up other molecules, restoring the bell curve shape of individual molecule’s kinetic energy.

Or so I understand, sorry I don’t have numbers and probabilities handy.

• daved46 says:

Ric, if you look up the mean time between collisions and the average emission time I think you’ll find it to be about 10,000 collisions per emission. So the small amount of CO2 in the atmosphere equilibrates with the atmospheric temperature

• I defer To Dave Burton’s conversation with William Happer, https://wattsupwiththat.com/2017/11/24/can-a-cold-object-warm-a-hot-object/#comment-2676749

It says there are about 1 billion collisions per emission!

That means LWIR emitted by the Earth’s surface, winds up heating nearby air (and itself) as the mean free path of the photons is quite short. So that energy stands a good chance of going back to the ground, trees, whatever. And it implies that the radiant energy that escapes the surface is from wavelengths not blocked by CO2 or H2O. Something I get reminded of every morning I scrape frost off my car.

• Toneb says:

“By what mechanism, exactly ?”

You need to consider a photons path-length to space.
By the time it gets to ~ 8km more can escape to space than be back-radiated.
Shine a torch into a thin mist.
The beam will be attenuated at a distance.
Shine a torch in fog.
The beam will be attenuated more quickly.

Yes more fog droplets.
But also accumulated collisions because of the path.
Which is why there is no lab experiment that can duplicate the GHE in Earth’s atmosphere.

The photons come back … more fog-droplet collisions – then the brighter the back-radiated (refected – same concept) the beam.

By that mechanism.

• You give an example of how radiation is reduced by mist and fog.
In what way is this a support of the gh-effect causing increasing levels of radiation?

“The photons come back”

WTF!

You are not shy at all!

Absorbed photons come back?

Lets talk about how you like to make things up.

Absorption of photons destroys the photon.

You have no clue.

64. Paul Bahlin says:

For commenters who offer the argument “warm objects instantaneously reject IR from colder objects, please consider that you have defined a reflection.

The logical end point of this fallacy is that albedo is a function of energy content.

There’s a noble prize waiting for you if you can prove that one.

• A C Osborn says:

So Laser Directing mirrors get as hot as the Laser Beam?
Or are already hotter than the beam?
Clouds are “White Hot”, Ice is also “White Hot” and both hotter than Sunlight.

• Paul Bahlin says:

Huh?

• “The logical end point of this fallacy is that albedo is a function of energy econtent.”

First, albedo is necessarily a consequence of geometry. Irradiation on a disc, distributed over a hemisphere, absorbed in a volume, is enough to reduce the power density to 0.7i5.

Obviously, the temperature will determine the amount of ice, snow, sand, and water surfaces.

So, yes, albedo is a function of energy content. Measured as emissive power, T^4.
But geometry is the main constraint-

65. Uncle Gus says:

Thank you, Willis Eschenbach!

You’ve left out the fact that the Greenhouse Effect ALSO works during the daytime. (In fact, according to that diagram, it delivers more radiant energy to the surface than the Sun does directly, since a whopping 105 W/m2 of sunlight is reflected by the Earth and clouds.)

It’s a subject that’s insanely complicated in detail, but almost moronically simple in general terms. All you need is the ability to visualise, and and a bit of faith that scientists from more than a century ago (who never even heard of Global Warming!) are NOT trying to con you.

• A C Osborn says:

Love it “Faith Based Science”.
The world was flat and the Sun and stars revolved around the Earth, do you still believe it?
Ulcers were caused by stomach Acid, still believe it? Never heard of Helicobacter pylori?
Do you believe in the Standard Model, the Big Bang, Dark Matter & Dark Energy, with no proof and no actual evidence?
Because their are plenty of Scientists that don’t and have Theories just as good to prove it.
http://earthsky.org/space/erik-verlinde-gravity-theory-no-need-dark-matter
http://www.dailymail.co.uk/sciencetech/article-5109023/Dark-energy-dark-matter-NOT-exist-study-says.html

You seem to be suggesting that Scientists are never wrong.

• Willis, thank you for writing this, I’ve wanted a decent post on the subject here to reference in the future. I apologize for not having written it myself years ago. You did a better job than I would have.

I guess we have nearly all the possible objections (and explanations) to what you describe. That could be a feature in its own right!

• Uncle Gus says:

All right, I’ll take some time on this one.

I never said that scientists were never wrong, I said that these particular scientists are not trying to con us from political motives connected with the Global Warming controversy (mostly because they were all dead long before it started…).

If you think someone is trying to put one over on you, you have to pick up on every detail. Otherwise, a certain amount of faith is necessary for the maintenance of sanity. Massive objects attract each other. That’s a simple principle put forward by Sir Isaac Newton, and we trust it. It might suit us to believe instead that gravity is caused by the Flying Spaghetti Monster with His Noodly Appendages, but why should we? There’s nothing glaringly wrong with Newtonian gravitation, and even Einstein only modified it slightly with his Special Relativity.

There’s nothing glaringly wrong with the Greenhouse Effect. “It contradicts the Second Law of Thermodynamics” only works if you don’t understand radiative thermodynamics (or think that other people won’t, and hope to make a name for yourself through an invented controversy). If you’ve understood the basic underlying concepts, you can visualise the phenomenon in general terms, and it makes sense. If it doesn’t make sense to you, it’s because you’re introducing some rubbish such as “two lots of photons can’t travel in opposite directions at the same time”. (Yes, they can.)

Climate scepticism, on the other hand, is a different sort of thing entirely. When it’s not purely and simply political, it’s based on established science and attacks the *conclusions* and the *methods* of the warmers, not their scientific principles. (Their *moral* principles, now that’s another thing…)

I don’t need to start believing nonsense, just to maintain my street-cred as a sceptic.

• Uncle Gus,

You say: …according to that diagram, it delivers more radiant energy to the surface than the Sun does directly

No it does NOT! The energy flow from the Sun that is absorbed by the surface is 169W/m^2. The energy flow from the surface to towards the atmosphere is exactly the same: 169W/m^2. The figures balance perfectly.

Take a closer look at the diagram…

Net upward sensible heat: 22W/m^2
Net upward latent heat: 76W/m^2
Net upward radiation: 392 – 321 = 71W/m^2
TOTAL: 169W/m^2

The mistake you have made is to look only at the 321W/m^2 downward component of radiation from the atmosphere and ignored the 392W/m^2 upward radiation from the surface. These two numbers cannot be dealt with separately. They are inextricably interlinked. You can’t have one without the other. You have cherry picked!!

For more details, please refer to my post of November 25, 2017 at 9:35 am.

• “The energy flow from the Sun that is absorbed by the surface is 169W/m^2. The energy flow from the surface to towards the atmosphere is exactly the same: 169W/m^2.”

You are claiming that a surface at 288K is emitting only 169W/m^2.

Do you pee your pants on a daily basis as well?

• Mack says:

Quoting Uncle Gus……”according to that diagram,it delivers more radiant energy to the surface than the sun does directly”.
No it does NOT! says David Cosserat..
Well, the diagram actually says this radiant energy from the “backradiation” is ABSORBED BY THE SURFACE.,David. So in Trenberth’s mind it is real and exists. It should cook bacon and eggs on the porch if left there overnight.
Incidently, Roy Spencer has doubts about the existance of these energy flows….
http://www.drroyspencer.com/2015/06/what-causes-the-greenhouse-effect/

• Uncle Gus says:

Actually, you’re right.

I was considering only the downward radiation, and ignoring the upward. In total, they will balance exactly in the long run – they have to, or the earth would quickly burn up or freeze.

And of course, all heating (except a minute amount of geothermal) comes from the Sun.

But the diagram does seem to imply that the Earth’s surface is heated more by IR from the lower atmosphere than by combined IR and visible light directly from the Sun. That is some efficient blanket!

To be honest, I don’t know how accurate that or any of the other analyses of radiative heat transfer in the atmosphere are, or indeed can be. I suspect that without a lot more direct measurement, all we can get is a sort of rule-of-thumb sketch. I’m sort of assuming we have that much because, well, otherwise we might as well all go down the pub…

66. A C Osborn says:

Willis, I am back with some more dumb layman questions.
You said that we can measure the LWIR coming from the CO2 in the Sky.
How do we do that?
How does the Measuring Device differentiate the CO2 from the rest of the Atmosphere between it and the CO2?
And how does it know it is LWIR?

• daved46 says:

We know it because we have tested the other atmospheric gases and know that only CO2 and Water will emit LWIR. So we measure the humidity calculate emissions from it and the rest is from CO2. As to how we know it’s LWIR you’d need to read the technical details for a given measurement device.

• tty says:

Actually all gases with more than two atoms in the molecule are GHG, and two-atom gases are too as a matter of fact, but only very weakly.

• A C Osborn says:

You miss the point completely.
You take a Sky Measurement, you get a Value, what is the value?, how do you know where it originates from?, and from what?
How do you measure the Humidity all the way up to the top of where it is with the measuring device that measures the Temperature?

• Dave Fair says:

The short answer, A C, is that satellites have sensors to detect whatever we want tracked.

• A C Osborn says:

No Satellites measure outgoing radiation, which shows more CO2 causes more atmospheric cooling.
They can’t measure Down Welling as it pointed away from the Satellite.

67. prjindigo says:

Can a child shoot an adult?

Its not the cold object that does the warming, it is the energy it releases.

• Uncle Gus says:

68. Ethan Brand says:

Willis: Very lucid explanation of a fundamental concept. When I engage in a discussion with someone, I have learned to quickly ferret out where the “bottom” is. That is, at what level is there agreement. Having a discussion at any level higher than that is generally a waste of time…if one is trying to resolve anything. At one point in my careers I was designing and making electrical power meters. I found that many of my customers, who ostensibly were trying to monitor electrical consumption, had no real understanding of the concepts of power and energy. Until that concept was understood, there was no point in engaging in any other conversation. Most of the time, I was successful in bridging that gap.

I cannot over state the importance of this general concept in communication. In the arena of AGW, I find the most fundamental/lowest level gap is the concept of the Scientific Method. In the realm of technical discussion, that’s pretty darn low (not sure you can get lower..:) )….unfortunately much lower even than the concept of radiation heat transfer.

Thanks for your excellent insights and writing skills…I always look forward to anything you write….

Regards,
Ethan Brand

• Uncle Gus says:

Beautifully put!

• Ethan
An important insight.
My work is with micro tomography systems – think hospital CT scanner but the size of a suitcase, sitting on a desk. We market these to scientists, mostly biologists. They often at first have no understanding of the nature of the technology, imagining it to have almost magical powers.

So you’re right – before any meaningful communication can occur one often needs to educate the client on the essential underlying science and principles, such as xray absorption and contrast (plus backprojection- reconstruction). Once that basis of understanding is reached, then more fruitful discussions can follow.

69. Sparks says:

“…If there is a block of wood between you and a block of ice, if you remove the wood, you’ll get colder because you will be absorbing less radiation from the ice than you were from the wood. You no longer have the wood to shield you from the ice.”

That statement is wrong.
You’ll get colder because the ice will be absorbing more radiation from your body than when the wood was there.

If you were to use a sheet of aluminum instead of a block of wood, the block of ice would be absorbing more radiation from your body through the aluminum sheet than through the block of wood.

The sheet of aluminum will become warmer, this is because of the block of ice absorbing radiation from your body.

For a so called “cold body to warm a hot body”, its energy potential would have to be quantified, if you give the cold body a value of 2C and the warm body a value of 5C the differential between the two values would be 3.5C. if you consider the differential between the aluminum sheet and your body, this would be 4.25C.

3.5C>2C therefor the sheet of aluminum has warmed by 1.2C. This is in fact the block of ice absorbing radiation from your body through the aluminum sheet and not the other way around.

• Uncle Gus says:

“The sheet of aluminum will become warmer, this is because of the block of ice absorbing radiation from your body.”

It’s sucking the heat out of your body!!!

Don’t get me wrong when I say that I absolutely love this. As the saying goes, it’s so bad, it’s not even wrong!

(And then not to make *any intelligible point at all* at the end of it all! Priceless.)

• Sparks says:

If you were to use a sheet of aluminum instead of a block of wood, the block of ice would be absorbing more radiation from your body through the aluminum sheet…

• Sparks says:

Uncle Gus
You’re giving me the impression that you believe that a persons warm body absorbs cold from a block of ice, are you for real?

I didn’t say “It’s sucking the heat out of your body!!!”
But a cold surface will draw the heat from a persons body, it’s called “heat transfer” not cold transfer.

Aluminum has less resistance than wood allowing the heat to leave the body to the block of ice more efficiently.

• “This is in fact the block of ice absorbing radiation from your body through the aluminum sheet and not the other way around.”

Nevertheless, the absorption is the transfer of energy. Less transfer, higher temperature of source, increased transfer, lower temperature of the source.

It is important to locate the source of energy, by measuring the higher density of energy, to determine the direction of the flow.

Co2 increases absorption. Increased absorption means increasing rate of transfer. Increasing rate of transfer means lower temperature of the heat source.

• Sparks says:

lifeisthermal Says;
“Co2 increases absorption”

Not at saturation, Do you understand that atmospheric Carbon Dioxide is above its limit of absorption?

Do you understand that Carbon Dioxide has no further effect once saturated?

lifeisthermal Says;
“Increased absorption means increasing rate of transfer “
And
“Increasing rate of transfer means lower temperature of the heat source “

Not applicable on planet Earth.

70. mkelly says:

In 1954, Hoyt C. Hottel conducted an experiment to determine the total emissivity/absorptivity of carbon dioxide and water vapor11. From his experiments, he found that the carbon dioxide has a total emissivity of almost zero below a temperature of 33 °C (306 K) in combination with a partial pressure of the carbon dioxide of 0.6096 atm cm. 17 year later, B. Leckner repeated Hottel’s experiment and corrected the graphs12 plotted by Hottel. However, the results of Hottel were verified and Leckner found the same extremely insignificant emissivity of the carbon dioxide below 33 °C (306 K) of temperature and 0.6096 atm cm of partial pressure. Hottel’s and Leckner’s graphs show a total emissivity of the carbon dioxide of zero under those conditions.

http://www.biocab.org/Overlapping_Absorption_Bands.pdf

Willis says: thermal IR emissivity is 1.0

What you are using for emissive is from water vapor not CO2. CO2 will not cause an increase in any temperature at thes concentrations or pressures.

• Uncle Gus says:

That actually makes sense!

I think it’s wrong, and I don’t have time to read it carefully and figure out why it’s wrong, but at least it’s out of Looneyland for a change!

(If it wasn’t such a departure from the accepted science on the subject, I probably wouldn’t be so sceptical. Does anyone better qualified than I am have anything to say?)

• mkelly says:

Dr. Hottel wrote the books on radiative heat transfer in combustion chambers. His charts are still used today.

It is not wrong. At the temperatures and partial pressures we talk about emissivity of CO2 is essentially zero.

I challenge anyone to use these charts and show that CO2 has an emissivity that not essentially zero at partial pressure of .0039 ppmV and 273K.

• Uncle Gus says:

Yes, it *does* say 33 degrees Centigrade! That’s hotter than I have my shower in the morning.

I still don’t believe it, but if true, think of the implications! CO2 effectively *transparent* to IR at the relevant temperatures. The whole Global Warming thing not a hoax or “settled science”, but based on a single unexamined (and fallacious, assumption)!

The irony!

Pity it’s BS…

• mkelly,

I have read the reference, but that only talks about the influence of CO2 in the overlapping bands with water. The point is that most of the CO2 influence is in the 10 micrometer band, where water is not active at all…

71. Roy W. Spencer says:

feeling a little spunky, were we, Willis? ;-)

72. Steven Mosher says:

thanks willis

watched these arguments for 10 years now. The cash analogy is pretty good. I may steal it someday.
the analogy, not the cash.

73. Mariano Marini says:

Here’s an interesting link. MIT scientists have built an incandescent light bulb whose envelope passes visible light but which reflects infrared radiation back at the filament. The result is that it takes less electricity to heat the filament to the desired temperature … a lot less electricity.

Why not fill a bulb with CO2 and compare (measure) the real contribution of CO2 in a “close” (sealed) word? This way we can know exactly the relation between filament and feedback temperature.
We could warm the filament at 50° and see how much at 400ppm of CO2 it will feedback.

• Mariano,

If you have a bulb with a diameter of around 50 km, you may measure a difference, if you can fix all other variables…

74. joletaxi says:

Hello from France(so You will excuse the bad english… I hope

I love Your exemple with money, because it is the central dilemna.

Me give 100 to You
but instead off refunding 75 to me, You decide to spend 75 buying beer
the net flow is 100 to You, and I just have to fall in tears.

do’nt forget that if CO2 is a very good “interceptor of radiative wave, it is also a very good emitter
sure, the bulk of the atmosphere will be warmed by collisions with CO2,but at equilibrum, the flux will be the same.
In the dry desert, the surface of the grond will be very hot at noon, and the atmosphere surrounding also by conduction, and convection
When the sun fall, the grond can freeze, very quickly
But it all depend of the capability off the limit layer to radiate to space.
The theory is that, when it warm beneat, it cool upside and the radiative capability fall?
I doubt of that.
First, if there is more CO2near the ground, , (that the way the theory works?) more radiative flux will be intercept, and thus, it will be cooler in the upper limit.layer
But If there is more CO2 near the ground, there is also more CO2 in the limit layer, and thus a better capability to radiate to space.
If it is hotter in the ground layer, the atmophere surrounding will expanse, and the t°of the adiabatique curve will change, so any consideration of the t° at the limit layer will be modified.

Just for the fun
what if there is only CO2 in the atmosphere, no water on the ground?

• Toneb says:

nr65:
” In the case of the Earth, the solar irradiance of 1,368 W/m^2 has a Stefan Boltzmann black body equilibrium temperature of 394 K, 121 C, 250 F. That’s hot. Sort of.”

OK

“The Earth’s albedo reflects away about 30% of the Sun’s 1,368 W/m^2 energy leaving 70% or 958 W/m^2 to “warm” the surface (1.5 m above ground) and at an S-B BB equilibrium temperature of 361 K, 33 C cooler (394-361) than the earth with no atmosphere or albedo.”

You forgot to divide by 4 as 1368 is TOA.
The Earth is a rotating oblate speriod.

So that’ll be 342 W/m2 x 0.7 (albedo)
=239 W/m2 available (on average) for each m2 of the sunlit surface of Earth.

That gives a S-B temp of 255K
But we live on a planet at 288K

33K warmer.
Why?

• Toneb says:

Wrong place

• Dave Fair says:

Because we are a water world, Toneb. H20 not only dominates in the radiative arena, but also gives convective cooling at the surface.

• KM says:

Toneb, you write:

“That gives a S-B temp of 255K
But we live on a planet at 288K

33K warmer.
Why?”

As mentioned by others, Earth’s surface is not a blackbody. A blackbody is an object that only exchanges energy with its surroundings through radiative heat transfer.

The surface also exchanges heat with the troposphere via conduction (and subsequently convection).

Earth together with its troposphere, however, is a reasonably good approximation of a blackbody. The tropopause just above the troposphere ensures that there is no convective heat flow between the troposphere and the stratosphere. Conductive heat flow at this boundary is also small.

The average temperature of the troposphere is around 255K. Coincidence?

Back to the question, why is the surface 33 K warmer than 255 K?
Going further, why is the temperature at the tropopause 33 K colder than 255 K?

This is due to the lapse rate. Hot air rises and cools off. cold air drops and warms up. The end result is a temperature gradient with 33K warmer than the average near the surface, and 33K colder than the average at the tropopause. The average, 255 K stays the same.

• gbaikie says:

–Just for the fun
what if there is only CO2 in the atmosphere, no water on the ground?–
I like fun.
Earth has 70% surface water and 30% land.
The average ocean surface is about 17 C and average land surface is
So without water surface, roughly earth should have an average temperature
of less than 10 C
[Because the 17 C ocean surface 17 C currently increases the average land
Air surface to 10 C].
Ocean warms land and land cools ocean.
Land surface air temperature can be much higher, but basically
ocean saves money and lands spends more money that it makes.
Land dweller are in debt to the ocean, and always have been..

75. Kip Hansen says:

Interesting analogy — the atmosphere, of course, is not “cold”, nor is the surface “warm” (or “warmer”) — they are only relatively in that relationship. It only confuses the issue to think in terms of cold and warm. It is not the temperature of the material between the two objects that is the primary factor — only the case that there IS something that interacts with radiant energy between them. If the interceding object were entirely transparent to radiant energy, there would be no effect — because the object in this case, the atmosphere, is made up of gases and particles that do interact with radiant energy, the atmosphere both absorbs and radiates that energy and becomes part of the energy flow equation.
The very principle of atmospheres acting as protection from the mind-numbing coldness (there is something that can be called cold) of space itself has long been know — anyone camping on the high, dry deserts of the American Southwest has experienced the speed and efficiency of sky “sucking” the heat out into space when the atmosphere is (relatively) thin and dry, denying us that protective blanket.
The same is true in the opposite case — on the tropical islands, where altitudes are low (and the atmosphere thick) and humidity is high, less of the energy delivered by the Sun during the day escapes to space during the night — thus we lie on our boats, praying for a breeze to cool the sweltering night.
The moon (and Mars) lacking an appreciable atmosphere is hot in the direct Sun and very cold (all my our measure) in the shade or on the darkside.
A blanket itself is not “warm” — it is simply fairly efficient at slowing energy flow through it — which is what the atmosphere does. The highly efficient silvered plastic emergency “space’ blankets used by Emergency Rescue squads is prime example — the allow very little body heat to pass through and reflect back a great deal of the radiant energy (TV shows often make the error of placing these around the victim with the shiny side out which makes a better picture, but is incorrect in use).

• “A blanket itself is not “warm” — it is simply fairly efficient at slowing energy flow through it”

An atmosphere is not “warm”, it simply increases the transfer of heat into it. The cause of increasing rate of transferred energy is reduced emissive power, observed as reduced T^4.

A blanket, or any other type of thermal insulation, causes reduced absorption of heat in the colder surroundings. Which is equal to reduction of the rate at which heat is transferred.

An atmosphere, and co2 in particular, causes increased transfer of heat, by adding heat absorbing molecules the its heat source.
This means, that an atmosphere acts exactly in the opposite way to how insulation affects temperature of a body.

Would you care to explain why your model is based on a violation of proven and applied physics?

76. johnosullivan says:

Willis (Anthony), Thanks. This is an excellent article. Kudos to you gentlemen.

• Toneb says:

” In the case of the Earth, the solar irradiance of 1,368 W/m^2 has a Stefan Boltzmann black body equilibrium temperature of 394 K, 121 C, 250 F. That’s hot. Sort of.”

OK

“The Earth’s albedo reflects away about 30% of the Sun’s 1,368 W/m^2 energy leaving 70% or 958 W/m^2 to “warm” the surface (1.5 m above ground) and at an S-B BB equilibrium temperature of 361 K, 33 C cooler (394-361) than the earth with no atmosphere or albedo.”

You forgot to divide by 4 as 1368 is TOA.
The Earth is a rotating oblate speriod.

So that’ll be 342 W/m2 x 0.7
=239 W/m2 available (on average) for each m2 of the sunlit surface of Earth.

That gives a S-B temp of 255K
But we live on a planet at 288K

33K warmer.
Why?

• Toneb says:

Doh

• The earth surface is not 33K warmer. The atmosphere is 33K colder.

This is caused by heat absorption=cooling of the source(earth surface)

Surface temp: (TSI/2pi*r^2)/(4/3pi*r^3)^2=383W/m^2=287K

77. nickreality65 says:

In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual.
 Galileo Galilei

The ONLY^3 reason RGHE theory even exists is to explain how the average surface (1.5 m above ground) temperature of 288 K/15 C (K-T balance 289 K/16 C) minus 255 K/-18C , the average surface (now ground) temperature w/o an atmosphere (Which is just completely BOGUS!) equals 33 C warmer w/ than w/o atmosphere.

That Δ33 C notion is absolute rubbish and when it flies into the nearest dumpster it hauls RGHE “theory” in right behind it.

The sooner that is realized and accepted the sooner all of us will have to find something better to do with our time and the taxpayers’ money. Maybe that’s what keeps RGHE staggering down the road.

The genesis of RGHE theory is the incorrect notion that the atmosphere warms the surface (and that is NOT the ground). Explaining the mechanism behind this erroneous notion demands some truly contorted physics, thermo and heat transfer, i.e. energy out of nowhere, cold to hot w/o work, perpetual motion.

Is space cold or hot? There are no molecules in space so our common definitions of hot/cold/heat/energy don’t apply.

The temperatures of objects in space, e.g. the Earth, Moon, space station, Mars, Venus, etc. are determined by the radiation flowing past them. In the case of the Earth, the solar irradiance of 1,368 W/m^2 has a Stefan Boltzmann black body equilibrium temperature of 394 K, 121 C, 250 F. That’s hot. Sort of.

https://science.nasa.gov/science-news/science-at-nasa/2001/ast21mar_1/

But an object’s albedo reflects away some of that energy and reduces that temperature.

The Earth’s albedo reflects away about 30% of the Sun’s 1,368 W/m^2 energy leaving 70% or 958 W/m^2 to “warm” the surface (1.5 m above ground) and at an S-B BB equilibrium temperature of 361 K, 33 C cooler (394-361) than the earth with no atmosphere or albedo.

https://springerplus.springeropen.com/articles/10.1186/2193-1801-3-723

The Earth’s albedo/atmosphere doesn’t keep the Earth warm, it keeps the Earth cool.

Bring science, I did. (6,200 views and zero rebuttals.)

http://writerbeat.com/articles/15582-To-be-33C-or-not-to-be-33C

http://writerbeat.com/articles/16255-Atmospheric-Layers-and-Thermodynamic-Ping-Pong

• Toneb says:

” In the case of the Earth, the solar irradiance of 1,368 W/m^2 has a Stefan Boltzmann black body equilibrium temperature of 394 K, 121 C, 250 F. That’s hot. Sort of.”

OK

“The Earth’s albedo reflects away about 30% of the Sun’s 1,368 W/m^2 energy leaving 70% or 958 W/m^2 to “warm” the surface (1.5 m above ground) and at an S-B BB equilibrium temperature of 361 K, 33 C cooler (394-361) than the earth with no atmosphere or albedo.”

You forgot to divide by 4 as 1368 is TOA.
The Earth is a rotating oblate speriod.

So that’ll be 342 W/m2 x 0.7
=239 W/m2 available (on average) for each m2 of the sunlit surface of Earth.

That gives a S-B temp of 255K
But we live on a planet at 288K

33K warmer.
Why?

• nickreality65 says:

Toneb

A sphere of radius r has 4 times the surface area as a disc of radius r. The incoming sees the earth as a disc the outgoing sees the earth as a sphere. In the bucket of warm poo model the incoming 1,368 discular W/m^2 is simply spread over the spherical ToA, i.e. divide by 4. Simple and dumb. K-T diagram.

The incoming is NOT actually spread evenly over spherical ToA to get an evenly heated 342 * .7 = 240. or what I refer to as the earth as a ball suspended in a warm bucket of poo.

The incoming strikes the ToA at an oblique angle so the perpendicular to ToA W/m^2 is 1,368 *cos latitude which ranges from 1,368 at the equator to zero at the poles.

The outgoing sees a sphere 24/7 with the heat leaving per Q = UAdT same as the walls of a house. Even Pierrehumbert recognized this in his 2011 paper.

Here’s an animated power point.

• Toneb says:

nr65:

“The incoming sees the earth as a disc ”

Yes, it does BUT we are calculating the energy budget for the whole Earth NOT just the sunlit side.
Hence a /4 factor is needed.

“Averaged over the entire planet, the amount of sunlight arriving at the top of Earth’s atmosphere is only one-fourth of the total solar irradiance, or approximately 340 watts per square meter.”

https://earthobservatory.nasa.gov/Features/EnergyBalance/page2.php

• KM says:

(I replied to this further up, but I see the same post is repeated here.)

Toneb, you write:

“That gives a S-B temp of 255K
But we live on a planet at 288K

33K warmer.
Why?”

As mentioned by others, Earth’s surface is not a blackbody. A blackbody is an object that only exchanges energy with its surroundings through radiative heat transfer.

The surface also exchanges heat with the troposphere via conduction (and subsequently convection).
Earth together with its troposphere, however, is a reasonably good approximation of a blackbody. The tropopause just above the troposphere ensures that there is no convective heat flow between the troposphere and the stratosphere. Conductive heat flow at this boundary is also small.

The average temperature of the troposphere is around 255K. Coincidence?

Back to the question, why is the surface 33 K warmer than 255 K?
Going further, why is the temperature at the tropopause 33 K colder than 255 K?

This is due to the lapse rate. Hot air rises and cools off. Cold air drops and warms up. The end result is a temperature gradient with 33K warmer than the average near the surface, and 33K colder than the average at the tropopause. The average, 255 K, stays the same.

• Mack says:

You’ve got the Sun going round and round the Earth, Toneb . The Sun never sets in space, and space is right there, just at the TOA….where satellites measure it as 1368w/sq.m. End of story…no division by 4 or anything. Yearly global average, non directional, covering whole globe at the TOA.

• Ben Wouters says:

Toneb November 25, 2017 at 9:35 am

You forgot to divide by 4 as 1368 is TOA.
The Earth is a rotating oblate speriod.

So that’ll be 342 W/m2 x 0.7
=239 W/m2 available (on average) for each m2 of the sunlit surface of Earth.

That gives a S-B temp of 255K

As any planet with one sun only one half of the Earth is illuminated, so better to divide by 2
(no concept of heat storage for BB’s)
Without atmosphere no clouds, so most probably a more moon like albedo (~0.11)
684 W/m^2 x 0,89 = 609 W/m^2
Dark side 0K ( ~3K if you insist)
SB gives (322K + 0K)/2 = 161K

Actual moon ~197K, actual Earth ~288K.
Why?

• Hugs says:

As any planet with one sun only one half of the Earth is illuminated, so better to divide by 2

The area perpendicular to sunshine is PI R², this flow is spread to half the surface 2 PI R² for half the time, so the factor is roughly four. Your largest mistake is to use surface temperature in S-B law. The Earth radiates and reflects a lot from colder atmosphere, putting TOA exit flow at 340W/surface m², but giving a much higher surface temperature due to the lapse rate.

By the way, what is the approximate surface temp as a function of amount of air in the atmosphere? The less air, the colder it is.

Or, if I make a hole 200m deep, how much that affects temp down there when conduction, convection and wind are not taken into consideration? I think it is about 1K. There’s a greenhouse.

• What’s RGHE? At least, what’s R in R… GreenHouse Effect.

• Tony says:

• nickreality65 says:

Radiative GreenHouse Effect. As opposed to CGHE Convective GreenHouse Effect which is how it actually works.

78. Paul Bahlin says:

Try this thought experiment…

Imagine a cool surface emitting, say, 3 watts per square meter of IR SMACK in the middle of the CO2 emission spectrum. Now focus this IR WITH a magical lossless 1 square meter lens into a 1 square millimeter beam and aim it at a surface emitting 300 watts per square meter.

Does the beam heat up that spot?

• Paul Bahlin, this thought experiment is about the question if it is possible to concentrate black body radiation with static lenses above 100% saturation. Well, it is not possible. For example, if you concentrate sunlight on an object by using as much mirrors and lenses as you like, the temperature of that object will never exceed the surface temperature of the sun.

• menicholas says:

Interesting.
So, if you had a 1 million mile wide lens and placed it near the sun and your lens was able to focus all the radiant energy from that side of the sun onto one tiny point of a solid object, that object could never be heated up to more than the 5000 or so degree temp of the sun?
I am not doubting it, but it sounds like a surprising result.
How do you know?

79. I explained this before. Forget all these analogies, well, the one with the blow torch and the candle is correct, forget all the others.

The easiest way to understand this is a simple industrial boiler such as is found in power plants. From cold start-up, with pipes and walls at ambient, we light the combustors. The blue natural-gas flame is at 4,000 degrees F. After equilibration the pipes and walls come up to the max for the particular metallurgy involved, typically 1,100 or 1,200F. Guess what? Blue flame is still at 4,000F, has not varied even a tenth of a degree.

Another easy one is binary stars, ubiquitous in our Galaxy. All have orbits with apogee and perigee. Guess what? The hotter star does not change its temperature AT ALL at perigee.

Go to engineering school, pass Heat Transfer and all related pre-requisite courses, then we can talk.

• mkelly says:

Michael please see my above post about Hottel charts and CO2 emissivity.

• tty says:

As a matter of fact radiative interaction between close binaries is a well established phenomenon.

80. I think the real greenhouse effect is our rapid heating by a high-energy, small-wavelength source and the slow cooling from our low-energy, long wave emissions. The rates of energy transfer are very different and fortuitously beneficial to our life on Earth.

81. Sparks says:

BTW you can not calculate electromagnetic energy budgets that way shown above.

If you have a (cold) object emitting 2W/m2 and a (warm) object emitting 5W/m2 You can not simply add both objects to get an electromagnetic energy budget of 7W/m2.

You have to work out what’s known as a differential between a (cold) object emitting 2W/m2 and a (warm) object emitting 5W/m2 which is known as potential.

This is done by adding together each object 2W/m2 and 5W/m2 and divide by 2
This will give you a differential value of 3.5W/m2 if there are other objects in the system we have to work out their differential values too, so lets say there were another objects with a potential of 4W/m2 and 6W/m2 we will get a differential value of 5W/m2

To work out our electromagnetic energy budget we then add both differential values of 5W/m2 and 3.5W/m2 and we get 8.5W/m2

The electromagnetic energy budget for this system is 8.5W/m2 with a potential of 15.5W/m2

82. at the molecular level depending on the angle of collision a hot molecule can cool a cold molecule and a cold molecule can heat a hot molecule.

this can easily be seen from the speed of balls on a pool table before and after a collision.

conduction is the elephant in the tea house overlooked in favor of sexy radiation.

83. Mr. Pettersen says:

I find this a bit frustrating that so many well educated people in here can struggle with this.
When we calculate T1-T2 we find the difference between the objects. We then can se how much warmer T2 can become.
The value T1 is the energy radiating from object 1 and since it’s already radiated away that energy has already left object 1
So we can find how much T2 can absorb but we can NOT find out how much energy that’s left in object T1. Remember once again that object 1 has already radiated away all the T1 value.

Removing T2 or adding a T3 will not change T1. T1 will always be T^4 of object 1
There will always be a T0 that any object can radiate to. (Space)
Remember the radiation window to space is not 100% closed so the surface can radiate to space.

And please don’t start counting photons. It’s not the number of photons that sets the temperature but the wavelength or frequency of the energy.
Using the money example it’s not the number of coins passed over the table, but the value of each coin that sets the balance of your account.

But I will say it’s nice to see this being discussed here at wattsupwiththat.

84. Not puny.

All the PE required to keep the entire mass of the atmosphere off the surface against gravity is involved.

That PE reservoir is the source of additional kinetic energy at the surface providing the observed surface temperature enhancement above S-B.

85. Dave in Canmore says:

If only everyone who disagreed with Willis first went through the examples he generously provided!

This thread does show why it’s good practice to be courteous- There’s nothing wrong with being incorrect, but being incorrect and indignant really makes you look foolish.

• Have done. Those examples are not adequate since they ignore conduction and convection as per my explanations above.

86. N. Jensen says:

Ric,

Thanks, do you have a reference/ link on this ?

It just seems terribly counter-intuitive to me, that 4 molecules should be able to warm 10.000.

So, I would like to know the reasoning behind, in detail..

Willis,

How many ‘objects’ are there in the atmosphere ?

And how do they transfer energy ?

AFIK the atmosphere is mostly made of molecules and atoms.

Also, is it not true that the molecules near earth are compressed by the weight of the atmosphere, and therefore warmer

• tty says:

“Also, is it not true that the molecules near earth are compressed by the weight of the atmosphere, and therefore warmer”

No. There is no relation between pressure and temperature. It is only changes in pressure that causes heating and cooling.

• tty,

Re the atmosphere, you say: There is no relation between pressure and temperature. It is only changes in pressure that causes heating and cooling.

Imagine that a suitable absorbing gas, initially at ambient temperature, is contained in a non-absorbing transparent container and that the gas is always at atmospheric pressure due to an escape valve in the lid.

I now aim a constant strength beam of EMR at the enclosure (e.g. from the Sun). Are you saying that the gas will not heat up to a steady-state level above ambient while remaining at the same pressure?

• tty says:

Your experiment only shows that a gas expands when heated and heats when absorbing EMR.

• What you say is true.
Now consider this: gravity adds CONSTANT compression of the atmosphere. It is a constant force, so it adds a constant amount of energy by constant compression.

• Hugs says:

But there is. It’s called the lapse rate and it not caused by the pressure gradient only by itself.

• Compression alone does not create ‘extra’ warmth at a planetary surface. You first need energy moving from an irradiated surface to the mass of an atmosphere via conduction. Then It is the surface variations in density that move energy to or from PE and KE via convection which ultimately produces that warmth once hydrostatic equilibrium has ben achieved and KE is being brought back from PE in descending columns of air.
The denser the atmosphere the more conduction can occur at a given level of insolation hence Venus having a very hot surface even after adjusting for distance from the sun.
Compression alone can generate heat on very large scales such as within stars and maybe gas giants but that is a phenomenon on a far larger scale.

• AndyG55 says:

Pressure allows the atmosphere to retain more heat.

The pressure/temperature gradient is a measured facet of all planets with atmosphere, up to a level of approximately 0.1 atm.

IRRESPECTIVE of atmospheric gas constituents.

• Correct.
There seems to be a universal rule that at densities commensurate with pressure of less than 0.1atm allow too little conduction and convection to significantly affect radiative fluxes.
Planets with such thin atmospheres tend to closely match the S-B equation.

87. Excellent primer, Willis, except that I would not have said this in quite the same way: “Can a cold object leave a warm object warmer than it would be without the cold object? While the answer is generally no, it can do so in the special case when the cold object is hiding an even colder object from view…”

The answer is not “generally no.” Counterexamples abound. One such counterexample is you, if you’re wearing clothes.

Moreover, instead of “even colder” I’ve have said “even colder, or less emissive, or more reflective, or less thermally conductive.” There are many ways in which the presence of a cooler object can make a warmer object warmer than it otherwise would be.

I assume that you’re attempting to educate some confused “sky dragon slayer(s),” who stubbornly persist in the erroneous belief that GHGs cannot warm the Earth because they think that the 2nd Law of Thermodynamics forbids a cold atmospheric gas from warming an already warmer planetary surface.

Good luck with that. Been there, done that, got frustrated. Like many folks at the opposite end of the opinion spectrum, evidence is irrelevant to slayers, because don’t want to understand it. Like the White Queen, they’d rather believe as many as six impossible things before breakfast than learn something which requires admitting, even to themselves, that they were wrong.

Here are some other folks who’ve also tried their hands at educating sky dragon slayers: Dr. Roy Spencer in July 2010 & April 2014, mark4asp on April 30, 2017 at 9:44 am, Frank on May 1, 2017 at 11:58 am, Ed Bo on April 30, 2017 at 9:02 am & May 1, 2017 at 10:38 pm, commieBob on April 30, 2017 at 8:55 am, Roger Sowell on April 30, 2017 at 9:09 am & May 1, 2017 at 6:28 pm, davidmhoffer on April 30, 2017 at 1:16 pm & April 30, 2017 at 2:02 pm, matthewrmarler April 30, 2017 at 8:05 pm, and MarkW on May 1, 2017 at 8:03 am..

An example I sometimes use is space blankets. They work. It works, in part, by reflecting IR from the body being warmed, back to the same body, which helps to warm it.

Space blankets actually work two ways. The more important mechanism is by blocking air movement (and thus convective and evaporative cooling). But the other way is by reflecting IR. That’s why they’re silvered, and that’s the feature which is somewhat analogous to the misnamed greenhouse gas effect.

The space blanket, itself, can be much colder than the body which it helps to warm, and it still works quite well.

GHGs added to the atmosphere are somewhat similar. They are colorants, which tint the atmosphere in the far infrared part of the light spectrum, which causes the air to absorb more LWIR radiation than it otherwise would.

The warm surface of the Earth glows in the far infrared, and those IR emissions represent radiant energy escaping from and thus cooling the surface. GHGs in the atmosphere absorb some of that energy, preventing it from escaping to higher altitudes or outer space. That recaptured energy warms the air a bit.

Those same GHGs also emit LWIR “back-radiation,” some of which makes it back to the surface, where it is absorbed, warming the surface a bit, which is one of several mechanisms by which warmer air warms the surface.

Clouds have a similar effect. On a cloudy night, the ground temperature will cool more slowly than on a cloudless night, even if the clouds, themselves, are much colder than the ground. The reason for that is that clouds reflect and re-radiate LWIR back toward the ground.

• Brett Keane says:

Your characterisation of the atmospheric gravito-thermal effect is as false as the warmista’s.

• What the blanket doesn´t do, is increasing absorption of heat in the colder surroundings.
Which is a proof of the flawed argument of the greenhouse-unicorn: that increasing heat absorption in the increasing amount of dry ice, cause increasing power density of its own heat source.

The blanket does the opposite of what co2 does, and what the atmosphere does. It insulates by reducing/preventing transfer of heat to the absorbing cold air.

• Thanks, nate, for the interesting reference.

• menicholas says:

Interesting to think of CO2 as colourants.
I have used another example of a colourant to counter the assertion that the amount of CO2, being what it is, could not possible have much effect on the entire atmosphere.
This seems at first glance plausible, in a horse sensey kind of way, but I happen to be aware of a very real example of something completely altering the optical properties of a fluid in even tinier concentration that CO2 exists in the air.
I have worked in the lake and wetlands management industry, and an oft-used product is a type of dye which is added to lakes and ponds to block light from penetrating the water and thus it inhibits the growth of unwanted algaes and aquatic plants.
There are a lot of brands of the dye that is used, and the remarkable thing is how small of an amount will dye an entire lake so dark it blocks all but a narrow band of visible wavelengths.
One quart of a product like Blue Lagoon, for example, will block light from getting to the bottom of an entire lake one acre in area and four feet deep. It will even do twice that volume but not quite as dark.
Most of the quart is not even the pigment.
As an aside, one might wonder what happens to the light?

• A C Osborn says:

If the “blue lagoon” prevents the light getting to the bottom of the lake it must be either reflecting or absorbing it.
If it is absorbing it, presumably it is making the upper depths warmer than the would normally be and the bottom colder.
Has anyone done any measurements to find out?

• What an interesting comment, menicholas! I had never heard of Blue Lagoon and products like it. Thank you for teaching me something.

Let’s do the arithmetic. Four acre-feet = 5,213,616 gallons. So 1 qt / 4 acre-feet = 0.1918 ppmv, blocks enough light from passing through 4 feet of water to prevent algae growth on the bottom. Impressive!

A column of the Earth’s atmosphere has about the same mass as a 30 foot column of water. So blocking the light through just four feet of water should require an even darker tint than blocking the absorbed shades of light through the Earth’s atmosphere.

• A C Osborn, a four-foot-deep pond is all “upper depths.” It matters not a whit whether the light is absorbed in the top one foot or by the dirt at the bottom.

Do you really wonder “ff it [the dye] is absorbing it [the light]”? And can’t you think of any better way to find out than with a thermometer?

• Quarts! Four acre-feet = 5,213,616 quarts.

and
s/ff/if/

Sigh.

• A C Osborn says:

daveburton, it is apparent from your condescending, dismissive and disparaging tone that you have been around Mr Eschenbach for much too long.
Or is it a prerequesite for being a CAGW adherent?
So let me tell you where I am coming from, I am just a lowly Engineer who did basic Thermodynamics 50 years ago.
But at that time I worked in a UK Metrology lab where they were measuring to 1 millionths of an inch and also in an Optics lab.
So I know about light and also heat transfer, which is a real problem when working at such fine measurements.
I was Trained to take measurements compared to Certified masters, not take anything for granted and not make assumptions and use Verification.
The absorptiuon of the light is an obvious conclusion, but I have not assumed what happens to the Light and the lake water, all I know is that the light has reduced at the bottom enough to inhibit algae and plants.
So in response to Menicholas’s muse “what happens to the light” I would test the temperature of the water at various depths before and after adding the dye to see if any areas changed temperatures.
If it is hotter at the top than before then that would strongly suggest that the dye was absorbing more of the incoming Radiation than before.
Obviously you have a superior method, presumably involving measuring the light itself, but that does not tell you “what happened to it”, just absence of it.
Please enlighten us, but without the condescension.

• Oh, fer Pete’s sake, A C. It’s dye, not glitter. Dyes absorb light, they don’t reflect it.

88. Kelvin Vaughan says:

If all warm things radiate heat, does oxygen in the atmosphere radiate at ambient temperature? If it does is it not a greenhouse gas too?

89. 1saveenergy says:

327 posts on this & 966 posts on Radiative Heat Transfer of CO2 so far.
all top quality Q & A + discussions, learning lots & that’s what we come here for, well done Anthony for hosting.
(who said science was boring…. or settled)

Would be really useful if all posts were numbered to stop getting lost, is that possible ?

90. N. Jensen says:

I’m not talking about changes in temperature or pressure.

I’m just talking about the fact, that the pressure at say 1 km above msl is higher than at 3 km above msl.

And that this in and of itself implies a higher temperature.

No matter how many CO2 molecules there are in the air.

Can we agree on that ?

• Sparks says:

The thicker a planets atmosphere is, (having more atmospheric mass) the higher its pressure will be. The higher the pressure, the more energy needed to move it around and the more energy that can be absorbed and stored in the system.
Venus’ atmospheric composition has nothing to do with its atmospheric pressure or temperature any more than Mars does, Their ratio of carbon dioxide may be similar, it is a thicker atmosphere on Venus absorbing more energy.
If you strip both Venus and Mars of their atmospheres, their planetary temperature will be equal, allowing for the distance of the sun.

• Density is the key. Thickness can also relate to depth.
The greater the density at the surface the higher the proportion of a given level of insolation that can be transferred to the atmosphere by conduction and the hotter the surface will become once convection creates a hydrostatic equilibrium.
Density is a product of mass and gravity alone because they alone create the pressure that forces molecules closer together.
Thus the surface temperature enhancement must also be a product of mass and gravity alone at any given level of insolation.

91. You should finally get rid of Kiehl & Trenberth energy balance figure and fluxes because they are obsolete and based on the wrong atmosphere. The balance by Stephens et al. is much better:

• Where is the component for the non radiative return of KE to the surface in descending air to offset sensible and latent heating of the atmosphere in thermals (rising air) ?
Stephens et al also miss that out and increase DWIR to offset that omission in order to achieve apparent balance.

• tty says:

Not only in descending air, also in descending rain and snow.

• Correct. As rain and snow fall they warm up with the surrounding air as they descend the lapse rate slope. Nonetheless they still are cooler than ambient at each point in the journey.

• tty says:

“Nonetheless they still are cooler than ambient at each point in the journey.”

Not always, but mostly (ever heard of freezing rain?). And nevertheless they carry a lot of heat down to the surface since they are way above absolute zero,

• Ok tty

There are exceptions but one should try to keep it simple. Sometimes rain falls through a freezing inversion layer on the way down but that is a local short lived event.

Radiative flux is not a conserved quantity. You cannot add 2 (or more) radiative fluxes from different sources and use the resulting algebraic sum to derive the sink temperature via S-B. It’s a complete bastardisation of physics and all diagrams, posts and comments assuming you can IN THE ENTIRETY OF THIS WHOLE BLOG SITE deserve to be thrown in the trash.

This is so fundamental I cannot believe it has not been understood properly.

• Ed Bo says:

You keep repeating this completely erroneous argument. If you had taken and understood the first few weeks of a decent thermodynamics course, you would not be trying to argue that.

Energy IS a conserved quantity. Radiative fluxes transport energy. The necessary corollary to the fact that energy is a conserved quantity is that you can add the energy inputs to any defined system or subsystem, and subtract the energy outputs, to calculate the change in the energy of that system or subsystem.

A decent introductory thermo course will make you do dozens, if not hundreds, of problems like this.

This concept is no more difficult than balancing your checkbook using the principle of “conservation of money”, adding the inputs and subtracting the outputs to calculate how much your bank balance has changed.

• Ed,
You need to obtain a refund from MIT. Radiative fluxes are not a conserved quantity. If you cannot grasp this simple concept, then you have no business making comments here. (hint: radiative fluxes are not measured in Joules)

• Ed Bo says:

SGW:

Radiative fluxes integrated over area and time are energy transfers. These most certainly can be added and subtracted, as indicated and required by the 1st LoT.

• Where on Earth did you get that misconception, Badger & SGW? Seriously, what is your source?

Ed, I am impressed. You are a most patient man.

• Dave Fair says:

There it is, aveollila! The risible 0.6 +/- 17.

Has anyone done any better measurements; say a figure less than 28 times the measured value.

• Kelvin Vaughan says:

How do the Watts relate to time? Are they Watts per second as in Joules?

92. Mike Jonas says:

Thanks, Willis, for an excellent article. As usual, Willis explains things clearly and simply.

It would have been helpful if the Second Law of Thermodynamics had been clearly stated up-front, so that we all knew exactly what the article was explaining. Unfortunately, in recent years the 2nd law has been re-stated in a way which renders it almost useless for general dicussion. The law as currently used says that entropy increases. That’s about as useless as “before present” being used to mean “before 1950” – it renders normal discussion almost impossible.

So, let’s go back to a meaningful version of the 2nd law: In the absence of work, there cannot be a net transfer of energy from a cooler object to a warmer object.

What “absence of work” tells you is that things very quickly get complicated if you don’t actively keep them simple. Willis actively kept his explanation simple, and his demonstrated conclusion was therefore clear: a cool object can make a warm object warmer than it would have been if the cool object had not been there. That’s all he set out to demonstrate, and demonstrate it he did.

What this means is that the 2nd law cannot be invoked to “prove” that CO2 (or GHGs) cannot warm Earth’s surface [*]. That’s all it means. It doesn’t mean that CO2 does or doesn’t warm the surface, only that one particular argument is invalid.

[*]_Sorry about the double negative, but it’s the best I could do.

93. Martin Mason says:

These discussions have been exceptional for me in that they’ve forced me to find out things for myself rather than accept what are often myths repeated on various blogs. The most shocking thing I’ve realised is that as sceptics we are a disorganised rabble who often disagree with each other more than we do the MMGW establishment. We know that they purvey poor science but we have no coherent response to their incorrect but coherent position. I truly believe now that the Lukewarmer position is a complete cop out like being agnostic rather than atheist. Radiation flux does not represent the transfer of heat and radiation received doesn’t always raise the energy level of a body, once this is understood everything falls into place.

• “Radiation flux does not represent the transfer of heat and radiation received doesn’t always raise the energy level of a body, once this is understood everything falls into place.”

Better thus:
Radiation flux does not adequately represent the transfer of heat when there are also non radiative processes going on.

94. “In the absence of work, there cannot be a net transfer of energy from a cooler object to a warmer object.”

Correct, but work is done against gravity in lifting the atmospheric molecules off the surface in the first place.
The presence of that work does allow a colder atmosphere to heat a warmer surface but only by non radiative means involving conduction and convection and only after the lifting of the atmosphere has been completed and hydrostatic equilibrium achieved.

Willis completely omits non radiative processes.

• Stephen
See my comment below at 1:11 pm.

• Thanks ptolemy2
Convection always reorganises to eliminate internal radiative imbalances.
Otherwise no hydrostatic equilibrium and no atmosphere.

95. For those who ask “How does a warm body know to ignore a colder photon?”

There are lots of resources on the web documenting quantum mechanic processes where:

“The electron can be excited only if hν corresponds exactly to an allowed electron
transition in the atom. “
http://www.ntec.ac.uk/Phys/pdfs/6-Emission%20Absorption.pdf

and

“The energy levels for all physical processes at the atomic and molecular levels are quantized, and if there are no available quantized energy levels with spacings which match the quantum energy of the incident radiation, then the material will be transparent to that radiation, and it will pass through.” http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

There are some heavy duty books on the subject for those who want to delve further into this.

In a nutshell, if the IR radiation from the earths surface progressing upwards, is not at one of the precise frequencies that CO2 responds too, then those photons of that frequency just carry on towards space unchanged. (disregarding broadening/wings/shoulders for now)

The range of spectral lines (frequencies/wavelengths) transmitted from the earth’s surface will be many due to the wide range of surface material, coverings etc varying from location to location. We can assume that it is going to be very difficult come up with a value with any accuracy.

The energy contained in a photon depends upon its frequency/wavelength. If the frequency is ‘correct’ it will be absorbed into a CO2 molecule. If the frequency/wavelength is incorrect, it will pass through it.

Just to add confusion, a photon excited CO2 molecule will ‘relax’ to its ‘ground’ state within a few nanoseconds, like all other excited molecules, the decay happens at an exponentially. However, researchers are finding some molecules can exhibint ‘non-exponential decay!: http://iopscience.iop.org/article/10.1088/1742-6596/538/1/012008/pdf

So, going with the theories of a couple of years ago, decaying to ‘ground’ state within a few tens of nanoseconds, thermal IR emitted from the earths surface, some excites a CO2 molecule, say 20nS later, the photon is re-emitted from the molecule and travels elsewhere. What effect does this have on any temperature of CO2 and surrounding gas?

So much to learn.

• tty says:

“The electron can be excited only if hν corresponds exactly to an allowed electron
transition in the atom.“

Quite true, which is the reason gases (with relatively simple, relatively isolated molecules) have line spectra instead of continuous “black body” spectra like solids. Even so there are lots of possible transitions even in fairly simple molecules.

Which you most certainly can’t do as soon as you are dealing with more than one isolated molecule.

Also note that even in the ideal case (=isolated atom in vacuum) the lines are a bit fuzzy due to the uncertainty principle

• Steve quoted, “The electron can be excited only if hν corresponds exactly to an allowed electron
transition in the atom…”
and tty agreed, “Quite true.”

Not true, if we’re talking about 15 µm LWIR. Electron transitions are much too high energy. 15 µm IR photons correspond to a CO2 molecular bending mode, not an electron transition.

The word “exactly” isn’t correct, either. Other sources of energy, such as interactions with nearby molecules and the motion of the CO2 molecule as a whole, can add to and subtract from the “exact” quantum of energy, creating pressure/temperature broadening and hence the “wings” of the spectral lines.
.

Steve also wrote, “a photon excited CO2 molecule will ‘relax’ to its ‘ground’ state within a few nanoseconds…”

Actually, in the case of 15 µm LWIR absorbed by CO2, the mean time for it to give up the absorbed energy by emission of another photon is on the order of one second!!!

Are you startled by that? So was I!

Atmospheric physicist Will Happer mentioned it in a UNC Physics Colloquium three years ago, and kindly explained it to me in a subsequent email exchange.

At 1 Atm and typical temperatures, that’s more than a hundred million times longer than the average time for a CO2 molecule to lose its absorbed LWIR photon’s worth of energy by collisional transfer to another air molecule. So, regardless of how much or little LWIR the CO2 molecules in the atmosphere are absorbing, they remain at almost exactly the same temperature as the other air molecules.

It also means that this lovely animated picture, from the NSF, which even illustrates the correct vibrational mode, is wrong more than 99.999999% of the time:

It also means that when a CO2 molecule in the atmosphere emits a LWIR photon, the energy to do was almost always acquired bu collision with another air molecule, rather than by absorbing a photon.

That implies that the amount of ~15 µm LWIR emitted by atmospheric CO2 depends only on the atmosphere’s temperature (and CO2 partial pressure), not on how the air got to that temperature. Whether the ground is very cold (and emits little IR) or very warm (and emits lots of IR) will not affect the amount of IR emitted by the CO2 in the adjacent atmosphere (except by affecting the temperature of that air).

• Thanks Dave for the interesting and educational email exchange!

• Hi daveburton! You have raised a point that, with few exceptions like Will Happer, is not understood by the majority of CAGW scientists and commentators. The radiative lifetime of a vibrationally excited CO2 molecule is of the order of 1 second, during which time approx. 10^9 to 10^10 collisions with the air molecules (mostly N2) will occur. So most of the energy of 667 cm^-1 photons emitted from a 288 K Earth surface and absorbed by a ground state CO2 molecule will be transferred during radiationless collisions to the non-radiating molecules of the troposphere. This is the mechanism by which greenhouse gases warm the troposphere.

In a previous posting in this thread, I gave a simple analogy of a black metal plate containing a 100 W heating coil (representing the solid and liquid surface of the Earth which absorbs incoming visible radiation from the Sun) and a similar black metal plate parallel to the first, but a passive absorber/emitter (with no internal heating coil) representing a layer of CO2 greenhouse gas. The bulk of the atmosphere consisting of non-radiating N2, O2 and Ar molecules can be represented by a large insulated tank of water or antifreeze connected by a well-insulated heat pipe to the passive plate. The rise of incoming Solar radiation during the daytime can be modelled by increasing the current in the heating coil in the first plate. The IR emitted from the first plate is then absorbed by the passive plate, which also warms up, but most of the heat energy is stored in the tank of water or antifreeze. There will be a lag in the temperature rise in the passive plate/storage tank, but if the current in the heating coil is shut off, simulating nighttime, the rate of cooling of both plates is reduced as heat is now transferred from the contents of the storage tank to the passive plate, which exchanges photons with the first plate. Another analogy could be an electronic power supply converting AC to DC with a single rectifier and a humongous filter capacitor which smooths out the ripple in the half-wave rectified DC output.

Prof. Happer correctly explained that the emission of IR from CO2 to outer space depends on the temperature of the emitting layer, which at 10 km is 220 K. However, this is strictly true only for frequencies close to 667 cm^-1 which are completely saturated all the way through the troposphere and into the stratosphere. Because there is a temperature inversion from 10 to 45 km due to absorption of incoming Solar UV and visible radiation by ozone in the stratosphere, doubling CO2 will cause escape to outer space to occur at higher altitudes where the temperature actually is higher. So the central 667 cm^-1 emission increases above 20 km, requiring less emission from the Earth’s surface for energy balance. Thus doubling CO2 will mean global cooling, not warming, when only these central frequencies are considered. Jack Barrett has run MODTRAN spectral calculations to 70 km, instead of truncating at 20 km (which on average is at 220 K), showing this increased stratospheric emission [see the section “The hard bit” at http://www.barrettbellamyclimate.com/ ].

Of course, doubling CO2 does result in net global warming, not cooling, because frequencies on the far wings of the CO2 absorption ditch are not totally saturated, even in the entire 10 km path length of the troposphere. This shows up as an increase in the area of the CO2 absorption ditch in the spectrum available at https://en.wikipedia.org/wiki/Radiative_forcing . To understand this net absorption, one must go to the integrated Schwarzschild Equation [see the section “Schwarzschild’s Equation” at Jack Barrett’s website].

The radiation intensity for a frequency in the wing of the CO2 absorption ditch is the sum of a Beer-Lambert absorption term and an emission term. Because the emission at 10 km occurs at 220 K instead of at 288 K, the emission term moderates the net absorption by a factor of 0.659 [the signal I = 1 – 0.659A , where A is the Beer-Lambert absorbance, and signal = 1 – A in the absence of the emission term]. This means that the literature value for climate sensitivity is too high, since it doesn’t include the factor 0.659. It is also too high because it was derived for absorption in a cloud-free sky. Clouds are composed of liquid water droplets or ice crystals which act as miniature Planck black bodies which absorb 100% of the IR emitted from the 288 K surface, and then re-emit at the lower temperature of the cloud top. Doubling CO2 beneath and inside the clouds will not increase the absorption because it is already 100%. So only the smaller path length from the cloud top to 10 km will have CO2 molecules that will increase absorbance on doubling CO2.

The increased absorbance on doubling CO2 occurs in sidebands centered at 618 and 721 cm^-1 [see the MODTRAN spectrum referenced above]. This occurs as photons emitted from the 288 K surface are absorbed by CO2 molecules in the v=1 first vibrationally excited state [for the energy level diagram showing the transition, see Diagram 3 in the section “Spectral transitions” at Jack Barrett’s website referenced above]. At 288 K, only 3.2% of CO2 molecules are in this v=1 excited state, and the percentage goes down as temperatures decrease with increasing altitude. The bottom line is that when the 62% of the Earth’s surface which is covered with clouds is considered, the climate sensitivity (not counting feedbacks) is not 1 K, but closer to 0.6 K. And water vapor feedback has been grossly overestimated. Even if it were 50%, this would raise climate sensitivity to 0.9 K, and increased cloud cover would be expected to bring this back down closer to 0.6 or 0.7 K. The literature value of 3 K is way too high. Therefore wrecking the economy to try to keep the effect of doubling CO2 to 2 K is unnecessary, wasteful, and foolish.

I don’t want to bore others with more details in this Forum, but if you contact me at rtaguchi@rogers.com , I can send you pdf files with extended discussions. Willis, too, might be interested, and I respect his intelligence and fair-mindedness.

• tony mcleod says:

rogertaguchi
“This is the mechanism by which greenhouse gases warm the troposphere.”

And the surface?

• Hi, RogerTaguchi. I’m glad you enjoyed the video of Prof. Happer’s colloquium. I made the video by combining his PowerPoint slides with an audio recording that I had made with my smartphone, which is far from ideal. But I think the result is serviceable.

I didn’t understand the black metal plates analogy, sorry.

I don’t think that the Effective Radiating Level (ERL) (or Effective Emission Height, or any of several similar names) is above the tropopause, except, perhaps, right at the 667 cm^-1 (15 µm) absorption peak. From what I’ve read, the ERL is within or below the tropopause over most of the 15 µm absorption/emission band. Various sources give the average ERL for CO2’s 15 µm band as being somewhere between 4 km and 12 km altitude, which is well below the ~20km altitude at which atmospheric temperature increases much with increasing height.

So the net effect of adding CO2 to the atmosphere and thus raising the ERL for the 15 µm absorption/emission band is to lower the average emission temperature, leading, as you noted, to net surface warming.

An interesting tidbit is the contrast between the calculated emission profile viewed at 20km, as shown in the Wikipedia article, and the actual satellite-measured emission profile.

Here’s the Wikipedia calculated profile:

Here’s a measured profile (part of Prof. Happer’s slide #16, for the Tropical western Pacific):

Look at the center of the CO2 “absorption ditch” — do you see the difference? The narrow spike in the middle of the measured spectrum is apparently emissions from CO2 in the warm but extremely thin stratosphere (above 20 km altitude).

You can also see it in Jack Barrett’s simulated spectra, looking down from 70 km (the last figure on his “hard bit” page†).

Note to future readers: the links on BarrettBellamyClimate.com are numbered pages, and the numbers occasionally change. So if the link doesn’t take you to the correct page, go to the main page, ctrl-F, and search for “hard bit”.

• A C Osborn says:

daveburton November 26, 2017 at 1:38 am
“So the net effect of adding CO2 to the atmosphere and thus raising the ERL for the 15 µm absorption/emission band is to lower the average emission temperature, leading, as you noted, to net surface warming.”

Why do you say that when the only Actual Measurements we have show that adding CO2 Inreases the Cooling of the Atmosphere and therefore of the Earth’s Surface.
Nasa have shown that inreasing CO2 increases cooling.
see the 2013 AGU Presentation starting at about 15 minutes on this Video.

I do not see the same correlation between Raw Surface Temps and CO2 increases.

• 1. Sorry, for some reason Wikipedia’s calculated 20 km spectrum graph didn’t show up in my 1:38am reply to RogerTaguchi (above). It should have been:

2. A C Osborn, thanks for the link.

(a) In that talk, Marty Mlynczak says that he’s discussing the thermosphere, above 100 km altitude. Cooling in the thermosphere/ionosphere does not imply cooling down in the troposphere.

(b) Here are his slides: https://fallmeeting.agu.org/2013/files/2013/12/PressConfMlynczakFinal.pdf

(c) For future reference, you can link directly to any starting point in a YouTube video using any of several syntaxes, like this (these all start that AGU video at the beginning of Dr. Mlynczak’s talk):

For example, here’s a direct link (using the “#t=19m49s” syntax) to the presentation immediately following M’s — which happens to have been by friend-of-WUWT Leif Svalgaard, on the topic of predicting solar cycles:

3. A C Osborn and The Reverend Badger, please note that in these “indented threads” there’s no way to tell who you’re talking to or what you’re taking about when you say things like, “Does this apply to…” or “Try talking about…”, unless you tell us. That’s why Willis always says, “When you comment, please QUOTE THE EXACT WORDS THAT YOU ARE DISCUSSING, so that we can all understand the nature of your objections.”

Try talking about e-m radiation, don’t use the word “photon” and don’t say “frequency/wavelength”, just “frequency”. When you do this the argument can take on a different “flavour” which MIGHT just help in the learning.

• Ed Bo says:

So Badger, the last hundred years of EMR physics has been a complete mistake, and you are the only super-genius who recognizes this?

• A C Osborn says:

Does this apply to all materials, or just CO2?

• Steve Richards November 25, 2017 at 1:04 pm

So, going with the theories of a couple of years ago, decaying to ‘ground’ state within a few tens of nanoseconds, thermal IR emitted from the earths surface, some excites a CO2 molecule, say 20nS later, the photon is re-emitted from the molecule and travels elsewhere. What effect does this have on any temperature of CO2 and surrounding gas?

So much to learn.

Indeed, the decay to the ground state in tens of nanoseconds is due to collisional deactivation and transfers energy to the surrounding molecules (about 10 collisions per nanosecond near the surface) thereby warming up the atmosphere. The radiative decay has a much longer characteristic time so doesn’t become the dominant process until higher in the atmosphere.

96. Mark - Helsinki says:

Energy Balance\resulting temperatures

At the fundamental level, how much kinetic energy is created from input, how much is retained after output
what happens to the KE retained, well we know, it makes the flipping weather, the oceans move, and so much more, including life, we are full of kinetic energy ourselves.

• Mark - Helsinki says:

But can we master this problem? of course not.

Temperature is just an average measurement of kinetic energy, right off the bat we lose accuracy.

But sciency jargon or whatever.

97. An elegant clarification, thanks Willis.

Radiation is far from being the whole story of heat movement in the atmosphere. The body of theory of Ilya Prigogine concerning nonlinear thermodynamics and dissipative structures is important, though often overlooked. Changes to the thermal structure of the atmosphere may well – according to principles of self-organisation – result in rearrangement of dissipative and oscillatory structures which could establish a new equilibrium with minimal or no overall change to thermal fluxes.

The Benard or convection instability is manifest in a situation in which a fluid layer is heated from below and kept at a fixed temperature above so as to create a temperature gradient in opposition to the effects of gravitational force. At small values of this gradient heat is transported from lower to upper regions by conduction [and radiation] and macroscopic motion is absent. Random motions of the molecules and a damping of convection currents characterise the state of the fluid. However, when the gradient exceeds a critical value a convective, macroscopic motion occurs generally in the form of rolls or hexagons (for variations see Koschmeider 1977). In short, out of an initial state that is completely homogeneous there arises a well ordered spatial pattern. Moreover, with further increases in the gradient the spatial pattern becomes oscillatory.

Kugler PN, Kelso JS, Turvey MT. 1 On the Concept of Coordinative Structures as Dissipative Structures: I. Theoretical Lines of Convergence. Advances in Psychology. 1980 Dec 31;1:3-47.

98. Mark - Helsinki says:

Most scientists hate to talk in simplistic terms, sadly, the like to sound oh so smart

• Mark
Essential chaos theory is much less complex than it sometimes looks. I’m a mere biologist. Hint – ignore the maths, just look at the pictures.

• tty says:

“ignore the maths, just look at the pictures”

Not recommended. You don’t necessarily have to do the mathematics but you need to understand the underlying principles and assumptions. Look at Mann and his hockey stick for a prime example of what can happen when you try to apply mathematics you don’t understand.

• ptolemy2, I just taught this subject to a class of biologists this week, differential equations in a bio class! For an example of order from chaos check out the cell cycle and cyclins.

99. Martin Mason says:

Stephen, I didn’t want to be so provocative as to say that any atmospheric model that omits the heat engine effect of the troposphere isn’t complete but you’re correct that it must.

How is it that the temperature at any point on earth can be determined by incoming solar radiation and surface emissivity with no reference to back radiation?

If radiation dominates how come the derivation of lapse rate does not include any radiation inputs?

If we have a downwelling heat source of 340 W/m2 why can’t we recover this at the surface as we can recover heat from solar radiation.

• Martin Mason

I assume you are addressing me but I don’t accept that any radiation other than external radiation dominates surface temperature.

Thus I raise the same questions as you.

As regards the apparent irrecoverability of DWIR as additional surface heat over and above S-B plus the mass induced surface temperature enhancement I suggest that any DWIR from the atmosphere is already merged into the surface temperature created by insolation, conduction and convection via the lapse rate slope.

As DWIR descends through lower levels of radiatively active material it distorts the lapse rate slope and reduces convective vigour so that warmed radiative material of any description finds itself at a lower height and warmer temperature than it otherwise would be so that it radiates more to space and the radiative capability for warming the surface is neutralised.

If there is to be a radiative balance with space all outgoing radiation must emanate either from the surface or from within the bulk of the atmosphere. If too much is going out from one source then the other adjusts accordingly. Could never maintain long term hydrostatic equilibrium otherwise.

• Paul Aubrin says:

“If we have a downwelling heat source of 340 W/m2 why can’t we recover this at the surface as we can recover heat from solar radiation.”
Very good question. The answer is simple: You need a GHG panel at 0K (-273.13°C).

• A C Osborn says:

Paul, it is much worse than that.
There is ZERO Empirical Data to show that there are 340 W/m2 hittng and “Warming” the surface.
There are simple tests where you can MEASURE the ACTUAL POWER of Sunlight hitting the surface.
Simple measurements with simple calculations that confirm the Satellite measurements of the Real Watts coming from the Sun.
see as an example
https://www.bing.com/videos/search?q=how+to+measure+the+power+of+sunshine&qpvt=how+to+measure+the+power+of+sunshine&view=detail&mid=1CE167A1295F871DFC441CE167A1295F871DFC44&FORM=VRDGAR
There are no such tests to show the Actual Power coming down as “Back Radiation”.
The closest they get is the sort of Experiment that Dr Spencer did and when the Target got COLDER and he said that proved that it was being warmed by the DWLIR because it would have been even colder if it wasn’t.
Does that sound like 340 Watts to you?

• Ben Wouters says:

Ferdinand Engelbeen November 26, 2017 at 11:09 am

There are several stations on earth measuring downward IR radiation, some of them already from the 1950’s:

Looking forward to something similar to the Solar Challenge
https://www.worldsolarchallenge.org/dashboard/timing
but this time using cars driven by backradiation panels, and let’s make it a night race ;-)

The pyrgeometers used in measuring backradiation calculate the radiation using emissivity 1.0 according a spec sheet I saw. Seems unrealistically high to me.

• Ben Wouters,

The problem is converting IR to power: conventional cells transform visible light to power, with some yield, but as IR photons have a lot less energy, that doesn’t work (yet).

Read some years ago the possibility to recover low temperature waste heat by an array of micro thermocouples on a chip, where one side was heated by waste water, the other side cooled with ambient air.
Problem here is that the surface at night is loosing more energy than what is coming in as backradiation. Thus the “warm” side is the car/panels, so you need a lot of mass/weight to give some power…

Radiation is only basic knowledge (of about 50 years ago…) for me, so I can’t directly discuss your point about an emissivity of 1 (for the atmosphere?).

• A C,
It’s even “worser” than you think. According to the Trenberth energy balance diagram, at the earth’s surface, backradiation from the atmosphere is over twice as powerful as the sun. Yet backradiation is powerless to heat up your pool at night as sunlight does during the day. Backradiation does not warm surfaces in the shade. Yet Trenberth assumes that 1 W/m2 of backradiation is equivalent to 1 W/m2 of solar radiation. Absurd.

The reason backradiation is powerless to perform work is simple. A colder atmosphere cannot transfer heat to a warmer earth surface.

The earth and atmosphere are purely passive. The only energy source available to heat the earth is the sun. And the sun has an average power of 341 W/m2 at the TOA. It is thermodynamic horse manure to indicate that this 341 W/m2 of solar energy gets magnified to 494 W/m2 at the earth’s surface.

• A C Osborn says:

Yes, they have less power, but how much less?
Why aren’t 340 LWIR watts = 340 sunshine watts?
This is the part I am trying to establish, I have seen at the bottom of this post that there is 1/40th of the energy for the radiation, so there is either 40 times as much dwi radiation or this whole back radiation thing does not work as advertised.
Everything I have seen up till now says it is not able to prove to work.
Thanks for the link to that CO2 ARM paper I remember it when it came out and it got a lot of stick at the time not least for having the last 5 years of data missing.
I may try and register with ARM to try and get some downloaded data as that paper only shows anomalies, which do not tell yo very much.

100. Berényi Péter says:

Willis, here are the five heat reservoirs in your picture.
1. sun
2. surface
3. troposphere
4. stratosphere
5. space
The sun is a tiny speck in the sky, the rest is some 186 thousand times larger. However, it is hot (5778 K), while space is cold (several kelvins), so radiative energy coming from the sun is millions of times larger than all the rest due to the fourth power in radiation law. Both heat reservoirs (1 &. 5) can be considered infinite on scales relevant to the climate system.

As for the rest, heat capacity of the surface (including oceans) is vastly larger than any one of your atmospheric layers.

About 30% of incoming solar radiation is rejected by the Earth system, it goes back to space without thermal interaction (albedo effect). That is, only 237 W/m2 gets thermalized in some part of the climate system and from then on it acts like heat (because it is heat), so “back radiation” has no meaning at all, only net heat transfer by radiation, which is an entirely different beast.

Albedo of the Earth system is strictly regulated. We know this, because annual integrated incoming solar radiation is exactly the same for the two hemispheres (due to a peculiar property of Keplerian orbits). Now, absorbed heat is also measured to be the same for the two hemispheres (for decades now, by satellites), in spite of the fact that clear sky albedo of the Southern hemisphere is much lower due to prevalence of oceans there (under clear sky conditions it reflects some 6 W/m2 less back to space than the Northern hemisphere). Not so under all sky conditions, which means there must be more clouds in the Southern hemisphere and by an amount required by an exact match.

This regulation is neither understood nor replicated by computational climate models. Therefore we have no idea what effect increasing level of atmospheric greenhouse gases may have on it, if any. However, temperature obviously depends on albedo in first approximation.

Another interesting notion is that radiative heat transfer from surface to troposphere is next to insignificant, it is 18 W/m2 compared to 98 W/m2 by convection. Therefore heat transfer between the surface and troposphere is hardly affected by greenhouse gases.

As there is no convective heat transfer to and from the other heat reservoirs, the rest of it is purely radiative.

But it is hard to tell, what effect of changing greenhouse gas levels may have on them. For example we do know the stratosphere is cooling lately (TLS: Temperature Lower Stratosphere).

However, it does not necessarily mean that the 147 W/m2 radiative heat transfer from here to space is decreasing, because outgoing heat radiation of a body depends not only on its temperature, but also on its emissivity. Which, for the stratosphere, is clearly increasing due to increasing greenhouse levels, because it equals to absorptivity under LTE (Local Thermodynamic Equilibrium) according to Kirchhoff’s Law, a defining feature of greenhouse gases.

101. Berényi Péter says:

Willis, here are the five heat reservoirs in your picture.
1. sun
2. surface
3. troposphere
4. stratosphere
5. space
The sun is a tiny speck in the sky, the rest is some 186 thousand times larger. However, it is hot (5778 K), while space is cold (several kelvins), so radiative energy coming from the sun is millions of times larger than all the rest due to the fourth power in radiation law. Both heat reservoirs (1 &. 5) can be considered infinite on scales relevant to the climate system.

As for the rest, heat capacity of the surface (including oceans) is vastly larger than any one of your atmospheric layers.

About 30% of incoming solar radiation is rejected by the Earth system, it goes back to space without thermal interaction (albedo effect). That is, only 237 W/m2 gets thermalized in some part of the climate system and from then on it acts like heat (because it is heat), so “back radiation” has no meaning at all, only net heat transfer by radiation, which is an entirely different beast.

Albedo of the Earth system is strictly regulated. We know this, because annual integrated incoming solar radiation is exactly the same for the two hemispheres (due to a peculiar property of Keplerian orbits). Now, absorbed heat is also measured to be the same for the two hemispheres (for decades now, by satellites), in spite of the fact that clear sky albedo of the Southern hemisphere is much lower due to prevalence of oceans there (under clear sky conditions it reflects some 6 W/m2 less back to space than the Northern hemisphere). Not so under all sky conditions, which means there must be more clouds in the Southern hemisphere and by an amount required by an exact match.

This regulation is neither understood nor replicated by computational climate models. Therefore we have no idea what effect increasing level of atmospheric greenhouse gases may have on it, if any. However, temperature obviously depends on albedo in first approximation.

Another interesting notion is that radiative heat transfer from surface to troposphere is next to insignificant, it is 18 W/m2 compared to 98 W/m2 by convection. Therefore heat transfer between the surface and troposphere is hardly affected by greenhouse gases.

As there is no convective heat transfer to and from the other heat reservoirs, the rest of it is purely radiative.

But it is hard to tell, what effect of changing greenhouse gas levels may have on them. For example we do know the stratosphere is cooling lately (TLS: Temperature Lower Stratosphere).

However, it does not necessarily mean that the 147 W/m2 radiative heat transfer from here to space is decreasing, because outgoing heat radiation of a body depends not only on its temperature, but also on its emissivity. Which, for the stratosphere, is clearly increasing due to increasing greenhouse levels, because it equals to absorptivity under LTE (Local Thermodynamic Equilibrium) according to Kirchhoff’s Law, a defining feature of greenhouse gases.

• Curious George says:

“annual integrated incoming solar radiation is exactly the same for the two hemispheres.” I thought that the Southern Hemisphere was getting more (we are closer to the Sun during the Southern summer). Link, please.

• Berényi Péter says:

Kepler’s second law
A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

It means angular speed is inversely proportional to the square of distance to the Sun. On the other hand radiation flux coming from the Sun is also inversely proportional to the same quantity. Therefore incoming radiation integrated for a given angle is constant. The angle between equinoxes is 180 degrees (not counting precession, which is slow anyway and gives a negligible contribution – 25 arc seconds between equinoxes).

Between the spring and fall equinoxes the Northern hemisphere gets as much incoming solar radiation at ToA (Top of Atmosphere) as the Southern one between the fall and spring equinoxes and vice versa.

Q.E.D.

But the point is, annual reflected, consequently absorbed radiation is also the same.

• Curious George says:

Peter, thank you, you are 100% right. Let’s get to numbers. The Northern hemisphere gets as much solar radiation in 186 days as the Southern hemisphere gets in 178 days. While the total is the same, the Southern hemisphere should be a little warmer – that’s what I meant.

• Berényi Péter says:

Annual mean temperature of the Southern hemisphere is in fact a bit lower, in spite of it is getting the same amount of energy input on annual bases.

This is why outgoing longwave radiation is not so balanced, the inter hemispheric difference is at least an order of magnitude larger (the Northern hemisphere radiates more to space).

To compensate for it, there is about 256 TW net heat transfer from the Southern hemisphere to the Northern one, mainly by oceanic currents. The difference comes out to be roughly 1 W/m^2.

• A C Osborn says:

Is the difference basically due to the absorption of so much more water in the SH?

• Berényi Péter says:

@A C Osborn November 27, 2017 at 4:22 am

Yes, this is why clear sky albedo of the Southern hemisphere is 1.75% lower than that of the Northern one. Deep water (like oceans) is very dark, except for shallow incidence angles, when water surface acts like a mirror. If the light ray is perpendicular to the surface, only 2% is reflected, while the rest is absorbed at some depth. On the other hand, small water droplets (like ones in clouds) are white.

This is how albedo is regulated to the extent annual average absorbed shortwave radiation be the same for the two hemispheres.

Some say regulation is done by the exact positioning of the ITCZ (Inter Tropical Convergence Zone), which is cloudy indeed.

However, it can’t be the whole story, because on average the ITCZ lies mostly in the Northern hemisphere, so the Southern one has to have abundant cloud cover elsewhere as well.

• A C Osborn says:

Thank you.

• menicholas says:

This diagram appears to show the ITCZ going right through the middle of the Sahara desert in Summer.
I do not think it is accurate

• Berenyi
Very illuminating and compelling.
I’ve always felt that normalising adaptation of a complex-chaotic (plus open and dissipative) system such as climate must be taking place.
Your discussion of unity of thermal budgets in both hemispheres is a strong pointer that this is indeed the case.

• Berényi Péter says:

There is a theoretical proof, that in reproducible closed (but not isolated) non equilibrium thermodynamic systems the MEPP (Maximum Entropy Production Principle) holds.

A system is reproducible iff microstates belonging to the same macrostate can’t evolve into different macrostates, that is, the system is kinda macrostate-preserving.

If the terrestrial climate system were of this kind, Earth would be pitch black as seen from space (it would have low albedo), because most of the entropy production happens, when incoming shortwave radiation gets absorbed and thermalized.

Earth is obviously not black, its albedo is about 30%.

Therefore it can’t be reproducible, that is, it must have at least two microstates belonging to the same macrostate, that can evolve into different macrostates in a short time.

Indeed it is irreproducible, for the climate system is chaotic, because there are turbulent flows in it. True, it is not turbulent below the Kolmogorov length scale, an order-of-magnitude estimate for which is about 1 mm in the atmosphere. However, the atmosphere contains some 5×10^27 cells of this size, with 2.5×10^16 molecules in each (1.5×10^17 degrees of freedom). With that many cells for example average energy per degrees of freedom (temperature) in a cell is expected to fluctuate by one part in ten million. Which, due to the butterfly effect, contradicts reproducibility.

BTW, we would need some 10^22 times more computing power to skip turbulence in computational climate models, which is impossible. So models have to be parametrized in this respect, but, unfortunately, there is no way to validate the parametrization. An Earth sized wind channel would be way too expensive and also, impossible.

The sum of this is that the chaotic nature of the terrestrial climate system not only makes exact calculations impossible, but also determines the color of Earth (light blue, as opposed to black). Also, theory of irreproducible systems (where Jaynes entropy can’t even be defined) is one of the few uncharted territories of classical physics, so we have no idea how albedo is regulated. The only thing we know, it is.

As albedo is a major player in determining surface temperature, the end result is it’s a bit early to construct computational models, because the theoretical foundations are lacking. And building expensive policies on non existent models is beyond insanity.

If only climate scientists would make actual science instead of being mindless advocates in a policy debate, it could turn out better. For example, someone should do experiments on an irreproducible closed thermodynamic system, obviously on one which is a small enough member of this broad class to fit into a lab setup. There are plenty of such systems and we may even learn something important by studying them. Furthermore, it is not a computational exercise for sure.

JOURNAL OF PHYSICS A: MATHEMATICAL AND GENERAL 38 (2005) pp 371–381
doi: 10.1088/0305-4470/38/21/L01
Received 22 December 2004, in final form 13 April 2005
Published 10 May 2005
Maximum entropy production and the fluctuation theorem
R.C.Dewar

• Curious George says:

Peter, a great explanation, thanks.

102. 1sky1 says:

There’s a crucial difference between the valid explanation given here in terms of the PRESENCE of a cold atmosphere, whose backradiation reduces RADIATIVE cooling of the surface, and the customary explanation of the so-called GHE, which attributes the ENTIRE increase in surface temperature solely to GHGs.

In fact, an atmosphere without any GHGs would still be warmed by thermal convection and conduction, reducing the cooling rate of the surface by those non-radiative mechanisms. And if water vapor were the sole GHG in the atmosphere, the surface temperature would not be greatly different from what is experienced now.

• 1sky1

The consensus view is that the reduction in cooling from the presence of GHGs accounts for the entire observed warming above S-B which is usually quantified as 33C.

You are correct that even a radiatively inert atmosphere would have conduction, convection and a surface temperature enhancement with radiation to space going out from the surface alone and the surplus surface KE being cycled up and down indefinitely and switched between KE and PE in convective overturning.

What happens if you then introduce radiative material of any kind, whether water vapour or not, is that some of the energy needed to go out to space comes from within the atmosphere so that less needs to go out from the surface.

In theory that would require a lower surface temperature but instead the lapse rate slope becomes less steep, the vigour of convection falls and energy is taken up by convection from the surface less fast so that the surface stays at the same temperature.

That is how one should integrate the thermal effects of radiative and non radiative processes.

• Brett Keane says:

Stephen Wilde
November 25, 2017 at 2:02 pm: Thankyou, Stephen.
Everything over 0K seems to radiate kinetically. If that has no effect, then what does that mean if anything for the spectra measured and touted by warmista?

• Berényi Péter says:

There were only very shallow convection in an atmosphere completely transparent in IR, only to make latitudinal heat transfer possible. Otherwise temperature would be uniform above a pretty low altitude.

However, between Earth’s surface and troposphere only 15% of heat transfer happens radiatively, but there is a vigorous convection driven by radiative cooling at the top of troposphere. If radiative transfer decreased inside it, heat flux would still remain constant, because convective transfer would increase driven by a greater temperature difference.

Therefore greenhouse gases only have an effect in the upper troposphere and lower stratosphere, where convection is thwarted.

• In an atmosphere completely transparent to IR all radiation to space would have to be from the surface and the vigour of convection would have to be at its fastest possible in order to get KE back to the surface in time to equalise radiation out to space with radiation in to space.
You would still have a temperature gradient along the pressure gradient due to the conversion of KE (sensible heat) to PE (non sensible heat) in uplift and the opposite in descent. Due to uneven surface heating and consequent density differentials in the horizontal plane an isothermal atmosphere would be impossible.
If you then add GHGs which provide an additional route for radiation to space from within the atmosphere then those GHGs also distort the adiabatic lapse rate slope to the warm side which reduces the vigour of convection which would otherwise be needed for convection to match surface radiation to space with radiation in from space. The GHGs are then assisting the shedding of radiation to space so that convection doesn’t work so fast.
Due to less vigorous convection the GHGs radiate to space more effectively from a lower warmer height along the lapse rate slope than would otherwise have been the case.
The net outcome is to neutralise any potential for surface warming from those GHGs.
That is why all planets with atmospheres can keep their atmospheres in hydrostatic equilibrium indefinitely regardless of the proportion of GHGs present.
The AGW radiative theory is fatally flawed.

• Berényi Péter says:

@Stephen Wilde November 28, 2017 at 11:26 am

Why would the atmosphere get colder upwards, if only the surface is cooling?
If it is not getting colder, no convection is possible.
In fact it should be getting colder at least at the adiabatic lapse rate, otherwise a parcel rising would get colder (and denser) than its environment, so the atmosphere would be stable against convection.

103. FYI: I’m on a physics forum trying to get some answers. Seems Postma is banned from commenting on WUWT, but is aware of this thread. Too bad, he should be able to contribute his view of things.

I used to think all this banning on all sides of the debates was a bad thing but about 10% of the regular contributors on here are repeatedly rude / insulting in comments so I am now thinking we could have a better debate if at least 100 more were banned.

• arfurbryant says:

TRB,

I agree wth you. In fact, I would like to suggest to Mr Watts that the best way forward with this particular topic – with is absolutely fundamental to the ‘CO2 = CAGW’ debate – would be to have a closed debate where, for example, a selection of around 8 people only are allowed to post whilst everyone else merely views. To be fair, I would try to arrange the posters to be equally representing the Warmist, Luke-Warmist, Sceptical and Denier groups. All chosen debaters would have to agree to debate without resorting to personal attacks, ridicule, sarcasm etc.
I for one would just love to see an intellectual debate between polite, knowledgeable and respectful protagonists with maybe one other person (ideally unbiased) as referee. Finalising the chosen debaters would be challenging but worthwhile, in my opinion.

As a starting premise or proposition, I would suggest: “There is no valid physical mechanism by which a change in atmospheric Carbon Dioxide can measurably affect the Earth’s average temperature.”

• J. Richard Wakefield, Mr. Postma does not “contribute” to discussions, he pollutes them. That’s why he’s unwelcome.

Anthony is very broad-minded about commenting at WUWT, and he welcomes commenters with a wide variety of viewpoints. (It’s a nice contrast to the heavy-handed censorship prevalent at most alarmist climate blogs.) But even Anthony’s great patience has its limit.

I strongly recommend that you stay away from Principia-Scientific (PSI). It is a disinformation site, run by nutjobs.

They mix truth and fiction, which just makes their web site even more dangerously deceptive than those hideous hoax/parody sites, because it makes the fiction harder to recognize.

“Falsehood is never so false as when it is very nearly true.”
– G.K. Chesterton

Much of the material on the PSI site is good articles which they have simply copied from other sources. But the material which they wrote themselves is mostly deceptive, or nuts, or both.

The four main authors at PSI are John O’Sullivan (CEO) [not to be confused with “the good John O’Sullivan,” who writes for National Review, and who the PSI John O’Sullivan occasionally impersonates], Joe Olson, Joe Postma, and Pierre Latour. In addition to the web site, they also have a book, entitled “Slaying the Sky Dragon.” So they are often referred to as “Sky Dragon Slayers,” or just “Slayers.”

Joe Olson and John O’Sullivan are the two most prolific author at Principia-Scientific, and Postma is #3. A google site search for Olson name finds 723 hits (down from 1760 in July). John O’Sullivan has 986 (down from 1030 in July), Joseph Postma has 236 (down from 453 in July), and Pierre Latour has 193 (down from 288 in July):

They used to have a 5th prolific author, Douglas Cotton, but he had a falling out with the others:

When I say those guys are nuts, I don’t mean they are just a little bit odd. I mean they are stark, raving, clinically insane. For example, this is a quote from a April 17, 2015 email from Joe Olson:

> “9/11 Conspircy Solved, Names, Connections, Details” and my interview
> with Dr James Fetzer, “Unequivocal 9/11 Nukes” are both on youtube.

Yes, you read that correctly. Joe Olson, one of the top two authors at Principia-Scientific, claims to know that the 9-11-2001 attacks were an “inside job” by the Bush Administration, and that the Twin Towers at the World Trade Center were destroyed with nuclear weapons. (If you care to do so you can search for those titles and find his youtube videos; if you are a masochist you can watch them, but I would not advise it.)

For the sake of your sanity, I strongly recommend that you stay away from Principia-Scientific, their web site, and their idiotic “Sky Dragon” book — and that includes Postma.

• arfurbryant says:

daveburton: [“When I say those guys are nuts, I don’t mean they are just a little bit odd. I mean they are stark, raving, clinically insane.”]

I consider your post to be utterly distasteful and not worthy of this blog or this debate. I am sure J Richard Wakefield is capable of rising above your personal opinion in making his own mind up.

Moderators – what were you thinking?

[distasteful, yes, but when somebody claims 9/11 was an inside job caused by nuclear weapons, it tends to support Mr. Burton’s opinion that their views on climate aren’t to be trusted.

Mr. Postma seems to be an equal-opportunity hater. read this https://climateofsophistry.com/2017/09/13/the-walking-braindead-flat-earther-science-denier-list/ -mod]

• SkepticGoneWIld says:

Right Dave,
Postma believes in the stark raving mad position that heat only transfers from warm objects to cold objects per the Second Law and radiative heat flow equation.

Postma has at least 7 peer-reviewed publications in legitimate science journals as well. And Dave has how many scientific publications?

• Many years ago I declined an invitation to join PSI because I found a certain lack of rationality.
The mass induced surface warming effect based on atmospheric mass and the gas laws via conduction and convection seemed too simple for them since it made them look as misguided as the AGW alarmists.

• Given his foul-mouthed abusive posts to anyone who disagrees with he should definitely not be allowed to post his trash on here!

EXCELLENT! Both highly relevant – well worth reading.

1. Radiative flux is not a conserved quantity. You cannot add algebraically the radiative flux intensities from two different sources and use the arithmetic result to derive the sink temperature via S-B.
2. Radiation from objects is electro-magnetic with a range of frequencies (with an upper limit) and NOT a stream of little ping pong ball like elementary particles (just stop it with the “photons”, you are confusing yourselves!).
3. If your mind is still boggling and you fancy a diversion down a different less traveled path but with the same destination try this;http://www2.ups.edu/physics/faculty/evans/Pictet%27s%20experiment.pdf

4. The ERL (Effective Radiating Level) is NOT a real physical location (level/layer) from which radiation zaps off to outer space. If you have even the slightest idea that it is please go and research the origin of the term.

105. Definitions:
Heat – is simply the transfer of energy from a hot object to a colder object. (hence a cold object by definition cannot transfer heat to a hotter object)

Temperature – (hot & cold) a measure of the average kinetic energy of the particles in an object.

Second Law – is about the quality of energy. It states that as energy is transferred or transformed, more and more of it is wasted. The Second Law also states that there is a natural tendency of any isolated system to degenerate into a more disordered state. In simple terms, entropy always increases.

Entropy – is defined as a state of disorder or decline into disorder.

Seems some here are thinking that hot mean heat. That IR radiation is a form of heat, which it is not when the IR is from a colder object than the warmer object interacting with the IR.

It’s the Watts wot done it! Radiative flux in WATTS/m2, people think its like their heaters, 2kW convector or WATT-eva. (Not Anthony’s fault, and probably not his parents’ either).

• arfurbryant says:

Superb post! Thank you.

106. Toto says:

Willis is a brave man to take this on. Willis is right, but this shows that the argument that global warming is simple physics is simply wrong. So many people do not understand this “simple” physics, so the physics is not so simple.

Figure 1 at the top and WE’s explanation of it are a very good start to explaining the Second Law of Thermodynamics. It explains that the law is about net flows. Flow is an amount of something over time, so the diagram could show the two boxes as radiators/absorbers of money over some time period like a day or year. It could even show coins going back and forth to represent photons; maybe that’s taking the analogy too far. Cash flow works to show how net flow is important. It also shows that there is no problem in a poor person sending money to a rich person; it does make the rich person richer.

However, the example does not explain the Second Law of Thermodynamics. There is nothing in the analogy which says the rich person has to radiate more money to the poor one, with a net cash flow of rich to poor.

The Second Law of Thermodynamics was formulated at first from observations, before they even knew what heat was. The theory of why it is so came later and it is not simple

107. WE, thanks. Unlike usual, skipped all comments to just say much appreciated post. Goes far to clarifying many here commented misconceptions. Well done, in layman’s English.

108. Paul says:

l am going back over 50 years in my memory but we had thermo course or courses first then had heat transfer courses and it all seemed simple and logical and followed in steps.

109. AndyG55 says:

In certain circumstances, a second object can slow the cooling of a warmer object.

Only H2O has the capacity to do this in our atmosphere.

110. Thought about this all night. The issue is the definition of the word “Warming.” “Warming” does not mean Cooling more Slowly, it means the Hot thing actually gets Hotter.

This is impossible, as, since the Hot thing gets Hotter from the Cool thing, the Cool thing will Get Hotter too, making the Hot thing also Hotter, Perpetual Motion.

This should do it for any sophisticated audience.

• Micheal Moon wrote, “Thought about this all night… “Warming” does not mean Cooling more Slowly, it means the Hot thing actually gets Hotter.”

Wrong.

Instead of thinking about it all night, you should have opened the door and checked the temperature outside.

On most nights, in most places, the temperature drops as the night progresses, then rises again during the day. If, thanks to GHGs in the atmosphere, or cloud cover, or anything else that retards cooling, the rate of cooling at night is slowed, then the temperature at dawn is warmer than it otherwise would have been.

So in this context “warming” and “cooling more slowly” are equivalent: they make air temperatures warmer than they otherwise would be.

Michael Moon wrote, “This is impossible, as, since the Hot thing gets Hotter from the Cool thing, the Cool thing will Get Hotter too, making the Hot thing also Hotter, Perpetual Motion.”

Wrong.

Have you never snuggled up with a pretty girl on a chilly night (“for warmth” you told her)?

Or… have you never built a campfire, or a fire with charcoal briquettes?

If you separate the burning coals, they soon cool too much to sustain combustion. But if you pile several coals together, near each other, they will keep each other hot enough to continue burning much longer.

https://www.wikihow.com/images/thumb/7/73/Create-a-Strong-Burning-Charcoal-Fire-Step-13.jpg/aid3802-v4-728px-Create-a-Strong-Burning-Charcoal-Fire-Step-13.jpg.webp

111. Gonna be a wise ass and point out that the microwave (IR) background radiation at 2.75 W/m2 is somewhat stronger than the ~2 W/m2 from increased CO2.

The burning question here is does earth radiation in the atmospheric window make space warmer than it would otherwise be?

• Does cosmic background radiation make the earth warmer than it would otherwise be?

112. And, if any of you would like to argue that Preventing Cooling, or even Slowing Cooling, is the same as Warming, then you have no place in this discussion…

113. Cooling more slowly is not the same thing as getting hotter. Are we clear? If not, go back to school…

114. Cooling more slowly involves the temperature going Down. Getting hotter involves the temperature going Up. Clear then?

115. One more time, going down more slowly is not the same thing as going up. Congratulate Willis on his idiocy several more times, or ruin this blog…

• Michael Moon, repeating the same wrong statement over and over does not make it less erroneous. In the context of the Earth’s climate (and in many other contexts) cooling more slowly is exactly equivalent to heating.

Everything on Earth is constantly gaining energy from some sources and losing it from others. It matters not a whit whether you increase the rate of energy gain or slow the rate of energy loss, if you do either then the thing gets warmer.

What’s more, accusing Willis Eschenbach(!) of “idiocy” is an expeditious way to prove beyond a doubt that you’re as dumb as HotWhoppers’ Miriam O’Brien (a/k/a Slandering Sou from Bundangawoolarangeera), who is the only other person I can think of who is foolish enough to say something like that.

116. From where comes a photon/EM wave? It comes from moving charge, as in a shock from static electricity. Yes, static electricity and light are virtually the same thing. So, photons, come from charge being forced to change direction, as in moving molecules bumping into each other. If one is moving faster than the other one, the resulting EM/photons cannot necessarily catch up to the faster moving one, hence, no heat transfer from cooler to warmer.

“There are more things between Heaven and Earth Horatio than are dreamed of in your philosophy…”

117. Just one more! If a cool thing could heat a hotter thing, the hotter thing, being hotter now, would heat the cool thing more, which now would get hotter, and heat the hotter thing a little more, and now we have positive feedback, everything keeps getting hotter, perpetual motion, no need to dig up more coal or oil or Nat-Gas, pretty sweet! Good luck with that…

• The precept here is that a cool thing can retard cooling of a hotter thing, not that it can make it hotter.

Why is that so hard for some people to understand?

• Toneb says:

RW:

Exactly.

“everything keeps getting hotter, perpetual motion”

Perhaps, just perhaps, YOUR Beliefs in the scince are therefore wrong….. and indeed the “problem”.
Just a thought you understand.

• Steven Mosher says:

They are in denial.

Yet they all claim to have minds open to evidence.
They all see the evidence but can always construct a path to ignore…misrepresent misunderstand or reject the evidence.

That relates directly to the character of all skeptics.

• Tim Folkerts says:

Michael & Toneb, surely you are familiar with the concept of a “convergent series”. This sort of feedback may or may not lead to some infinite runaway result. Can you somehow prove the series diverges? Or are you simply picking one answer more or less at random?

• Toneb says:

“This sort of feedback may or may not lead to some infinite runaway result. ”

Err – I don’t remember saying anything about a “runaway result”.
Just that the GHE is real.

• No, Steven Mosher, it relates to the character of all mankind. Well, most of mankind, anyhow — certainly most climate activists.

It is amazing to me how evidence, even overwhelming evidence, so rarely causes people to waver in their opinions. E.g., when I show climate alarmists good news, like the fact that the 30% increase in atmospheric CO2 over the last 70 years has caused no significant increase in the rate of sea-level rise, they should be glad.

Wouldn’t you think that people who claim to be concerned about climate change would be relieved at such excellent news?

Strangely, they rarely are. Usually, good news upsets them. They’re apparently so emotionally wedded to their dystopian nightmare that they’ve come to want it to come to pass. No matter how conclusively it is proven, they reject the best scientific evidence, and cling doggedly to dystopian fantasies. They’d rather be right than safe. They’d rather believe their children are doomed than accept that they were wrong. So when you show them good news, they get mad, and respond with angry insults!

That’s pathological. I wonder if it is connected in some way to the strange popularity of horror movies. Maybe it’s the cats.

“When my information changes, I alter my conclusions. What do you do, sir?”
– John Maynard Keynes (paraphrased)

• Sparks says:

Steven Mosher

“That relates directly to the character of all skeptics.”

I’m skeptical of that, I’m an engineer, I have a BSC, I am qualified in electrical engineering, computer maintenance and networks, 25+ years programing experience, I owned a professional web design company and built an office/workshop servicing computer and office equipment for local business before even google existed, I have built schools, collages, homes and businesses in an engineering capacity. I have also worked for about 10 bands as a crew member, involving the production and the setting up of equipment including the programing and servicing of robotic lighting systems in hundreds of venues around the UK and Ireland, I am also qualified in horticulture I hold a hfe cert in numeracy. I have studied astronomy, solar physics and planetary mechanics, I have no criminal record, I am personally very polite, well mannered and have an awesome sense of humor and my mum says I’m good looking. Two years ago I ran into a burning building 3 times in an attempt to save a dogs life.

For the past 3 years I have worked as an electronics engineer for a security company and have been working on a project designing and building a portable security system, it’s kinda top secret…

I’m not sure that I can even begin to do myself any justice in a quick comment, but you would be the last person I would ever ask for a character reference.

• Sparks, you neglected to mention that your excrement is odorless.

• Tim Folkerts says:

• So, Sparks, do you have a significant other in your life?
Asking for a friend . . .

• AndyG55 says:

“have been working on a project designing and building a portable security system, it’s kinda top secret…”

DOH !!!!! Not any more it isn’t …

• menicholas says:

I think we can all agree that the people here commenting, at least those that agree with whatever position we hold on some issue or another, are the bestest of the bestest that humanity has to offer.
But not the ones that we disagree with, nosirree Bob…nope. They are rotten to the core…and probably dress silly, have bad hair, and stinky breath.
I can pert near guarantee it.

• Toneb says:

MM:
Apologies – In haste
I see you are on the “physics” side of the discussion.

• nate says:

Michael,

No perpetual motion, no more than adding insulation to my house causes that. Keep it simple. Insulation is a cool thing, it slows heat transfer from the warm house to the cold outside. Keeping all else equal (furnace heat output to the house held fixed), the house will warm.

Find the fault with this.

• Uncle Gus says:

The Sun is the furnace.

The GHGs in the atmosphere are the insulation.

(You’ve got to be *very* specific with these bozos, Nate. And even then it rarely makes any difference…)

• A C Osborn says:

Uncle Gus.
Yes it is a shame that those same Insulating GHGs are between the Furnace and the Earth isn’t it?
Half the time the biggest GHG ie Water droplets and water vapour are cooling the Earth, are they not?
Or have you never noticed how much colder it is on a Cloudy day?

Or is it a shame, would we want the same daytime temperatures as the Moon?

• nate says:

AC,
“Or have you never noticed how much colder it is on a Cloudy day?
Beg to differ. It can go either way.

This time of year cloudy days can be warmer—indicative that warm moist air has moved in aloft. Late in the day, and at night, clouds block radiative cooling to space, keeping temps elevated.

• Michael Moon November 25, 2017 at 10:00 pm
Just one more! If a cool thing could heat a hotter thing, the hotter thing, being hotter now, would heat the cool thing more, which now would get hotter, and heat the hotter thing a little more, and now we have positive feedback, everything keeps getting hotter, perpetual motion, no need to dig up more coal or oil or Nat-Gas, pretty sweet! Good luck with that…

Not if it’s a convergent series in which case a new steady state is reached.

• Paul Bahlin says:

It actually is convergent. If you do the math, it reduces to a really nice Taylor series expansion of
1/(1-h).
where h is the percentage of upwelling that is returned to the surface.

You can easily show that the energy returned to the surface is

[(1/(1-h)-1] of the incoming. So if h=0 nothing returns and if it is 1, the planet blows up.

118. Willis,

You are abysmally ignorant. You are not stupid, but you do not belong in the deep end of this pool, seriously. You are doing more harm than good with your profound misunderstanding of radiation and Heat Transfer. I beg you, withdraw from these questions, Mann will mock you and the entire blog..

• Rainer Bensch says:

• Mike Jonas says:

Michael Moon. Those are strong words, and unprettily abusive. But it is you who has missed the point. After the title Can a cold object warm a hot object?, Willis’ opening words were:
Short answer? Of course not, that would violate the Second Law of Thermodynamics … BUT it can leave the hot object warmer than it would be if the cold object weren’t there.

All of your comments have been misdirected, because they have addressed the first part of the above where you seem to have failed to notice that Willis agrees with you (“Of course not“), whereas virtually the entire article addressed the second part (“it can leave the hot object warmer than it would be if the cold object weren’t there“).

119. Thank you Willis, for an excellent and easy to understand explanation of a challenging subject. However, I think two conditions should be added to make the explanation more stringent.

Firstly, the question should be “Can a Cold Object Warm a Hot Object by means of radiation only
Without this condition, the answer would be “Yes, of course, just take on some clothes and you will soon notice the effect of warming from insulation”.

The second condition is that you have to take into account the effect of reflection. You describe the interaction as if all the bodies were perfect IR absorbers and perfect IR emitters. I know it is to make the description simple, but here I think you are oversimplifying.

A cold object can warm a hotter object by either shielding the warm object from an even colder object, or by being painted in a more reflective surface.

I think the situation easiest to understand is if both the cold and the warm object is rigged in a way so that they initially have a fixed temperature. Say that object A is 40 Celsius which is 20 Celsius warmer than object B because it is heated from an internal source, for instance a lightbulb.

The distance between A and B is so large that only heat exchange by radiation need to be taken in account.
Can we then do something with object B, so that the temperature in A increases?
Yes, we can for instance paint B in a more reflective paint.

Another way to increase the temperature in A is to place another object with a temperature of 30 C between A and B, but with enough distance from A to avoid any heat exchange other than radiation.

/Jan

• arfurbryant says:

Jan: “[A cold object can warm a hotter object by either shielding the warm object from an even colder object…”]

Seriously? When have you ever put hot coffee into a Thermos flask to find the coffee gets warmer with time?

• nate says:

Arfurbryant,

Many people here are giving their own examples, such as you have here, substituting them for the original post. The problem is when these examples remove key factors from the original post, that make all the difference.

Your thermos example: you need to compare the coffees temp over timme in the thermos to what it would be if the coffee was in a regular cup. Answer: it will be warmer in the thermos.

• arfurbryant says:

Nate,

Thanks for your input. I agree that examples used should be clear. Words are important in this debate.

Hence, Jan’s statement is just plain wrong. You are obviously aware that insulation does not make the warm object warmer. It does, however, retard the rate of cooling of that warm object. No argument with that.

But to imply that my ‘example’ somehow is not correct or even relevant is also not true.

In this context, you should be saying (as Willis did in the first paragraph of the OP) that a cold object can make an object warmer than it would have been if the cold object was not there. But instead you choose to use the phrase ‘warmer in the thermos’. You mean warmer in comparison with no thermos! To be clear, you should say that it is cooling slower than if there was no thermos. The coffee will NEVER be warmer.

Am I being pedantic? Yes. Am I wrong? No.

Does Willis answer his own question “Can a cold object warm a hot object?” Yes in the short answer but he then subtly changes the goalposts by answering a different question with a BUT in the second part of the answer.

Adding a cold object (insulation in this context) in NO WAY makes the warm object even hotter.

Cooling more slowly is not the same as warming.

CO2 is not an effective insulator and it is not in sufficient quantities to insulate, even if it was effective. Adding CO2 at the rate of just over one ppm per year should not be compared to any form of insulating material such as steel, blankets, overcoats etc.

I hope this clarifies my point.

Regards,

Arfur

• Arfurbryant,

I do not think you read my example before you responded so I repeat it here:

“I think the situation easiest to understand is if both the cold and the warm object is rigged in a way so that they initially have a fixed temperature. Say that object A is 40 Celsius which is 20 Celsius warmer than object B because it is heated from an internal source, for instance a lightbulb.

The distance between A and B is so large that only heat exchange by radiation need to be taken in account.

Can we then do something with object B, so that the temperature in A increases?

Yes, we can for instance paint B in a more reflective paint. “

Do you disagree with this?
/Jan

• nate says:

OK, however. in the GHE, there is a continuous heat source, the sun. Again this source of heat is conveniently left out of many examples. In your coffee example that is equivalent to having an immersion heater supplying a steady power to the cup. In that case the coffee WILL BE warmer in the better insulated thermos, as compared to a regular cup.

• arfurbryant says:

Hi Jan,

Yes I did read your comment and the part that I objected to mostly was the phrase “A cold object can warm a hotter object by either shielding the warm object from an even colder object…”

As to your further point about object A being made warmer by either the addition of reflective paint on object B or by adding a third object C at intermediate temperature (30C) well, YES, I completely disagree with you!

In your examples, there is NO WAY that object A can be made warmer by either painting object B or by adding object C.

Kind regards,

Arfur

• Arfurbryant

Imagine a poor homeless person freezing in the streets a winter night. His body temperature has fallen to 35 C and he is desperate because unless he recover his body temperature he will die. Then he finds some thick good insulating winter clothes.

Should he bother to take them on?

According to you, a colder object, the clothes, can never help him recover his body temperature.

Of course, he will be warmer because the insulation help him lose less heat and more of the energy from his body functions will then stay in his body and warm it up.

This is similar to object A, which will warm up because more of the energy from its inner energy source, the lightbulb, will stay in the box when some of the radiation is reflected back.

/Jan

• arfurbryant says:

Hi Jan,

You are wrong.

The homeless person will not recover his body temperature by putting more clothes on. All that would do is make him ‘feel warmer’ by reducing his rate of cooling. Unfortunately for him, as far as his body (core) temperature is concerned, all the extra clothes do is to delay his impending death from hypothermia.

By the way, this is not to say putting the clothes on is a bad idea! It would buy him (or her) more time…

Seriously, just try sleeping outside tonight with a really thick sleeping bag around you. Take your body temperature before you get into the bag and take it again in the morning. Without adding any form of external or internal heat, you will find your body temperature is colder in the morning. If you start to feel cold during the night, simply add another sleeping bag and again take your temperature in the morning. You will still be colder in the morning…

Kind regards,

Arfur

• Arfurbryant

Unfortunately for him, as far as his body (core) temperature is concerned, all the extra clothes do is to delay his impending death from hypothermia.

Do you really mean this?

Well, you are wrong, and I honestly think you are far off now, but let me try one more:

Imagine, as a thought experiment, that the new warm clothes are perfectly insulating. No heat escapes from his body after he takes them on.

Since he is not dead, his body functions must burn some calories, which create some heat, where do you think that heat will go?

/Jan

• arfurbryant says:

Hi Jan,

Yes, i really meant that. People who sleep outside will have a lower body temperature in the morning, irrespective of how much clothing they have. If you doubt me, please speak to either a physicist or a medical professional. Don’t take my word for it.

The trouble with ‘thought experiments’ is that people don’t usually think them through.

The direct answer to your question (where does the heat go?) is this:
If the insulation is truly 100% then there will be no heat loss from inside the insulation but, equally, there will be no heat gain by the body. (Can I assume there is an air gap between the body and the insulation?)

However, the person will die relatively quickly because he/she will be unable to breathe once the air inside the 100% insulation is exhausted (and the O2 is replaced by CO2). In addition, the body functions you refer to will use up stored energy, which is of a finite amount and is irreplaceable without compromising the 100% insulation. Once the stored energy is depleted he/she will die.

Either way, ‘thoughtfully’ covering a homeless person with 100% insulation is not going to do him any favours and his body core temperature will not increase at any point.

Sorry if you disagree but please ask someone else. I see no point in discussing this further but I genuinely wish you all the best in the future.

Kind regards,

Arfur

• Goodbye Arfur

I had never though that I would ever meet a person who think that putting on insulating clothes in a cold day will not make you warmer.

Now I have.

All the best

Jan

• Jan Kjetil Andersen, after patiently, but fruitlessly, trying to explain a little bit of basic science to arfurbryant, wrote, “I had never though that I would ever meet a person who think that putting on insulating clothes in a cold day will not make you warmer.”

In The Road Less Traveled, psychiatrist Scotty Peck defined mental health as dedication to reality. A mentally healthy person is a person who tries very hard to understand the world clearly. He is willing, even anxious, to revise his internal “map” of reality, when new information conflicts with previously held opinions.

Just like the leftists who blather on about “my truth” and “our realities,” as if reality were a matter of perspective, the slayers are not mentally healthy. Evidence, logic, and and even frequently repeated personal experience — as when they put on a coat every day before going outdoors in the winter — are irrelevant to slayers like arfurbryant, Michael Moon, skepticgonewild, “the bad John O’Sullivan,”† Joe Olson, Joe Postma, Pierre Latour, et al.

The “slayer” nonsense is such pure, refined ignorance that it inspired me to wax poetic, which is rare for me. With apologies to the late, great Ogden Nash, I give you:

Sky Dragons
The Second Law they twist and shove,
to slay their dragons from up above.
But this I know by actual test:
Use a blanket and you’ll shiver less.

† “The bad John O’Sullivan” is the Chief Slayer, i.e., CEO of PSI. He should not be confused with “the good John O’Sullivan,” who writes for National Review, and whose identity the bad John O’Sullivan has been known to borrow, when convenient.

“The bad” John O’Sullivan and his comrades cannot be trusted. Here’s a particular infuriating example of their dishonesty:

In 2015 O’Sullivan & Pierre Latour blatantly lied on the PSI web site, and also in emails, about Dr. S. Fred Singer’s views. The web site article was entitled, “Singer Concurs with Latour: CO2 Doesn’t Cause Global Warming.”

After several people objected, in email O’Sullivan wrote, “Fred Singer has now come over to PSI’s view that CO2 can only cool. I suggest you contact him.”

So I did. I forwarded that to Prof. Singer, who was then 90 years old, and asked him:

Dear Dr. Singer,
Please confirm or deny this allegation.
Warmest regards,
Dave

Prof. Singer replied succinctly:

denied SFS

That’s what I expected. I forwarded it to O’Sullivan & the other slayers, plus some of the people who had been trying to persuade PSI to remove the dishonest article, including Dr. Singer’s friend, Lord Christopher Monckton. But O’Sullivan still refused to to remove the dishonest article from the PSI web site.

Prof. Singer elaborated in a subsequent email:

Friends, There is a sure way to smoke out deniers like PS [Principia Scientific]
Just ask them if GH models violate the 2nd Law of Thermo …
These people just won’t accept the existence of DWR (downwelling IR from atm to sfc)
— even if measured empirically
No point wasting more time — as Jo said
Fred

Jo Nova weighed in, and appealed to O’Sullivan (and cc’d the other three), to do the right thing:

John, this is a simple publishing issue. Claiming that Singer supports PS when he doesn’t, is using his name to promote your group at the expense of Singer and the skeptic movement as a whole. The simple request to take down the article, or correct it, should have been apologetically complied with immediately.

This ongoing fracas is damaging the skeptics as a whole, and wasting much time. We fight opponents with billions – let’s focus on the real enemy.

Jo

Joe Postma defended O’Sullivan’s and Latour’s dishonesty. He wrote:

Except it’s not an expense to Singer or the skeptic movement… Because PSI’s position of criticizing the alarmist greenhouse effect isn’t a negative. Unless you’re trying make it so or make it appear like it is so.

Which is strange, confusing, duplicitous behaviour in this skeptic movement.

If a scientist says that climate sensitivity might be indistinguishable from zero, it’s well within reasonable bounds of inference to say that it ipso facto agrees with PSI’s position that this is the result of there being no radiative greenhouse effect, whether the scientist agrees with that or not. It doesn’t harm the scientist unless you’re thinking you need approval from the alarmists…you don’t…

Eventually, after Lord Monckton threatened legal action, PSI finally did edit and tone down the article, making it less flagrantly false. It now says “Singer Converges on ZERO Climate Carbon Forcing.”

• Sorry about that. that mess is the result of typing <blockquote> when I intended to type </blockquote> Sigh.

[fixed-mod]

• Thank you for the support Dave, your poetic verse is hilarious

/Jan

• Thanks Dave. That was quite hilarious.

Please explain how 341 W/m2 of solar insolation at the TOA gets magically multiplied to 494 W/m2 at the earth’s surface when the sun is the ONLY source of energy to heat the earth.