# 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.

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Joel O'Bryan
November 24, 2017 7:22 pm

“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
November 24, 2017 7:30 pm

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

reallyskeptical
November 24, 2017 8:49 pm

If people are still unclear on this, Rabbit has a great explanation of this here, like only the Rabbit could do it: http://rabett.blogspot.com/2017/10/an-evergreen-of-denial-is-that-colder.html

AndyG55
November 25, 2017 2:54 pm

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
November 26, 2017 2:03 am

“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
November 26, 2017 2:50 am

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
November 26, 2017 8:59 am

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).

Tony
November 26, 2017 5:28 pm

As for Eli’s thought experiment, here’s a reasonable comment to start from:

Though for full context you could read from a little earlier. It’s gone through overall in quite some detail from thereabouts onwards. And is of course thoroughly refuted.

Not up for discussion here, obviously. Dusted and done, etc…

Tim Folkerts
November 26, 2017 7:39 pm

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
November 27, 2017 12:24 am

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

Tony
November 27, 2017 12:29 am

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.

November 28, 2017 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.)

Leo Morgan
November 24, 2017 9:55 pm

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
November 25, 2017 1:11 am

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
November 25, 2017 1:11 am

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.

ferdberple
November 25, 2017 9:03 am

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.

co2islife
November 24, 2017 7:23 pm

“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
November 24, 2017 10:38 pm

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.

November 25, 2017 1:58 am

David Ramsay Steele

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

co2islife
November 25, 2017 6:57 am

LOL, great story

lemiere jacques
November 25, 2017 12:12 am

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
November 25, 2017 7:00 am

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
November 25, 2017 7:45 am

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
November 25, 2017 12:24 pm

thank you. I thought I was the only one 🙂

November 25, 2017 2:46 pm

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
November 25, 2017 4:50 pm

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
November 26, 2017 7:49 am

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
November 26, 2017 10:49 am

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
November 26, 2017 11:11 am

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

Crispin in Waterloo
November 27, 2017 11:39 pm

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.

wildeco2014
November 28, 2017 4:45 pm

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
November 25, 2017 7:52 am

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.

November 25, 2017 8:27 am

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 .

November 25, 2017 12:28 pm

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.

co2islife
November 25, 2017 4:31 pm

CO2 is 0.00004 or 0.04% of the atmosphere. Is it plausible that “activating” 1 out of every 2,500 molecules in the atmosphere can actually result in a material temperature change?
How to Discuss Global Warming with a “Climate Alarmist.” Scientific Talking Points to Win the Debate.
https://co2islife.wordpress.com/2017/01/16/how-to-discuss-global-warming-with-a-climate-alarmist-scientific-talking-points-to-win-the-debate/

Crispin in Waterloo
November 26, 2017 7:51 am

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.
November 26, 2017 10:26 am

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.!

Tom Halla
November 24, 2017 7:24 pm

Nice succinct argument.

co2islife
November 24, 2017 7:27 pm

If I take Cold Sodium and mix it with cold water, I get a very hot reaction.
https://youtu.be/NTFBXJ3Zd_4

Joel O'Bryan
November 24, 2017 7:37 pm

That was a hydrogen explosion you heard.

2Na(s) + 2H2O(l) –> 2NaOH (l) + H2(g)

Joel O'Bryan
November 24, 2017 7:44 pm

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

Urederra
November 25, 2017 2:43 am

nope

co2islife
November 25, 2017 7:06 am

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.

November 24, 2017 7:58 pm

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

Catcracking
November 24, 2017 8:25 pm

of course, agree.

co2islife
November 25, 2017 7:04 am

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
November 25, 2017 2:23 pm

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.

Ferdinand Engelbeen
November 26, 2017 12:03 am

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.
November 25, 2017 12:53 am

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
November 25, 2017 7:02 am

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.

November 24, 2017 7:28 pm

This is essentially the same argument Roy Spencer made seven years ago. It is nicely rebutted here. http://principia-scientific.org/images/stories/pdfs/Pierre_commentsadded.pdf

November 24, 2017 7:34 pm

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

November 24, 2017 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.

November 24, 2017 8:09 pm

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.)

reallyskeptical
November 24, 2017 8:45 pm

“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.

November 24, 2017 8:46 pm

“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”?

November 24, 2017 8:53 pm

“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
November 24, 2017 9:21 pm

“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
November 25, 2017 1:22 am

“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
November 25, 2017 3:48 am

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
November 25, 2017 4:01 am

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
November 25, 2017 4:50 am

“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
November 25, 2017 8:15 am

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
November 25, 2017 9:12 am

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.

November 25, 2017 11:42 am

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 :
http://cosy.com/Science/1DeqDiagram.jpg
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
November 26, 2017 1:42 pm

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
November 24, 2017 8:49 pm

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
November 25, 2017 1:34 am

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

A C Osborn
November 25, 2017 1:52 am

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

Jaakko Kateenkorva
November 25, 2017 3:15 am

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
November 25, 2017 3:35 am

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
November 25, 2017 9:23 am

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.

Editor
November 25, 2017 7:56 am

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

November 24, 2017 7:55 pm

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
November 24, 2017 8:28 pm

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.

November 24, 2017 8:41 pm

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
November 25, 2017 6:36 am

“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
November 25, 2017 2:49 pm

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
November 24, 2017 8:47 pm

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.

November 24, 2017 8:56 pm

“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).

commieBob
November 24, 2017 11:28 pm

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
November 25, 2017 9:53 am

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
November 25, 2017 1:38 pm

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
November 25, 2017 3:20 pm

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
November 26, 2017 12:01 am

“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.

Ferdinand Engelbeen
November 26, 2017 6:43 am

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
November 26, 2017 7:20 am

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
November 26, 2017 7:54 am

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.

Ferdinand Engelbeen
November 26, 2017 3:42 pm

Crispin,

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

RayG
November 24, 2017 8:47 pm

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

Nick Stokes
November 24, 2017 8:49 pm

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.

November 24, 2017 8:58 pm

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
November 24, 2017 9:01 pm

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.

Nick Stokes
November 24, 2017 9:29 pm

“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
November 25, 2017 1:36 am

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
November 25, 2017 7:01 am

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
November 25, 2017 3:28 pm

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.

Ferdinand Engelbeen
November 26, 2017 6:50 am

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
November 26, 2017 2:27 am

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
November 26, 2017 4:32 am

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
November 26, 2017 4:36 am

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
November 26, 2017 5:42 am

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.

Bill Treuren
November 24, 2017 7:58 pm

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.

November 24, 2017 8:03 pm

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?

November 24, 2017 11:04 pm

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
November 24, 2017 11:54 pm

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.

November 25, 2017 3:08 pm

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.

Brett Keane
November 24, 2017 8:05 pm

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.

Paul Aubrin
November 25, 2017 11:30 am

“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.

November 25, 2017 6:48 pm

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
November 24, 2017 11:58 pm

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
November 25, 2017 5:42 am

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
November 25, 2017 11:35 am

“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/

eyesonu
November 24, 2017 8:07 pm

Willis,

Nice condensed and simple straight forward explanation.

Dave Fair
November 25, 2017 12:19 am

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.

KM
November 24, 2017 8:11 pm

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. 🙂

Catcracking
November 24, 2017 9:13 pm

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
November 24, 2017 9:35 pm

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 10:34 pm

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
November 25, 2017 12:23 am

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

November 25, 2017 2:30 am

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
November 25, 2017 3:56 am

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
November 25, 2017 5:40 am

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
November 25, 2017 5:53 am

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
November 25, 2017 7:34 am

“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.

http://www.dinosaurtheory.com/temperature_atmosphere.jpg

tty
November 25, 2017 8:48 am

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:

http://learningweather.psu.edu/sites/default/files/Lesson3/absorptivity.png

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
November 25, 2017 9:18 am

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
November 25, 2017 10:37 am

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.

November 25, 2017 3:23 pm

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
November 25, 2017 4:43 pm

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
November 26, 2017 8:07 am

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.

Anne Ominous
November 24, 2017 8:12 pm

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
November 25, 2017 5:01 pm

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
November 27, 2017 2:37 am

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.

November 24, 2017 8:18 pm

“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
November 24, 2017 10:14 pm

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
November 25, 2017 6:52 am

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

I’m afraid it seems that some do.

Toneb
November 25, 2017 12:39 am

“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
November 25, 2017 2:43 am

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

November 25, 2017 4:03 am

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
November 25, 2017 4:45 am

“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
November 25, 2017 8:53 am

“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
November 25, 2017 5:06 pm

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
November 26, 2017 3:43 am

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
November 26, 2017 4:51 am

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.

John
November 24, 2017 8:29 pm

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.

November 24, 2017 8:35 pm

“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.

November 24, 2017 9:00 pm

“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.

Windchaser
November 24, 2017 9:42 pm

“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
November 24, 2017 9:45 pm

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.

Toneb
November 25, 2017 12:43 am

“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.”

Roy Spencer has …

http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/

Editor
November 25, 2017 6:50 am

“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
November 25, 2017 9:07 am

“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
November 25, 2017 10:39 am

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

Paul Aubrin
November 25, 2017 11:59 am

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
November 25, 2017 5:10 pm

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.

Ferdinand Engelbeen
November 26, 2017 7:25 am

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
November 25, 2017 12:48 pm

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
November 25, 2017 1:08 pm

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
November 25, 2017 6:07 pm

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.

November 24, 2017 8:37 pm

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
November 25, 2017 4:52 am

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.

donald penman
November 24, 2017 8:54 pm

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.

November 24, 2017 9:02 pm

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

jIM a
November 24, 2017 9:46 pm

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
November 25, 2017 5:09 am

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
November 24, 2017 9:44 pm

“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).

donald penman
November 24, 2017 9:30 pm

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

Extreme Hiatus
November 24, 2017 9:32 pm

“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.

November 25, 2017 4:07 pm

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.

eyesonu
November 24, 2017 9:35 pm

Willis,

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

Tony
November 25, 2017 5:22 am

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

Charles Gerard Nelson
November 24, 2017 9:36 pm

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

November 25, 2017 5:22 am

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.

Julian Flood
November 24, 2017 9:40 pm

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 25, 2017 10:04 pm

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
November 25, 2017 4:10 am

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

Nicholas Denman
November 24, 2017 9:58 pm

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
November 25, 2017 9:57 am

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.

angech
November 24, 2017 10:08 pm

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.

November 24, 2017 10:35 pm

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

angech
November 24, 2017 11:42 pm

So I cannot ask for a hot cup of coffee?

November 25, 2017 1:19 am

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.

November 25, 2017 5:09 am

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
November 25, 2017 4:26 pm

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.

November 26, 2017 2:21 pm

“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.

November 26, 2017 2:31 pm

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

Crispin in Waterloo
November 27, 2017 10:47 pm

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.

Gary M
November 24, 2017 10:09 pm

Gabro
November 24, 2017 10:10 pm

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.

Charles Gerard Nelson
November 24, 2017 10:57 pm

why?

November 24, 2017 11:22 pm

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
November 27, 2017 10:34 am

angech
November 24, 2017 10:18 pm

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
November 24, 2017 10:35 pm

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
November 25, 2017 3:23 am

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.

November 25, 2017 4:26 am

Black holes are not real?

Tony
November 25, 2017 4:38 am

Dolphins are not real?

angech
November 26, 2017 11:13 pm

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

Tony
November 27, 2017 1:51 am

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
November 27, 2017 2:19 am

(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)

November 24, 2017 10:27 pm

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.

Sparky
November 25, 2017 4:18 am

You want to try nested steel greenhouses.

November 25, 2017 12:16 pm

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
November 25, 2017 12:37 pm

“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
November 25, 2017 12:58 pm

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.

November 25, 2017 4:24 pm

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

gbaikie
November 25, 2017 5:58 pm

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.

November 24, 2017 10:32 pm

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
November 25, 2017 1:09 am

“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 …

November 25, 2017 1:22 am

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
November 25, 2017 1:44 am

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

Ed Bo
November 25, 2017 3:10 pm

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.

Rick C PE
November 24, 2017 10:57 pm

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.

November 25, 2017 12:21 am

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.

November 24, 2017 11:16 pm

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
November 25, 2017 2:46 am

“No rational “skeptic” denies…”

No true Scotsman denies…

November 25, 2017 6:16 am

‘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
November 25, 2017 6:26 am

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