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:

all_flows_net_flows

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.

my energy budget large

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:

two way radiation equation

with the following variables:

two way radiation 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:

two way radiation equation expanded

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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joelobryan

“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).

joelobryan

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

reallyskeptical

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

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. !

tjfolkerts

“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

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.

tjfolkerts

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

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

https://climateofsophistry.com/2017/10/06/slayers-vindicated-by-additional-independent-researchers/#comment-31140

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…

tjfolkerts

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

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

Tony

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

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

Bob Shapiro November 28, 2017 at 11:46 am

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

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

TANSTAAFL …

w.

Leo Morgan

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

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

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

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

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

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

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

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

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

co2islife

“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

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.

David Ramsay Steele

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

co2islife

LOL, great story

lemiere jacques

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

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

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

thank you. I thought I was the only one 🙂

Aphan

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

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

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

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

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

Crispin in Waterloo

DMacKenzie

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

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

Considered this:

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

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

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

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

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

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

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

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

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

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

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

Case 2.

Crispin in Waterloo:

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

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

Case 2.

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

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

Read the post for the long version.

w.

wildeco2014

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

co2islife

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

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

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

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

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

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

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

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

co2islife

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

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.

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

Nice succinct argument.

co2islife

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

joelobryan

That was a hydrogen explosion you heard.

2Na(s) + 2H2O(l) –> 2NaOH (l) + H2(g)

joelobryan

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

Urederra

nope

co2islife

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

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

Catcracking

of course, agree.

co2islife

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

J Richard

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

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

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

Crispin,

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

Gary Pearse.

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

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.

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

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

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

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

w.

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

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

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

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

w.

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

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

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

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

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

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

Regards,

w.

reallyskeptical

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

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

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

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

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

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

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

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

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

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

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

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

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

reallyskeptical

“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

“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

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

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

“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

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

Bigfoot:

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

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

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

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

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

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

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

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

A C Osborn

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

Jaakko Kateenkorva

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

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

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.

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

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

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

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

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

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

daved46

“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

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

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

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

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

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

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

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

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

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

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

w.

commieBob

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

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

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

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

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

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

Crispin in Waterloo,

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

tjfolkerts

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

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

Crispin,

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

RayG

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

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

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

Catcracking

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

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

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

menicholas

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

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

menicholas

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

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

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

LdB

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

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

Perhaps on re-reading your post, that is what you are saying.
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

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

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.

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

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

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

Aphan November 24, 2017 at 8:03 pm

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

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

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

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

w.

w.

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

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

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

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

https://climateandstuff.blogspot.com/2013/03/does-thermal-radiation-travel-from-cool.html?m=1

commieBob

Aphan November 24, 2017 at 11:04 pm

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

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

commiebob,

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

Brett Keane

Bare planet. Add atmoaphere.
Radiating object. Add object radiating less.

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

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

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

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

Bare planet. Add atmoaphere.
Radiating object. Add object radiating less.

I don’t understand that.

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

True. And … so what?

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

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

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

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

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

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

Regards,

w.

Paul Aubrin

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

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

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

commieBob

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

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

“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

Willis,

Nice condensed and simple straight forward explanation.

Thanks.

Dave Fair

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

Interesting, but please allow me to play the devil’s advocate here.

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

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

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

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

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

Let the flaming begin. 🙂

Nah, no flaming. You say:

Problems:
1. The atmosphere is not opaque.

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

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

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

w.

Catcracking

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

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

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

KM November 24, 2017 at 9:35 pm

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

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

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

1. Clear skies

2. Clouds

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

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

w.

KM

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

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

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

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

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

richard verney

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

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.

http://clivebest.com/blog/wp-content/uploads/2010/01/ir-spectra-earth.png

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

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

“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

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

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

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

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

You state:

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

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

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

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

The GHG religion is disproven by all these other experiments.

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

Crispin in Waterloo

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

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

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.

Thanks, Anne.

w.

Crispin in Waterloo

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.

From Wiki-read-with-care:

“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

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.

“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

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

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

I’m afraid it seems that some do.

Toneb

“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

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

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

Tony

“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

“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

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

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

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

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

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

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

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

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

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

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

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

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

w.

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

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

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

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

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

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

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

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

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

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

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

w.

Windchaser

“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

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

“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/

http://www.drroyspencer.com/2010/07/experiment-to-test-the-temperature-influence-of-infrared-sky-radiation/

“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.

More reading material on this:

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

“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

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

Paul Aubrin

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

Count to 10

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

J. Richard Wakefield

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

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

Tony mcleod

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

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

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.

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

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

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

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

jIM a

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

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

“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

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

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

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

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

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

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

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

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

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

eyesonu

Willis,

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

Tony

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

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

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

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

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

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

JF

Julian Flood November 24, 2017 at 9:40 pm

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

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

Good to hear from you, Julian, long time.

w.

Hi de hi.

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

Talking of warming…

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

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

JF

Hugs

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

Nicholas Denman

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

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

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

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

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

angech

So I cannot ask for a hot cup of coffee?

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

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

mkelly

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

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

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

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

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

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

Must proof read!!

Crispin in Waterloo

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

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

Gabro

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.
comment image

why?

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

Curious George

Link, please.

angech

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

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

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

Black holes are not real?

Tony

Dolphins are not real?

angech

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

Tony

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

(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)

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.

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

w.

Sparky

You want to try nested steel greenhouses.

The Reverend Badger

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

“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

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

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

gbaikie

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,
Likewise the steel has 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.

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

“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 …

http://www.drroyspencer.com/2010/08/help-back-radiation-has-invaded-my-backyard/

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

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

Ed Bo

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

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

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

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

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

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

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

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

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

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

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

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

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

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

“No rational “skeptic” denies…”

No true Scotsman denies…

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

Tony

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

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

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

Tony

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

Tony, you embarrass yourself.

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

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

Tony

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

I Came I Saw I Left

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

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

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

tty

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

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

Dave Fair

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

co2islife

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

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

I Came I Saw I Left

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

co2islife

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

I Came I Saw I Left

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

co2islife

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

Ed Bo

ICISIL:

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

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

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

Excuse me? You just restated exactly what I said.

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

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

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

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

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

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

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

I’m sorry Writing Observer-

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

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

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

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

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

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

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

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

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

co2islife

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

Ed Bo

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

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

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

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

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

AndyG55

“and it is just a metaphor”

No, Its a FALLACY !!

Dennis Sandberg

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

Don K

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

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

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

Hi Willis! Your argument is correct for two emitters with independent energy sources (e.g. two glowing light bulb filaments, two Suns, etc.). However, the Sun does not emit any significant far-infrared radiation at 667 cm^-1 which can be absorbed by CO2 in the atmosphere. The 288 K surface of the Earth, however, emits far-infrared radiation whose Planck black body curve peaks near 667 cm^-1. CO2 is such a powerful absorber that 667 cm^-1 radiation emitted from the Earth’s surface is almost 100% absorbed within metres of the surface. By Kirchhoff’s Law that a good absorber is a good emitter, CO2 will emit almost 100% at 667 cm^-1 (parabolic dish antennas are not only good radio receivers, but are also good radio emitters/radiators). But the excited CO2 molecules formed on absorption of 667 cm^-1 photons can also lose their excitation energy during radiationless collisions with the main molecules of the air (N2, O2, Ar) which cannot and do not re-emit any significant amount of infrared (IR) because their molecules do not possess any permanent electric dipole moment. The energy does not go away, but ends up as translational and rotational kinetic energy of the departing molecules. The heat capacity at constant pressure for linear molecules like N2, O2 and CO2 is 7k/2 per molecule, where k is Boltzmann’s constant. Since N2 and O2 together outnumber CO2 by 1,000,000:400 = 2500:1, almost all of the absorbed energy ends up warming N2, O2 and Ar molecules (and is stored as enthalpy, heat content, H). This is the mechanism of the greenhouse effect, and has nothing to do with back-radiation. Back-radiation does exist, but it can be thought of as energy flow that just balances an equal energy flow for two surfaces at thermal equilibrium, with no net change in the temperature of either. Some might argue that excited CO2 molecules can also be formed on collision of ground state CO2 molecules with high-energy (fast-moving) air (N2, O2, Ar ) molecules, and that powers the back-radiation. But this energy would come from the air molecules, whose average kinetic energy (temperature) must decrease; no, at thermal equilibrium there can be no net warming of the Earth’s surface by back-radiation alone. Here is an analogy that might make this more clear: suspend two flat parallel black metal plates in a blackened vacuum chamber cooled with liquid helium to 4 K, close to the 3 K of the cosmic background microwave radiation. If one metal plate contains a 100 W heating coil, and the other is a passive radiator without a heating coil, at thermal equilibrium 50 W will be emitted from the outside surfaces of the two plates, for a total of 100 W outward toward the 4 K inner walls of the enclosing vacuum chamber. But the space between the two plates will have 50 W exchanged back-and-forth, so there is no net heating or cooling of the plates. If the two plates are moved together until touching, the total output will still be 100 W, with 50 W outward from each surface. But the back-radiation will, like the forward radiation it balanced in the previous gap, disappear with no change in temperature or output. Note that when the two plates are separate, 50 W is emitted from both sides of the plate with the heating coil, and 50 W from both sides of the passive plate, for a total of 200 W emission. How can this be powered by the 100 W heating coil? The Law of Conservation of Total Energy is not violated because the 50 W back-radiation from the passive plate to the plate with the heating coil just balances 50 W in the opposite direction, so the back-radiation photons can be considered to be photons initially emitted from the powered plate and then reflected back to the power source. This is not the same as photons reflected from a mirror, because a mirror does not emit in the forward direction, so the mirror analogy is wrong for CO2 in the atmosphere which absorbs and re-emits in the forward direction.

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

Patrick MJD

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

Richard111

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

The Reverend Badger.

There are clearly 2 schools of thought.

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

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

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

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

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

Good luck with it.

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

mkelly

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

mkelly,

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

mkelly

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

The Reverend,

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

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

Which proves that Willis is right…

http://hockeyschtick.blogspot.com/2014/11/why-cant-radiation-from-cold-body-make.html?m=1

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

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

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

Aphan:

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

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

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

Something wrong with the theory?

A C Osborn

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

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

A C Osborn,

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

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

Mark - Helsinki

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

No need for a large article

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

Done

Mark,

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

Simple standard physics says…

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

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

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

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

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

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

Bingo!

Dave Fair

David, a simplified discussion and a question:

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

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

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

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

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

Dave Fair

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

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

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

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

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

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

wildeco2014

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

The Reverend Badger

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

Ed Bo

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

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

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

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

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

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

Ed Bo

Adrian:

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

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

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

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

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

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

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

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

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

Ex falso, quodlibet.

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

Mark - Helsinki

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

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

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

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

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

All about passive vs non passive transfer of kinetic energy

Mark - Helsinki

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

Mark - Helsinki

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

4TimesAYear

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

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

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

Mark - Helsinki

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

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

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

RAH

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

Patrick MJD

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

RAH

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

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

Mark - Helsinki

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

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

Any claims otherwise are either utterly delusional or dishonest.

Mark - Helsinki

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

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

Mark - Helsinki

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

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

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

tty

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

Mark - Helsinki

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

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

Mark - Helsinki

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

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

Mark - Helsinki

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

We’re dealing with ghosts here

arfurbryant

Can A Cold Object Warm a Hot Object?

No. End of discussion about radiation.

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

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

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

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

To Willis’s figures:

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

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

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

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

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

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

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

The science is far from settled!

Kind regards to all.

Arfur

daved46

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

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

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

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

arfurbryant

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

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

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

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

arfurbryant

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

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

menicholas

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

menicholas

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

menicholas

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

Richard111

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

menicholas

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

Richard111

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

tty

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

menicholas

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

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

Richard111

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

Esa-Matti Lilius

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