## Simple Experimental Demonstration that Cool Objects Can Make Warm Objects Warmer Still

August 26th, 2016 by Roy W. Spencer, Ph. D.

I continue to receive emails (not to mention the hundreds of sometimes nasty blog comments) objecting to what I just expressed in the title of this article. So, I thought it would be useful to propose a simple experiment that demonstrates the concept.

This is a considerably simpler task than my recently proposed experiment to measure the warming effect of adding CO2 to the atmosphere (which I now believe was not possible, at least as I originally proposed it). This experiment would be easy enough for high school students to perform, and maybe even junior high students. It probably does require a good multi-probe temperature monitoring and data logging device. I use the Extech SD200 three-probe temperature sensor. Alternatively, a \$50 handheld IR thermometer might be used in a pinch, if you are careful.

Background

One of the supposed arguments against atmospheric greenhouse gases keeping the Earth’s surface warmer than if those gases were not present is the claim that, since the atmosphere is colder than the surface, it would violate the 2nd Law of Thermodynamics for a cold object (the atmosphere) to increase the temperature of a warm object (the Earths surface).

The Wikipedia entry for the 2nd Law of Thermodynamics includes the following statement from Rudolph Clausius, who formulated one of the necessary consequences of the 2nd Law (emphasis added):

“Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.”

The statement by Clausius uses the concept of ‘passage of heat’. As is usual in thermodynamic discussions, this means ‘net transfer of energy as heat’, and does not refer to contributory transfers one way and the other.

The italicized words are important, and have been ignored by my critics: while it is true that the net flow of heat must be from higher temperature to lower temperature, this does not mean that the lower temperature object cannot (for example) emit radiant energy in the direction of the warmer object, and thus increase the temperature of the warmer object above what it would otherwise be.

But, this statement I just made will lead to endless arguments and objections (watch the comments, below), with hand waving qualitative statements about absorbing and emitting molecules and photons and entropy and perpetual motion and such.

So, let’s envision a simple experiment that will mimic what happens in the climate system, using visible light to heat a warm surface which then cools through infrared radiation toward a cold surface.

Energy Balance of the Global Climate System

The sun’s energy that is absorbed by the Earth’s surface raises its temperature. The surface then loses energy through both (1) non-radiative loss to the atmosphere (conduction, evaporation, and convection), as well as (2) infrared radiative loss to the atmosphere and to outer space.

What is interesting is that the clear atmosphere is mostly transparent to the sunlight passing through it and warming the system, but it is not transparent to the infrared radiation trying to escape back out to cool the system. This is the basis of the atmospheric “greenhouse effect”.

Now, in order for the climate system to maintain a roughly constant average temperature, there must be energy balance: the rate at which the earth-atmosphere system gains energy from the sun must match the rate at which the system loses infrared energy to the cold depths of outer space, which has a radiating temperature of almost absolute zero. If you can reduce the rate at which it cools to outer space, the climate system will increase its temperature until it emits enough infrared energy to restore radiative balance. This is the basis for global warming theory: increasing carbon dioxide in the atmosphere reduces the rate of IR energy loss to deep space, resulting in some warming. (The warming is actually in the lower atmosphere, while the upper atmosphere cools).

The Experiment

We can mimic these radiative processes by continuously heating one surface with halogen light bulbs (which more closely approximates the solar spectrum of light than incandescent bulbs). The hot surface will then radiatively cool toward a second surface which is chilled (e.g. with dry ice inside a cooler). The heated surface will also lose heat through conduction to the surrounding air, too, but we will reduce that effect with Styrofoam insulation….what we are looking for is a radiative effect.

The following cartoon shows the basic setup.

The heated surface is painted black to absorb as much visible light as possible and raise its temperature. The chilled surface is painted with Krylon white #1502 which has an infrared emissivity close to 1.0 (allowing it to efficiently absorb IR energy from the heated surface) while the white color also reflects visible light and so avoids heating from the lamps.

At some point, energy balance will be reached when temperatures stabilize. (Of course, eventually all of the dry ice will be used up…so there is limited time to do the experiment…maybe an hour or two). I suggest putting the heated surface on top so any heated air goes upward and away from the experimental setup. Similarly, the chilled surface will have chilled air spilling down the sides.

Now, if we simply insert a piece of room-temperature cardboard in between the heated surface and the chilled surface, we should see an increase in the temperature of the heated surface despite the fact that we just used a cooler (room temperature) object to make a warmer object even warmer still, in apparent violation of the 2nd Law of Thermodynamics. The cardboard can probably just be laid on top of the cooler. Or, a sheet of Styrofoam might work just as well, if not better. The temperature of the air between the heated and chilled surfaces could be monitored with the third probe from the SD200 to answer any objections that the intervening cardboard is somehow reducing the mixing of air between the hot and cold surfaces (which shouldn’t happen anyway, if the heated surface is above and the chilled surface is below).

The intervening cardboard (or Styrofoam) sheet mimics what the atmospheric greenhouse effect does, at least in terms of energy flows (but it’s a solid surface, rather than a gas, so maybe it’s more analogous to the greenhouse effect of thick cirrus clouds, which completely block the transfer of infrared light).

I don’t know just how much the observed temperature increase in the heated surface will be when the cardboard sheet blocks its view of the chilled surface. Maybe 1 deg. F, maybe 10 deg. F. But it should be observable. The effect will be greater the bigger the temperature difference that can be maintained between the heated and chilled surfaces, and the closer you can get them together so the heated surface “sees” mostly the chilled surface, instead ot the room-temperature surroundings with which it is also exchanging infrared radiation.

Now, this experiment does not prove that gases can do what the cardboard has done; that is a separate issue that is much more complicated to demonstrate with an experimental setup. It only answers the 2nd Law violation claims some have made against a cool object (here, the cardboard sheet) causing a heated object to be warmer than if the cool object was not present, which is what the Earth’s greenhouse effect does.

NOTE TO COMMENTERS: I intend to delete any comments which include personal insults.

NOTE TO READERS OF COMMENTS: Some commenters here throw around technical terms and make grand assertions and detailed arguments which I consider fallacious. I do not have time to counter them all every time they arise, although I have addressed virtually all of them in other posts over the years.

### 314 Responses to “Simple Experimental Demonstration that Cool Objects Can Make Warm Objects Warmer Still”

1. Nabil Swedan says:

“This is the basis for global warming theory: increasing carbon dioxide in the atmosphere reduces the rate of IR energy loss to deep space, resulting in some warming. (The warming is actually in the lower atmosphere, while the upper atmosphere cools)”

This statement is absolutely true. The question is that does this valid statement supported by observation can only be explained by the greenhouse gas effect and rdaiative forcing? The answer is no. There are other ways to explain it through thermodynamics.

Here is another way to explain it:

Swedan, N. (2015) Anthropogenic and Natural Forcings as Functions of Emission Time. Development in Earth Science (DES) 3, 1-7.

• Roy Spencer says:

pdf?

• Nabil Swedan says:

DOI: 10.14355/des.2015.03.001

By the way, more than one publisher expressed interest in this paper including the Russian Academy of Sciences.

• DHMacKenzie says:

Nabil, I read your paper. You have applied the equation for dry adiabatic lapse rate, which applies in the troposphere, to the mesopause which cools by radiation. In the troposphere parcels of gas moving up and down result in reasonably close correspondence to basic thermodynamics pressure temperature relationships, those being larger than temperature changes due to radiative effects at low altitude. The opposite is true in the mesopause where radiative effects dominate and convection and pressure gradient are nearly irrelevant. Are you sure your paper was peer reviewed ? It doesn’t pass mine.
Sorry Dr. Roy, I am way off topic here….

• Nabil Swedan says:

Yes, it passed vigorous peer-review for more than one publisher expressed interest. The paper is a summary of my book titled Global Warming Calculation and Projection which has had over 14,000 review. What is your concern? I would be pleased to answer.

• Ball4 says:

Nabil, I read your paper too. Can you reveal the reviewers? I understand if not.

Your paper uses the 9.8 ideal lapse rate dry air in the troposphere as the T change method (from delta MW your .005% for 100ppm CO2), not radiation. You seem to apply ideal LR theoretically as if it changed in the dry mesosphere where the lapse is shown constant (isothermal) in standard atm. over z ~8km. So the differentials would be zero. Also the local p and density are practically nil in mesosphere compared to troposphere. Can you comment?

The derivation of ideal dry LR derivation doesn’t involve radiation (look closely the infinitesimal slab is assumed constant T(z) so no radiation effect). So of course you find no radiation effect in the idealized thick slab. Not there to begin with. Comments?

• Nabil Swedan says:

1) The mesopause is not a point, it is a region as you can find from reference [16]. Its pressure is not zero but approaches zero.

2)The lapse rate equation is derived based on applying the first law of thermodynamics for a rising air parcel. No radiation term is present in the derivation. Therefore, do not expect any radiation terms within the atmospheric air. They simply do not exist based on derivation of the equation and its observed validation. This subject is discussed under Discussion and Conclusions, first paragraph.

3) The last paragraph of the manuscript on page 3 clearly states that the lapse rate equation applies to the lower atmosphere only. However, the differential dT applies throughout the atmosphere because variation occur infinitesimally with time. This is a good mathematical exercise for you to demonstrate.

• Ball4 says:

1) Mesosphere is indeed a region including ~8km of constant T(z) in the standard for a LR 0 not 9.8, yet Nabil writes “it can be demonstrated that the differential of air temperature of the lapse rate equation with respect to variations in the concentration of carbon dioxide emissions applies throughout the atmosphere.” This does not answer how your differential of T wrt CO2 can be nonzero when the T(z) is constant Nabil.

2)Because you agree radiation is excluded in dry LR derivation, any conclusion drawn is absent of radiation so “do not expect any radiation terms within the atmospheric air” can not possibly be an accurate conclusion. More correct would be “do not expect any radiation terms within the dry LR air”

3) Sure dry LR is for Zo to Ztrop. but you apply it in that eqn. just before that paragraph to “at the mesopause, Z‐Z*” = hm, avg. mesosphere height. This is inconsistent writing.

Also, in eqn. 1) you note “ε =Emissivity of the atmospheric air, dimensionless, approximately equal to unity.” This implies very low transmission like L&O of 0 but the surface can easily be seen from space, the L&O surface does have ε = near unity (0.96-0.97) no transmission, some reflection. Global median, the emissivity of atm. air is actually measured around 0.8. Meaning equatorial tropics 0.95 and dry arctic regions 0.7.

• Nabil Swedan says:

When you do the mathematical exercise you will find that dT decreases with the increase in CO2 regardless of its absolute value. This is not dT with respect to height. Give it some thinking.

• Ball4 says:

Yes. Then your paper needs to be improved to take into the calculus consideration the constant 0 LR in mesosphere and 9.8 in troposphere. And to get up to mesosphere the constant 0 LR and decline in temperature with increasing z of the stratosphere (& T decline as CO2 increases). Or better yet improve it by removing mesosphere entirely and just use troposphere.

Added CO2 in the mesosphere will take the constant 0 LR and make it negative, your paper does not show that in the calculus, you write about “dT decreases with the increase in CO2” but in & above the stratosphere dT increases with the increase in CO2 & your paper calculus does not as it relies on changes only in Cp.

Think about the wet LR. In that case you have to include Cp of condensed water along with Cp of air & increased CO2. Think about that too as it more meaningful in the midlatitude tropics (6.5 LR).

• Ball4 says:

Better: take into calculus increase in T wrt z but decrease in T wrt added CO2 in the stratosphere.

• Nabil Swedan says:

The approach you suggest was used in the book in 2009(Global Warming Calculation and Projection); it made little difference, not much. Of course room for improvement exist.

2. Norman says:

Dr. Spencer

An excellent set up. The only sad thing is that unless those who complain about the 2nd Law violation for GHE do the experiment themselves, they will not accept whatever results anyone else will get (they will have some excuse to explain the temperature rise). They will not accept results and yet they will not take the time to do the experiment themselves.

On the next thread you post they will say the same thing as always.

• Gordon Robertson says:

@Norman…”…they will not accept whatever results anyone else will get (they will have some excuse to explain the temperature rise)”.

We don’t need an excuse, the work was done by a genius, Rudolf Clausius, circa 1850.

Of course, others will claim his work is old which is tantamount to claiming the data sets put out by UAH are influenced by big oil.

You don’t need this experiment, Norman. Grab a steel bar on one end and apply a propane torch to the other end. See how long you can hang onto the bar. The heat moves almost instantaneously from one end to the other and please don’t tell me heat travels from the cool end and warms the flame in the torch.

• Norman says:

Gordon Robertson

Heat travels from Hot to Cold, energy travels in both directions.

You could do this experiment to verify. If you have on bar at 50 C and put it against a bar at 0 C measure the rate of temperature loss of the 50 C bar and the temperature gain of the 0 C bar.

Now put a 50 C bar up to another bar at 50 C and what happens?

No heat flow but energy is still moving about. The atoms or molecules composing the material have not stopped moving. They are still jiggling about their fixed points in the matrix. Since a surface will emit a wide spectrum of IR it can be logically concluded that molecules at the surface have many different energy states. Some are jiggling faster than others. When two 50 C bars are in contact the surface molecules are still jiggling and passing energy back and forth. The reason nothing changes is because the two opposing energy exchanges are taking place at the same time. It is an equilibrium state but the molecules have not stopped their motion and energy exchange just because the temperatures are the same. Hope that makes sense to you.

3. Tim Folkerts says:

I’ve done a similar experiment (conceptually the same but with a few differences in the set-up).

It worked.

4. Dr. Spencer says below which is 100% correct and what I am basing my theory on as to why/how the climate may change.

Now the climate will change if the energy balance is changed and how can that be accomplished? I say first and most telling would be through albedo changes associated with extreme prolonged minimum solar conditions that will result in making the terrestrial items that determine albedo increase the albedo. Those terrestrial items being cloud cover, snow cover ,and sea ice coverage on a global basis to increase.

Secondly the amount of solar energy coming into the climatic system weakening somewhat would change the energy balance of the climatic system and thirdly an increase in major volcanic activity would do likewise.

This is how the climate can change.

So why not GHG effect? Because I think the strength of this depends upon the climate. Why? Because CO2 always follows the temperature never has led it, at least not yet.

Energy Balance of the Global Climate System

The suns energy that is absorbed by the Earths surface raises its temperature. The surface then loses energy through both (1) non-radiative loss to the atmosphere (conduction, evaporation, and convection), as well as (2) infrared radiative loss to the atmosphere and to outer space.

What is interesting is that the clear atmosphere is mostly transparent to the sunlight passing through it and warming the system, but it is not transparent to the infrared radiation trying to escape back out to cool the system. This is the basis of the atmospheric greenhouse effect.

Now, in order for the climate system to maintain a roughly constant average temperature, there must be energy balance: the rate at which the earth-atmosphere system gains energy from the sun must match the rate at which the system loses infrared energy to the cold depths of outer space, which has a radiating temperature of almost absolute zero. If you can reduce the rate at which it cools to outer space, the climate system will increase its temperature until it emits enough infrared energy to restore radiative balance. This is the basis for global warming theory: increasing carbon dioxide in the atmosphere reduces the rate of IR energy loss to deep space, resulting in some warming. (The warming is actually in the lower atmosphere, while the upper atmosphere cools).

5. Nabil Swedan says:

“The italicized words are important, and have been ignored by my critics: while it is true that the net flow of heat must be from higher temperature to lower temperature, this does not mean that the lower temperature object cannot (for example) emit radiant energy in the direction of the warmer object, and thus increase the temperature of the warmer object above what it would otherwise be.”

Objects radiate at any temperature. However, can radiation from a cold object reach the surface of a warm object? This is very difficult to prove experimentally in the laboratory. It is easier to experiment at the surface of the earth instead.

An object between surface and outer space will of course raise surface temperature. We see that all the time in a cloudy day in winter. Clouds are droplets of water, they are an object. Are greenhouse gases an object? Based on physics, they are part of the air mixture and behave as such. This air mixture is in intimate contact with the surface and therefore no radiation can occur between air and surface. Radiation occur between two objects separated apart.This is not the case between air layers “slabs” or “slabs” and surface. The whole radiative forcing concept is against basic physics. Air “slabs” do not radiate among themselves, they can only exchange heat by convection.

• Christopher Game says:

Nabil Swedan says that contiguous slabs of atmospheric air cannot radiate between one another, and that radiation cannot pass between the ground and the contiguous atmosphere. He claims basic physics says so.

Textbooks of radiative transfer do not at all support what Nabil says. They say the opposite. They say that contiguous slabs of atmospheric air always radiate between one another, and that radiation always passes between the ground and the contiguous atmosphere. No gap is needed for radiative transfer between objects. Slabs of atmospheric air are included in the concept of objects for this context.

• Nabil Swedan says:

Not engineering reference books that are based on successful applications. If you are referring to climate textbooks taught in some schools particularly in the west, then you are right. However, most of the world do not agree on these climate text books and do not teach radiative forcing in their schools. They all agree on engineering reference books. In engineering, there is only convection between atmosphere and surface and atmosphere and atmosphere layers.

• Tim Folkerts says:

A Heat Transfer Textbook, 4th edition
“This book is an introduction to heat and mass transfer oriented toward engineering students. It may be downloaded without charge.”

And it has as section on radiation of gases for furnace design. (Note also it is from MIT).

http://web.mit.edu/lienhard/www/ahtt.html

• George Hong says:

thanks for the book. there is a strange experiment 1.3 about being able to feel cold based on an absence of “back radiation”. the room is emitting in all directions but if you open the fridge there is no back radiation from that direction and you can actually feel it. don’t know how that is supposed to work. as soon as i open the door cool air comes out. somebody said the same about the night sky. turn your face towards the night sky and you should feel the cold. maybe my internal thermometer is broken.

• Christopher Game says:

cites
“Perry,R.H.,andGreen,D.PerrysChemicalEngineersEngineers Handbook. SixthEditionMcGraw‐Hill,U.S.A.(1984)Editors
HaroldB.CrawfordandBeatriceE.Eckes,p.4‐52to4‐66.”

I do not have a copy of the 6th edition. So I will cite the 8th edition. It does not have the same break-up into sections of text. In the 8th edition, radiative transfer is considered in Section 5, edited by Hoyt C. Hottel, James J. Noble, Adel F. Sarofim, Geoffrey D. Silcox, Phillip C. Wankat, and Kent S. Knaebel.

In Perry’s ‘Chemical Engineers’ Handbook’ 8th edition, 2008, edited by James O. Maloney, McGraw-Hill, USA, on page 5-15, under the heading ‘HEAT TRANSFER BY RADIATION’, sub-heading ‘GENERAL REFERENCES’, I find the following entries, amongst others.

Thermal Radiation Heat Transfer and Properties, Wiley, New York, 1992. Goody, R. M., and Y. L. Yung, Atmospheric RadiationTheoretical Basis, 2d ed., Oxford University Press, 1995. Hottel, H. C., and A. F. Sarofim, Radiative Transfer, McGraw-Hill, New York, 1967.

I have studied Goody & Yung, and to some extent Hottel & Sarofim. They disconfirm your proposal that contiguous slabs of atmospheric air cannot radiate between one another, and that radiation cannot pass between the ground and the contiguous atmosphere.

On page 5-35 of Perry’s Handbook, 8th edition, I read: “Reconsider a generalized enclosure with N volume zones confining a gray gas. When the N gas temperatures are unknown, an additional set of N equations is required in the form of radiant energy balances for each volume zone.” The term “volume zone” includes the notion ‘slab of atmospheric air’. The method of volume zones, otherwise called ‘slabs’, is used by the reference you cite, admittedly in a different edition. This disconfirms your proposal that contiguous slabs of atmospheric air cannot radiate between one another, and that radiation cannot pass between the ground and the contiguous atmosphere.

• Nabil Swedan says:

If you are referring to furnace sizing, you are talking about a glowing air at 1800 degrees C or more. Normally product of combustion such as water and CO2 are used in the design. This is just a method of sizing the furnace not a greenhouse gas effect, and it is not the only method. The glowing air radiate of course, and we see that in a kitchen stove. The atmosphere on the other hand is at ambient temperature or less, there is a big difference.

• Ball4 says:

Nabil, The atm. also glows at ambient temperature (night and day) just your lying eyes cannot detect that glow. IR telescopes do detect the atm. glow so they are built at great expense above as much atm. as possible.

• Christopher Game says:

Nail Swedan, the volume zone principle holds at all temperatures above absolute zero.

As you rightly observe above at August 26, 2016, at 12:58 PM:

Volume zones, or ‘slabs’, of gas are objects in this context. You are right that the radiation from gases is far stronger at very high temperatures. But it is still importantly large at tropospheric temperatures, and occurs at all atmospheric temperatures. To make it vanish, the temperature would need to be taken to absolute zero, which is not feasible. Gases absorb radiation at all temperatures.

• Nabil Swedan says:

Christopher Game,

Design of all equipment and buildings I know of discard radiation between ambient air and equipment. 340 w/m2 of greenhouse gas effect at night is too huge not to be considered in the design. No engineer ever account for it in the design at night. Only proponents of the greenhouse gas effect do for they are not held accountable.

• Ball4 says:

If there was no 340 at night, they would have to consider THAT in the building designs. And building designs do consider air ambients for the location; Antarctic hut HVAC design and equatorial HVAC are going to be NOT the same.

• Christopher Game says:

Nabil Swedan, I read your 340 W/m2 as the disputed radiation from atmosphere to ground. It is associated with radiation from ground to atmosphere. The latter is normally a relatively little amount greater, for example in this case say about 350 W/m2. The heat transferred is then about 10 W/m2 in round numbers. Though myself not an engineer, I can guess, subject to your correcting me, that 10 W/m2 may be neglected by an engineer for building design?

The downwelling radiation from atmosphere to ground varies rather greatly from time to time and from place to place. It is very often correlated in magnitude with the simultaneous and collocated upwelling radiation from ground to atmosphere. Mostly the radiation upwelling from ground to atmosphere exceeds that downwelling from atmosphere to ground by such relatively small amounts. The correlation is remarkably close.

Nevertheless, there occur at various times and places important departures from that close correlation. They are associated with so-called temperature inversions: the “lapse rate” is negative just above the ground. Nocturnal inversions are quite common in some conditions a little before sunrise. In the polar winter night, with no sunrise shortly pending, polar inversions can be very strong, so that the atmosphere is temporarily and locally radiatively heating the ground. This latter occurs when the local effective average temperature of the atmosphere exceeds the local effective ground temperature, always in accord with the second law of thermodynamics.

Overall on average, the usual situation dominates, so that the ground radiatively heats the atmosphere by about 10 W/m2. In round numbers on global average, 60 W/m2 of ground radiation penetrates the atmosphere and so escapes directly to space.

Most of the cooling of the ground is by immediate conduction with secondary convection, and by evaporation. The ground may also be cooled by hail or snowfall. These kinds of cooling are what you are emphasizing. On global average they may be about 50 W/m2, about five times the radiative cooling of ground to atmosphere.

A primary effect of an increment in atmospheric carbon dioxide is to slightly reduce the above-mentioned 60 W/m2 that escapes directly to space. This slightly increases the above-mentioned excess 10 W/m2 of heating of atmosphere by ground. There are many important and not fully known secondary effects.

Warmophobics emphasize and exaggerate the effects of added atmospheric carbon dioxide, and emphasize and exaggerate the human contribution to the addition. There are many motivations for warmophobia. It is a very harmful group delusion and political ideology.

• Nate says:

Nabil,
I dont understand your point. Do molecules not radiate and absorb EM waves? If I shine IR light through a slab of CO2 or air, does the slab not absorb some of this, heat up and radiate?

If gas molecules cannot radiate than I dont understand how a HeNe laser can work.

• Nabil Swedan says:

The Beer-Lambert equation is a Law. Atmospheric air molecules absorb solar radiation as heat based on this law. Heat is then convected throughout the atmosphere. This is the global pictures. Do some molecules reach to the point of excitement and radiate as a laser? May be may be not. In engineering, this “laser” effect, if exists, is never accounted for in air related calculations, never. Greenhouse gas effect is not accounted for in industrial application, never. In a kiln application, there is more carbon dioxide from one end to the other than the entire carbon dioxide in a column of air. Yet, the 340 w/m2 of greenhouse gas effect is never accounted for in the calculations. Only few schools of the west teach this effect; this is not enough. You need the world to agree with you, and you have a long way of convincing a head of you.

• Norman says:

Nabil Swedan

Wow! Where did you take your College Chemistry courses for you Chemical Engineering degree?

They must teach you some really crazy ideas at that University. I have not read where air molecules absorb solar radiation except ozone and UV. The atmosphere is transparent to visible light waves which are what most Solar radiation is (some IR but bulk is visible).

In your Chemistry classes did you do any spectrometer work?
If you had you would know that radiation can penetrate through a gas. What class did you take in engineering that informed you that IR cannot go through air??? Have you every used a TV remote that uses IR as its signal carrier? It makes is all the wave across the room with no problem.

I will look up the mean-free path of a 15 micron photon in the atmosphere.

• Tim Folkerts says:

Nabil, don’t confuse a model (“slabs” of air) with reality. Radiation in the atmosphere is from one molecule to another. Molecules are not “touching” and can radiate to other molecules many meters away.

• Gordon Robertson says:

@Nabil Swedan…”An object between surface and outer space will of course raise surface temperature”.

No it won’t, only solar radiation can raise the temperature. If anything else could we’d have been in serious danger of Hansen’s tipping point a long time ago.

You are also presuming that radiation is the only means of lowering surface temperature. Convection will do it just as well as Lindzen describes in his paper on the greenhouse effect. Heat can effectively be leeched from the surface through conduction and convection.

Lindzen has claimed that without convection the Earth’s surface could reach 72 C. Convection and conduction are already keeping the surface cooler by 60C on average if he’s right.

Look what happens in winter on the Canadian and US prairies when the tilt of the Earth’s causes reduced radiation from the Sun. They fall well below 0C. Only the Sun can recover the atmospheric warming the following spring. Nothing in the atmosphere will warm the prairies.

At the same time, warm ocean current keep the coast of southwestern Canada above 0C most of the winter. We should stop this metaphor about the GHE and look at the oceans as the source of atmospheric warming.

• Ball4 says:

“No it wont, only solar radiation can raise the temperature.”

Has Gordon not experienced rising surface temperatures over night? Happens quite often actually, you know like when a night time cloud passes as shown in NOAA ESRL data that Dr. Spencer used in the last post. In fact his water temperature was shown higher observing the high cirrus cloud than the water not observing the cloud last summer.

• Nabil Swedan says:

Gordon,

In our colloquial sense, clouds keeps the surface warmer in winter. Surface radiation is slower under clouds than under clear sky in a winter night. We feel warmer with clouds overhead as a result.

6. Christopher Game says:

Nabil Swedan says that contiguous ‘slabs’ of atmospheric air cannot radiate between one another, and that radiation cannot pass between the ground and the contiguous atmosphere. He claims basic physics says so.

Textbooks of radiative transfer do not at all support what Nabil says. They say the opposite. They say that contiguous ‘slabs’ of atmospheric air always radiate between one another, and that radiation always passes between the ground and the contiguous atmosphere. No gap is needed for radiative transfer between objects. Slabs of atmospheric air are included in the concept of objects for this context.

7. A C Osborn says:

Of course a gap has to be involved.
Otherwise “Radiation” would have nothing to do with it.
It would be Conduction.
If Dr Spencer believes in Back Radiation warming the Earth’s Surface as per the NASA heat budget diagrams, perhaps he can explain why it can do no work, unlike Solar Radiation and Infra Red Radiation etc.
Solar Stills Cool things at night not warm them as they do in the day time.

• Roy Spencer says:

“back radiation” is part of the infrared radiation flow you mention. So I don’t know what your objection is. Downwelling IR radiation acts to increase the temperature gradient between the lower atmosphere and upper atmosphere (e.g. Manabe and Strickler, 1964), by allowing solar heating to warm the lower atmosphere to a superadiabatic state, leading to convective instability, and thus actually HELPS to do thermodynamic work.

• Chic Bowdrie says:

I realize you are the one with the degree and 30 years experience, but that comment is badly worded IMO. Radiative flow does not mean heat flow. Usually air must be at a higher temperature for heat to flow to air at a lower temperature. So downwelling IR makes the temperature gradient greater than it would otherwise be, but no heat flows downward unless the temperature of the above is greater than below. (There might be an exception as explained in the next paragraph.) This is the condition known as an inversion, correct? Solar heating does not need help from DWIR to create a superadiabatic state, although the more back radiation the greater the adiabatic state will be. But the Point being there is no heat flow from back radiation AKA DWIR unless the air above is warmer than below.

Actually since I am a supporter of the gravitothermal effect, I think there is a range where the temperature gradient is less than the environmental lapse rate, but not yet negative, and some heat would flow from cooler air to warmer air. This would have to be proven experimentally, of course, and would be difficult because of the need for an extremely long and well-insulated apparatus, as you have alluded to many times.

8. gbaikie says:

–I dont know just how much the observed temperature increase in the heated surface will be when the cardboard sheet blocks its view of the chilled surface. Maybe 1 deg. F, maybe 10 deg. F . But it should be observable. —

If used a mirror [regardless of it’s temperature] it should cause higher increase in temperature, than the cardboard or a sheet of Styrofoam. Oh, also assuming one is going to paint either with same flat white paint as aluminum sheet on top of cooler.
So a mirror could be the reflective side of aluminum foil tacked/taped onto the cardboard

— The effect will be greater the bigger the temperature difference that can be maintained between the heated and chilled surfaces, and the closer you can get them together so the heated surface sees mostly the chilled surface, instead ot the room-temperature surroundings with which it is also exchanging infrared radiation.–

It seems if have CO2 ice touching the aluminum sheet rather have air space between them, the aluminum sheet should be
closer to the -78.5 C temperature of dry ice.
Also to limit convectional, you turn it over sideways so one doesn’t descending and rising convection- still have some air convection but it reduces the effect.

But if instead go with the vertical, you try leaving the lid off the cooler, putting water with dry ice, and see if water vapor smoke [similar to clouds] increases the temperature [by reflection of water droplets].

9. doctor no says:

Roy, another excellent post.
Have you considered collecting these and assembling them as topics/practical experiments
that could be incorporated in a climate-related course?

10. Rob JM says:

Thermodynamics does not apply to quantum scale interactions.
Nothing else needs to be stated.

Additionally one form of heating, such as GHG or gravitational warming does not have a monopoly, since they are not mutually exclusive.

The existence of other warming mechanisms does however diminish the effect of GHG, thus further invalidating the notion of dangerous manmade warming.

Water vapour positive feedback on the other hand does clearly violate the 2nd law, which is why no evidence for existence can be found. If evaporation caused warming then evaporation would warm all by itself, ie any reaction that can happen must happen.

It should also be noted that gravitational heating also explains the faint young star paradox quite well, ie thicker atmosphere producing warmer world with less solar energy.

• Gordon Robertson says:

@Rob JM…”Water vapour positive feedback on the other hand does clearly violate the 2nd law…”

Of course it does, if positive feedback of any kind existed in our atmosphere we’d have been toast long ago.

There is no reason to believe that the current concentration of CO2 in our atmosphere, 0.04%, has changed significantly, even going back before the industrial revolution. The IPCC admitted in the 2001 review that ACO2 is only a small fraction of that 0.04% figure based on a concentration of 390 ppmv.

There has been no catastrophic warming with 0.04%.

The only evidence that it has changed came from ice core proxies in Antarctica where the concentrations of pre Industrial CO2 levels were cherry picked (according to Jaworowski). Circa 1930, scientist like Kreutz measured CO2 concentrations in the atmosphere (according to Beck) which exceeded 400 ppmv.

We saw what happened with the tree ring proxies of Mann et al over 1000 years. They started showing cooling when real data was showing warming.

11. Dan Pangburn says:

Roy, It should work.

The cold plate will certainly see condensation, possibly frost (if dry ice is used) which I see no problem with. Heating the heated aluminum sheet from below means the plates must be spaced a distance apart which allows lots of radiation interaction with the environment blurring the results. If the top plate was heated from above, the plates could be moved close together, minimizing environmental effects but people who need a demonstration to understand this stuff probably also wouldnt understand that it doesnt matter how this plate is heated as long as it is heated at a constant rate.

Instead of dry ice, it should work with just water with ice floating which for sure makes a constant temperature.

As to scale, IMO the setup could be much smaller than implied by Styrofoam cooler.

12. Mike Flynn says:

It’s a matter of observation that the Earth’s surface cools at night. The surface is emitting more EMR than it is absorbing. Winter is cooler than Summer for the same reason.

The Earth has cooled for four and a half billion years for the same reason.

Now we have people apparently seriously suggesting that a bottle of water, or a concrete block left on the surface will be hotter each day, and after a hundred years or so, have heated by whatever the current belief happens to be. For convenience say 1.5 C.

The contention seems to be that reducing the CO2 concentration in the air would then result in the test object reducing its temperature by say 1.5 C over another 100 years or so.

If anybody complains that it is impossible to measure such slow rates of temperature increase or decrease experimentally, then I would have to say that your hypothesis is invalid, as it cannot be falsified experimentally.

Pouring money into a cure for unfalsifiable hypothetical problems doesn’t seem terribly cost effective to me.

Cheers.

13. Mike Flynn says:

Even simpler test to “prove” cold objects can raise temperature of warmer.

Take Thermos flask (smaller is better, here.)

Poke wires through stopper.

Solder 12 V 25 W (or bigger. Bigger gives more spectacular result!) globe to wires so globe sits inside flask when stoppered.

Seal flask with stopper. Attach 12 V supply to wires outside flask.

Measure current through globe. Wait for either noise inside flask, or sudden cessation of current through globe due to glass melting and subsequent destruction of filament due to excessive heating beyond design limits.

Indicates why things like power transistors dissipating reasonable power internally need adequate heat sinks. The transistor junction gets hot, and being surrounded by cooler material, increases its temperature beyond melting point. Too well insulated, obviously.

Luckily, the Earth’s nighttime surface is a big heat sink, radiating to a 4K environment. Even when overlaid by much GHG, it manages to dissipate all the daytime heat. Whew!

Simple physics. Still doesn’t support global warming.

Cheers.

• Gordon Robertson says:

@Mike Flynn…”Even simpler test to prove cold objects can raise temperature of warmer”.

Mike, with all due respect, inserting a fragile heat source into a thermos bottle, which represents a confined space and has reflective glass walls, and allowing the heat to build up inside the thermos till it shatters the glass, is hardly an example of a heat being transferred from a cold object to a warm object.

As you correctly pointed out, a power transistor requires a heat sink to increase the surface radiation to cool the device. You’ll notice that the heat sink is general mounted outside an amplifier with the transistors on the outside. If you insert the heat sink and transistor into a confined space with reflective glass walls, you can expect the transistor to heat up as well.

That’s not due to a cold object heating the transistor is strictly due to a lack of convection.

• Mike Flynn says:

Gordon,

I agree. I was trying to point out the silliness of the idea that insulators provide heat. My point is that some would claim that the atmosphere works the same way as the Thermos flask – although to a much reduced degree. The walls of the Thermos are at a much lower temperature, and many would say this demonstrates the fact that a hotter body (the lamp filament) can have its temperature raised by back radiation from a cooler body – that of the Thermos, or, in the case of the Earth, the atmosphere. Of course, it doesn’t, as alluring as the idea seems at first sight!

Likewise, people claim that the atmosphere allows more total energy in than it allows out, thus resulting in a build up of energy from the Sun retained by the Earth, resulting in increased surface temperatures. People might claim that the increased temperature cannot be actually measured, even though an increase of say 1 C / century is only 0.01 C per annum, which appears well within the measuring range of current thermometry. Nor can this effect be demonstrated in the laboratory, even using CO2 at concentrations of 1 million parts per million, rather than the 400 ppm or so in the atmosphere.

Altogether somewhat unsatisfactory, if claiming that an insulating atmosphere raise the temperature of the surface – particularly on the night side.

The Earth has cooled since its creation, a fact accepted by most, but denied by some. Insulation provided by the atmosphere would appear to be a less than adequate mechanism for suddenly raising the surface temperature of a body which has been steadily cooling for four and a half billion years.

Cheers.

• RICHARD says:

Hmm Mike,, i guess if you sealed the Sun it would explode which would seem to be a better explanation of your experiment above.

We are talking about energy that is being reflected off a surface.

14. fonzarelli says:

It’s all very simple… If surface temps are not warmer at the end of the night than they were at the beginning of the night, then there is no violation of the 2nd law of thermodynamics.

15. An Inquirer says:

My basic question, Dr. Spencer, is why do you spend time on this subject? Certainly you have many opportunities to make valuable contributions elsewhere. How much is gained by talking to brick walls?

• wert says:

Not all of us are brick walls.

16. Steve Fitzpatrick says:

Hi Roy,

Not sure why you keep trying to convince people who are either incapable or simply unwilling to understand the concept of blackbody radiation. As I have said before, ’tis tilting at windmills, nothing more. If you want to ‘prove’ how a cooler object can warm a warmer object, you can use a vacuum chamber and four metal plates (painted black, of course). One plate is heated electrically with constant power ( say 100 watts. Two identically sized plates, located close to the heated plate but on either side, are held at constant temperature by circulating constant temperature water (say 10C) within the plates. After the power is on for a while, the heated plate will warm to an equilibrium temperature. Since the plates are in a vacuum, heat transfer between plates is almost 100% radiative, not by convection or conduction.

After the temperature of the heated plate is constant, a fourth thin plate, neither heated nor cooled, is slipped between the heated plat and one of the cooled plate, but touching neither. When the system reached a new equilibrium, the heated plate will be warmer than before the fourth plate was added, even though the added fourth pate is cooler than the warm plate. .The heated plate will warm, even though the added (4th) plate is

• Steve Fitzpatrick says:

is cooler, because all plates radiate according to the Stefan-Boltzmann equation,.

• Gordon Robertson says:

@Steve Fitzpatrick…”Not sure why you keep trying to convince people who are either incapable or simply unwilling to understand the concept of blackbody radiation”.

For one, blackbody radiation is a reference to a totally theoretical cavity resonator. It is meant as an idealization of broadband EM and applies basically to very hot bodies like stars.

For another, radiation must obey the 2nd law when it comes to heat transfer. Show me any blackbody problem that deals with heat transfer in the real world.

When you have two very hot independent heat sources close to each other, there is little doubt that heat can be transferred both ways. In real life, on Earth, that seldom happens and I seriously doubt that any good work has been done to quantify the transfers.

In the Earth’s surface/atmosphere model, you have a warmer body heating a cooler body via EM radiation. There is no reason to accept, based on physics, that the cooler atmosphere can transfer heat back to the warmer surface. If you have proof of that I’d like to see it.

• Steve Fitzpatrick says:

Gordon,

You really don’t understand what you are talking about. You are exactly the kind of person who would be well served to learn a lot more than you currently know, so that people like Roy would not waste their time refuting your nonsense.

There is never net radiative transfer from a cooler to a warmer object, and nobody has suggested that. What Roy has said is that materials emit infrared based on their temperature, independent of where that infrared is going. The photons don’t ‘know’ the temperature of an object they will ultimately encounter, just as they don’t ‘know’ anything about the arrival of photons from another radiating object. Blackbody photon emission is always independent of emission from elsewhere, and independent of target temperature. A fellow named Albert figured out that all radiation was quantized and emitted and absorbed as individual ‘packets’ of specific energies around the turn of the last century. The photons you see from stars could have been emitted many millions of years ago; those photons did not know when they were emitted they would ultimately be absorbed by the retinal cells in a creature (you) which did not exist when they were emitted. The requirement of causality demands that photon emission be independent of target… or target temperature.

Blackbody radiation applies to things other than stars. The four plate experiment I described would show that. The IR emission from gases like CO2 in the atmosphere is non-directional…. some goes towards space, some towards Earth. CO2, water vapor and other IR active gases do not know to emit only in the direction of space. There is no difference between ‘back radiation’ and outward radiation from CO2 in the atmosphere, there is only random radiation in all directions, based on temperature.

So do you believe or not believe that adding an unheated plate between the heated plate and one of the constant temperature plates, in the experiment I described above, will cause the temperature of the heated plate to rise? If you believe the heated plate will warm, then please explain why you think that happens. If you believe the added plate will not cause the heated plate to warm, then would you like to wager \$50,000 on that?

• Nate says:

Yes indeed.

Gordon, Nabil, anybody else, I think you should seriously consider Steve’s well framed arguments and questions, and point by point make a reasoned argument, or agree with them.

As a point of reference, in cryogenics work and textbooks, it is absolutely essential to understand the principles behind the 4 plate example he gives. In fact if the number of plates is increased you get whats called ‘super insulation’, which is widely used, in e.g. MRI machines.

• Dan Pangburn says:

Multi-layer insulation (MLI) is also widely used on artificial satellites as part of temperature control.

• gbaikie says:

” A fellow named Albert figured out that all radiation was quantized and emitted and absorbed as individual packets of specific energies around the turn of the last century. ”

was born we were building steam engines and correctly plotting
the position of objects in our solar system- though it’s true that Albert’s theory explained the oddity of Mercury’s orbit.

Roughly Albert explained why one need certain amount of energy
to emit light or X-rays. Or infinite watts does not cause x-rays. Or watts of energy doesn’t equal watts of energy- or thousand people handing hands does not create visible light.
So any material which given enough intensity of energy will emit light or X-rays- but all the back radiation of earth doesn’t. Though one person can pedal a bike which causes a generator to make a light glow white.
So something 300 K emits a portion of radiation that something 200 K doesn’t.

–So do you believe or not believe that adding an unheated plate between the heated plate and one of the constant temperature plates, in the experiment I described above, will cause the temperature of the heated plate to rise? If you believe the heated plate will warm, then please explain why you think that happens. If you believe the added plate will not cause the heated plate to warm, then would you like to wager \$50,000 on that?–
Hmm check it out:
–One plate is heated electrically with constant power ( say 100 watts. Two identically sized plates, located close to the heated plate but on either side, are held at constant temperature by circulating constant temperature water (say 10C) within the plates. After the power is on for a while, the heated plate will warm to an equilibrium temperature. Since the plates are in a vacuum, heat transfer between plates is almost 100% radiative, not by convection or conduction. —
Heat a plate by 100 watts tells you nothing- the plate could emitting gamma rays or be hotter than the sun.

• Steve Fitzpatrick says:

Absolute, nonsensical rubbish. You are simply wrong, and you can’t appreciate that because you don’t understand what you are talking about. It is a bit like when someone is dead,: they don’t know it, but everyone else does. I sincerely suggest that you learn more and write less.

• Gordon Robertson says:

@Steve Fitzpatrick…”You really dont understand what you are talking about. You are exactly the kind of person who would be well served to learn a lot more than you currently know, so that people like Roy would not waste their time refuting your nonsense”.

You’d be well served to put away the emotion and drop the ad homs. Emotion reveals an investment in a theory.

“The photons dont know the temperature of an object they will ultimately encounter, just as they dont know anything about the arrival of photons from another radiating object”.

You seem to be suffering the same delusion that EM of any frequency and intensity WILL be absorbed by atoms, no matter what. You are wrong, Bohr said so decades ago. Atoms absorb in specific frequency ranges and even at that the EM will only warm the atom if it can raise it’s energy level.

Tell me why the EM from a colder object should raise the atomic energy in the atoms of a warmer object? The 2nd law says they don’t.

• Steve Fitzpatrick says:

As I said, you really just don’t understand. My comments have no ‘ad homs’, they are directed only at pseudo-scientific nonsense. Lots of materials absorb and emit IR over a very wide wavelength range, or if you prefer, a wide photon energy range. Liquid water is a strong absorber from about 1 micron to >1 cm wavelength, so absorbs all of the infrared wavelengths emitted by CO2 and other IR emitters in the atmosphere. Sure, emission from CO2 gas corresponds to an energy state transition, but that doesn’t keep warm water on the ocean’s surface from absorbing the IR photons emitted by cooler CO2 gas in the atmosphere…. water absorbs all IR wavelengths. Liquid water also emits at photons that same wavelength, and some these are absorbed by CO2 in the air. Both are emitting and absorbing photons at the same time. The net IR flux is (and must be) from warmer toward cooler, but that doesn’t mean CO2 doesn’t ‘back radiate’…. it radiates randomly in all directions, including some back toward Earth.

I’d like to know what you predict will happen in the four plate experiment described above.

• Ball4 says:

Gordon, “You seem to be suffering the same delusion that EM of any frequency and intensity WILL be absorbed by atoms, no matter what.”

It is not a delusion to suffer from, that is exactly what Planck found from profuse proper testing and later proved by theory. As Steve writes, objects (water, gas, solid) absorb/emit EMR at all frequencies at all temperatures all the time. Plug any frequency and any temperature into Planck ideal law and you get a nonzero intensity. Might be below today’s equipment to measure but tomorrow’s? Who knows.

Despite that single atoms and polyatomic molecules do have quantum jumps in their electronic energy levels! Hint: the atoms translational KE is NOT quantized.

• mpainter says:

B4:”…what Planck found from profuse proper testing and later proved by theory. ”
###
Surely this is backwards. Surely you mean to say “explained by theory” or some such phrasing other than “proved”. Theories explain, they do not prove. Observations, not theories, “prove”, insofar as they generally confirm convincingly.

• Ball4 says:

mp, Planck first found his formula from much testing in his lab and also labs of many others; the theory was somewhat available in pieces so that guided him, only after he had his empirical results was the analytical complete theory worked out from 1st principles & it then supported his experimental results. Achieved the status of a law.

• Gordon Robertson says:

@Steve Fitzpatrick…”After the temperature of the heated plate is constant, a fourth thin plate, neither heated nor cooled, is slipped between the heated plat and one of the cooled plate, but touching neither. When the system reached a new equilibrium, the heated plate will be warmer than before the fourth plate was added, even though the added fourth pate is cooler than the warm plate. .The heated plate will warm, even though the added (4th) plate is cooler…”

What exactly are you trying to imply here, that the cooler plate warms the warmer plate through EM radiation to the hot plate? If so, that’s a flagrant violation of the 2nd law.

It’s well understood in electronics that power transistors require heat sinks to increase the surface of radiation in order to cool the transistor. The plate in your experiment is acting to interfere with the radiative properties of the atoms in the heated metal.

Radiation intensity drops off as the square of the distance. Before you insert the plate, the plates surrounding the heat plate will absorb EM from the heated plate and rise to the temperature of the heated plate. They will enter a state of thermal equilibrium.

If you insert a plate in between, the radiative field is reduced and becomes more intense. The inserted plate will very quickly rise to the temperature of the heated plate and it will likely be hotter than the surrounding plates. Therefore a new state of equilibrium will be required requiring the heated plate to become hotter.

Let’s get one thing straight. An electrically heated plate emits radiation based on the current through the plate. If the plate gets hotter, the resistance will increase and the current will reduce, cooling the plate, not warming it.

I claim this experiment is nothing more than another thought experiment.

• Steve Fitzpatrick says:

The two outer plates are held at a constant temperature (I said 10C), they will not equilibrate with anything, they are held at constant temperature. The heated plate is supplied with a constant rate of energy (100 watts), and will rise in temperature until radiative transfer from the heated plate to the two constant temperature plates equals 100 watts. When a thin additional plate is placed between the heated plate and one of the cool plates, the added plate will reduce the net radiative loss on one side of the heated plate and so cause the heated plate to warm. The system will reach a new equilibrium when the total net radiative loss from the heated plate returns to 100 watts. At that new equilibrium, the two outside plates will remain at 10C, the heated plate will be warmer, and the added thin plate will be at a temperature between the heated plate and 10C. Using Stefan-Boltzman, you can calculate what the new equilibrium temperatures for the heated plate and added plate will be… yes, Stefan-Boltzman will include ‘back radiation’ from cooler to warmer surfaces in the calculation, even though the NET radiative flux is always from warmer to cooler.

• Steve Fitzpatrick says:

By the way, it is relatively simple to design a control system (analog or digital) which will hold the electrical power flowing into the plate at a constant 100 watts over a broad range of temperatures, independent of how warming changes the resistance of heating element(s) in the plate. You may not be familiar enough with controls to understand this is not difficult to do, but it is not.

17. Gordon Robertson says:

@Roy…”The statement by Clausius uses the concept of passage of heat. As is usual in thermodynamic discussions, this means net transfer of energy as heat, and does not refer to contributory transfers one way and the other”.

I don’t think it’s fair to claim a quote is from Clausius when the first part is from Clausius and the second part, quoted above, came from whoever created the wiki entry. Clausius did not state the second part or imply it.

As you know, anyone can create a wiki article and if it’s wrong, someone has to challenge it by changing it. Then a wiki moderator makes a ruling on whether the change is applicable.

One of the moderators on Wikipedia was William Connolley, a contributor to realclimate, who had a field days panning anything attributed to Fred Singer. He apparently hated Singer and he took his disdain of skeptics so far that he was released from the wiki. Wiki articles on science have to be read very carefully.

Clausius mentioned nothing about ‘net transfer of energy as heat’. He mad it clear that ‘all’ the heat must be transferred from a warmer object to a cooler object unless compensation is applied externally to force a transfer from a cooler object to a warmer object.

The reference to contributory transfers make no sense whatsoever because in a system without compensation there is no two way transfer.

Even moreso, claiming heat is transferred

His analysis, like that of Carnot is based on a heat engine. A heat engine can transfer heat in one direction only unless provisions are made to transfer heat from a colder object while immediately replacing the heat removed.

• Gordon Robertson says:

sorry…part of my reply got lost. I started to say, “Even moreso, claiming heat is transferred…”eat is transferred as

That was meant to read, claiming heat is a net transfer of energy as heat is misleading. In radiative transfer, heat is transferred via EM. Clausius address that situation and claimed the same, that heat can NEVER be transferred from a cold body to a warm body without compensation.

Heat transfer in a solid, like an iron bar, does not require EM to transfer heat, the heat is transferred directly from atom to atom, and that heat is the kinetic energy inherent in the atom. Climate alarm has forced us to consider only radiative transfer, but even at that, radiative heat transfer must obey the 2nd law as laid out by Clausius.

• Ed Bo says:

Gordon:

Clausius is very clear that the compensation he is talking about is the (greater) energy transfer from hot to cold – both in radiative and conductive heat transfer.

I’ve quoted you his exact words many times, and they are as clear as clear can be. Yet you completely fail to understand them!

• Gordon Robertson says:

@Ed Bo…”Clausius is very clear that the compensation he is talking about is the (greater) energy transfer from hot to cold both in radiative and conductive heat transfer”.

Sorry…he is specific about the compensation claiming it must be supplied externally and that, on its own, heat can NEVER be transferred from a colder to a warmer body.

• Ball4 says:

NEVER? No. This is an instance for micro thermo. to locally conflict with macro.

Gordon, consider in a cooler iron bar contacting a warmer iron bar where a slower vibrating atom in the warm bar might get an increase in KE from a faster vibrating atom in the cool bar. Micro KE flows cold to hot. Statistically by M-B distribution that will happen often but always net of that micro action, macro KE overwhelms it statistically & macro KE increases in cool bar.

Might be easier visualize in a gas.

• Ed Bo says:

Gordon:

I have quoted Clausius’ exact words to you multiple times, and you seem completely incapable of understanding them. For the benefit of others, at least, here is his key explanation of exchange in heat transfer:

“In this case we have not a simple transmission of heat from a colder to a warmer body, or an ascending transmission of heat, as it may be called, but two connected transmissions of opposite characters, one ascending and the other descending, which compensate each other.”

It is the transfer from hot to cold that compensates for the transfer from cold to hot. There is NO mention of “external” compensation in this or in the entire section I have quoted you repeaatedly and at length. He is NOT talking about refrigeration systems here, where a work input is required to achieve a NET heat transfer from cold to hot. He is talking about basic heat transfer alone.

• Kristian says:

Ed Bo says, August 28, 2016 at 9:20 AM:

It is the transfer from hot to cold that compensates for the transfer from cold to hot. There is NO mention of “external” compensation in this or in the entire section I have quoted you repeaatedly and at length. He is NOT talking about refrigeration systems here, where a work input is required to achieve a NET heat transfer from cold to hot. He is talking about basic heat transfer alone.

Yup. You’re quite right, Ed Bo. Robertson here has simply misunderstood on a fundamental level, it seems, what Clausius is actually saying.

18. Bill Hunter says:

To ensure temperature is the only difference between the removable cardboard and the cooled aluminum plate, I would recommend they both have a similar surface and color.

also engineering experience teaches that radiant barriers (reflective ceilings) are quite effective in reducing downwelling heat and quite ineffective at reducing upwelling heat. . . .because of convection eliminating the effect.

But I think your experiment is worth doing to demonstrate, if you can, that the heated black plate will get warmer in the presence of a colder object.

If you flip the experiment upside down to mimic the greenhouse effect you may not be able to get enough to measure.

• Gordon Robertson says:

@Bill Hunter…”…engineering experience teaches that radiant barriers (reflective ceilings) are quite effective in reducing downwelling heat and quite ineffective at reducing upwelling heat. . . .because of convection eliminating the effect”.

When you talk about convection, you are talking about the transfer of heat via atoms (molecules) of air. That is far different than the transfer of heat via radiant transfer.

There is no upwelling or downwelling heat in radiative transfer. The heat remains in the respective bodies, increasing and decreasing as emitted EM is absorbed. Unless, convective heat transfer in association with heat conductance from a body to the air is in effect.

• Bill Hunter says:

thats all very true but what difference does it make?

• Gordon Robertson says:

@Bill Hunter…”…thats all very true but what difference does it make?”

The premise held by many here, including Roy, is that heat can be transferred radiantly from a cooler body to a warmer body, supposedly slowing its rate of cooling.

• Ball4 says:

Yes Gordon, a premise substantiated by Dr. Spencer atm. test last summer. KE in the atm. was reduced and KE on the surface increased.

• Bill Hunter says:

Gordon failing to prove the greenhouse assertion does not prove that the aasertion is false. All it should do is increase the uncertainty. as dr syun akasofu says to understand manmade climate variation one must first understand natural climate variation.

• mpainter says:

B4, I believe you are referring to the outdoors experiment with the two coolers filled with water, specifically to the incidental atmospheric warming (as registered by the site thermometer) as the cirrus passed overhead.

You seem to forget that ice radiates in the wavelengths of the “atmospheric window”, which is the explanation for the warming by cirrus of the surface. The question is this: should radiation by clouds be defined as part of the GHE? It is, after all the primary means by which the atmosphere warms the surface.

• Bill Hunter says:

Mpainter, do you have a reference for the radiating frequencies of ice and clouds? It would be helpful for me.

• Ball4 says:

No forgetting mp, that ice radiates is well known, the details are on the net vs. wv, water droplets. The important big deal is Dr. Spencer’s test showed added LW from the cooler cirrus was detected several inches down in his unshaded warmer water by thermometer reasonably after the clouds appeared and the temperature became higher than the shaded water over the test time. Dr.Spencer provided an analysis showing the test findings were reasonable.

This means the LW was absorbed in the top several microns and deeper, the added radiant energy turned into KE was NOT immediately evaporated as many assert. The LW radiant energy was detected by test to be higher water temperature (higher KE) slightly more than the shaded water, when both were cooling all night.

How you want to partition the energy is up to you, get on it.

• Tim Folkerts says:

Gordon: “The premise held by many here, including Roy, is that heat can be transferred radiantly from a cooler body to a warmer body, supposedly slowing its rate of cooling.”

Ball4: “Yes Gordon, a premise substantiated by Dr. Spencer atm. test

sigh … First Gordon offers a strawman, and then Ball4 incorrectly rebuts. So to put things a bit straighter.

1) No one (with a decent understanding of thermodynamics) says that heat, Q, is ever transferred from from a cooler body to a warmer body.

2) Rather, the claim (backed by innumerable experiments and well-established theory) is that photons radiated by ANY body (hot or cold) can be absorbed by ANY OTHER body (hotter or colder).

3) The energy of the absorbed photons must be accounted for when looking at the energy balance of the receiving body (conservation of energy!). The energy of those absorbed photons will make the receiving body warmer than it would be if it didn’t receive those photons and absorb that energy.

4) It is this flow of photons (this “back-radiation”) that Dr Spencer has substantiated, not any heat, Q, (net flow of energy) from cold to warm.

• gbaikie says:

“1) No one (with a decent understanding of thermodynamics) says that heat, Q, is ever transferred from from a cooler body to a warmer body.”
Or mirror temperature as little to do with the radiant
energy it transfers

“2) Rather, the claim (backed by innumerable experiments and well-established theory) is that photons radiated by ANY body (hot or cold) can be absorbed by ANY OTHER body (hotter or colder). ”
If count a mirror as absorbing the energy rather than re-radiating or reflecting it, but absorbing energy suggest
“(of a solid) hold (molecules of a gas or liquid or solute) as a thin film on the outside surface or on internal surfaces within the material.”- Google

Or if something absorbs energy then it’s no longer transmitting energy from some source, rather it absorbs the energy and becomes a source of energy which radiates according to how energetic [warm] it becomes.
So sunlight is 6000 K when it’s reflected or re-radiated it’s still 6000 K, but when something absorbs the sunlight it’s heated and at this warmer state radiates according to
this higher temperature.
Or visible light of sun is converted [absorbed] and IR light is radiated from the objects warmed by the sunlight.
Or a better mirror tends to adsorb a smaller amount of the energy as compared to what it’s reflecting.

• Ball4 says:

Tim, in Dr. Spencer’s test last summer the “heat can be transferred radiantly from a cooler body to a warmer body, supposedly slowing its rate of cooling. premise is indeed correct shown by the thermometers in the water containers. And it agrees with all your points, Gordon’s premise was actually proven beyond reasonable doubt, if and only if, you read KE properly according to Clausius’ defn. for heat.

This is a perfect example of the confusion caused by the heat term.

The KE in the cooler cirrus cloud was transformed into LW EMR which was absorbed by the warmer water on the surface not under the shade transformed back into added KE in the somewhat opaque water causing the temperature readout to be higher than the shaded water temperature. Both water tubs cooled all night.

No violation of entropy in 2LOT, universe entropy went up in the experiment. This is about a simple, clean test as you can find and it is in situ! Not a lab.

• mpainter says:

Should radiation by clouds be included in the GHE?

• mpainter says:

Test

19. Gordon Robertson says:

@Roy…”…with hand waving qualitative statements about absorbing and emitting molecules and photons and entropy and perpetual motion and such”.

You have made the rules about insults which is your right. However, the statement above is not only insulting, it’s patronizing as well. I have been studying in the field of electronics most of my life and I have done in-depth studies on atomic theory. I’m no expert but I do have a decent understanding of how atoms interact based on the Bohr model and quantum theory derived in the 30s.

In electrical engineering, you have to go even deeper to understand how semiconductors work. They are not intuitive devices at a deeper level.

You don’t appear interested in discussing that theory which is the basis of understanding heat. You seem to be introducing thought experiments of your own design while implying people like myself, who have spent years studying the basics of atomic theory, are stupid.

You dismissed my theories on energy a while back without presenting your own understanding of energy. If you don’t understand what energy is, and no one does, you have to at least be specific about what kind of energy you are talking about.

Maybe all the different forms of energy have a common root but that remains to be discovered. Meantime, scientists have defined specific forms of energy with very different properties.

From what I have seen, you have failed to distinguish between electromagnetic energy and thermal energy. They have very different properties. Talking about energy transfer between bodies is not good enough. You have to specify what heat is by definition and how it is transferred.

You seem to have bought into the popular notion that heat is energy in motion. That is simply not true. Kinetic energy is energy in motion and kinetic energy is generic. That is, it can apply to all forms of energy like mechanical, gravitational, nuclear, thermal, electromagnetic, and so on.

Heat is kinetic energy but heat is not electromagnetic energy. Heat is defined as the kinetic energy of atoms. EM in transit after emission from an atom is not associated with atoms. Heat must be associated with atoms therefore the transfer of heat via EM applies to the heat in certain atoms and the heat in another body of atoms.

The heat goes nowhere, all it does is diminish in one body as EM is emitted and increase in the other as the same emitted EM is absorbed.

That’s why the wiki article is just plain wrong, at least the part that wrongly interprets the statement by Clausius.

• Tim Folkerts says:

“You seem to have bought into the popular notion that heat is energy in motion. That is simply not true. “

You seem to have bought in to the notion that you understand heat, but that all the textbooks and professors and misinformed. The notion that heat, Q, is a process that transfers energy is “popular” because it is useful and agrees with all modern usage of the word “heat”. The idea you keep referring to is “internal energy, U.

• Gordon Robertson says:

@Tim Fokerts “The notion that heat, Q, is a process that transfers energy is popular because it is useful and agrees with all modern usage of the word heat. The idea you keep referring to is internal energy, U”.

Clausius used the term Q as well, as in S = {integral of}dQ/T.

Recognize it? It’s entropy. Clausius described it in words as the sum of infinitesimal changes of heat over a process, into or out of a system, at the temperature, T, at which the infinitesimal change takes place.

Now go look at the utterly obfuscated modern definitions of entropy. Most people describing entropy today have not the slightest idea what it means. They get lost in the equations and cannot explain the reality.

Clausius defined heat as the kinetic energy of atoms. What else could it be? If you want to accept an obfuscated abstraction of heat that is your right.

David Bohm, a physicist and a friend of Einstein, claimed that equations without a reality to back them are garbage.

• Kristian says:

From the hyperphysics site:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heat.html

This example of the interchangeability of heat and work as agents for adding energy to a system can help to dispel some misconceptions about heat. I found the idea in a little article by Mark Zemansky entitled “The Use and Misuse of the Word ‘Heat’ in Physics Teaching”. One key idea from this example is that if you are presented with a high temperature gas, you cannot tell whether it reached that high temperature by being heated, or by having work done on it, or a combination of the two.

To describe the energy that a high temperature object has, it is not a correct use of the word heat to say that the object “possesses heat” – it is better to say that it possesses internal energy as a result of its molecular motion. The word heat is better reserved to describe the process of transfer of energy from a high temperature object to a lower temperature one. Surely you can take an object at low internal energy and raise it to higher internal energy by heating it. But you can also increase its internal energy by doing work on it, and since the internal energy of a high temperature object resides in random motion of the molecules, you can’t tell which mechanism was used to give it that energy.

In warning teachers and students alike about the pitfalls of misusing the word “heat”, Mark Zemansky advises reflecting on the jingle: “Teaching thermal physics is as easy as a song: You think you make it simpler when you make it slightly wrong.”

Zemanzky’s plea:
Don’t refer to the “heat in a body”, or say “this object has twice as much heat as that body”. He also objects to the use of the vague term “thermal energy” and to the use of the word “heat” as a verb, because they feed the misconceptions, but it is hard to avoid those terms. He would counsel the introduction and use of the concept of internal energy as quickly as possible.

Zemansky points to the First Law of Thermodynamics as a clarifying relationship. The First Law identifies both heat and work as methods of energy transfer which can bring about a change in the internal energy of a system. After that, neither the words work or heat have any usefulness in describing the final state of the sytem – we can speak only of the internal energy of the system.

(My emphasis.)

• Gordon Robertson says:

@Kristian…”The First Law identifies both heat and work as methods of energy transfer which can bring about a change in the internal energy of a system”.

If you read Clausius, he does an excellent analysis of the relationship between internal energy, heat transferred from or into a body, and the work done by the body or done on the body.

Your statement in another thread about the 2nd law applying only to macro processes is absolutely correct according to Clausius. He claimed the internal energy is not important in understanding the relationship between heat and work which he claimed to be equivalent.

The only place I disagree with you and the author in your article is that Clausius himself states clearly that heat is the internal energy of the atoms in a body. He demonstrates that by analyzing atoms constrained to a lattice in which they vibrate. He claimed the vibrations represent work and heat since they are both equivalent.

In a gas like steam, super heated water molecules are raised to a high energy level. They also have a higher temperature and temperature is a measure of thermal energy. When the steam is applied it does work. Heat is converted to mechanical energy.

I think far too much is made of the semantical argument between heat and thermal energy. Who cares what name is attached to the kinetic energy that does work, call it foobar for all I care.

When super-heated steam does work in a steam locomotive, it is the molecules of water that are doing the work. It’s their kinetic energy that is driving the pistons that drive the wheels. That kinetic energy is heat, thermal energy, or foobar.

I don’t care what it’s called I just know it can be transferred in one direction only, without compensation.

• Kristian says:

Robertson, you write:

Yes, I’ve read Clausius. His use of the term “heat” is outdated. In fact, within the field of thermodynamics it’s been outdated for well over a hundred years, although confusion still exists to this day, hence Zemansky’s plea. Clausius “heat” term is now replaced by the term “internal energy”. For the reasons stated in the hyperphysics quote above. This fact, however, doesn’t mean that Clausius was somehow wrong. We have simply clarified some important thermodynamic principles, making them more physically precise and internally consistent, thus easier to work with.

• Ball4 says:

According to Mark Zemansky Kristian misuses the term internal energy. See MZ original pdf by google string: zemansky use and misuse of heat pdf

Kristian should read the last two paragraphs. U is system internal energy, Q is constituent kinetic energy change to system by virtue of a temperature difference, W is constituent kinetic energy change by work done to/from the system.

That is Kristian’s misuse of the heat term also as per Kristian’s hyperphysics link and MZ paper, and Clausius own defn.

• mpainter says:

Kristian,
I don’t see how you can discard the Clausius definition of heat without discarding Claudius altogether. It’s a very big change to make “heat” a transfer process rather than a property of matter which can be transferred.

• Kristian says:

Ball4 says, August 27, 2016 at 8:49 AM:

According to Mark Zemansky Kristian misuses the term internal energy. See MZ original pdf by google string: zemansky use and misuse of heat pdf

Kristian should read the last two paragraphs. U is system internal energy, Q is constituent kinetic energy change to system by virtue of a temperature difference, W is constituent kinetic energy change by work done to/from the system.

Ball4 obviously suffers from a serious case of a-diabaticism. He simply cannot and will not get his mind to grasp the simple thermodynamic concept of “heat” [Q], which has been well-established in the field now for way more than a hundred years.

I’m not sure what childhood trauma brought this phobia upon our friend here, but it sure must’ve been a quite severe one … The correct physical definition of “heat” has been shown and explained to him at least several tens of times at this point, but still he ALWAYS returns to his confused “caloric” ramblings.

This is what Zemansky says about heat and internal energy, Ball4, from his own book “Heat and Thermodynamics” (fifth edition), pp.78-79:

“We therefore give the following as our thermodynamic definition of heat: When a system whose surroundings are at a different temperature and on which work may be done undergoes a process, the energy transferred by nonmechanical means, equal to the difference between the internal-energy change and the work done, is called heat. Denoting this difference by Q, we have

Q = U_f – U_i – (-W),
Q = U_f – U_i + W

where the convention has been adopted that Q is positive when it enters a system and negative when it leaves (just the opposite of the sign convention for W). The preceding equation is known as the mathematical formulation of the first law.

(…)

A convenient way to remember the sign convention is to write the first law in the form

U_f – U_i = Q – W,

and to think of a system as an energy “bank”, in which positive heat is regarded as a deposit and positive work as a withdrawal.”

… and pp.80:

4-5 Concept of Heat

Heat is energy in transit. It flows from one part of a system to another, or from one system to another, by virtue of only a temperature difference. When this flow has ceased, there is no longer any occasion to use the word “heat”. It would be just as incorrect to refer to the “heat in a body” as it would be to speak of the “work in a body”. The performance of work and the flow of heat are methods whereby the internal energy of a system is changed. It is impossible to separate or divide the internal energy into a mechanical and a thermal part.”

This is a pretty conclusive statement, I would say.

• Ball4 says:

Kristian, your MZ clip is entirely consistent with what I wrote and explicitly shows you misuse the term internal energy (U) right here:

“The performance of work and the flow of heat are methods whereby the internal energy of a system is changed.”

Change in internal energy U: delta U=Q+W

MZ consistently applies Clausius’ definition of heat in his 1st memoir up front p. 18: “We…shall assume generally that a motion of the particles (in a body) does exist, and that heat is a measure of their (kinetic energy).”

So Kristian also misuses the term “radiant heat”. There is no KE in EMR. Correctly, there is radiant energy in EMR

• Gordon Robertson says:

@Kristian…”Heat is energy in transit. It flows from one part of a system to another, or from one system to another, by virtue of only a temperature difference. When this flow has ceased, there is no longer any occasion to use the word heat”.

Far too generalized, Kristian, and vague. We generally call energy in transit ‘kinetic energy’. Clausius claimed heat is that kinetic energy.

You have to call it something, leaving it as plain ‘energy’ is vague. You can have kinetic energy related to an electrical field. Does it have the same properties as the KE related to thermal energy?

As far as heat flowing from one part of a system to another, that may be true within a conductor, or even as a mass of atoms in convection, but it is not true in radiation. Heat does not ‘flow’ between bodies via radiation, EM does the flowing, heat remains with the atoms in either body.

However, since the 2nd law restricts heat transfer to one direction, without compensation, the meaning is clear. EM can only transfer heat in one direction via radiation.

I realize there are well founded theories that claim otherwise but I think those theories have in many cases been stated completely out of context. I’ll go so far as to claim many have not been tested. I mean, how do you measure the effect of a star at 2 million C surface temperature on a star at 1 million C surface temperature?

Where else in high temperature physics has this been done?

• Gordon Robertson says:

@Kristian…”Ive read Clausius. His use of the term heat is outdated”.

Either that or the modern usage is wrong. His usage makes eminently more sense than what Zemansky et al present. His usage is clear while that of Zemansky is vague and far too generalized.

Consider your apparent stance on this, that back-radiation from the atmosphere cannot raise surface temperature. That falls directly in line with the theories of Clausius and the 2nd law he proposed. He proposed it because Carnot had claimed there were no losses in a heat engine and that is the premise of AGW.

We’re talking perpetual motion, one of the reasons the 2nd law was presented. Agw IS BASED LARGELY ON THE 1ST LAW AND IT’S MAJOR FAULT WAS THAT IT WOULD ALLOW PERPETUAL MOTION IN CERTAIN INSTANCES.

CAPS not intended. I hit the cap button by mistake and I’m too lazy to correct it.

• Kristian says:

Gordon Robertson says, August 27, 2016 at 6:14 PM:

You have to call it something, leaving it as plain energy is vague. You can have kinetic energy related to an electrical field. Does it have the same properties as the KE related to thermal energy?

You don’t understand. The 1st Law is basically a statement of the conservation of energy. It says that a thermodynamic system contains a certain amount of energy, associated with its temperature. This amount (or base fund/storage) is called the system’s “internal energy”, denoted by U. The system can lose or gain internal energy, its U might go up or down from one state to another, in two ways only: 1) through a transfer of energy as a result of a temperature difference (in standard thermodynamics, this is called a transfer of HEAT), denoted by Q, or 2) through a transfer of energy as a result of WORK being performed, denoted by W. The W includes all energy transfers into and/or out of the system that aren’t a direct result of a temperature difference. So basically you have HEAT [Q] … and everything else, work [W]:

U_f – U_i = Q – W

Q is simply the net amount of energy transferred into or out of the system from the initial to the final state simply as a result of a temperature difference between the system itself and its surroundings. The standard thermodyanmic definition of this particular quantity [Q] is “heat”. In fact, Q describes both the amount of energy transferred this way AND the mechanism by which the amount was transferred.

However, it doesn’t matter HOW that amount of energy was thermally transferred. The only thing that matters is that is WAS thermally transferred. There are several different ways of transferring heat into or out of a thermodynamic system, so-called “heat transfer mechanisms”. You have ‘conduction’, ‘evaporation/condensation’, ‘radiation’ and ‘convection/advection’.

Radiation can also be a heat transfer mechanism, Robertson. For sure, not all radiation is a transfer of heat, like some people seem to assume. But heat can also be transferred via radiation.

When you sit next to a fire, the fire warms you by transferring radiant heat to you. You don’t have to put your hand into the fire in order for it to heat you. It does so from a distance. Like the Sun. The Sun warms us by transferring radiant heat to us. That’s not to say that the electromagnetic energy it sends out is ITSELF heat. No, it’s SW and LW electromagnetic energy. And that’s it. Furthermore, you do not experience being heated (as in ‘warmed’) by it until it is absorbed by your body and converted into molecular kinetic energy. But the net thermal transfer of energy from the Sun to you is what we call HEAT [Q] (in this case, radiant heat), and the increase in molecular kinetic energy inside of you as a result of absorbing and thermalising the energy thus transferred is what we call (a rise in) INTERNAL ENERGY [U].

That’s the distinction.

Why do we call the energy transferred to you as ‘heat’ ‘internal energy’ once it resides inside of you?

Because, as Zemansky very clearly points out:
“[Heat] flows from one part of a system to another, or from one system to another, by virtue of only a temperature difference. When this flow has ceased, there is no longer any occasion to use the word “heat”. It would be just as incorrect to refer to the “heat in a body” as it would be to speak of the “work in a body”. The performance of work and the flow of heat are methods whereby the internal energy of a system is changed. It is impossible to separate or divide the internal energy into a mechanical and a thermal part.

… and, paraphrased:
“Surely you can take an object at low internal energy and raise it to higher internal energy by heating it. But you can also increase its internal energy by doing work on it, and since the internal energy of a high temperature object resides in random motion of the molecules, you cant tell which mechanism was used to give it that energy.

And that’s the whole point. Since we cannot know what part of the random motion of a body’s molecules, giving it a temperature, is due to “heating”, and what part is due to “work being done”, we can’t just call it “HEAT”. Just as we can’t just call it “WORK”. That’s why, at the point where the energy transferred to the body via the fundamentally different mechanisms of ‘heat’ and ‘work’, has ended up inside the body, all we can say is that the amount of ‘molecular kinetic energy’ in the body has gone up. The body’s base fund/storage of “internal energy” has increased.

Yes, the term “internal energy” is pretty vague. But that’s only because it describes a reality that’s pretty vague. The distinction between heat and work is simply erased once the energy enters the system …

• Ball4 says:

“you do not experience being heated (as in warmed) by it until it is absorbed by your body and converted into molecular kinetic energy.”

Very good progress Kristian, you have advanced by studying Zemansky from your days of ” Because heat isnt kinetic energy…(Clausius) use of the term “heat” is outdated…now replaced by the term “internal energy”.”

Now a little more tweak of turning radiant heat into the more physically meaning radiant energy. But if not, then your understanding has to be the same as the readers: “Thats not to say that the electromagnetic energy it sends out is ITSELF heat. No, its SW and LW electromagnetic energy” i.e. radiant energy.

And make sure MZ’s undefined, unreliable concept of thermal energy in one post might mean heat (KE, Q) and another post mean internal energy (U). The reader can’t be the one who figures it out. The best way to think of thermal energy is it is the shortened term for U from therm [odynamic intern]al energy to thermal energy.

• Tim Folkerts says:

“Now go look at the utterly obfuscated modern definitions … “

At some point you have to at least *consider* the idea that any sense of obfuscation is due to YOUR poor understanding, and consider that maybe thousands of people who have studied this more than you have might actually know more than you do!

Statistical mechanics is actually a very elegant theory that encompasses classical thermodynamics and then provides further insights and predictions. You might just as well say that since you find relativity confusing, Einstein and generations of physicists are wrong.

• Gordon Robertson says:

@Tim Folkerts…”At some point you have to at least *consider* the idea that any sense of obfuscation is due to YOUR poor understanding, and consider that maybe thousands of people who have studied this more than you have might actually know more than you do!”

Of course. But my understanding comes from reading Clausius and he does what many modern scientists fail to do, explain the theory in words. When many modern scientists try to use words, it comes out sounding obfuscated.

This does not apply only to physics. I was trying to learn C++ programming and I could not grasp what was meant by a ‘class’. I understood programming fairly decently but the authors I read, many of them, obfuscated the meaning ridiculously. They talked around the subject rather than making a statement.

Finally I came across a book by the author of C++, Bjarne Stroustrup. He stated in one sentence, “a class is a user-defined type”.

Bingo!! I knew what a type was, like int, char, real, struct, etc. They are all predefined types in many languages. The class allows you to define your own type.

Why could more than a dozen authors not explain that? It’s obvious they did not know themselves. They could use a class but they could not explain it.

Clausius is extremely clear about heat. Nothing is left to the imagination. I have read many modern authors who are very ambiguous about it, leaving me to suspect they really don’t know what they are talking about. With entropy it gets even worse.

• Kristian says:

Robertson,

Have you read Maxwell and his take on “heat” and “radiation”? If you trust Clausius, you should also trust Maxwell.

The principle is very simple: The radiant heat between two objects at different temperatures is the NET of all the thermally emitted radiative energy flowing between the two. Mathematically, you can see this net flow as the sum of two oppositely directed radiance ‘fluxes’. In reality, however, the net flow is rather the probabilistic average of ALL photon paths and frequencies moving through ALL points in threedimensional space between the two objects (this came after Maxwell). You can basically see it as a “photon cloud” with a certain energy distribution between the warmer and the cooler object. The macroscopically detectable ‘movement’ of energy through a radiant energy field such as this, is the ‘radiant heat’, and it always goes down the intensity gradient, from hotter to colder. If there were no temperature difference between the two objects, then there would be no detectable macroscopic movement of energy through the field, even though energy would still move about chaotically/randomly at the quantum (microscopic) level.

• Gordon Robertson says:

@Kristian…”Have you read Maxwell and his take on heat and radiation? If you trust Clausius, you should also trust Maxwell”.

I am a big fan of Maxwell, especially considering he is Scottish, as am I. By DNA at least.

You do realize that he was a mathematician, albeit a rather brilliant one. Typically, when he attended school in Scotland, they referred to him as ‘Daftie’ Maxwell. Daft, of course, being the Scottish vernacular for dumb, or stupid.

I agree with you on just about everything you say in your posts but I cannot accept the term radiant heat, even if Maxwell used it. To do so is to confuse EM and thermal energy. Radiation measured from a hot body is not heat. You might regard it as potential heat but it is not converted to kinetic energy till it is absorbed by atoms. If there are no atoms about it will travel through space doing nothing.

Planck used the term ‘heat rays’, which I think is erroneous, although we can excuse him for that owing to the era in which he lived and worked. He did exemplary work on the electromagnetic spectrum to tease out the notion of quantum energy levels but his heat rays suggest he thought EM carried heat.

Heat is not a property of EM, which is defined based on its wavelength and intensity. In fact, in the equation E = hf, Planck’s constant, h, associates EM to its frequency. Planck has to be aware of that and I think he used heat in ‘heat’ to clarify the source of the IR.

Consider the iron bar I go on about. EM is not involved in the transfer of heat in an iron bar and the 2nd law applies there. EM is not involved in the convective transfer of heat or in conductance, so why should it make a difference in radiative transfer? It’s nothing more than a transport mechanism.

The only way the second law can apply in radiative transfer is if EM from the cooler body has no effect.

Mind you, I am referring to the case where the warmer surface warms a dependent source like the atmosphere. I am not talking about two stars in close proximity. I don’t know what would happen there.

• Ball4 says:

Gordon, “EM is not involved in the transfer of heat in an iron bar”

That is too strong Gordon, the iron atoms are emitting EMR – they glow in the visible at certain T – and those rays come from the interior per Planck not the exact surface, so EMR is also emitted and absorbed internally by the iron atoms as well as their transfer of KE (the heat).

• mpainter says:

Hall of mirrors. Undoubtedly EM is emitted by atoms bound in the crystal lattice, or in liquid state, but this originates as kinetic energy.
Something is peeping around the corner, but which way is the corner?

• Ball4 says:

Look in the mirror.

• DHMacKenzie says:

Gordon, start here
http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node136.html
Most of your concerns with Dr.Roy’s blog post will dissolve. Heat flow is from hot to cold, no violation of 2nd law. All good….

• Gordon Robertson says:

@DHMackenzie…”Heat flow is from hot to cold, no violation of 2nd law”.

I’m trying to reply but Roy’s filters are cutting me off.

Where in your link does it state that? All I saw in the example was a highly abstracted application of Kircheoff, likely taken completely out of context.

• DHMacKenzie says:

Gordon, those are simple examples of radiative heat transfer calculations for a basic heat transfer course. Someone who doesn’t “get” those calcs simply isn’t able to carry on a knowledgeable conversation about radiation heat transfer from ground to sky. Just trying to help you make your posts sound like you know what you are talking about. Cheers.

20. Gordon Robertson says:

@Roy…”…if we simply insert a piece of room-temperature cardboard in between the heated surface and the chilled surface, we should see an increase in the temperature of the heated surface…”

Roy…here’s why you are wrong.

In a solid iron bar, if one end is heated, the heat, as kinetic energy, transfers automatically from the hotter end to the cooler end. There is no transfer of heat in the opposite direction, just as when electric current travels down an iron bar from negative to positive there is no reverse current flow from positive to negative.

Any EM radiated by your cardboard insert would have no effect on the hot plate simply because the EM from the cooler body does not have the required frequency and intensity to affect the much higher level energy electrons in the hotter body.

• Tim Folkerts says:

just as when electric current travels down an iron bar from negative to positive there is no reverse current flow from positive to negative.

Surely as an electrical/electronic engineer, you have come across the idea of “drift velocity”. The average speed of electrons in a current-carrying wire is a few cm/second.

On the other hand, the “Fermi velocity” gives the speeds of the conduction electrons, which is on the order of 1,000,000 m/s. Electrons in a wire are moving every which way at about 1,000,000 m/s. Electrons are flowing “backwards” at high speed all the time. Slightly more are moving “forward”.
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi.html#c1

So yes, there are indeed “currents” moving both ways; electrons moving toward he negative terminal of a battery and electrons moving toward the positive terminal. There is just a slightly larger current heading the “right” way.

• Gordon Robertson says:

@Tim Fokerts…”Surely as an electrical/electronic engineer, you have come across the idea of drift velocity. The average speed of electrons in a current-carrying wire is a few cm/second”.

Yes, I have, that’s why I used the word ‘current’ rather than electrons. An electric current is made up of charges that travel independently of electrons but from electron to electron.

One of my high school electronics teacher demonstrated that decades ago. He took one of those old wooden rulers with the grove running along it’s length and set up steel balls along the groove so they were touching. He tapped the end ball closest to him and the ball at the other end shot off the ruler. The rest did not move. The force he applied moved from ball to ball, causing the end one to shoot off.

Please don’t try to tell me based on the Fermi level that electrons under an electric field are going both ways. That’s like claiming the electrons in the beam of a cathode ray tube are being emitted by the phosphors on the screen at the same rate approximately that they are being emitted by the cathode. Electrons under an electric field move only one way.

If an electric current is done from a valence shell to a valence shell there would not be room in the shells to accept electrons trying to go the other way. In semiconductors, they pass that off as holes moving the opposite direction to electrons but even Shockley, who invented that concept, explained that the holes are illusion used to demonstrate a concept in p-type silicon, not a reality.

• Tim Folkerts says:

“Please dont try to tell me based on the Fermi level that electrons under an electric field are going both ways. “

Should I also not try to tell you that molecules in the atmosphere are moving in all directions at 100’s of m/s, even when there is a breeze of a few m/s blowing in one direction?

• Gordon Robertson says:

@Tim Folkerts…”Should I also not try to tell you that molecules in the atmosphere are moving in all directions at 100s of m/s, even when there is a breeze of a few m/s blowing in one direction?”

Come on, Tim, we are talking specifically about electrons constrained to move through a conductor under the influence of an electric field. The electrons are drawn to the positive field and repelled by the negative field. Electrons are in fact driven through a conductor in one direction only.

You have to be careful with this. In electrical engineering theory the convention is still that ‘current’ flows from positive to negative. They claim that based on a theoretical positive test charge that some claim has mass. With that +ve test charge as the basis, electrons can be regarded as flowing against the field, although, having worked in electronics where the convention is the opposite, I regard that theory as nonsense.

If you ever see a reference to electrons flowing against the field, please consider they are talking from the conventional +ve to -ve current flow.

In semiconductor theory you often see references to hole current and holes are treated as real. However, Shockley, who was responsible for developing much of that theory, claimed the hole is an imaginary unit put forward to help clarify semiconductor theory.

• Ball4 says:

“the heat, as kinetic energy, transfers automatically from the hotter end to the cooler end. There is no transfer of heat (as kinetic energy) in the opposite direction”

Gordon, consider the vibrations of the constituent iron atoms of your bar. Literally this would mean they do not vibrate about their quantum location both ways. Obviously there is a problem in your wording which Tim is correctly pointing out.

There is KE vibration in the opposite direction; net KE transfer is one way by 2LOT. Just like incoherent photons travel both ways between absorbing bodies and the bodies’ net constituent KE transfer is found from 2LOT.

• Gordon Robertson says:

@Ball4…”Gordon, consider the vibrations of the constituent iron atoms of your bar. Literally this would mean they do not vibrate about their quantum location both ways”.

It’s maybe easier for me to visualize this because I was not educated from the quantum perspective. The vibrations have nothing to do with quantum theory and they are not abstractions as in spin. According to Clausius, the atoms actually vibrate mechanically, like a harmonic oscillator. From that he concluded that work is being done and since heat and work have an equivalence, the vibrations represent heat as well.

When a steel plate is heated, the vibration increase overall and the bar expands. It’s mechanical, not quantum. Think of the atoms being joined by tiny springs. That won’t prevent energy being transferred along the springs from atom to atom.

I have inadvertently used quantum theory while doing problems in chemistry involving atomic energy level. Also, I have used the theory while doing problems with covalent boding theory where the valence electron orbitals have orbital spaces based on probability.

You don’t need any of that for the 2nd law. It’s enough to understand that energy is transferred from atom to atom via their bonding electrons. After all, when the atoms vibrate, that vibrations has to stress the covalent bond, and if it exceeds the bonds abilities the bond will break.

Before it breaks, energy is transferred from a higher electrical potential to a lower potential as in an electric circuit, or from a higher level of thermal energy to a lower level as with heat transfer in a conductor.

It’s a mechanical problem at the atomic level that doesn’t really require quantum theory. At least, not to the level of abstraction, as in specifying spin.

• Ball4 says:

Well, that seems to be agreement iron atoms in the lattice vibrate both opposite toward the hot end and forward toward the cool end as Tim was pointing out.

• Toneb says:

“In a solid iron bar, if one end is heated, the heat, as kinetic energy, transfers automatically from the hotter end to the cooler end. There is no transfer of heat in the opposite direction, just as when electric current travels down an iron bar from negative to positive there is no reverse current flow from positive to negative.”

Eh??

What has that got to do with the GHE?

A heated bar transfers it’s heat via conduction to the cool end.
Different thing.

In conduction 2 adjacent molecules that have differing kinetic energy transfer that energy from the most energetic to the least energetic. It can only be one way.
In radiative transfer, the cooler object radiates some energy that is received by the cooler one’s “reservoir” and consequently it’s cooling is slowed.

• Gordon Robertson says:

@Toneb…”In conduction 2 adjacent molecules that have differing kinetic energy transfer that energy from the most energetic to the least energetic. It can only be one way”.

I was referring to an iron bar that has no molecules, only iron atoms. Molecules are a reference to two or more atoms joined by covalent bonds.

The kinetic energy to which you refer is thermal energy. Kinetic energy is a generic term used to describe energy in motion. In this instance it describes thermal energy.

The 2nd law applies to heat transfer of any kind, according to Clausius. I was demonstrating that it applies in an iron bar therefore it has to apply in radiation. The mistake that’s made, is confusing the transporting agent, the EM radiation, for the thermal energy.

Thermal energy in radiation does not move from body to body. When an atom cools, it’s valence electrons drop down to a lower energy state and emit EM. That represents a reduction of thermal energy in one body. The emitted EM can travel to a cooler body and be absorbed by an atom’s valence electrons, forcing them to a higher energy level. That atom increases in thermal energy.

However, thermal energy can only move from the hotter body to the cooler body, as in the iron bar.

• Toneb says:

“However, thermal energy can only move from the hotter body to the cooler body, as in the iron bar”

It is the NET flow of energy that cannot move from the cooler to the warmer.
That the warmer one get WARMER at the expense of the cooler ones energy.

Photons emitted by the cooler body are received by the warmer one.
All objects above 0K radiate, so if the cooler and the warmer are in proximity they will both receive each other’s photons. It’s just that the flow of photons will be GREATER from the warm to the cold.
The warmer body cannot know that photons impinging on it come from a cooler one and reject them!
Ergo the warmer body it is increasing it’s energy reservoir and cools more slowly than otherwise.
What does not happen is that the NET flow of energy is to the warmer body from the cooler.
The presence of the cooler body slows the warmer body’s cooling (presuming that there is a cooler sink surrounding the both).

It s analogous to a large water tank draining water at 10 gal/min. If it contains 100 gals then it will completely drain in 10 mins.
However if a small tank drips 1 gal/min into the bigger – the bigger is still losing 9 gal/min and will take 11 mins to drain. The net flow of water was from the small tank to the large one BUT the larger took longer to empty.

• Gordon Robertson says:

@Toneb…”Photons emitted by the cooler body are received by the warmer one”.

Why are you talking about photons, I specified an iron bar where the atoms are constrained in a lattice and heat transfer is via valence electrons?

Can you please move away from the radiative model and consider the iron bar? The 2nd law applies to it, heat can only be transferred from the hot end to the cooler end and IR exchange is not involved.

Clausius claimed the 2nd law held for all three forms of heat transfer so you need to step back and consider heat from the perspective. It has nothing to do with a two way IR flow. Heat is about atoms, not IR per se.

The transfer of heat must be between atoms. Visualize that and show me how that can take place via radiation. The atoms don’t transfer through the medium, what is going on?

If you see that clearly, the unidirectional aspect of the 2nd law makes sense without resorting to net flow of IR.

• Tim Folkerts says:

Gordon says “The emitted EM can travel to a cooler body and be absorbed by an atoms valence electrons, forcing them to a higher energy level. That atom increases in thermal energy.”

You seem stuck in a Bohr Model mentality, with valence electrons jumping up to a higher level, and then unable to jump up that amount again. The molecule gains energy from a high energy photon or a high voltage electron, then falls back to the ground state in one or more quantized jumps. An idea typified by diagrams like this:
http://www.daviddarling.info/images/hydrogen_spectrum.gif

With molecules or solids, the energy of IR photons (like the 15 um IR band) is absorbed by a completely different mechanism — vibrations of the molecule or solid (“phonons”). Here the molecule can gain multiple units of energy and simply vibrate with higher quantized energy.

As such, even a “hot” molecule that is already vibrating can absorb additional energy from additional photons to vibrate with a larger amplitude — whether such photons came from a hotter source or a colder source.

• Gordon Robertson says:

@Tim Fokerts…”With molecules or solids, the energy of IR photons (like the 15 um IR band) is absorbed by a completely different mechanism vibrations of the molecule or solid (phonons). Here the molecule can gain multiple units of energy and simply vibrate with higher quantized energy”.

Phonons don’t apply to conductors of heat, they are a quantum physics equivalent of valence electron heat transfer that apply to insulators. Insulators provide a far more complex problem.

I have used an iron bar to remove the complexity. I am trying to demonstrate the second law by showing that heat in an iron bar can be transferred in one direction only. It is the same with radiative transfer.

Why do you persist in talking about molecules when I have deliberately specified an iron bar consisting of pure ferrous atoms? And why do you insist on talking about molecular vibrations as if molecules are not atoms bound together by valence electrons?

Of course a hot ‘atom’ can absorb more energy provided the applied energy exceeds it’s current energy level. That is not the case with IR from a colder source.

• Toneb says:

*correction*

… warmer one’s reservoir …

• Nate says:

So Gordon, We won’t see a rise in the temperature of the heated surface?

Prove it. All previous evidence suggests you will be wrong. But go ahead and try.

• Gordon Robertson says:

@nate…”So Gordon, We wont see a rise in the temperature of the heated surface?”

Thus far, this is a thought experiment. The 2nd law says it won’t work. I am going with the tried and true 2nd law.

I don’t need to prove anything. Clausius did it for me over 150 years ago and I confirmed heat transfer in a physics lab back at university when I measured the heat transfer in a bar of steel.

Please note, in the lab, we were not required to measure heat transfer from the cold end of the bar to the warm end. In those days in thermodynamics, if you had suggested that, you’d have been laughed out of the lab. Or expelled.

• Tim Folkerts says:

As noted way up at the top, I have done an experiment quite similar to this. It worked.

• Nate says:

Gordon,

If you are unwilling to accept that the experiment (and similar ones) has been tried by many, and it works, then you need to try the experiment to prove or disprove for yourself. Otherwise your protest that it will not work is pointless.

What’s fascinating and bewildering to me is your whole attitude about established science. The science that we are discussing with you is well established, empirical science, tried and tested by many, and applied successfully to many situations, e.g. weather prediction models, cryogenics (I mentioned above). If it is wrong then these applications would not work.

For me, if I’m presented with some established science that doesn’t make complete sense to me, but is outside of my field of expertise, then I can explain this in two ways:

1) There is a gap in my knowledge about this. People who have made this field their life’s work understand this and if I learned more, most likely it will make sense to me.

2) The established science in this field in this is wrong, all of the experts (such as Roy and the National Academy of Sciences) are wrong and I am right.

Can you explain why you go with #2??

21. Gordon Robertson says:

@Roy…the thing I find most frustrating about your experiments is that your UAH data proves you wrong. There has been no warming since 1998 and since your data sets started in 1979, according to your 33 year report, there has been little or no true warming despite significant increases in atmospheric CO2.

AGW is often referred to as the extended greenhouse theory. If it’s so wrong, why are you not at least questioning the greenhouse theory itself? Lindzen has and his explanation is significantly different than yours. His doesn’t dwell on radiation, he goes into convection heavily.

Gerlich and Tscheuschner, two scientist who work in the field of thermodynamics, have literally demolished the theory. It was attacked first by Woods circa 1909 when he discovered through experiment that greenhouses warmed due to a lack of convection and not due to radiation per se.

In fact, Woods invoked the radiative laws which describe the reduction in radiative intensity with the square of the distance from the source. He claimed surface radiation would be so weak a few feet above the surface as to render it useless as a heating agent.

If you want to do an experiment, turn on a 1500 electric stove ring till it’s cherry red. You know it will cook the skin on your hand if you hold your hand an 1/8th inch away but if you move your hand back a foot the radiation wont harm your skin. If you move your hand 4 feet away you likely wont even feel it. That’s how quickly radiation intensity dissipates.

In your current experiment, the radiation from your cool source will be ineffective a 1/4 inch or less above the surface.

• Bill Hunter says:

Saying the UAH data proves Roy wrong is an argument from ignorance that assumes there is no other reason for why no warming has been observed.

• Gordon Robertson says:

@Bill Hunter…”Saying the UAH data proves Roy wrong is an argument from ignorance that assumes there is no other reason for why no warming has been observed”.

Get with the program, Bill, we are talking about GHGs in the atmosphere causing the atmosphere to warm. Since UAH began issuing data sets in 1979, their 33 year report has claimed little or no true warming over that period.

Show me how that truth can be represented as ignorance when Roy’s own data shows his radiation theories don’t work.

• Toneb says:

Gordon:

“Gerlich and Tscheuschner, two scientist who work in the field of thermodynamics, have literally demolished the theory. It was attacked first by Woods circa 1909 when he discovered through experiment that greenhouses warmed due to a lack of convection and not due to radiation per se.”

First of all, yes we know that the “Greenhouse effect” as it pertains to the atmosphere is badly named.
Doesn’t make the theory wrong my friend.

Also The Gerlich & Tscheuschner paper has been extensively rebutted.
Here for one….
http://www.worldscientific.com/doi/abs/10.1142/S021797921005555X

“In this journal, Gerhard Gerlich and Ralf D. Tscheuschner claim to have falsified the existence of an atmospheric greenhouse effect.1 Here, we show that their methods, logic, and conclusions are in error. Their most significant errors include trying to apply the Clausius statement of the Second Law of Thermodynamics to only one side of a heat transfer process rather than the entire process, and systematically ignoring most non-radiative heat flows applicable to the Earth’s surface and atmosphere. They claim that radiative heat transfer from a colder atmosphere to a warmer surface is forbidden, ignoring the larger transfer in the other direction which makes the complete process allowed. Further, by ignoring heat capacity and non-radiative heat flows, they claim that radiative balance requires that the surface cool by 100 K or more at night, an obvious absurdity induced by an unphysical assumption. This comment concentrates on these two major points, while also taking note of some of Gerlich and Tscheuschner’s other errors and misunderstandings.

If you disagree with the rebuttals, would you care to find any other Climate scientists supporting their, err, *science* (and thereby falsify the empirical science found in ALL text books on the subject)

Bad papers do get through (unaccountably) sometimes – however peer-review does not stop when a paper is published.
It is utter tosh.

• Ball4 says:

The G&T paper published in a journal with this policy: “Our peer review process is confidential and identities of reviewers cannot be revealed.”

Gordon: “Gerlich and Tscheuschner, two scientist who work in the field of thermodynamics”

Can Gordon back that up with proof?

As I have pointed out to Gordon, G&T write out the Clausius definition of heat but fail to follow this definition repeatedly, for example: “ground’s heat radiation”. KE is not found in radiation. Correctly the ground’s constituent particle KE reduces as the particles emit EMR.

• Gordon Robertson says:

@Ball4…”Can Gordon back that up with proof?”

Read the paper and supply an intelligent rebuttal, never mind the ad homs. Their expertise is thermodynamics speaks for itself as opposed to the several rebuttals that expose the ignorance of the authors to even basic physics.

Meantime, here’s a very interesting article on the abject failure of Canadian climate models and how much money we have spent in the world on such scientific nonsense.

• Ball4 says:

Ok, Gordon can not back up his assertion.

The extensive rebuttals have already been published Gordon. Blogs have filled up with the debate. The paper fails to accomplish the title. That G&T cannot consistently stay with the defn. of heat from Clausius (that they agree with) – as those who properly work in the field of thermodynamics would do – is evidence that Gordon will not be able to back up his assertion.

• Gordon Robertson says:

@Toneb…”Their most significant errors include trying to apply the Clausius statement of the Second Law of Thermodynamics to only one side of a heat transfer process…”

The thing most disconcerting in your reply is that you don’t seem to have read the paper yourself and understood it. You seem to have been too busy appealing to authority.

I read that rebuttal when it first came out from Eli Rabbett who uses that nym to cover his real persona of Halpern.

I spotted the error in that comment before G&T issued a rebuttal. Halpern et al made the same egregious error as many climate alarmists by presuming IR radiation is thermal energy. They were referring to a model presented by G&T in which bodies emitted IR and presumed the IR was heat. They then concluded that G&T were guilty of ignoring the IR from the cooler body.

G&T responded that IR and heat cannot be arbitrarily summed. You sum one or the other in a system but you cannot sum IR radiation fluxes and apply that as heat transfer.

The other rebuttals take a similar tack. One of them attacks only the G&T version of a planet without an atmosphere as opposed to a planet with an atmosphere AND oceans.

22. Kristian says:

Spencer, you say:

This is the basis for global warming theory: increasing carbon dioxide in the atmosphere reduces the rate of IR energy loss to deep space, resulting in some warming.

Yes, that’s the hypothetical “greenhouse warming mechanism”, illustrated here:

Only problem is, we don’t observe this mechanism (the raising of the effective radiating level (+Z_e) to force sfc warming (+T_s) by keeping T_e constant) to be operative and effective in the real Earth system, even as the atmospheric content of CO2 and water vapour has gone up significantly since the mid 80s. Because what does the data show?

There has been no observable “enhancement” of any “radiative GHE” since at least 1984-85. More ASR, not less OLR, is the direct cause of the radiative imbalance at the ToA. And thus, of the net accumulation of energy inside the Earth system leading to warming. The Sun, not CO2.

• Ball4 says:

Incorrect on OLR Kristian, Dr. Spencer is correct in the first clip you use, OLR has decreased slightly in the 11 year period studied in latest 2016 paper from the CERES team on principle investigator Loeb’s website. Have you still not shown your method replicate’s the CERES team results?

• Kristian says:

Ball4,

1) “Principal CERES investigator” Norman Loeb has very clearly stated that the observed decrease in OLR between ~2002-03 to ~2012-13 was due to the change in ENSO states over the period, that is, not due to some “enhanced rGHE”.

2) Tropospheric temps also went down over the same period. In fact, the OLR simply appears to track TLT over time. Which is to be expected.

3) CERES EBAF Ed2.8 and Ed4 are close to identical over the period in question, as shown before. Ball4 knows this full well.

• Ball4 says:

1) Sure, ENSO affects global, annual OLR as confirmed by Tmedian in UAH v6Beta 5. Loeb: “a consistent decrease in LW
TOA flux that is mainly associated with ENSO variability”.

2) Kristian hasn’t yet correctly shown this in UAH v6Beta5.

3) Yes, the CERES team data quality publication explains the small differences. Kristian avoids answering: “Have you still not shown your method replicates the CERES team results?”

• SkepticGoneWild says:

Ball4,

I don’t know why you go on and on regarding CERES. CERES data quality has huge problems. I refer you to two papers:

1. Earthâ€™s energy imbalance and implications (James Hansen)
http://pubs.giss.nasa.gov/docs/2011/2011_Hansen_ha06510a.pdf

Hansen’s paper is an eye opener. Read Section 14.6.1 Some quotes:

“The notion that a single satellite at this point could measure Earthâ€™s energy imbalance to 0.1 W mâˆ’2 is prima facie
preposterous. Earth emits and scatters radiation in all directions, i.e., into 4Ï€ steradians. How can measurement of
in all directions? Climate change alters the angular distribution
of scattered and emitted radiation. It is implausible that changes in the angular distribution of radiation could be modeled to the needed accuracy, and the objective is to measure the imbalance, not guess at it.

“The precision achieved by the most advanced generation
of radiation budget satellites is indicated by the planetary energy imbalance measured by the ongoing CERES (Clouds and the Earthâ€™s Radiant Energy System) instrument (Loeb et al., 2009), which finds a measured 5-yr-mean imbalance of 6.5 W mâˆ’2
(Loeb et al., 2009). Because this result is implausible, instrumentation calibration factors were introduced to reduce the imbalance to the imbalance suggested by climate models, 0.85 W mâˆ’2 (Loeb et al., 2009).”

The CERES satellites cannot measure the alleged energy imbalance, so a phony “calibration” factor is applied to obtain the result that the models suggest. That is science? So the cause of the imbalance, the enhanced greenhouse effect, cannot be measured.

The second paper is a CERES publication, which still indicates major errors in measuring the alleged TOA energy imbalance.

• Kristian says:

SkepticGoneWild,

Thanks. And you’re of course very right about the accuracy issue.

However, you’re shooting at the wrong target here. We’re not discussing the accuracy of CERES EBAF, because we’re not dealing with absolute quantities. We’re dealing with anomalies, the change over time. And anomalies are a matter of precision, not of accuracy.

It is a well-known fact that the CERES measurements are highly inaccurate, but at the same time precise to an impressive degree. Ad we’re only interested in the precision here …

• Ball4 says:

SKW, “The notion that a single satellite..”

There is not just ONE satellite, Loeb 2016 lists results for three. There are more planned to fly.

The calibration is not phony (skw term), read the CERES data quality paper critically. Every instrument is calibrated, even mercury thermometers. Especially these instruments are not calibrated to models, read the paper link you posted. There are 9+ causes to any temporary TOA imbalance, you mention only one.

• SkepticGoneWild says:

Ball4,

I never said there was only one satellite. Neither did Hansen in his paper. You did not read the paper properly.

Hansen pointed out MULTIPLE issues with orbiting satellites, including the fact that the the ARE calibrated to read what the climate models suggest:

“The precision achieved by the most advanced generation of radiation budget satellites is indicated by the planetary energy imbalance measured by the ongoing CERES (Clouds and the Earths Radiant Energy System) instrument (Loeb et al., 2009), which finds a measured 5-yr-mean imbalance of 6.5 W m−2 (Loeb et al., 2009). Because this result is implausible, instrumentation calibration factors were introduced to reduce the imbalance to the imbalance suggested by climate models, 0.85 W m−2 (Loeb et al., 2009)

• Nabil Swedan says:

That is a very long ENSO between 2002 and 2013. ENSO cycle is much shorter. I have no doubt in my mind that it is a decreasing trend of LW and time will tell. See this work that preceded that of Norman Loeb et. al (2016):

McGee Steve (2013). Is the Earth energy in deficit? Unpublished work in a scientific journal, discussed and published on Climate Etc. on November 28, 2013, http://judithcurry.com /2013/11/28/is-earth-in-energy-deficit/

Steve found a decreasing trend of about 0.05 w/m2 annually in agreement with my theoretical calculations.

• Kristian says:

Nabil Swedan says, August 27, 2016 at 6:33 PM:

That is a very long ENSO between 2002 and 2013. ENSO cycle is much shorter.

Swedan, you need to distinguish between the continuous natural ocean/atmosphere process dubbed “ENSO” and single ENSO events. A single ENSO event certainly couldn’t last from 2002 and 2013, but ENSO, the continuous process, could and would indeed.

There was a clear preponderance of positive ENSO ‘states’ between 2002 and 2007, and, conversely, a clear proponderance of negative ENSO ‘states’ between 2007 and 2012:
https://okulaer.files.wordpress.com/2016/08/nino3-4.png
(NINO3.4 SSTa, NOAA OIv2)

This very much affected the trend in TLT over the 2002-2013 period. It went down. And when TLT went down, the OLR naturally followed.

I have no doubt in my mind that it is a decreasing trend of LW and time will tell.

Not sure what you mean by this.

Here’s OLR through the global ToA (CERES EBAF) vs. global TLT (UAH) from 2000 to 2016:
https://okulaer.files.wordpress.com/2016/08/uahv6-vs-ceres-olr-gl1.png

The OLR clearly and evidently follows the TLT. If the TLT goes down over time, then so does the OLR. If the TLT stays flat over time, then so does the OLR. And if the TLT goes up over time, then so does the OLR.

IOW: Total OLR at the ToA is primarily simply a radiative effect of tropospheric temperatures.

• Nabil Swedan says:

Kristian,

One observation is not enough to make a big conclusion. A climate related observation must be examined from different angles to make an educated conclusion based on my experience. If you research different topics you will find that trend of the temperature of the mesopause is decreasing. Not one paper but many. Here is one:

“Long-term midlatitude mesopause region temperature trend deduced from quarter century (19902014) Na lidar observations
C.-Y. She 1,2 , D. A. Krueger 1 , and T. Yuan 2”

There are so many papers consistently indicating reduction in the temperature of air in the stratosphere and mesosphere. When the temperature of the mesopause has a decreasing trend, we know that Long wave radiation at the TOA must have a decreasing trend. This is physics. When the decreasing trend of LW at TOA is measured by two unrelated entities, at least one has no vested interest, then it makes sense to accept the decreasing trend of LW at TOA. Finally, when the observed radiation trend of LW at TOA is calculated theoretically as a function of greenhouse gases, then the conclusion is final and undisputed: Long wave radiation at the top of the atmosphere must have a decreasing trend and the cause is GHG. The overall energy balance yields to surface warming. This is observations supported by math and physics. There cannot be a better scientific approach. Anthropogenic Global Warming is a fact, period.

• Kristian says:

Are you pulling my leg, Swedan?

• Nabil Swedan says:

No, I am serious based on available facts. You do not have to agree, prove me wrong! The debate is more enjoyable with people who disagree with me. With those who agree, there is nothing to talk about, and there is no learning.

• Ball4 says:

Nabil, I am not at all familiar with Climate Forecast System Reanalysis in that link, I would go with CERES data as that is used in the various balance papers that get discussed so often & is not a forecast. TFK09 has no imbalance at TOA, Stephens 2012 has TOA 0.2 inward surplus +/- 0.4.

Note the CI and also in Loeb 2016 CI ranges 0.1 to 0.44. Your 0.05 (not sure where you got that) thus is not meaningfully within the observational accuracy. Note also this net does not say anything about how OLR has been changing as does Loeb 2016 paper.

• Ball4 says:

0.6 inward surplus +/- 0.4, geez.

• Nabil Swedan says:

The -0.05 w/m2 is meaningful for it is on annual basis. To compare it with yours per decade, then multiply by 10 and it becomes -0.5 w/m2 per decade as opposed to -0.1 to -0.44 per decade of Loeb (2016). It is close to the upper limit. The CFSR was the best available at the time. I agree with you from reading the paper of Loeb (2016) that the CERES is what I will be watching for from now on.

Please note that the decreasing trend during the day is -0.73 to -0.89 w/m2 per decade based on Loeb 2016. This is not far off from that of Steve McGee of -0.6 w/m2 per decade, who appear to have done the work with the presence of outgoing short wave, or during the day. They are in agreement as far as I can tell. Very interesting and fundamental to understanding what is going on with the climate.

• Ball4 says:

oh, ok. Yes the Loeb decreasing OLR trends from the various instruments are meaningful after calibration compared to their confidence intervals. Glad they are doing CIs now as in Stephens 2012.

• Toneb says:

“Only problem is, we dont observe this mechanism (the raising of the effective radiating level (+Z_e) to force sfc warming (+T_s) by keeping T_e constant) to be operative and effective in the real Earth system…..”

That’s because it’s NOT observable as there is an average effective height of emission for water vapour, N2O, ozone and all the other ghgs, as well as CO2.

• Kristian says:

Apparently, Toneb has no deeper knowledge whatsoever of the completely unsubstantiated hypothesis he’s promoting.

T_e is directly associated with total OLR at the ToA, Toneb. T_e is also directly associated with Z_e. If Z_e (Earth’s effective radiating level to space, basically its apparent planetary “blackbody surface” as seen from space) moves up, then T_e (and hence the total OLR at the ToA) is supposed to stay unchanged while T_s rises. As illustrated in this diagram:

However, we don’t see any of this happening in the real world. In fact, quite the contrary.

• Ball4 says:

Kristian, your linked figure is up to tropopause so doesn’t illustrate what Dr. Spencer correctly writes top post: “increasing carbon dioxide in the atmosphere reduces the rate of IR energy loss to deep space, resulting in some warming. (The warming is actually in the lower atmosphere, while the upper atmosphere cools).”

• Toneb says:

Or touche…

Apparently Kristian has reading comprehension difficulties.

I quote you again ….

“Only problem is, we dont observe this mechanism ….”

You said OBSERVE my friend.

AS I said, the effective emission height in Earth’s atmosphere is not observable, due to the various constituent GHG’s, and to boot the varying vertical path length of those gases due to ongoing weath around the planet (eg convection).

• Kristian says:

Toneb says, August 27, 2016 at 11:03 AM:

I quote you again …

“Only problem is, we dont observe this mechanism …”

You said OBSERVE my friend.

Yes. And we don’t observe it. So what’s your problem?

Thing is, we WOULD’VE observed it if OLR stayed flat as TLT went up over time. We don’t.

AS I said, the effective emission height in Earths atmosphere is not observable, due to the various constituent GHGs, and to boot the varying vertical path length of those gases due to ongoing weath around the planet (eg convection).

Who’s got difficulties with his/her reading comprehension here? I never stated nor implied that we can observe the effective radiating level itself, Toneb. I stated that we cannot observe the “greenhouse warming mechanism”. For the “greenhouse warming mechanism” to (potentially) work, T_e (OLR) cannot be observed to go up with the TLT. It is.

• Ball4 says:

Again, Kristian writes “(OLR) cannot be observed to go up with the TLT. It is.”

Again, not according to CERES team 2016 paper which shows OLR slightly reducing over ~11 years. Kristian has not yet shown his method replicates the CERES team results.

• Kristian says:

Toneb says, August 27, 2016 at 11:03 AM:

Haha. Says Toneb.

• Toneb says:

“Yes. And we dont observe it. So whats your problem?”

I’ll repeat the quote for the hard of comprehension….

“Only problem is, we dont observe this mechanism (the raising of the effective radiating level (+Z_e) to force sfc warming (+T_s) by keeping T_e constant) to be operative and effective in the real Earth system, even as the atmospheric content of CO2 and water vapour has gone up significantly since the mid 80s. Because what does the data show?”

So, you spout a strawman in order to rubbish AGW theory and I call you out.
Simples.

• Kristian says:

Toneb says, August 27, 2016 at 1:17 PM:

Ill repeat the quote for the hard of comprehension …

“Only problem is, we dont observe this mechanism (the raising of the effective radiating level (+Z_e) to force sfc warming (+T_s) by keeping T_e constant) to be operative and effective in the real Earth system, even as the atmospheric content of CO2 and water vapour has gone up significantly since the mid 80s. Because what does the data show?”

So, you spout a strawman in order to rubbish AGW theory and I call you out.
Simples.

Too funny! Toneb, you’re evidently unable to comprehend the first thing about what’s being written, and then you try to blame your shortcomings on me. Try to read up on your own favourite hypothesis and how it describes its “warming mechanism” before making a fool of yourself in spouting meaningless ‘objections’ like you do here …

Again, for the slow learners: Between 1984/85 and 2016 we do not observe the OLR through the ToA to remain flat over time as TLT go up. We rather observe the OLR to follow TLT over time.

That’s AGW out the door, Toneb. Simples.

The strawman work is all yours: “We cannot observe the effective radiating level.” LOL!

• Ball4 says:

“We rather observe the OLR to follow TLT over time.”

Again, Kristian hasn’t yet shown that with UAH v6Beta5. CERES team shows the opposite of Kristian’s results (that don’t have CIs) in Loeb 2016 (with CIs). OLR decreased.

• Kristian says:

Ball4 says, August 28, 2016 at 4:22 AM:

Hi, troll.

Again, Kristian hasn’t yet shown that with UAH v6Beta5.

Yes I have. Several times, even.

CERES team shows the opposite of Kristian’s results (…)

No, they don’t. They show exactly the same as my results. Which has already been shown you on multiple occasions.

(…) Kristian’s results (that dont have CIs) (…)

Yes, they do. Loeb et al., 2016 are simply using a newer version. But every official version of the data is calibrated and validated. You’re obviously ‘misunderstanding’ the concept of a straightforward version upgrade.

OLR decreased.

Yes, because the TLT decreased. Over the particular period in question. But significantly not from 2000 to 2016, and not from 1984/85 to 2016.

How many times do you need to have this information spoon-fed to you, Ball4? All you need to do is have a look at this plot:
https://okulaer.files.wordpress.com/2016/08/uahv6-vs-ceres-olr-gl1.png

• Ball4 says:

“But significantly not from 2000 to 2016, and not from 1984/85 to 2016.”

That assertion is not persuasive until Kristian fully discloses the method like any published author and show Kristian CI & Kristian method replicates Loeb 2016 results. Your UAH anomaly work is suspect when looked at critically, not persuasive as you also do not show your work, thus it is a suspect curve from assertion only. I cannot even try replicate it.

Become persuasive Kristian, show us the work, you know – like a published author does. The light of day is a wonderful thing.

• Kristian says:

Seriously, what are you on?

• Ball4 says:

Advancing good, replicable, tested science, in the light of day Kristian.

Drag out your work, shed some light on it. Doesn’t do anyone else any good hidden away in your computer. If you disagree with CERES team, it is ok, they would like to know why, me too. Show them, show me, show the peanut gallery. Free the work! Make progress, you already have shown some, keep going.

• Kristian says:

Well, I have to ask, because I’m being stalked by this person that simply can’t get over having lost an argument, and who consequently, inanely and in monomaniacal fashion, at every opportunity he gets, rather starts pointing at plots such as this …:
https://okulaer.files.wordpress.com/2016/06/uahv6-tlt-trop-x.png

… and says: “But, hey guys, look at the data inside this particular segment …:
https://okulaer.files.wordpress.com/2016/08/uahv6-tlt-trop-x2.png

… The OLR is going down, isn’t it? You know, the red crooked line behind the blue crooked temperature line. And therefore the rest of the plot doesn’t matter. Kristian is wrong, and I am right after all. Yuhuu!”

Seriously, only a person with mental issues can go on and on like this, seemingly without ever so much as considering the glaring fundamental problem with (and thus, quite frankly, the utter stupidity of) his own argument …

It’s right there, for all to see, in the very plots above.

Also, here’s CERES EBAF Ed2.8 OLR (black) vs. Ed4 (red and blue) over the period of my stalker’s choice:
https://okulaer.files.wordpress.com/2016/08/olr-30-30-paper.png

This has been shown to him already on multiple occasions. Does he care? Nope.

23. Lewis says:

Dr. Spencer,

I take slight exception to the terminology. It seems the cardboard acts much like a lid on a pot on the stove, it only slows the dissipation of heat, it does not heat the pot.

The cardboard is a barrier to the cold, basically reflecting the heat of the warmer surface. For all intents, it is another insulator. So it only reflects or slows the loss of heat it does not, of itself, heat the heated surface.

In effect, it is a blanket on the bed, with you under it.

On a related subject. It seems the effect of the GHG’s is to raise the temperature of the earth/atmosphere some amount, at which radiative equilibrium occurs.

The question, which should be simple for the scientists to answer, is how much of a temperature rise is required for that equilibrium to occur?

• gbaikie says:

“The cardboard is a barrier to the cold, basically reflecting the heat of the warmer surface. For all intents, it is another insulator. So it only reflects or slows the loss of heat it does not, of itself, heat the heated surface.”

With this experiment, the assumption is the cardboard because
it’s expected/assumed to slow down the loss of black painted aluminum plate above it, that the black painted aluminum plate will rise in temperature.
And if it does, then the “colder cardboard” is causing the hotter black painted aluminum plate to get warmer. Or warmer [but colder than black painted aluminum]
cardboard, warms more than the even colder white painted plate [which is cooled by the CO2 ice].
So if experiment worked, the 20 C cardboard warms the +70 C
black paint aluminum plate.

Or what is warming the black painted plate is the heat from
the halogen floodlights and cardboard is suppose to reduce
the loss of heat of black painted plate [more than cooled white painted and cooled by CO2 ice, plate].

AND this is suppose to *not* be related to a reduction of convection heat loss of black painted aluminum plate- or the hoped for warming is suppose to be caused by radiant energy
exchanges which are suppose to be different if the cooler
plate is not as cold

-In effect, it is a blanket on the bed, with you under it.-
Except a blanket [or clothing in general] is about blocking convectional heat loss rather than blocking radiant heat loss.

–On a related subject. It seems the effect of the GHGs is to raise the temperature of the earth/atmosphere some amount, at which radiative equilibrium occurs.

The question, which should be simple for the scientists to answer, is how much of a temperature rise is required for that equilibrium to occur?–

Well this is simple and it’s also complicated- depending on
what is meant.
The earth land surface in sunlight does not get much above 70 C and ocean surface temperature does not get much above 35 C. You could call that an equilibrium temperature.
Or you could be concerned about surface air temperature- and that could be more complicated.
The highest air temperature recorded in the world was in 1913
in Death Valley. You could consider that as related to the equilibrium temperature.
Or you could be talking about average night and day temperature. 15 C is average temperature of air surface temperature of Earth. Continental US average temperature is about 13 C. And greenhouse gases according to GHE theory is
suppose to increase this temperature. And one can imagine runaway effects from greenhouse gases.

24. Bryan says:

Roy

Colder objects do not always make warmer objects warmer still.
Even though they radiate to the warmer object and the warmer object absorbs their radiation.

For example

A hollow metal shell at 80C has a lump of ice placed inside and closed.

The metal shell will drop in temperature much faster than without the ice even though it absorbs the radiation coming from the ice.

• gbaikie says:

–The metal shell will drop in temperature much faster than without the ice even though it absorbs the radiation coming from the ice.–

And this also true if it’s water rather than ice with the energy needed to melt ice.

But by adding water or ice to sphere, one is increasing the thermal mass of the object. Therefore once metal sphere reaches
0 C, it will cool slower and compared to sphere without the
ice or water.

Let’s say one has a sphere which is 1 meter in diameter and it can opened in middle and can be sealed again. So it has clamp at it’s equator which can seal and make it air tight.

So have southern hemisphere on the ground, and place insulation
at it’s bottom- in it’s interior. And top of this insulation you put a block of ice. Then you put northern hemisphere on top
and clamp it shut.

So hollow sphere walls is made of 1 cm thick iron.
Outside volume of sphere [not including the clamping part]
cubic meter. Or one has .04 cubic meters of iron that has the
mass of 7874 times .04 [ Iron density near 7.874 g/cm3].
So iron sphere wall has total mass of 314.96 kg. Call it
315 kg. And have clamp be made of iron and weighs 85 kg.
Total iron mass: 400 kg.
Insulation- a block foam with curved bottom which sits on bottom of sphere- roughly 30 cm cubic block. Or if like
make it out of aerogel:
“Aerogel is a synthetic porous ultralight material derived from a gel, in which the liquid component of the gel has been replaced with a gas. The result is a solid with extremely low density and low thermal conductivity.”- wiki
It’s mass is insignificant and blocks all conduction of heat.
Place 30 cm cube of water onto top of insulation [encase in plastic bag]- 27 liter of water or 27 kg of water [or about 7 gallons of water]

And finally put sphere on concrete ring- so it doesn’t roll around- say, 30 cm interior diameter. Though have pegs so there is 1 inch [or more] air space at bottom except where sitting on the pegs.

So put the sphere in a desert and at night and heat iron of hollow sphere to 80 C. Could put boiling water into it, and then remove water. Place insulation and block of ice in sphere, then clamp it shut.
Compare to a heated sphere which doesn’t have block of ice in
it.

How long does it take for heated sphere of 80 C to cool down to night temperature of around 10 to 20 C if lacks the 30 cm cube of water?

If did same thing at location on Moon at night, how long would take to cool the same temperature as the desert night air [10 to 20 C]

• gbaikie says:

Put above iron sphere on lunar surface. Have it it’s pegged stand, on flat plain at equator. Don’t bother with insulation
and put in sphere a block of ice at -150 C and is 27 kg of ice.
Assume iron and stand and ice is -150 C when the sun comes up.
-150 C = 123 K
So sunlight is above horizon and it would be warming a hemisphere and remain throughout the day time heating same amount of area of the sphere- though part of sphere which is heated changes- other variation of the clamp at it’s equator [which not going to describe and I will ignore- or you imagine that it’s not there].
So it’s disk is 1 meter diameter and lit hemisphere would have uneven heating. The area at zenith to Sun will be hottest and can be warmed to 120 C.
The sphere has no air in it, and block of ice will become gas long before 0 C.
It’s long day, by time of sunset, the sphere should have uniform temperature of about 100 C. Or sphere will full of steam and hot water at somewhere around 100 C and will have pressure of somewhere around 15 psi.
So we check on it near sun down. The ground around it will be cooled to temperature below 100 C and sphere will be warmer than the ground.
Or if sun is at 10 degree above horizon it will be warming the same amount of the sphere as it was at noon. The level
ground will be receiving a lot less sunlight as compare sunlight at noon. Or ground at zenith is heated to 120 C
ground and when sun is at 30 degrees above horizon will be about 45 C and about 30 hours later when sun closer to 20 degrees above horizon the ground will be about 10 C.
See graph:
http://staff.diviner.ucla.edu/science.shtml

So before sun goes down, the ground will be fairly cold and the sphere will be fairly hot.
And days later when nearer the middle of the night on the Moon, the ground will remain colder than the sphere.
How much warmer requires some math- or above doesn’t require math.

• gbaikie says:

water at 120 C has pressure of 2 bars:
http://www.engineeringtoolbox.com/saturated-steam-properties-d_457.html
Can assume water never goes above 120 C, or pressure does not
exceed 30 psi.
And can assume sphere can withstand far more than 30 psi pressure.
Question is there any amount water which could be added that would prevent the water from reaching say 100 C.
Or it has 27 kg of water and could hold as much as 480 kg
of water [filled completely with water] though since it
starts as ice, the limit would be about 440 kg of ice at -150 C.

• Gordon Robertson says:

@Bryan..”The metal shell will drop in temperature much faster than without the ice even though it absorbs the radiation coming from the ice”.

Clausius stated very clearly that heat can NEVER be transferred on its own from a colder body to a warmer body. Who said radiation from the ice is absorbed by the warmer body?

• Ball4 says:

“Who said radiation from the ice is absorbed by the warmer body?”

Planck. Because that’s what his referenced tests demonstrated.

• Bryan says:

Gordon Robertson

Assuming for simplicity that conduction and convection have been eliminated in this case by experimental set up

Radiation from the warmer shell will be absorbed by the ice .
Radiation from the colder ice will be absorbed by the shell.
The intensity of the radiation from the shell is much greater than that from the ice.
Heat is transferred from the shell to the ice
The temperature of the shell drops( it loses internal energy)
The temperature of the ice increases(it gains internal energy)

25. Bryan says:

Instead of thinking about an insulating layer as being a heat source it is more helpful to identify the insulating property of the barrier.

A layer that radiates will send some energy back to the heat source which will be absorbed.
This energy will slow the rate of heat transfer.

Convection

As well as hot air rising cold air falls.
Once again some of the internal energy of the rising parcil of air is returned to the region of the heat source.

Conduction

In a metal object (say a poker in a fire) the lattice atoms vibrate.
The ones at the hot end faster than the ones at the colder end.
By ‘collision’ energy is transferred from hotter to colder atoms.
But the faster vibrating atom does not lose all its kinetic energy to the colder in the collision.
This recoiling atom returns some energy back to the hotter direction .

So is ‘The Greenhouse Effect’ nothing more than radiative insulation?

If so the grand title of ‘An Effect’ seems overdone.

• Gordon Robertson says:

@Bryan…”In a metal object (say a poker in a fire) the lattice atoms vibrate.
The ones at the hot end faster than the ones at the colder end.
By collision energy is transferred from hotter to colder atoms”.

It’s not collision at all, the atoms are constrained in a lattice. Unless you overheat the poker and break some bonds there are no atoms free for collision.

Thermal energy is transferred atom to atom via valence shell electrons.

With radiation it’s similar. Atoms at the body boundary are free to emit EM to the atmosphere. Perhaps that can take place even deeper than the boundary since the spacing between atoms is relatively immense. Once the EM is radiated, the atoms cool and that EM can be absorbed by a cooler body where the process is reversed.

That’s how heat is transferred radiatively.

26. Steve Case says:

I dont know just how much the observed temperature increase in the heated surface will be when the cardboard sheet blocks its view of the chilled surface. Maybe 1 deg. F, maybe 10 deg. F. But it should be observable.

In other words, you didn’t actually do the experiment. A YouTube of such a set-up would be great.

As I understand it, the back radiation from CO2 is around 15 and a black body that has a peak radiation of 15 is about the temperature of a block of dry ice, -109F, and isn’t going to warm anything warmer than that. But it will impede the rate of cooling compared to an absence of any other back radiation i.e., the absolute zero of outer space. But the actual warming comes from the sun. In the case of your experiment which you didn’t actually do, the heating comes from the two lamps (radiation) and ambient air (conduction) between the two aluminum sheets. Looks like they are about a foot or so apart from your drawing. So you’re claiming that the loss of (radiation) from the ambient air a foot or so away is going to make a measurable difference. I think you ought to actually do the experiment and measure how much that difference is.

• Steve Case says:

I see that the website blocks ASCII code that’s not on the keyboard and although [ALT+0181] yields the micron symbol on the reply page it disappears when you click on [Submit Comment]

so my post should have read:

“… the back radiation from CO2 is around 15 microns and a black body that has a peak radiation of 15 microns …”

• Gordon Robertson says:

@Steve Case…”I see that the website blocks ASCII code thats not on the keyboard and although [ALT+0181] yields the micron symbol on the reply page it disappears when you click on [Submit Comment]”

Try finding the html equivalent of [ALT+0181] and see if that posts. it does for a smiley.

• Tim Folkerts says:

“As I understand it, the back radiation from CO2 is around 15 and a black body that has a peak radiation of 15 is about the temperature of a block of dry ice, -109F, and isnt going to warm anything warmer than that. “

No, not quite. CO2 radiates at several IR wavelengths, including around 15 um, 4 um and 2.5 um. These wavelengths tell you NOTHING per se about what temperature could be reached when CO2 is used to warm something. The spectrum of the radiation from any object depends on the temperature of that object, and the temperature of the radiating object determines the maximum theoretical temperature that could be reached using radiation from that object.

So if CO2 is at -40 C, it (by itself with no other input power) could theoretically warm something up t0 -40C. CO2 at +100 C could theoretically warm something to +100 C.

• Gordon Robertson says:

@Tim Fokerts…”So if CO2 is at -40 C, it (by itself with no other input power) could theoretically warm something up t0 -40C. CO2 at +100 C could theoretically warm something to +100 C.”

In an atmosphere that was at least 90% CO2, not in an atmosphere that is 99% N2 and O2 with all CO2 at 0.04%.

Only in climate models, where CO2 has been given an arbitrary and unrealistic ability to warm the atmosphere could your claim be true.

• Ball4 says:

Doesn’t warm the atm. Gordon, you did not comprehend the top post: “This is the basis for global warming theory: increasing carbon dioxide in the atmosphere reduces the rate of IR energy loss to deep space, resulting in some warming. (The warming is actually in the lower atmosphere, while the upper atmosphere cools).”

• Gordon Robertson says:

@Steve Case…”But it will impede the rate of cooling body’s ability to cool.compared to an absence of any other back radiation”

What’s your reasoning? How can a cooler body in proximity to a warmer body slow the cooling of the warmer body?

It would have to surround it as a sphere but then it would be absorbing IR solely from the warmer body. That does not represent a case where IR emissions from the cooler body is interfering with the warmer body’s ability to cool.

Why should it? Can you give a reason based on the atomic structure and not simply as a thought experiment?

• Steve Case says:

Can you give a reason based on the atomic structure and not simply as a thought experiment?
In a word, no. The thought experiment says if a body cools by radiation, and some of that radiation is returned, then it would follow that the cooling would be retarded.

• Norman says:

Gordon Robertson

I know you have an extreme passion in posting numerous times the definition of heat based upon the writings of Clausius.

Yes we know, heat is both internal kinetic energy and the transfer of that kinetic energy.

Roy’s set-up is not a thought experiment, it is a valid experiment you could perform. I think that might be a more useful thing to do than try to prove to people your definition of heat is the only valid one to use.

Do Roy’s test, see what happens, then explain to the bloggers what you believe is taking place and why.

• Steve Case says:

Roys set-up is not a thought experiment,

It is until it’s done.

I wish he had done it before he posted it and what he thought the results would be.

• gbaikie says:

He said he guessed the result would be that cardboard would cause the black painted aluminum plate to be:
“Maybe 1 deg. F, maybe 10 deg. F. But it should be observable.”
And suggested he would heat black painted aluminum plate
Or 1 to 10 F is about .5 to 5.5 C.
I doubt cardboard will increase black painted 75 C surface to 75.5 C.
If cardboard was a mirror surface, I think it might increase
75 C to 77 C- which I think is actually firmly quite measurable.
Or at least I pretty sure mirror surface will heat more than paper.
But all a mirror warming the black paint surface would prove is that clouds could warm the surface. Which few people argue about.
I think also if mirror warms the same as cardboard, it provide a clue that neither mirror or cardboard is radiantly warming- or would seem to point to strong possibility of it involving convectional warming.
But the point I see of Spencer posting about it before doing
it, is to provide opportunity for comment and make suggestions which might help him do a better experiment.

• gbaikie says:

Oh, maybe he meant 75 F.
So make it 76 to 85 F.
No, that doesn’t make any sense- or what is he calling
room temperature

• Norman says:

Steve Case

Roy has already done actual experiments and posted the results. For those that do not understand how the GHE works his own testing will not convince them. They need to do the testing themselves, it is the only way, then they have to think about what is taking place and try to describe it.

• Steve Case says:

Maybe 1 deg. F, maybe 10 deg. F. But it should be observable.

Hmmm A real simple set-up would be to hold a thermometer several inches over a block of dry ice, and then over a block of wood and take note of any difference.

• Kristian says:

Norman says, August 27, 2016 at 7:21 PM:

Yes we know, heat is both internal kinetic energy and the transfer of that kinetic energy.

No, Norman. “We” specifically do not know that. Because it’s not the case. I know that you yourself have a seriously hard time grasping the simple thermodynamic concept of “heat”. That’s why you simply do not get my argument against the “back radiation” explanation of the so-called “GHE”. You shouldn’t let yourself be even more confused, then, by the ramblings of people such as Robertson on this particular issue.

“If you can reduce the rate at which it cools to outer space, the climate system will increase its temperature until it emits enough infrared energy to restore radiative balance.”

That was a fun experiment. Here is another. Let’s drive from, say the countryside near Las Wages, to somewhere in Deep South country, and compare the day/night temperature differences.

Before departing Las Wages, I am waging a bet that the countryside in the Deep South will not cool at night to the same degree that the Nevada countryside does.

After picking up my millions of winning, I will send some to you, Dr. Spencer.

• Ball4 says:

Chad, your one drive is weather not climate. To get climate, you will need to drive from LV to all global thermometer field locations annually and make it a whole career of driving.

I’m on it.

• Ball4 says:

Good. Report back in 30 years.

Having been born in the forties, still working, and wondering what on earth to do upon retiring, now I am re-purposed!

I will circulate a sign-up form for co-drivers to facilitate this investigation, but any prospective volunteers need to be aware that my wheels do not possess the benefits of air conditioning; but, there is a solution for that dilemma.

While traveling at sea level under the oppressive heating situation (i.e. 14.7 psi atmospheric level) of the air column directly above us, we will merely ascend the nearest 8,000 ft. hill to cool, because there the atmospheric pressure of the air column above us will only be @ 11 psi.

Ah yes, the benefits of 5 psi difference for thermoregulation cannot be overstated.

• Gordon Robertson says:

@Chad Jessup…”…we will merely ascend the nearest 8,000 ft. hill…”

You should read some stories from Everest from about 20,000 feet up. The solar radiation is constant but the weather and temperatures change remarkably due to convection.

On the South Col, between Everest and Lhotse, in May, they can be subjected to 100 mph winds with temperatures at -30C.

Climbers on both K2 and Everest, stranded high on the mountain because they left it too late in the day to summit, have barely survived the cold, even in summer months. One of them on a US expedition was rendered a medical basket case for months afterward.

• mpainter says:

You can run, but you can’t hide. You, too, shall perish miserably, along with the rest of us, my friend.

G.R. – I am familiar with the physics of your subject. I was sarcastically making fun of the Huffman, et al types.

• Mike Flynn says:

“M. Martins has recently added to our knowledge by making observations on the heating of the soil at great elevations. He finds on the summit of the Pic du Midi the heat of the soil exposed to the sun, above that of the air, to be twice as great as in the valley at the base of the mountain. ‘ The immense heating of the soil,’ writes M. Martins, ‘ compared with that of the air on high mountains, is the more remarkable since, during the nights, the cooling by radiation is there much greater than in the plain.'”

– From Tyndall.

People talking about temperature at elevation often confuse air temperature with surface temperature. The thinner the atmosphere, the less impediment to the Suns rays, the fiercer the radiation.

Google “nose guards mountaineers” images. Exposed skin, at altitude, under clearish skies, gets badly sunburnt really quickly. If the air is cold, and there is wind, you won’t even realise how badly you’ve been burnt, until much later.

Insulation protects you from the sun, just as a fireman’s heavily insulated suit protects him from radiant heat.

The Earth sits in Nature’s freezer of around 4 K. There’s a heat source about 150 million kilometres away. The side facing the Sun gets hot, the unheated side gets quite cold. Thank god for the insulating atmosphere. Cooler during the day, warmer during the night. Stay on the surface, and get into the shade, or don some insulation if the Sun gets too fierce. Build a fire if it gets too cold.

Global warming due to insulation? I think not – hypothetical experiments of dubious construction notwithstanding.

Cheers.

M Flynn –“Exposed skin, at altitude, under clearish skies, gets badly sunburnt really quickly.” True, but that is due to UVB not visible light. I was once in SE Asia where one day the sun was not visible the entire day due to heavy cloud cover, yet I was severely sun burnt, because earth’s atmosphere is transparent to ultraviolet light.

That was a very painful lesson in atmospheric physics.

Oops. Should have stated that UV which gets past the ozone.

28. Kelvin V aughan says:

If you placed air with 0.08% CO2 in between, instead of the cardboard, how much hotter would the aluminum get?

• Gordon Robertson says:

@Kelvin Vaughan…”If you placed air with 0.08% CO2 in between, instead of the cardboard, how much hotter would the aluminum get?”

Great point. And you were generous to double the concentration of CO2 in the actual atmosphere.

• Norman says:

Kelvin Vaughan

I hate to “pop” your great idea but if you look at the set-up there is already CO2 between the aluminum plate and the dry ice…the experiment is set up in the air.

29. Mlton Hathaway says:

I believe it was on this blog that I once saw a reference to a classic experiment that involved two parabolic reflectors facing each other. A candle flame placed at the focal point of one reflector would create a hot spot at the focal point of of the other. But what I found really interesting is that a cube of dry ice placed at the focal point of one reflector would create a cold spot at the focal point of the other.

Can anyone provide a link for this experiment? I found some links, but all used a single mirror, and only exhibited a very small effect.

Anyway, if it is true that one can ‘focus cold’ with parabolic mirrors, then it seems like this discussion is ended. All one has to do is replace the dry ice with water ice and see if the temperature of the cold spot increases.

• gbaikie says:

here you go:
http://www.exo.net/~pauld/physics/coldspot.html

The short version:
“The block of ice is almost 100 degrees celsius cooler than room temperature. It gives off less thermal radiation than the room around it does. This lower level of thermal radiation is imaged by the mirror at the location of the visible light image of the block of dry ice. Less radiation arrives at this point, and so the air at that point is cooler, although only by a fraction of a degree.”

30. mpainter says:

What do you get when you have ten physicists together arguing about thermodynamics?

• Ball4 says:

Which is the reason proper test labs exist mp.

• Mike Flynn says:

M Ball,

Exactly. That which cannot be verified by repeatable experiment remains assumption. Cargo Cult Scientism in some cases. Correlation is not an experiment, neither are model outputs.

The heating properties of varying concentrations of CO2 in a volume of air are yet to be demonstrated by repeatable experiment. Until that time, they remain in the same category as thoughts about phlogiston, the ether, the indivisible atom, (or the thoughts of Maxwell, Planck, Bohr, and others about the nature of radiation, before experiments showed they were mistaken, and Einsteins Nobel Prize was well deserved.)

Cheers.

• Norman says:

Mike Flynn

The GHE is verified by actual measurements. You point an IR detecting instrument upward and it detects IR radiation, not only does it detect the downwelling flux but it also can determine the value of that flux.

The verifiable experiment is one you could do on your own. Take an IR lamp and move it close to a surface and measure the temperature increase of the surface. Logically you would conclude the IR lamp was the heat source. Now you can logically conclude that the surface is absorbing the IR energy and converting it into kinetic energy that you can measure with a thermometer. So you have measured downwelling flux of IR and you prove that IR is absorbed by a surface.

If that is not verifiable science what do you consider science to be?

• Ball4 says:

My own experiments? I’ve done the simple ones, especially the IR thermometer. I rely on Prof. Tyndall’s more complicated experiments (thoroughly replicated and honed) reported in 1861. They show the heating properties of varying concentrations of CO2 in a volume of air and were demonstrated repeatedly as he was so surprised. College thermo. lab courses repeat them annually now, only some of the students pass as evidenced by certain blog postings.

Also the experiments of Dr. Spencer are quite reliable, replicable if you think needed.

• Mike Flynn says:

I’m talking about the non existent (to date) experiment which shows that increasing the concentration of CO2 between the heat source and an object raises the temperature of the object.

Conversely, lowering the concentration of CO2 between a heat source and an object supposedly reduces the temperature of the object.

Tyndall showed that CO2 absorbs energy, reducing the amount reaching an object from a heat source. This results in a drop in temperature – in accordance with the later radiative transfer equations.

Cheers.

• Norman says:

Mike Flynn

Why wouldn’t the Earth qualify as the experiment?

You have downwelling IR. IR is composed of photons that each carry real energy that will convert to kinetic energy when absorbed by a surface.

• Ball4 says:

“Im talking about the non existent (to date) experiment which shows that increasing the concentration of CO2 between the heat source and an object raises the temperature of the object.”

That is exactly the testing Prof. Tyndall performed so it is existent. Tyndall’s experiment showed CO2 absorbs energy, reducing the amount reaching an object from a heat source thereby increasing the temperature of the air chamber measured by his thermometers.

This has also recently been performed in situ on the real atm. with the same results as Tyndall’s over a long time period.

• mpainter says:

Norman, have you altered your views? Are you ready to partition the “back radiation absorbed by the surface”?

• Norman says:

mpainter

Already been done by myself and Ed Bo. No point in continuing trying to explain my view to you is there?

But just for the benefit of trying.

I will refer to this global energy budget diagram since I believe it is most current. The values on the page are derived from measured values.

The effects of GHE are already considered in the partition of energy.

If I am correct from previous discussions with you on this topic, you were of the opinion that the 340.3 DWIR would heat the ocean surface and cause a greater evaporation rate?

The current evaporation heat loss of 86.4 W/m^2 has this flux already factored in.

The enhanced GHE with doubling CO2 is supposed to result in an increase of DWIR by 3.7 W/m^2
https://wattsupwiththat.com/2016/08/22/willis-and-i-walk-the-planck/

So add to the Current value of 340.3 and you get 344 W/m^2 DWIR

Now I am thinking you have a very good point! Intelligent. Thanks.

It would be very difficult to determine how much warmer the ocean surface would get at this point. If it was purely radiative it would increase temperature of the ocean surface by 0.67 C. But with this increase in surface ocean temperature the other effects would also increase lowering this maximum temperature but by how much? You have radiation at 30% so it would seem the ocean increase in temp would amount to 0.2 C. Also you point out it is ocean’s that determine air temperature the surface temp would then only rise 0.2 C but even Roy’s UAH graphs so an increase. But looking more closely at Roy’s temperature graphs it looks like before the El Nino of 1998 the temperature was meandering around -0.2 (lower some years, higher others) and after the El Nino it seems to meander around 0.1 which would be a 0.3 C increase which would about match what you are saying.

Thanks for pushing the point. It is a very good one. Take care and have a good day!

• mpainter says:

Norman, right, my view is that “back radiation absorbed by the surface” should be added to the insolation _if_ one believes that all radiation absorbed by the surface is accountable in a partition of energy loss. Adding the insolation and the “back radiation absorbed by the surface” of the KT diagram gives the sum of 494 W that needs to be partitioned in such an accounting.

But, if one accepts the principle that at least one third of the energy loss at the surface is latent, there is a BIG problem reconciling the 494 W with the well-constrained 80W of latent heat loss, as partitioned at the surface. How to reconcile these two figures should be the preoccupation of the AGW proponents. But they ignore it. Hmmm.

• Tim Folkerts says:

“But, if one accepts the principle that at least one third of the energy loss at the surface is latent, there is a BIG problem reconciling the 494 W with the well-constrained 80W of latent heat loss”

I’ve never heard of the “At Least One Third Of The Energy loss At A Surface Is Latent” Principle. And I can’t think of any reason why I should accept such a principle.

The numbers in the diagram add up. Energy is conserved. The Second Law is upheld.

Counted one way (net IR flux), then evaporation is way more than 1/3 of the energy loss. Counted another way (two-way IR fluxes), evaporation is way less than 1/3. But the way we choose to divvy up the numbers doesn’t change the underlying physics. There is no need to “reconcile” anything.

• gbaikie says:

— Tim Folkerts says:
August 28, 2016 at 12:17 PM

But, if one accepts the principle that at least one third of the energy loss at the surface is latent, there is a BIG problem reconciling the 494 W with the well-constrained 80W of latent heat loss

Ive never heard of the At Least One Third Of The Energy loss At A Surface Is Latent Principle. And I cant think of any reason why I should accept such a principle. —

I have don’t know how much is latent, but if accept that larger amount of energy is converted to to latent that should alter the cartoon.

What is known is that more than half of all sunlight reaching
the Earth surface occurs in region called the tropics. And the tropics is 40% of surface area of Earth. And about 80% of tropics is ocean area.
It rains a lot in the tropics and in addition a vast amount moisture is transported out of the tropics.
It’s also known the ocean surface temperature is constrained
by evaporation- preventing ocean surface temperature from exceeding 35 C. One also has many clouds systems in the tropics which acts as engines which draw up moisture from the surface.
One can have rain daily in the tropics, and one has intense rainfall events in the tropics.

• mpainter says:

Tim, yes indeed, there is a great discrepancy between the partitioning of the energy flux at the surface and the KT type diagrams. This was very much discussed in an exchange between me and Ed Bo on the preceding post (Desert Rocks…).

You see Tim, latent energy loss at the surface is constrained by observations: worldwide precipitation totals. This is 80-84 W/sq m, according to the various KT type energy budget diagrams. See if you can figure out the problem, and it’s a big one.

• Ed Bo says:

Tim:

For some ungodly reason, mpainter thinks it is improper to first subtract the downwelling LWIR from the upwelling LWIR to get the net LWIR when computing percentages of various types of heat transfer.

He thinks the only proper way to do it is to first add the downwelling LWIR to the solar insolation.

He also claims to have a source that says a majority of energy losses from the ocean surface are evaporative. Despite my repeated requests, he has refused to provide any kind of link or even citation, so we could see how that source calculates this figure.

mpainter said to you: “See if you can figure out the problem.” The only real problem is that he does not understand the associative property of addition and subtraction as well as most 10 year olds do.

• mpainter says:

Ed Bo says:
August 24, 2016 at 5:07 PM
mpainter:
###

Above begins the lengthy exchange on the preceding post (Desert Rocks, Nevada Surfrad station) if anyone is interested. Don’t know why Ed Bo broke off there and came here. I think he is looking for help.

The partition study, for which I had no link, partitions ocean energy loss at 60% – 30% – 10% for latent energy, radiation, and conduction, respectively. Kiehl, Trenberth use 50%, 39%, and 11%, roughly, in their partition scheme for the whole globe. If one accepts the KT scheme, the other follows naturally. Or so it seems.

• mpainter says:

Norman, you have touched on another mental gymnastics feat of the AGW proponents who warm the surface with back radiation and then refuse to allow that the surface has absorbed that same back radiation when you proceed to partition energy loss.

• Tim Folkerts says:

Why even state percentages???

IF everyone agrees that evaporation accounts for ~ 80 W/m^2 of energy transfer, then everyone agrees. Dividing this number by various other numbers doesn’t add any other information.

• mpainter says:

Tim, there’s no issue concerning the amount of latent energy loss at the surface. The whole issue is the partition of the surface emission of total radiation absorbed by the surface, as shown in the KT diagram. The KT diagram dodges the problem by simply not partitioning the energy loss attributable to back radiation absorbed by the surface.

For good reason: 494 W of radiation would produce three times as much latent energy emitted by the surfface as actual.

• Ed Bo says:

Tim:

Amen! mpainter has spent most of the last week arguing that the way K&T partition the surface energy balance is completely ridiculous because he does his calculations in a different order than they do.

I have been trying to tell him for this whole period that the K&T split, amounting to 50% evaporative, 39% radiative, 10% conductive over land and sea (when considering each as a NET value) is completely consistent with something he saw on the internet once (but cannot find now) that split this 60%, 30%, 10% for the sea alone. He has been telling me that I have no idea what I am talking about.

Now HE says here in this thread that they are consistent. Amazing! And he says that I am the one looking for help…

• Norman says:

mpainter

I will again you this more updated global budget.

With this one you have:

Incoming Solar absorbed by surface: 163.3 W/m^2
Upwelling IR: 398.2 W/m^2
Downwelling IR: 340.3 W/m^2
Evaporation: 86.4 W/m^2
Other: 18.4 W/m^2

You need to realize that all these are based upon actual measurements.

So if you take the Incoming Solar and Downwelling IR you get a surface flux of 503.6 W/m^2

So now you want this partitioned?

Okay if you do it this way then your IR loss would now be 398.2/503.6 or 79.07% from the surface. Evaporation would be 86.4/503.6 or 17.16% and other would be 18.4/503.6 or 3.65%

I hope you realize that the figure you have of 60, 30 and 10 is arbitrary based upon what values you use. In the case of the 60,30,10 they were taking the Net IR as it is the only energy loss.

So in that partition you would (numbers will vary because of an updated global budget). 57.9/163.3 = 35.46 for radiation

NOTE: in the partition you have to balance the energy coming in with the energy leaving or you will get heating or cooling. In the current global budget they have an NET 0.5 watt increase. The incoming energy is slightly greater than the outgoing.

Continuing the partition as was done by your source at some point. Based upon outgoing divided by incoming.

86.4/163.3 = 52.91% for evaporation
18.4/163.3 = 11.27%

So how you partition things really does not matter because the values are not based upon how you partition the numbers but are based on observational measurements over a period of time.

So if you partition it rigidly on actual energy IN and OUT you get
Evaporation: 52.91%
Other: 11.27%

If you partition it with your choice nothing changes in the 10 year measured values. They will still be the same but your partition % will be different.

Evaporation: 17.16%
Other: 3.65%

The measured values are not changing only the way you partition the % of surface energy loss.

I am not sure why you think nonradiative values would increase if you partition your way, all it does is change the % of the outgoing loss. It can’t change the actual values.

I hope that makes some sense to you. I think I did it as clearly as possible. If not I guess I have time to work another angle. I hope this one will end it though.

• mpainter says:

It’s quite simple. The back radiation is absorbed as kinetic energy and thus it becomes a whole new game because the kinetic energy is eventually returned to the atmosphere as latent, conducted and radiant energy. Partition that.

But you can’t, because the results are absurd if you use the accepted partition ratios.
Very big problem for the KT type energy budget diagrams.

• Norman says:

mpainter

Sorry but no. You are wrong. Your thinking is flawed and you can’t understand the reality that the values in the diagrams are actual measured values taken over long periods of time. No increase of latent heat, the latent heat is measured values with the Backradiation taking place. If you removed the backradiation the latent heat effects would be much lower because the surface would be much cooler.

You will not understand the flaw in your own thinking but consider that they are real measured values of what is going on. The diagrams are not the flaw, the flaw is in your own reasoning but no one can see their own flaws.

• mpainter says:

Diagram says “333 W absorbed by the surface”. Tell me what that means to you Norman. Let’s see who has the flaws.

• Norman says:

mpainter

I will try again but for some reason you are incapable of understanding it, stuck somewhere in a belief that can’t collapse.

You still do not understand that all the flows are measured and the effects taken into consideration but that is the reality. If you go to CERES web page and make global energy fluxes you can see how they arrived at their results. Lots of work though, you make grids and add all the flows up and divide by the number of grids to get the overall averages.

Here will be another approach for you, maybe you will see what everyone is telling you, probably not.

Okay using your approach to add solar flux and DWIR as one total flux of 503.6 W/m^2.

You have 503.6 W/m^2 reaching the surface -398.2 leaving the surface as Upwelling IR which leaves you with 105.4 W/m^2 to be removed by the other heat transfer mechanisms.

Take away the 18.4 W/m^2 that is conduction and convection combined and now you have 105.4 W/m^2 incoming – 18.4 leaving with the imbalance at 87 W/m^2.

If you attempt to think as you are currently thinking (which is absurd and wrong) and want to for some reason increase the evaporation loss then you would have to have cooling. The sea surface shows some warming (slight) but no cooling at this time. If you take your 87 remaining energy imbalance and subtract the 86.4 attributed to evaporation you are left with 0.6 W/m^2 excess (which makes sense since the water temp is slowly going up). If you added even 10 W/m^2 to evaporation for 96.4 W/m^2 because you falsely think the DWIR is adding more energy to the surface of the water than radiation is removing from that surface (why you persist in this notion is beyond those that can think logically as so many have tried to explain to you).

So with your idea say more evaporation because of the DWIR flux.

You have 87 watts after all other loses are taking into the equation and if you increase the evaporation to 96.4 (just 10 W/m^2 more) you end up with a -9.4 W/m^2. This would lead to a noticeable cooling of the ocean surface which is not taking place by any measurement I have seen.

I really do not understand how you logically keep coming to your absurd and illogical conclusions even when it is shown quite clearly why your thinking is flawed and incorrect.

• mpainter says:

Norman, your calculations show that your partition ratios are latent ~ 15%, radiation ~ 81%, and conduction ~ 4%, based on approximately 500 W/sq m surface emissions.

This is not the partition ratios of studies that have been published in the literature.
My point is that the KT diagram does not square with these published studies.

Norman,You say:

“If you attempt to think as you are currently thinking (which is absurd and wrong) and want to for some reason increase the evaporation loss then you would have to have cooling. The sea surface shows some warming (slight) but no cooling at this time.”

###

Norman, my aim is not to decrease evaporation loss, or dispute evaporation figures. My aim was to point out the revealing discrepancies between the KT diagram and studies that partition surface energy loss.

I used the word “constraint” in my discussions referring to latent energy loss; my position is that latent energy loss acts as an observational constraint in partitioning surface energy loss. You have wholly misunderstood my comments, if you even read them. It’s your inability to read, or grasp, my comments that lead you to interminable and fruitless discussions.

• mpainter says:

Ed Bo made some ill-informed and snarky comments on the previous thread. Here he mis-states my position and tries to bolster his self-image, it seems. I urge readers not to take Ed Bo’s comments here at face value. For those who are interested, see the previous thread, referenced above.

• Kristian says:

Norman, as long as you remain unable (or unwilling) to appreciate the elemental difference between the solar flux and the “DWLWIR”, you will always make these same analytical mistakes and end up confusing yourself into thinking that certain surreal notions are in fact true.

You say, August 29, 2016 at 7:51 AM:

You still do not understand that all the flows are measured and the effects taken into consideration but that is the reality.

As has been repeatedly pointed out to you: In reality, these opposite flows are specifically NOT MEASURED, they are only COMPUTED! The only thing actually measured (as in ‘physically detected’) is … the unidirectional HEAT flux.

Okay using your approach to add solar flux and DWIR as one total flux of 503.6 W/m^2.

No! You cannot add a HEAT flux and a conceptual ‘component flux’ and expect to get a physically meaningful result, Norman.

You have to wake up! The “DWLWIR” is NOT an extra added flux of energy to the surface. It is only there to begin with because of the larger ‘component flux’ from the surface up. It is completely dependent on this. The two cannot and must not be separated. And so, the “DWLWIR” is ONLY part of the surface heat LOSS. Mathematically (and ‘physically’) it SUBTRACTS from the “UWLWIR” to give the ‘net LW’ (radiant HEAT) flux.

IOW, your budget makes absolutely no physical sense, and ESPECIALLY it makes no thermodynamic sense.

The way you can think of this is through Clausius’ principle of “compensation”: Whenever some photons originating from a cooler body are absorbed by a warmer body, there are ALWAYS more photons simultaneously being emitted by the warmer body to the cooler. This ensures that you will NEVER be able to DETECT a rise in “internal energy” [U] in the warmer body. Simply because it never happens. And since it NEVER occurs that the “internal energy” of a warmer body faced with a cooler one at any point in time increases even incrementally, even for a split second, then it makes fundamental sense to say that – macroscopically/thermodynamically – the cooler body never actually TRANSFERS energy to the warmer body. The exchange of energy between the two is synchronous and instantaneous, and the probabilistic average of all the microscopic absorp tion and emission events, that individually cannot be accounted for, is a macroscopic NET MOVEMENT, a TRANSFER, of energy from the warmer to the cooler body only.

Yes, in a way it’s all semantics, but it’s still important to make the point, because what you do when you place the “DWLWIR” on the INPUT side of the energy budget of the surface, is that you make it seem, and hence confuse yourself into believing, that the energy from the cooler atmosphere to the warmer surface actually DOES in fact increase the surface “internal energy”, that the surface absorp tion of the “DWLWIR” constitutes a separate and independent process or event from the surface emission of the “UWLWIR”, and that the former process/event actually contributes to CAUSING the latter to happen, because seemingly, according to your logic, the surface needs to absorb both the solar flux (165 W/m^2) and the “DWLWIR” (345 W/m^2) in order to be ABLE to emit 398 W/m^2 worth of “UWLWIR”, from a temperature of 289K, rather than one of a mere 232K (from 165 W/m^2 alone).

Which is turning an actual temperature EFFECT – the “DWLWIR” – into what might be perceived as a temperature CAUSE. Causality is turned right on its head …

• yes, it is true that the DWIR is an “effect” in the sense that it only happens as a result of the surface of the Earth heating up and, in turn, the atmosphere, too.

But that does not negate the fact that the existence of DWIR increases average surface temperature above what they would have been if DWIR did not exist. Maybe it IS semantically misleading to say that DWIR represents an energy “source” that further *heats* the surface….that’s why I prefer to phrase it as a process that reduces the rate of *cooling* of the surface. But semantic distinctions still don’t change the equations which are used to quantitatively describe the resulting temperatures and energy fluxes.

• mpainter says:

Yeah, Norman, can’t you do anything right?☺

• Norman says:

mpainter

YOU: “This is not the partition ratios of studies that have been published in the literature.
My point is that the KT diagram does not square with these published studies.”

If you look at the partition studies in literature they are using NET radiation not breaking it up individually as you want to do. If you want to get the partition that is achieved in your literature then you must use a NET radiation flux and then it works. If you want to split the radiation up as you try to do then the partition is different. Not so hard to grasp. Wake up and think and it will make sense to you. There is no flaw in the budget diagrams or the partitions you refer to. The flaw is in your approach and that is the only flaw. You refer to partitions that use NET radiation then for your own purpose you use individual radiant flows and say there is a mistake when really none exists except in your own flawed thought process.

• Norman says:

Kristian

YOU: “You have to wake up! The DWLWIR is NOT an extra added flux of energy to the surface. It is only there to begin with because of the larger component flux from the surface up. It is completely dependent on this. The two cannot and must not be separated. And so, the DWLWIR is ONLY part of the surface heat LOSS. Mathematically (and physically) it SUBTRACTS from the UWLWIR to give the net LW (radiant HEAT) flux”

I disagree and I have challenged you to explain how your view is logical or consistent with established atomic physics but to date you have not. The two fluxes are separate and real regardless of what your beliefs are.

The molecules of the Earth’s surface are vibrating the same as they were before a DWIR and emitting the same IR. The process: Molecules are excited by internal heat and jiggling of surrounding molecules, they move to a higher energy state, they emit a photon of IR and go to a lower energy state. The amount of energy being emitted by the surface is based upon its temperature. The DWIR is real, it is based upon H20 and CO2 molecules being excited by collisions with other molecules then radiating away an IR photon and dropping down to a lower energy state.

Unless you can explain why another flux would inhibit or suppress the process of radiation emission it is bogus and faulty thought process. You are just plain wrong in your understanding of radiation physics and should read up on it again. I can’t help you. Look at a physics textbook and read about radiation and how it is generated.

• Norman says:

Kristian

YOU: “Yes, in a way its all semantics, but its still important to make the point, because what you do when you place the DWLWIR on the INPUT side of the energy budget of the surface, is that you make it seem, and hence confuse yourself into believing, that the energy from the cooler atmosphere to the warmer surface actually DOES in fact increase the surface internal energy, that the surface absorp tion of the DWLWIR constitutes a separate and independent process or event from the surface emission of the UWLWIR, and that the former process/event actually contributes to CAUSING the latter to happen, because seemingly, according to your logic, the surface needs to absorb both the solar flux (165 W/m^2) and the DWLWIR (345 W/m^2) in order to be ABLE to emit 398 W/m^2 worth of UWLWIR, from a temperature of 289K, rather than one of a mere 232K (from 165 W/m^2 alone).”

I do not confuse myself at all thinking the DWIR increases surface internal energy. The DWIR is less than the UPIR so why would it increase the internal energy of the surface? Where do I make such a claim in any of my posts?

Internal energy is based upon the total number of joules the body has. If you are removing 398 and adding 340 you are losing joules, the surface internal energy is not going up. Why do you in any way think that is my conclusion?

In my thinking the surface does not “need” to absorb both solar and DWIR, it just does. There are photons from both sources and a surface that can absorb them. What physics do you have to support the notion the surface will not absorb both fluxes?? None that I know of.

Mike Flynn, the king skeptic, already set up an experiment to demonstrate two way radiation. You have holes drilled in two plates, one heated the other not and have detectors behind the plates that pick up photons and count how many go through. Taking the cross section of the holes you can determine the larger flux that makes up these photon counts.

• mpainter says:

Norman, enough of your confusion. “Net” is in principle, additive. ALL energy accounted for in the partition of surface emissions is net by definition and thereby is meaningful. Energy emitted by the oceans is complete down to SST approaching 0 C. There is your “Net” energy flux, via meridional overturning circulation. All radiation absorbed by the surface is eventually emitted as one or another form of energy. This truism escapes you, somehow. Nothing to be done.

• Tim Folkerts says:

One quick analogy to point out the logical flaw in mpainter’s argument here.

Suppose I run a business with gross sales of \$1,000,000 and expenses of \$900,000 for a net income of \$100,000. And I have to pay \$20,000 in taxes.

I could say I pay 20% of the net income as tax. I could also say I pay 2% of the gross income. Both are equally correct.

Now suppose someone has heard that “tax rates are supposed to be about 20% of income”. That doesn’t make the 2% figure wrong. That doesn’t mean that I would have to pay \$200,000 in taxes as soon as I utter the words “my gross income is \$1,000,000”.

It just means different people are looking at different ratios and different ways to state the same thing.

• mpainter says:

Regarding the flaw in my logic, I had assumed that thermodynamics was a well defined and firm set of principles.
A complete innocent, I never imagined that it was a an analogy to tax and accounting problems.

That was the flaw in my logic.

• Norman says:

mpainter

I guess I have to find the Official mpainter dictionary to be able to communicate with you.

YOU post this definition: “Norman, enough of your confusion. Net is in principle, additive.”

How so?

Definition of Net: “Full Definition of net
1
: free from all charges or deductions: as
a : remaining after the deduction of all charges, outlay, or loss compare gross
b : excluding all tare
2
: excluding all nonessential considerations : basic, final ”

With a Net it is how you define additive. With radiation the UPwelling IR is removing energy from the surface. The DOWNwellingIR is adding energy to the surface. The amount of radiation the surface is getting from radiation is the sum (additive) of the energy into the surface (DWIR) 340 W/m^2 plus the negative energy leaving the surface or in simpler terms minus this flux of 398.2 W/m^2.

You can do it either way to get net radiation:

positive 340-398.2 = -58.2
or positive (because it is adding energy to the surface)
340 + (-398.2) = -58.2

Either way the surface is losing 58.2 W/m^2 via radiation. If not for GHG the surface would be losing 398.2 W/m^2

No GHG in atmosphere:
0-398.2
or
0 + (-398.2)

With either preference you are still going to lose energy at a much higher rate than you gain until you reach a new cold equilibrium.

Not that you will comprehend this post but never-the-less it is most interesting to try and understand how your brain processes information.

• mpainter says:

Norman says:”Not that you will comprehend this post but never-the-less it is most interesting to try and understand how your brain processes information.”

###

This is how my brain works:

“The net force is the vector sum of all the forces that act upon an object. That is to say, the net force is the sum of all the forces, taking into account the fact that a force is a vector and two forces of equal magnitude and opposite direction will cancel each other out.”

• Ball4 says:

Now mpainter is an expert on Newton’s law.

Picture two garden hoses single stream pointed at each other, same pressure, 10feet apart, same flow rate: neither pointer gets wet.

Conclusion: water has mass

——

Picture two flashlights in a dark room, the two beams pointed at each other, both photon streams clearly reach the other & “paint” the holders. Two streams!

Conclusion: Incoherent photons have no mass, do not interact or if you prefer, incoherent EMR waves have no mass, do not interact. Two streams!

• Norman says:

mpainter

Finally you point to a way to demonstrate a concept. You understand forces and vectors. Great.

Okay instead of radiation which seems to confuse several people what if we use some positively charged electric plates.

Two situations for you.

First situation is to have a negatively charged ball between two negative charged plates. One plate has a negative charge of 398 units, the opposing plate has a charge of 340 negative units. Which way will the negative charge ball move? How many units of force will it experience in the direction it moves?

Now the other situation. Now you have a similar negative charged ball but only one negatively charged plate with 398 units of negative charge. Will the ball move away faster than the one in the first situation?

• mpainter says:

Well, Norman, you flubbed it again. You forgot to add solar. I’ll show you the proper procedure once more (sigh).

There’s that word “Net”. I know that you love it.

That “Net” is what you now partition. For example, you subtract UWIR from this and that is your first partition of surface energy loss. The remainder of the radiation “absorbed by the surface” is latent energy and conductive energy loss, not yet partitioned.

Any questions?

• mpainter says:

“Net” is additive in science. That’s the principle my definition illustrates.
Lest you miss the point.

• Norman says:

mpainter

Your post is making the exact same conclusion I had already posted so what is the flub?

There are different ways to partition the energy flows. You can’t really say one is wrong or right. They give different partition % but one does not seem more valid then the other.

One way takes the net radiation as a loss and uses only solar input as energy received by the surface, the other three are all losses, IR is counted as a loss, evaporation and convection + conduction.

By the way adding a minus is the same as subtracting a plus they equal the same thing.

I am really not sure what your complaint is anymore. I think all possible modes of expressing radiation have been determined and all the outcomes lead to the same results.

No GHG the Earth is much colder (not sure about he -18 C though, they get that from a reflection of 30% but I think a lot of the Earth’s albedo comes from clouds)

https://en.wikipedia.org/wiki/Albedo#/media/File:Albedo-e_hg.svg

If Earth had no water and a nitrogen/oxygen atmosphere it might have a similar albedo as the moon of 0.12. The incoming solar would then be 300 W/m^2 and the Earth would have an equilibrium temperature of -3.45 C. Considerably warmer than the -18 C that uses the full albedo (of which a large part is clouds caused by a GHG)

• Ed Bo says:

mpainter:

You say: “But you cant, because the results are absurd if you use the accepted partition ratios.”

You keep talking about these “accepted partition ratios” without ever showing a source as to who proposed these ratios and how they derived them, or who it is who “accepts” them.

I have been prodding you for a week for some source, and you are completely unwilling or unable to provide one. At this point, I can only conclude that you are, as the kids say, “making [stuff] up”!

• Norman says:

mpainter

Time to end the ridiculous nature of this long debate over nothing.

I found information on both ways of partitioning surface fluxes.

Here is the one you are always referring to.
http://onlinelibrary.wiley.com/doi/10.1002/eco.1633/full

Now look here they are doing it the way you like. And it shows a much different partition percent because they are not adding the radiation fluxes in this one as they do in the one you are talking about. Also note they do state it is NET radiation in the first one. Repeat again so it can finally find a way into your brain, NET again NET adding the radiation fluxes together and then doing the percentages. Net, net or net net.

Think NET once more net.

http://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php

This is the one you like. Note ….no NET. NO added radiation fluxes before the percentages are calculated. Again NO NET, not added, not net radiation.

Can you finally understand these things or will you go to the next thread and bring it up again and again?

NET in one
NO NET in the other.

Same exact energy fluxes, just different ways to show it.

• mpainter says:

I’m tired of your snark, Norman. If you don’t like my comments, then ignore them. In fact, I would prefer that.

• Norman says:

mpainter,

I actually like your comments. Sorry for the snark. I guess I did go to far with the “net” thing in that post.

• Norman says:

“too far” not to far sorry for my poor grammar!

31. Norman says:

Mike Flynn

http://www.jstor.org/stable/108724?seq=32#page_scan_tab_contents

Page 31 and 32 of this paper from Tyndall will confirm what Ball4 is saying.

Experiments have been done and verified. It will not convince you.

32. Kelvin Vaughan says:

You haven’t popped my bubble as I said 0.08% CO2.

33. RICHARD says:

“If you can reduce the rate at which it cools to outer space, the climate system will increase its temperature”

Interesting,

If i put a tea cosy around my teapot i can make the tea hotter than when i poured in the hotwater.

• Norman says:

RICHARD

Yes you can if you continue to apply heat to the water in the teapot. That is what is missing in many people’s thought process on the issue. They neglect to understand that the surface in question has an influx of energy (with Earth system it is an on/off type influx) with Roy’s experiment it is a continuous heat source.

The climate system will increase in temperature (to a point, new equilibrium) if you reduce the rate it is cooling to outer space and leave the incoming energy the same. It is a First Law understanding. If a system has an increase in energy either by increase in input or decrease in output the situation is the same, the body warms up.

• RICHARD says:

Norman-

the heat remains the same from the source- ie the kettle, you are not warming the kettle, you are stopping the teapot from cooling as quickly with the tea cosy, you are not making the teapot hotter.

• wrong, Richard.

You can make an open pot of water on the stove, which is not quite boiling, start to boil by putting a lid on it.

You are increasing the temperature of the water by initially reducing the rate of heat loss.

Norman is correct. It is basic 1st Law thermodynamics.

I suspect that even cooks who have never taken physics understand this.

• RICHARD says:

Roy , we are not on top of the sun !

• Tim Folkerts says:

Sure we are “on top of the sun” and receiving energy from it. Or are you suggesting that the earth does not receive energy from the sun???

• RICHARD says:

Tim, we are looking at reflected energy , not direct energy,

it’s the same as the blanket and human body analogy,

we need to measure the energy reflected from the human body on a wall- if we could , not block the energy directly from the body, if you could put a blanket on the wall of reflected energy from the body you would only be slowing down the energy loss.

• RICHARD says:

Roy, – the source is the electric kettle, once the hot water has left the kettle into the teapot, a lid or tea cosy is not going to make it boil.

• RICHARD says:

Roy-

“I suspect that even cooks who have never taken physics understand this”

I suspect that tea makers in Japan realize that when the hot water has left the kettle and entered the teapot you cannot make it hotter with a lid or tea cozy.

• Ed Bo says:

Richard:

Do you seriously not understand that a body with a steady energy input, like electric current through a resistor to a tea kettle (or the sun shining on the earth), inhibiting energy loss to a colder ambient environment, like the lid on the teapot will make the body hotter?

Little kids understand this!

• RICHARD says:

ed-

“like the lid on the teapot will make the body hotter?

hotter than what?

the initial temp of the water entering the pot.

No!

the lid will slow down that initial temp loss. It wont make it hotter than the temp from the water being poured in.

A slight difference me thinks.

I like to heat the tea pot with boiling water before i put the teabag in befroe actually making the tea in the pot. Heating the pot is making my next stage of pouring in for the actual tea more pleasant and what is happening is not getting the initial heat loss. I can not go beyond this in making it hotter .

• RICHARD says:

obviously when the hot water has left the kettle i cannot make that water any hotter.

• RICHARD says:

i can only hope to slow down its cooling,

• Richard, I can only assume you are purposely diverting attention from the issue being discussed.

The energy analogy to the climate system is this: The pot of water is the climate system, the stove give an energy source, like the sun, the lid is the atmosphere. The stove (sun) is always on. Since the lid (atmosphere) is at a warmer temperature than deep space, it can cause the temperature of the Earth’s surface to rise.

No one is talking about adding water or taking water away from the pot, which would be like adding or removing big chunks of Earth.

• RICHARD says:

Roy,

I see the kettle as the source- ie the sun,

the hot water being poured obviously energy from the sun

the tea pot the earth as it wee, the lid or the tea cosy what every you want it to be regarding your atmospheric gasses etc.

So the hot water hits the teapot

can this hot water be made hotter or are you just slowing down the cooling?

• OMG…you think the kettle is the sun?? and water being poured is energy from the sun?

Dude…I’m afraid I cannot be any clearer than what I have already stated. Maybe you should go back and read more and assume less…

• RICHARD says:

Roy, lets try it this way then

The kettle is always on,

the water leaving is hot and constant.

the hot water hits the teapot and is constant,

can this constant hot water that has left the source be made hotter ?

• I have no idea what you mean by “water is leaving is hot and constant”, or by “the hot water hits the teapot and is constant”.

I’m beginning to think you are purposely wasting my time with ambiguous objections, when the issue is very simple.

A fixed amount of (non-boiling) water sitting in a pot and being heated by the stove. If you somehow insulate the pot (add a lid, or put a blanket around it), the water temperature will rise.

It’s as simple as that. There is no “water leaving”. I don’t know where you got that impression from the discussion.

• Ed Bo says:

Richard:

The analogy that everyone else is using is this:

The electrical power into the kettle (or the power in the flame under the pot) is like the sun’s power to the earth.

The water in the kettle or the port is the earth (with its thermal capacitance.

The lid of the kettle or pot is like the “greenhouse gases” in the atmosphere, acting as a barrier between the hot water and the coolder ambient.

The question is this:

With a steady power input to the kettle or pot, is the steady state temperature of the water in the kettle or pot (it’s not getting poured out) greater with the lid on than with the lid off?

• Thank you, Ed. Sometimes I think I’ve entered an alternate universe…

• RICHARD says:

“Richard:

The analogy that everyone else is using is this:

The electrical power into the kettle (or the power in the flame under the pot) is like the suns power to the earth.

The water in the kettle or the port is the earth (with its thermal capacitance.

The lid of the kettle or pot is like the greenhouse gases in the atmosphere, acting as a barrier between the hot water and the coolder ambient.

The question is this:

With a steady power input to the kettle or pot, is the steady state temperature of the water in the kettle or pot (its not getting poured out) greater with the lid on than with the lid off?”

and yet the true analogy is-

the kettle is is the sun
the energy from the sun is the hot water being poured, the earth is the teapot!

• Richard, for some reason you cannot grasp the simple example we are discussing, and want to change it into something else.

So, why not make:
1) the lid be the sun,
2) the water be the cold depths of outer space,
3) the kettle be the Sahara Desert, and
4) when we pour the water it’s Niagra Falls.

Then we can figure out how many angels can dance on a hot stove when they are wearing asbestos boots.

• RICHARD says:

Roy , lets make it really simple,

once again help me here, once the hot water has left the kettle can this hot water be made hotter without going back into the kettle and its heating coil.

yes or no?

• No, it cannot.

Does putting a lid on a warm pot of water being continuously heated by the stove result in the water temperature going higher than if the lid wasn’t there?

Yes, or no?

• RICHARD says:

No, it cannot.

thank you Roy,

that is all i needed to know.

“Does putting a lid on a warm pot of water being continuously heated by the stove result in the water temperature going higher than if the lid wasnt there?

if you blocked the Sun magically would it blow- probably

if you blocked a volcano would it blow certainly – probably

blocking the actual source is a recipe for disaster,

is the earth the source or a reflection of energy from the source- yes or no?

• Well, this is interesting…

But you refuse to do the same, and then pose another yes-or-no question.

• RICHARD says:

(snip) no more questions from you till you give me the same courtesy I extended to you. -Roy

• RICHARD says:

Roy,

I was very polite, that was unfair to snip.

• Ed Bo says:

Richard:

The earth and its atmosphere have no mass transfer (that isn’t utterly trivial) with the rest of the universe. So the analogy you keep trying to foist on us, with outgoing water constituting mass transfer, is just not appropriate.

You keep dodging the key question about Roy’s analogy. Why?

• RICHARD says:

and i even gave you a “got you” on my last comment you snipped!

The fact you continue to avoid answering a simple yes-no question, as I did for you, means you are no longer welcome here. -Roy

34. Johann Wundersamer says:

So if earth is packed in a CO2 / Styrofoam holder summer warmth holds longer; winter cold stays longer.

What’s achieved.

• no, winter is warmer, too, because the winter atmosphere is still warmer than the absolute zero temperature of deep space.

35. While complicated , I don’t think the experiment shows anything which could not be computed with the tools presented in Incropera , et al “Intro to Heat Transfer” .

I differentiate the issue of whether various materials can retard heat flow including thru “back radiation” — which is an “of course” — from the issue of how the interior of a ball can have a higher internal thermal energy density than that calculated for its spectrum as seen from the outside and the spectrum of its heat source .

That is , how higher temperatures at the bottoms of atmospheres relative to their tops , that is their orbital gray body temperatures , 3% in the case of the Earth , 125% in the case of Venus , — or calculated “colored body” temperatures — can be explained by a spectral phenomenon .

I don’t see that this sort of experiment addresses that at all .

• if you read the article, you would know it does not address it and was not meant to address it. I’m only addressing the 2nd Law argument some make that a cool object cannot make a warm object warmer still. You are correct, though, that using some basic assumptions and the relevant equations one would get a qualitatively similar result. But the people I am trying to reach apparently don’t accept well-known engineering and physics concepts, or computer calculations.

• “the people I am trying to reach apparently dont accept well-known engineering and physics concepts, or computer calculations.”

Can’t argue with that . Too true .

But it’s Hansen’s et al’s claim that some spectral effect could cause us to runaway to Venus that I see as the endlessly proffered boogeyman .

A large part of my sense of physics was inculcated by experiencing the brilliantly simple cheap experiments of the PSSC curriculum . And I think one of the most elegant experiments in the understand of radiant transfer was Ritchie’s 1830s experiment simply using hot water : http://cosy.com/Science/Ritchie'sKirchhoffExperiment.html .

There’s an as yet unmet need for brilliantly simple YouTubes quantitatively demonstrating these relationship .

• RICHARD says:

Bob,

He’s up against Harvard on that one-

“Such a runaway greenhouse effect is thought to have happened in Venus’s early history (the surface temperature of Venus exceeds 700 K). It cannot happen on Earth because accumulation of water vapor in the atmosphere results in the formation of clouds and precipitation, returning water to the surface”

http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap7.html

• don’t know who wrote that, but it’s not very well phrased. First of all, a “runaway greenhouse effect” is never possible…temperatures eventuall reach the point where infrared emission causes sufficient cooling to result in energy equilibrium and temperatures no longer rise.

Yes, water is an important part of cooling the Earth’s surface…but that short-circuits only about 60% of the greenhouse heating of the Earth’s surface. Read Lindzen’s 1990 BAMS paper, “Some Coolness Concerning the Greenhouse Effect”, or Manabe and Strickler (1964).

36. Anthony Mills says:

Following Bob Armstrong’ comment,any basic engineering text will derive the radiative heat transfer between a hot black surface of area A at temperature T1 to cooler black surroundings at T2 as

Q=sigma*A*(T1^4-T2^4) W

An increase in T2 reduces Q,the rate of heat loss from surface 1.Whether the cooler surroundings are “warming” surface 1 is a issue of semantics.Heat transfer engineers would not make such a statement .

• Ball4 says:

Anthony’s surface T2 is Dr. Spencer’s cooler surface, the ice.

“An increase in T2 reduces Q, the rate of heat loss from surface 1”

Yes Anthony when Dr. Spencer inserts the warmer cardboard surface than ice (but still cooler than T1) in view of T1.

And the data when performed convincingly show in a simple experimental demonstration that: Cool Objects Can Make Warm Objects Warmer Still.

Also right, Anthony, heat transfer engineers would not make such a statement because there are only radiant energy transfer engineers who would make such a statement as shown by Dr. Spencer experiment now posted to be the physical truth.

37. Christopher Game says:

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38. Larry R. Shultis says:

“…emit radiant energy in the direction of the warmer object, and thus increase the temperature of the warmer object above what it would otherwise be.”

Question: how do you determine “what it would otherwise be”, that is a temporal measurement and if the temperature is being augmented as in an open or closed system such as a planetary system, that would be near impossible analyze.

Take a two piece bar, as a single bar, of metal and take it to some uniform temperature and graph its temperature over time as it cools. Then take the same bar to the original temperature and separate the two pieces so that the total surface area will be greater. Graph the temperatures of the two pieces over time. Do the same for different separation distances. Will the radiation flow between the two pieces change the cooling rates of the bars with respect to their separation distances or add energy to each other with the result that temperatures would be higher than they would have been in some way at the same cooling times? I would suspect that the separated bars would cool faster than the single bar but that the separated bars would have the same cooling rates independent of the separation distances because the exchanged radiation will not add energy to the hot bar since the received intensities will be less than the emitted intensities and thus would be similar to cooler bars radiating toward hotter bars.

That is the simplest system I can think of where an isolated system might give some usable data.

39. MikeN says:

Can anyone describe a simple at home experiment that demonstrates temperature increase leads to higher CO2 levels?

• David Appell says:

MikeN says:
“Can anyone describe a simple at home experiment that demonstrates temperature increase leads to higher CO2 levels?”

Sorry, I got your request backwards.

I do not think you can find a home experiment that shows what you are looking for — effects are too small.