Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still

July 23rd, 2010 by Roy W. Spencer, Ph. D.

Probably as the result of my recent post explaining in simple terms my “skepticism” about global warming being mostly caused by carbon dioxide emissions, I’m getting a lot of e-mail traffic from some nice folks who are trying to convince me that the physics of the so-called Greenhouse Effect are not physically possible.

More specifically, that adding CO2 to the atmosphere is not physically capable of causing warming.

These arguments usually involve claims that “back radiation” can not flow from the cooler upper layers of the atmosphere to the warmer lower layers. This back radiation is a critical component of the theoretical explanation for the Greenhouse Effect.

Sometimes the Second Law of Thermodynamics, or Kirchoff’s Law of Thermal Radiation, are invoked in these arguments against back radiation and the greenhouse effect.

One of the more common statements is, “How can a cooler atmospheric layer possibly heat a warmer atmospheric layer below it?” The person asking the question obviously thinks the hypothetical case represented by their question is so ridiculous that no one could disagree with them.

Well, I’m going to go ahead and say it: THE PRESENCE OF COOLER OBJECTS CAN, AND DO, CAUSE WARMER OBJECTS TO GET EVEN HOTTER.

In fact, this is happening all around us, all the time. The reason why we might be confused by the apparent incongruity of the statement is that we don’t spend enough time thinking about why the temperature of something is what it is.

How Cooler Objects Make Warmer Objects Even Hotter

One way to demonstrate the concept is with the following thought experiment, which I will model roughly after the Earth suspended in the cold of outer space. Even my oldest daughter, a realtor who has an aversion to things scientific, got the right answer when I used this example on her.

Imagine a heated plate in a cooled vacuum chamber, as in the first illustration, below. These chambers are used to test instruments and satellites that will be flown in space. Let’s heat the plate continuously with electricity. The plate can lose energy only through infrared (heat) radiation emitted toward the colder walls of the chamber, since there is no air in the vacuum chamber to conduct the heat away from the plate. (Similarly, there is no air in outer space to conduct heat away from the Earth in the face of solar heating.)

The plate will eventually reach a constant temperature (let’s say 150 deg. F.) where the rate of energy gain by the plate from electricity equals the rate of energy loss by infrared radiation to the cooled chamber walls.

Now, let’s put a second plate next to the first plate. The second plate will begin to warm in response to the infrared energy being emitted by the heated plate. Eventually the second plate will also reach a state of equilibrium, where its average temperature (let’s say 100 deg. F) stays constant with time. This is shown in the next illustration:

But what will happen to the temperature of the heated plate in the process? It will end up even hotter than it was before the cooler plate was placed next to it. This is because the second plate reduced the rate at which the first plate was losing energy.

(If you are unconvinced of this, then imagine that the second plate completely surrounds the heated plate. Will the heated plate remain at 150 deg., and not warm at all?)

Since the temperature of an object is a function of both energy gain AND energy loss, the temperature of the plate (or anything else) can be raised in 2 basic ways: (1) increase the rate of energy gain, or (2) decrease the rate of energy loss. The temperature of everything is determined by energy flows in and out, and one needs to know both to determine whether the temperature will go up or down. This is a consequence of the 1st Law of Thermodynamics involving conservation of energy.

Note that the above example involving 2 plates, one hotter than the other, is apparently where the greenhouse effect deniers (sorry, I couldn’t help myself) would claim the “physically impossible” has occurred: The presence of a colder object has caused a warmer object to become even hotter. Again, the reason the heated plate became even hotter is that the second plate has, in effect, “insulated” the first plate from its cold surroundings, keeping it warmer than if the second plate was not there.

The only way I know of to explain this is that it isn’t just the heated plate that is emitting IR energy, but also the second plate….as well as the cold walls of the vacuum chamber. The following illustration zooms in on the plates from our previous illustration:

What happens is that the second plate is heated by IR radiation being emitted by the first plate, raising its temperature. The second plate, in turn, cannot cool to the temperature of the vacuum chamber walls (0 deg. F) because it is not in direct contact with the refrigerant being used…it can only lose IR at a rate which increases with temperature, so it achieves some intermediate temperature.

Meanwhile, the cooler plate is emitting more radiation toward the hot plate than the cold walls of the vacuum chamber would have emitted. This changes the energy budget of the hot plate: despite a constant flow of energy into the plate from the electric heater, it has now lost some of its ability to cool through IR radiation. Its temperature then rises until it, once again, is emitting IR radiation at the same rate as it is receiving energy from its surroundings (and the electric heater).

As we will see, below, in the case of the Earth being heated by the sun, the vacuum chamber “wall” (outer space) is close to absolute zero in temperature. Putting anything between that (essentially infinite) heat sink and the Earth’s surface will cause the surface to warm.

Examples are All Around Us

Examples of objects with lower temperatures causing objects with higher temperatures to become even higher still are all around us.

For instance, in terms of these most basic heating and cooling concepts (energy gain and energy loss), the same thing happens when you put a blanket over yourself when it is cold. The blanket stays cooler than your skin, but it nevertheless makes your skin warmer than if the cooler blanket was not there. Even though the direction of flow of heat never changes (it is always from warmer to cooler objects), a cooler object can still make a warm object even hotter.

It doesn’t matter what the mechanisms of energy transfer are….if the presence of a cooler object keeps a warmer object from losing energy as rapidly as before, the warm object will become even hotter.

But if you insist on another real-world example involving infrared radiation, rather than heat conduction, let’s use clouds at night. Almost everyone has experienced the fact that cloudy nights tend to be warmer than clear nights.

The most dramatic effect I’ve seen of this is in the winter, on a cold clear night with snow cover. The temperature will drop rapidly. But if a cloud layer moves in, the temperature will either stop dropping, or even warm dramatically.

This warming occurs because the cloud radiates much more IR energy downward than does a clear, dry atmosphere. This changes the energy budget of the surface dramatically, often causing warming — even though the cloud is usually at a lower temperature than the ground is. Even high altitude cirrus clouds at a temperature well below than of the surface, can cause warming.

So, once again, we see that the presence of a colder object can cause a warmer object to become warmer still.

Extending the Concept to the Atmosphere

As mentioned above, in the case of the cold depths of outer space surrounding the Earth’s solar-heated surface, ANY infrared absorber that gets between the Earth’s surface and space will cause the surface to warm.

This radiative insulating function occurs in the atmosphere because of the presence of greenhouse gases, that is, gases that absorb and emit significant amounts of infrared energy…(mostly water vapor, CO2, and methane). Clouds also contribute to the Greenhouse Effect.

Kirchoff’s Law of thermal radiation says (roughly), that a good infrared absorber is an equally good infrared emitter. So, each layer of the atmosphere is continuously absorbing IR, as well as emitting it. This is what makes the Greenhouse Effect so much more difficult to understand conceptually than solar heating of the Earth. While the sun is a single source, and most of the energy absorbed by the Earth is at a single level (the surface of the ground), in the case of infrared energy, every layer becomes both as source of energy and an absorber of energy.

It also helps that our eyes are much more sensitive to solar radiation than they (or even our skin) are to infrared radiation. It’s more difficult to conceptualize that which you can’t see.

Our intuition begins to fail us when presented with this complexity. The following illustration shows some of these energy flows: just the IR being emitted upward and downward by different atmospheric layers. If I included arrows representing the IR energy being absorbed by those layers, too, it would become hopelessly indecipherable.

As a result of the atmosphere’s ability to radiatively insulate the Earth’s surface from losing infrared energy directly to the “cold” depths of outer space, the surface warms to a higher average temperature than it would have if the atmosphere was not there. The no-atmosphere, global average surface temperature has been theoretically calculated to be around 0 deg. F.

This, then, constitutes the basic mechanism of the Greenhouse Effect. Greenhouse gases represent a “radiative blanket” that keeps the Earth’s surface warmer than it would otherwise be without those gases present.

In fact, research published in the 1960s showed that, if the current atmosphere suddenly became still – with no wind, evaporation, and convective overturning transporting excess energy from the surface to the upper atmosphere – the average surface temperature of the Earth would warm dramatically, from 0 deg. F with no greenhouse gases, to about 140 deg. F. That the real world temperature is much lower, around 59 deg. F, is due to the cooling effects of weather transporting heat from the surface to the upper atmosphere through convective air currents.

Weather as we know it would not even exist without the greenhouse effect continuously destabilizing the vertical temperature profile of the atmosphere. Vertical air currents associated with weather act to stabilize the atmospheric temperature profile, but it is the greenhouse effect that keeps the process going by warming the lower atmosphere, and cooling the upper atmosphere, to the point where convection must occur.

What About Kirchoff’s Law?
One of the statements of Kirchoff’s Law is:

At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.

Many well-meaning people think that one of the consequences of Kirchoffs Law of radiation is that an individual layer of the atmosphere that absorbs infrared energy at a certain rate must also emit energy at the same rate. This is NOT true.

The rate of emission becoming the same as the rate of absorption occurs in the very special case where (1) the temperature has reached thermal equilibrium, and (2) that equilibrium is the result of only those two radiative flows, in and out of the object.

Interestingly, this condition of a layer emitting the same amount of IR as it is absorbing is virtually never met anywhere in the atmosphere. This is because of the vertical, convective flows which are also transporting energy between layers.

In the global average, air below about 5,000 feet in altitude is absorbing more infrared energy than it emits, while air above that altitude (up to the top of the troposphere, the 80% of the atmosphere where weather occurs) is losing infrared energy faster than it is gained.

The reason why these two regions stay at roughly a constant temperature, despite very different rates of infrared loss and gain, is convective heat transport by weather: air heated by sunlight absorbed at the Earth’s surface has its excess energy transported to the upper troposphere, where a lack of water vapor (Earth’s main greenhouse gas) allows that energy to escape more rapidly to space.

The 2nd Law of Thermodynamics: Can Energy “Flow Uphill”?
In the case of radiation, the answer to that question is, “yes”. While heat conduction by an object always flows from hotter to colder, in the case of thermal radiation a cooler object does not check what the temperature of its surroundings is before sending out infrared energy. It sends it out anyway, no matter whether its surroundings are cooler or hotter.

Yes, thermal conduction involves energy flow in only one direction. But radiation flow involves energy flow in both directions.

Of course, in the context of the 2nd Law of Thermodynamics, both radiation and conduction processes are the same in the sense at the NET flow of energy is always “downhill”, from warmer temperatures to cooler temperatures.

But, if ANY flow of energy “uphill” is totally repulsive to you, maybe you can just think of the flow of IR energy being in only one direction, but with it’s magnitude being related to the relative temperature difference between the two objects. The result will still be the same: The presence of a cooler object can STILL cause a warmer object to become even hotter.

Anyway, that’s my story, and I’m sticking to it. Until someone convinces me otherwise.

So, let the flaming begin! No, really, have fun…but if you want your comments to remain available for others to read, please keep it civil.

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SUGGESTION FROM ROY (7:50 a.m. Monday, July 26): If you want to add intelligently to this discussion, you need to actually read (1) what I have said, and (2) what others have said. Chances are, your point has already been made and discussed.


350 Responses to “Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still”

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  1. Andrew says:

    So basically, the misunderstand is that it is incorrectly thought that it is being said that the cold atmosphere adds heat to the surface. A cold body cannot add (net) heat to a warmer body, but it can inhibit the loss of heat from that body.

    Do I have this about right?

  2. Johan says:

    I find your example very intriguing. I made a poor attempt to translate this to “first principles”. So, the only energy input in the system is the electric current (motion energy of the electrons, I suppose). This motion energy is transferred to motion energy of the charges (protons, electrons) in the first metal plate, thus raising its temperature. But some of the motion energy of the atoms in the metal is converted to electromagnetic radation (photons), mostly in the infrared region of the spectrum. In equilibrium, and without the second plate, the total energy input (motion of the electrons) equals the increased motion of the atoms in the metal (higher temperature) + the heat (IR radiation) “radiated” out to space. But with the second metal plate, and total energy input being equal, part of the radiated heat from the first plate is absorbed by the second plate. This adds motion energy to the atoms of the second plate (inreasing its temperature), but part of the absorbed energy is radiated out to space, but also back to the first plate, where this absorbed photon energy leads to a higher temperature of the first plate. Until a new equilibrium is reached of course, but I won’t try to put that in a formula. Your oldest daughter is a lot smarter than I am for sure! (and my apologies for everything I got wrong in the above, I’m not a physicist obviously).

    • I tend to stay away from discussing what happens on the molecular level. Most of what I presented relies on peoples common sense and experience.

      • Anonymous says:

        Fair enough. It so happens I’m a great admirer of Feynman, who as you most probably know maintained that the statement “all things are made of atoms” contains the most information in the fewest words. For what it’s worth, I think you’re doing a splendid job of explaining complicated matters to the interested layman.

  3. Steve Fitzpatrick says:

    Roy,

    Thanks. I grow tired of explaining it over and over to “die-hard skeptics”. Now I can just suggest they read this post!

    • Anonymous says:

      I think you should read the post from the physicist Joss below.

      One of the problems that has come to light is that Trenberth was using 2D equations (used for 2D models of the atmosphere) while calculating the energy budget of the atmosphere using 3D models. This gave more warming than if he had used the correct 3D equations. In the 2d equations, there is back ratiation, but NOT in the 3D equations as the back radiation is cancelled out.

  4. Chuck L says:

    Dr. Spencer, Thank you for the clear and understandable discussion of two misconceptions about the laws of thermodynamics as related to the greenhouse effect. Can you discuss the other point which is always discussed about CO2′s heating the atmosphere is logarithmic, i.e. the doubling of CO2 causes a decreasing amount of heating as the total amount of CO2 increases in the atmosphere. For the record, I do not think man-made CO2 is the main reason that global temperatures have warmed.

    Thank you.

    • I looked at this before. While the point is true, it is already taken into account in the theory of radiative forcing. Unless I am mistaken, I don’t think this factoid affects how the IPCC has reached their conclusions.

  5. adriaan says:

    Dear Dr. Spencer,

    I think you have made several mistakes in this model. First of all geometry issues with a plate in a circular space pose some complicated problems. If a second plate is close to a heated plate, the second plate will assume the same temperature as the heated plate, and start to emit as if it were the heated plate. Take the infinitesimal small distance between two plates, one heated and the other not, they will assume the same temperature when the distance becomes zero. So your model is wrong with 160F for the heated and 100 for the cold plate. The heated plate will not heat up, the cold one will heat up until it is radiating exactly the same amount and spectral profile of radiation. And that is also overlooked: the spectral distibution of the radiation emitted by a plate at 100 and 160 F are different, with longer wavelengths more abundant in the colder plates’s spectrum, containing a lower total energy. So during the time that is needed to reach equilibrium, the hot plate may get hotter, but the total energy flow will remain constant. At the end, the second plate is as hot as the emitter originally was (In absence of the specific heat of each plate, which neglected in my reasoning).

    Take a light bulb, run it with specified voltage and measure power output and color temperature. Enclose it with a 100% reflective sphere, at 0F. Run it again and measure output and color temperature. In this case you have 100% back radiation with exactly the same spectral distribution. Your bulb will not be burning brighter and more white (ie at higher temperatures).

    But I may be wrong. I have always read your blog with the utmost interest for your challenging ideas.

    Thanks for that, and please keep up the good work!

    • (1) geometry of the vacuum chamber (which are usually made cylindrical, but there are box-shaped ones, too) does not affect the qualitative point being made, nor does the shape of the object being heated.
      (2) the second plate will not reach the same temperature as the first plate…even if the two are bolted together! That’s because the heater is in the first plate. The plates would have to be perfect heat conductors, which nothing is.
      (3) yes the wavelength distribution at which the two plates radiate will be somewhat different. But this does not affect the conclusion, either: A hotter plate emits more IR than an identical, cooler plate. Period.
      (4) I’m not sure what your light bulb conundrum has to do with the issue at hand.

      • Jere Krischel says:

        For #2, what if the first plate simply had more mass? Since it isn’t a perfect conductor, is the assertion that more mass given a constant surface area will increase the temperature of the plate?

        • we are getting lost in details…the actual temperature the 2 plates equilibrate to will depend upon many factors, including the mass of the plates, their thermal conductivity, their specific heat capacity, their geometry, the relative positions of the two plates to each other, etc.

          These are all things that some engineers deal with on a daily basis…building thermal models. And yes, they must include “back radiation” from cooler objects to warmer objects to get good answers.

  6. JohnWho says:

    Dr. Spenser:

    You said “The plate will eventually reach a constant temperature (let’s say 150 deg. F.) where the rate of energy gain by the plate from electricity equals the rate of energy loss by infrared radiation to the cooled chamber walls.”

    My question – Is 150 deg. F the maximum temp that plate could reach based on the amount of electricity being used to heat it?

    If so, then clearly some other factor is required to make that plate hotter than 150 deg. F.

    If not, then what is the hottest that amount of electricity can heat that plate? In your example the only heat source (the way I understand it) is the “electricity”. Until that plate reaches a temp higher than the highest temp provided by the heat source (electricity), is it being heated by any of the cooler objects or are they simply keeping it from cooling?

    I hope I’ve explained my thoughts clearly.

    • If I understand your question, then no, the maximum temperature the plate could reach for a given energy input has not been reached. The more you can reduce the rate of energy loss to the cold walls, the hotter the plate will get. Yes, the surrounding objects act to control the rate at which the plate can lose energy. I have no idea what happens if you can keep the plate from losing energy at all….I suspect the heater wire melts (Ha!)

      • Anonymous says:

        The ‘core’ of the heater, the electrical wire elements, are glowing at a far higher temperature than what you measure for the heater on the outside. The wires of the heating coil can reach temperatures of up to 1000 C.

        So. Yes, you are right that the wire element will melt if heat is not removed from the heater coil.

        But, sadly, the heating of the second plate is not hotter than the heating wire of the heating core, your argument falls flat, as heat is indeed flowing from the hot centre core to the second plate.

  7. Bomber_the_Cat says:

    What you say is essentially correct. A hot object will cool down more slowly in a warm room, than in a cold room. This is because heat flows in both directions (this applies to conduction as well as to radiation). However, the NET flow of heat will always be from the hot object to the cold object, at a rate proportional to the difference in their temperatures.
    Because this subject is potentially very confusing, it is important to be rather precise in the language used to describe this process. In your example of the plate in a vacuum chamber, to say that the heated plate ‘cools down more slowly’ when the second plate is introduced is not precise and will confuse. The heated plate will continue to lose heat at the same rate as before. In fact, since it increases its temperature from 150 deg. to 160 deg., it will lose heat even faster than before ( the rate it loses heat, by radiation, is proportional to the 4th power of temperature). The reason it gets warmer, is not because it loses heat less quickly, but because it is getting back radiation from the second plate; this represents an additional heat input to the first plate.

    The diagram of radiation coming from the Sun is not helpful. Is it implying that no infra-red radiation comes from the Sun? It also implies that the greenhouse effect depends on the temperature of gases in the atmosphere.

    Kirchov’s Law pertains to blackbody/greybody emmisivity. This is not relevant to greenhouse gas theory. If it were, then oxygen and nitrogen would be much more important than any trace gases, because they are at the same air temperature, have similar emissivities and are much more abundant. The emmisivity of gases is very very low. This is why an infra-red camera in a police helicopter is not ‘whited-out’.

    • Anonymous says:

      The greenhouse effect depends on a sub-molecular vibrational property of the CO2 molecule. CO2 molecules absorb infra-red energy in quantum packets called photons. The photons absorbed are those in the 14 to 16 micron wavelength band. This absorption band is significant because the Earth emits it’s own radiation in a spectrum typical of a blackbody at, say, 300 deg.K. The peak emittance of a body at 300 deg.K is at about 10 micron, i.e. close to the CO2 absorption band. The reason that CO2 absorbs these photons and not those at other wavelengths is that the photons between 14 and 16 microns have precisely the right amount of energy to raise the CO2 molecule to a higher energy state. It thus ‘captures’ that photon and raises itself to what is called an ‘exited’ state. It subsequently releases that quanta of energy as another photon, at the same wavelength, and returns to ‘ground’ state. It does this very quickly, within nanoseconds. The direction of re-emitted photon is spherically random. Sometimes it will be re-emitted spacewards and sometimes the photon will be re-directed back to Earth. The latter keeps the surface warmer than it would otherwise be.
      If the CO2 molecule collides with another gas molecule before it emits its captured photon, the energy is converted instead into kinetic energy (heat) and the molecule is said to be ‘thermalised’. It can no longer emit the photon (and therefore doesn’t contribute to the greenhouse effect). If it emits its photon first, before collision, this obviously does not result in any temperature change, since the energy has ‘gone’. The rate of absorption/emission by this mechanism is not temperature dependent. If the molecule doesn’t absorb a photon it cannot re-emit it! The rate at which the CO2 emits its photons therefore, is dependent on the rate at which it receives them in the first place, not on its own temperature. The back-radiation to Earth by this mechanism is characterised by having wavelength of about 15 microns. The energy It is not distributed over the spectrum that would be expected from a blackbody at the gas temperature.

      I see from the later postings that you are having difficulty convincing people that the addition of the ‘passive’ plate in your model will result in the first plate becoming warmer. You are quite correct of course. And as you say, there are many everyday examples all around us. Some people just don’t get it.

      • Rob says:

        “If the CO2 molecule collides with another gas molecule before it emits its captured photon, the energy is converted instead into kinetic energy (heat) and the molecule is said to be ‘thermalised’. It can no longer emit the photon (and therefore doesn’t contribute to the greenhouse effect).”

        But it heats the atmosphere! It transfers the absorbed light into the kinetic energy of another molecule (such as N2 or O2, which are transparent in the IR), which is a net heat gain. The point is that light from the sun is absorbed by the surface of the earth, which re-radiates it as longer wavelengths, where it is absorbed by CO2 or CH4 or H2O. This absorption can result in re-radiation in a random direction, thereby transfer the absorbed photon, or the photoexcited greenhouse gas can transfer its energy by collision, causing a net increase in the temperature (kinetic energy) of the atmosphere.

        Sunlight heats the ground which radiates in the IR region to CO2 which either re-radiates this energy or transfers it to other IR transparent molecules, resulting in a net conversion of light into heat. Without the greenhouse gas, the IR radiation from the surface of the earth is lost back into space.

  8. I agree that the concepts can get muddled with ambiguous words like “heating” and “cooling”. I’m sure I could have been more precise.

    Yes, I realize the IR properties of gases is different. But, to the extent something is an infrared absorber — whether a solid or CO2 or water vapor molecules scattered through the atmosphere — it is also an infrared emitter. Depending upon the wavelength, this can be very weak or very strong.

    And, yes, the greenhouse effect DOES depend upon the temperature of the gases in the atmosphere, in the sense that it changes the temperature. Generally speaking, the rate of absorption is not temperature dependent, but the rate of emission is very temperature dependent.

  9. Mark Hugo says:

    I really don’t like arguing with Dr. Spencer.

    I think he has it “all together”. But I would like to point out that Dr. Elsasser’s “Heat Transfer By Infrared Radiation In the Atmosphere” is sort of the “modern technological basis” for atmopsheric energy analysis.

    Like Einstien’s 1905, “On the Electrodynamics of Moving Bodies”, the work of Elsasser is not “unique”, but rather a concise summary of the 123 references he cites at the end of his 107 “masterpiece”.

    Perhaps the most impressive result from Dr. Elsasser’s work is the “Atmospheric Radiation Chart”, from which day to day heat up and cool down rates (and temperature profiles) can be calculated.

    On page 23 he says this in regard to the usage of the charts and their derivation: “It may be noted that since the flux in the carbon dioxide band is equal, at any level, to a definite fraction of the black body radiation corresponding to the temperature at that level both in the upward and downward direction, the RESULTANT flux of th carbon dioxide vanishes in the approximation of the chart. This is a fair approximation to the truth in the lower atmosphere (for the upper atmosphere, see section 12/of the 107 page paper by Elsasser)..”

    Now granted, Dr. Elsasser does say it’s “in the realm of the approximation” for his “day to day” work. So one has to ask “how close is the approximation”?

    However, when one considers the work of Plass and others (about 1955 vintage), and Dr. Elsasser’s section 12, which show CO2 as a (on balance) UPFLUX agent in the Stratosphere, and thus a net COOLING AGENT…I for one, am mystified where the “tropospheric warming” effect of CO2 comes from.

    Perhaps someone needs to precisely analytically exposite the reason CO2 is, in contemporary times, a net downflux agent?

    Respectfully:

    Mark Hugo, P.E.

    • In Hugh Elsasser’s quote you provided, he only mentions that the atmospheric CO2 layer as SOURCE of IR has no net directional preference, which is true. But his statement is a non sequitor, because he does not address the effect of the layer as an ABSORBER. BOTH are necessary to say anything about temperature.

  10. After reading the paper by Gehrlich published in the International Journal of Modern Physics in March 2009, I am struck by Dr. Spencer’s inability to explain away the fact that the principal method of heat exchange in the atmosphere isn’t radiation, but convection. In fact, Wood’s experiment in 1909 showed that greenhouses (the ones made out of glass) don’t work due to radiation filtering at all—but entirely due to convection. (Read the paper to understand Wood’s experiment.)

    No one is doubting that perhaps radiation has something to do with the temperature of something or other. What those who claim the Greenhouse Effect is bogus are saying is that radiation is irrelevant because its effect is too small to be measured. Only convection really matters when talking about heat being transferred from the ground to the atmosphere. This is why, as Mark shows, increasing CO2 in the atmosphere is measured to COOL the earth, rather than raise it.

    Those who wish to defend the Greenhouse Effect must stop talking about radiation and start talking about convection, or rather, why convection isn’t so important as everyone knows it is. As long as the earth and the atmosphere transfer heat through convection, we need to consider this method of heat transfer.

    • Jonathan, I really don’t think you read (or understood) what I wrote. No one is claiming convection isn’t important, and I do discuss its role. Yes, a real (glass) greenhouse shows the importance of convective heat transfer…I even say what the average effect is in degrees of surface temperature for the atmosphere with and without convection.

      If you want to object to something specific I said, then do so. But don’t assume you know what I said based upon your preconceptions.

      • Anonymous says:

        Dear Roy,
        Your inappropriate choice of an analogy that presides in a vacuum chamber where only radiation, of the 3 modes of energy transfer, occurs cannot pass without comment. It is either deceitful sleight of hand or a clumsy choice of thought experiment. I choose the latter because I hold you in esteem.

        I’m extremely grateful to you as one of the ‘folk’ you have kindly entertained by emails. You deserve respect for openly addressing this now. However, I am most frustrated at your continued avoidance of my emailed first question to you:

        Why are you giving a ‘free pass’ to the unphysical concept of ‘back radiation’ when no serious scientist entertains the notion of ‘back convection’ or ‘back conduction’?

        As we all know from Lindzen and Choi (2009) ERBE satellite data shows no evidence of any back radiation signal and Earth is not a vacuum chamber.

        There are now a growing number of eminent international scientists coming out to disagree with you on this incongruity of thinking. Is it not time to apply Occam’s Razor and dump a hypothesis that requires a ‘free pass’ to uphill radiation but not conduction and convection, and is proven to have no real world atmospheric signal.

        Please kindly address my question and show us in the laws where radiation transport has different rules to conduction and convection and hereinafter try to forego disingenuous vacuum-based analogies as it ill behoves you.

        • Your inappropriate choice of an analogy that presides in a vacuum chamber where only radiation, of the 3 modes of energy transfer, occurs cannot pass without comment. It is either deceitful sleight of hand or a clumsy choice of thought experiment.

          But radiation IS the only mode of energy transfer between the Earth and outer space. I use it to make a basic point regarding *radiation*. It is not meant to be a model of all the inner workings of the climate system.

          … I am most frustrated at your continued avoidance of my emailed first question to you:
          Why are you giving a ‘free pass’ to the unphysical concept of ‘back radiation’ when no serious scientist entertains the notion of ‘back convection’ or ‘back conduction’?

          Hmmm…how can I argue against such iron clad logic? :)

          As we all know from Lindzen and Choi (2009) ERBE satellite data shows no evidence of any back radiation signal and Earth is not a vacuum chamber.

          (1) Lindzen and Choi (2009) did not specifically address back radiation.
          (2) I never claimed the Earth is a vacuum chamber…but it is located in one! :) In case you missed this detail, the second bar in the experiment represents the IR-absorbing atmosphere placed between the heated Earth and the vacuum of outer space.

          There are now a growing number of eminent international scientists coming out to disagree with you on this incongruity of thinking. Is it not time to apply Occam’s Razor and dump a hypothesis that requires a ‘free pass’ to uphill radiation but not conduction and convection, and is proven to have no real world atmospheric signal.

          I actually a fan of Occam’s Razor, but not if it requires us to ignore things we know that happen, and which cannot be explained without the concept of back radiation.

          Please kindly address my question and show us in the laws where radiation transport has different rules to conduction and convection and hereinafter try to forego disingenuous vacuum-based analogies as it ill behoves you.

          My post already addresses this…did you read it?

        • Anonymous says:

          Roy it would appear that option one, “deliberate slight of hand” be the most likely in this case.

          “But radiation IS the only mode of energy transfer between the Earth and outer space.”

          You know well enough that this poster is referring to your inappropriate analogy that resides in a vacuum when what we are actually discussing is whether “back-radiation” can actually occur in a fluid where there is convection. Whether or not back radiation can occur in a vacuum is neither here nor there with regards to this discussion, (although a word of advise, when you discover a perpetual motion device, the trick is usually to keep it under your hat until the patent has been obtained!).

          As you are well aware, this is a discussion about whether “back-radiation” can occur in the Earths atmosphere where convection will overwhelm any “back-radiation” in your analogy if it were set in the atmosphere (which is clearly why you have set it in a vacuum). You then deceptively and falsely state, “But radiation IS the only mode of energy transfer between the Earth and outer space” while deliberately ignoring over 600 km of atmosphere.

          So I ask you this, what is the point in discussing anything with you if you’re not even going to be honest ? ? ? ?

          “Back-radiation” in the atmosphere is overwhelmed by convection and is refuted by simply lowering a lit match to the back of your hand. Then try the same test with your hand over the match and notice the difference.

          So it is clear why you always use analogies set in a vacuum when discussing atmospherics.

          Just as it is clear why you pretend not to notice when someone points that out to you Roy.

  11. Annabelle says:

    Surely it should be possible to set up a real experiment to test the “thought experiment”?

    • Of course it would be possible.

      • Anonymous says:

        Dr. Spencer,

        Are you aware of any physical experiments done with regard to GHG since the Woods experiment in 1909? (MythBusters does not count) And if not, what is preventing someone from carying out such an experiment?

        I feel that not doing physical experiments in this field is scientific malpractice by omission.

        JT

  12. Peter D says:

    Where could I buy a house heating system based on this principle. From your comments, I would need very little input energy to heat up a decent house in winter. I could cut my heating bills dramatically. Why don’t builders use this system?

  13. John Marshall says:

    Energy radiated from the earth’s surface is adsorbed by the atmosphere. This radiated energy is in one direction, upwards, but energy re-radiated by the atmosphere is radiated in all directions so very little gets back to the surface. Most gets back to space. Your plate experiment looks reasonable but the second plate will reflect some energy back to the heated plate because it is not a black body radiator and so cannot adsorb all radiation falling onto it. Some must escape and in so doing will increase the heated plate temperature slightly, which will radiate more energy again. It does not act as if it is manufacturing energy out of nothing like a PPM which the 2nd law states is impossible!
    The Trenberth paper’s graphic showing the so called heat budget is not in balance so violates both laws. Temperature measurements of the atmosphere show standard adiabatic cooling with no heat input from re-radiation. CO2 does adsorb IR but will loose it just as quickly as Kirchoff tells us. It will also transfer energy to the other atmospheric gasses through conduction so heating them up, it cannot fail to do so as this is one way in which it will loose the energy.
    The theory of GHG’s fails through violating the laws of thermodynamics never mind all those thought experiments.

    • Energy radiated from the earth’s surface is adsorbed by the atmosphere. This radiated energy is in one direction, upwards, but energy re-radiated by the atmosphere is radiated in all directions so very little gets back to the surface. Most gets back to space.

      How can this be, John? You say a layer of the atmosphere emits as much downward as it does upward, but “little gets back to the surface”?

      Your plate experiment looks reasonable but the second plate will reflect some energy back to the heated plate because it is not a black body radiator and so cannot adsorb all radiation falling onto it. Some must escape and in so doing will increase the heated plate temperature slightly, which will radiate more energy again. It does not act as if it is manufacturing energy out of nothing like a PPM which the 2nd law states is impossible!

      First of all, atmospheric layers do not “reflect” IR energy…only absorb and transmit. Second, even if there is reflection, it will cause the temperature of the warmer object to rise even more. There is no energy “manufactured”.

      The Trenberth paper’s graphic showing the so called heat budget is not in balance so violates both laws.

      Yes, the Kiehl & Trenberth energy budget DOES balance…go add up the numbers yourself.

      Temperature measurements of the atmosphere show standard adiabatic cooling with no heat input from re-radiation.

      The temperature profile of the *troposphere* shows that convective heat transport is important. As I state in my post, it is radiation that destabilizes the troposphere so convection can occur in the first place.

      CO2 does adsorb IR but will loose it just as quickly as Kirchoff tells us.

      And a blanket over your body loses heat to its surroundings as fast as it gains it from your body…but it still makes your body warmer, doesn’t it?

      It will also transfer energy to the other atmospheric gasses through conduction so heating them up, it cannot fail to do so as this is one way in which it will loose the energy.

      Actually, since air is a very poor conductor of heat, the role of conduction in the atmosphere is very minor.

      The theory of GHG’s fails through violating the laws of thermodynamics never mind all those thought experiments.

      Not, it does NOT violate these laws! John, I think you did not read what I wrote.

  14. Jack Savage says:

    Am I alone in wondering how this (fundamental!) point of physics seems to be a matter of a certain amount of debate in the climate world?
    It is a minefield out there for us laymen, that is for sure.
    Thank you for an exposition which, although the latter half of it goes over my head, has cleared up a matter (Global Warming breaks the Second Law of Thermodynamics shock horror) which has been puzzling me.
    Whilst I am quite content with the idea that there seems to be no evidence out there that we are all going to fry any time soon, I was never comfortable with the idea frequently put forward that changing the constitution of our atmosphere, no matter how slightly, would have absolutely no effect on the physics of the temperature of the planet.
    To carry forward the blanket analogy, even the slightest difference in the textile mix in the blanket would have an effect, although possibly infinitesimal, on how warm it kept the sleeper.

  15. Woodsy42 says:

    Thank you for that explanation, it accords with everything I learned at school years ago. I had noticed in the blogosphere that many people keep repeating the mantra ‘heat cannot travel from cold to hot’. It can’t of course, but people don’t seem to understand that radiation travels both ways while ‘heat’ represents the ‘net difference’.
    But I’m not a high powered physicist so is it fair to try and summarise in very simple language thus:
    Adding a second bar in your experiment means that the energy travels around between the objects and some is ‘held’ in the cooler one, therefore less energy is lost from the system (the jar). This makes the ‘balance’ of energy in the jar a bit higher than otherwise because the energy hangs around longer. Basically it’s acting as energy storage.
    Even so I am not convinced as an AGW argument because:
    (i) Unlike your experiment our atmosphere is not full of iron bars or plates. Your illustration works for any heat retaining molecule in the atmosphere and is not specific to CO2.
    (ii) The opposite happens when the heat is turned off, just as your system will COOL when the cold second piece is first added.
    (iii) While accepting that the 2-bar system results in more energy being retained I don’t see why that must necessarily mean a higher temperature.
    Interesting discussion though :-)

    • Even so I am not convinced as an AGW argument because:
      (i) Unlike your experiment our atmosphere is not full of iron bars or plates. Your illustration works for any heat retaining molecule in the atmosphere and is not specific to CO2.

      Actually, the small amount of greenhouse gas molecules do virtually all of the IR absorption and emission, but they act as little heater/coolers that very rapidly heat or cool the rest of the air around them. This happens extremely rapidly, explained through the kinetic theory of gases.

      (ii) The opposite happens when the heat is turned off, just as your system will COOL when the cold second piece is first added.

      Just as when the heat is turned on, when the heat is turned off, the hotter bar will cool more slowly than if the 2nd bar was not there.

      (iii) While accepting that the 2-bar system results in more energy being retained I don’t see why that must necessarily mean a higher temperature.

      If more thermal energy is retained in a mass with a certain heat capacity, then yes, the temperature MUST be higher. Wouldn’t want to violate the 1st Law of Thermodynamics, you know. Stiff penalty.

  16. Peter Hsu says:

    Dear Dr. Spencer:

    I am a big fan of your site, but this article left me wondering. Just some questions:

    Shouldn’t the “Chiller Chamber” walls be at absolute zero (-459.67 degrees Fahrenheit) as opposed to 0 degrees F.?

    Would not bodies (in your example the Chiller Chamber) at 0 degrees F would still be radiating energy towards the heated bar (and thereby raise the temperature of said bar)?

    Your everyday example of the blanket leaves me slightly cold.

    I’ve always thought that blankets reduce heat loss by interfering with air convection (cooling mechanism) around the body, and trapping moister warm air, and hereby reducing perspiration, another cooling mechanism.

    If you believe that the back radiation provided by a wool blanket is enough to keep you warm (or even raise your temperature), would not a lead blanket would keep you even more toasty. Surely lead’s greater mass will provide even more back radiation than wool.

    Yours respectfully

    Peter Hsu

    • Shouldn’t the “Chiller Chamber” walls be at absolute zero (-459.67 degrees Fahrenheit) as opposed to 0 degrees F.?

      It doesn’t matter what temperature they are at to illustrate the basic concepts….just that they are colder than the 2 bars.

      Would not bodies (in your example the Chiller Chamber) at 0 degrees F would still be radiating energy towards the heated bar (and thereby raise the temperature of said bar)?

      Yes, and I alluded to this in the article.

      I’ve always thought that blankets reduce heat loss by interfering with air convection (cooling mechanism) around the body, and trapping moister warm air, and hereby reducing perspiration, another cooling mechanism. If you believe that the back radiation provided by a wool blanket is enough to keep you warm (or even raise your temperature), would not a lead blanket would keep you even more toasty. Surely lead’s greater mass will provide even more back radiation than wool.

      Yes, the specific mechanisms of heat transfer with a wool blanket around your body are mostly different (radiation plays only a minor role). But maybe you are missing the point…anti-greenhouse people make the generalized claim that a cooler object placed near a warmer object can not result in a rise in temperature of the warmer object. I’m showing that, energetically, that is not true.

      • coturnix19 says:

        the greenhouse people are right… you forget that when you place ‘cool’ object near hot object, you actually remove even colder object (the walls) that had been there in the first place. So you are saying right thing but formulate it wrong way. GHPeople say wrong thing, but formulate it right way =)

      • coturnix19 says:

        to clear things, lets do it this way: suppose the heated plate has temp of 150F and walls have temp of 0F. Would the plate warm if another plate, with temp of -100F were placed between warm plate and walls? Energy flux from the cool plate to the hot plate would be times lower than from the walls that were there before screened by cold plate. Therefore, the net flux would be towards cool plate, and warm plate would be cooled by a really cold object. So you see, cold object can only cool warm object (at least in given set-up, i am not trying to understand how non-blackbodiness affects this, i am not up to it yet), your fallacy was that you place a warmer object than there had been before.

        So.

        Can cool object placed next to warm object warm the warm object even more? No. Unless you put it in place on even colder object that ‘dbeen there before.

        In case of atmosphere, cold upper atmosphere screens lower one from even colder (~3K) space above.

        • Can cool object placed next to warm object warm the warm object even more? No. Unless you put it in place on even colder object that ‘dbeen there before.

          EXACTLY! There is no fallacy in what I presented. You, too, are in fact agreeing with me.

  17. Lionell Griffith says:

    One small point. ALL the energy that is said to be back radiated from the second surface came from the first surface. If that is the case, the first surface no longer has that energy. ALL that is happening is the energy the back radiated energy simply replaces the energy that was lost. Its rather like 2 – 1 = 1 for the first radiative event and 1 + 1 = 2 for the back radiative event. Nothing is added to the system and therefor no temperature increase.

    What is being presumed for ALL back radiation increases temperature arguments is that 2 – 1 = 2 and 2 + 1 = 3. If that were the case, one could draw infinite energy from a system built according to Dr. Spenser’s thought experiment. One must go a bit more deeply into the matter than simple superficial “common” sense arguments.

    Rather than all the verbal hand waiving and rhetorical word twisting I suggest performing an actual experiment to see what actually happens. For example, a mirror reflecting light that has been reflected from a surface back to the surface does NOT increase the brightness of that surface and, therefor, does not increase its temperature. Try looking into a mirror, does your face warm because of the back radiation increasing its temperature?

    Back radiation does not add, it simply replaces that which was previously lost. The so called greenhouse effect that is presumed to work by back radition is nothing but BS!

    • One small point. ALL the energy that is said to be back radiated from the second surface came from the first surface. If that is the case, the first surface no longer has that energy. ALL that is happening is the energy the back radiated energy simply replaces the energy that was lost. Its rather like 2 – 1 = 1 for the first radiative event and 1 + 1 = 2 for the back radiative event. Nothing is added to the system and therefor no temperature increase. What is being presumed for ALL back radiation increases temperature arguments is that 2 – 1 = 2 and 2 + 1 = 3. If that were the case, one could draw infinite energy from a system built according to Dr. Spenser’s thought experiment. One must go a bit more deeply into the matter than simple superficial “common” sense arguments.

      Lionell, what you are forgetting is that while all of this is happening, energy is still being pumped into the first bar with electricity. You are talking about conservation of energy, which is not being violated as you imply. I believe what your example means is that if we turned OFF the heater, then added the second bar, the temperature of the first bar could not go up, which is true…it would just slow the rate of cooling of the first bar. But critical to the whole experiment is that, like the sun heating the surface of the Earth, there is energy being continuously pumped into the system from outside.

      For example, a mirror reflecting light that has been reflected from a surface back to the surface does NOT increase the brightness of that surface and, therefor, does not increase its temperature. Try looking into a mirror, does your face warm because of the back radiation increasing its temperature?

      Actually, yes, if the mirror is replacing something dark, it will cause your face to warm slightly. Imagine you are surrounded by darkness, and your face is illuminated by a point light source. If a mirror is placed so that some of the light reflected off your face is reflected back again toward your face, it will brighten your face and warm it slightly.

  18. Joss says:

    “The 2nd Law of Thermodynamics: Can Energy “Flow Uphill”?
    In the case of radiation, the answer to that question is, “yes”. While heat conduction by an object always flows from hotter to colder, in the case of thermal radiation a cooler object does not check what the temperature of its surroundings is before sending out infrared energy. It sends it out anyway, no matter whether its surroundings are cooler or hotter.”
    ________________________________________

    Roy, I do appreciate your efforts but I am sorry to disagree with most of the above description dealing with the second law and the energy going “uphill”.
    This never occurs in your experiment because, in fine, both the first and second plate are heated by the electric device which is an external source of heat.

    Once again, you go on with your statement about the “photon that does not stick its finger outside before leaving”.
    IMHO, this is not true.

    Indeed, Clausius was a smart guy. He said it all in a single sentence. :

    “Again as regards the ordinary radiation of heat, it is of course well known that not only do bodies radiate to cold, but also that cold bodies radiate to hot; nevertheless the general result of this simultaneous double exchange of heat always consist, as is established by experience, in an increase of the heat in the colder body at the expense of the hotter.”
    Rudolf Clausius “ The Mechanical Theory of Heat “ 1879.

    Here the crucial words are “simultaneous double exchange”.

    This means that your descriptions using incident photons and asking questions about what they do (or do not) are somewhat irrelevant.

    Dealing with photons, or photon streams, or “infrared energy” necessarily implies a sequential, step-by-step description of the process, which is inappropriate when dealing with a “simultaneous double exchange” in thermodynamics.

    In this case, it is much safer to use an electromagnetic wave description of this “simultaneous double exchange”. This would avoid bizarre (and silly ?) questions about photons requesting permissions, looking around etc. (which circulate here and there).
    Actually, due to radiative interaction, two bodies in empty space, know about each other because of the electromagnetic field.
    And yes, electromagnetic waves do know where they are going to and where they come from !
    I guess, photons too. But who knows ? until we are able to lay down the correct equations for the photons (implying momentum conservation, too).

    Explaining thermodynamics with photons as is usually done by climatologists, is misleading.

    Gerlich and Tscheuchner state it differently :

    “It cannot be overemphasized that a microscopic theory providing the base for a derivation of macroscopic quantities like thermal or electrical transport coefficients must be a highly involved many-body theory. Of course, heat transfer is due to interatomic electromagnetic interactions mediated by the electromagnetic fi eld. But it is misleading to visualize a photon as a simple particle or wave packet travelling from one atom to another for example. Things are pretty much more complex and cannot be understood even in a (one-)particle-wave duality or Feynman graph picture.”

    All the best (a french physicist)

    • Clausius was a smart guy. He said it all in a single sentence. : “Again as regards the ordinary radiation of heat, it is of course well known that not only do bodies radiate to cold, but also that cold bodies radiate to hot; nevertheless the general result of this simultaneous double exchange of heat always consist, as is established by experience, in an increase of the heat in the colder body at the expense of the hotter.”
      Rudolf Clausius “ The Mechanical Theory of Heat “ 1879.

      Yes, if you put a warm body next to a cold body as an energetically closed system, the temperature of the warmer body cannot increase. But what you are forgetting here, Joss, is that there is an external source of energy that is being continuously pumped into the warmer body. Clausius’s statement only applies to the former case. Because solar energy is being continuously pumped into the Earth’s surface, the addition of a “radiatively insulating” layer around the Earth will cause the surface to become warmer than if the layer was not there.

      • Vangel says:

        I must make a confession here. Although I am an engineer by training I have not really solved any thermodynamics problems for 30 years so I am quite rusty and may miss something.

        But I think that you are missing the argument that was being made. Let us look at the Clausius sentence again:

        “Again as regards the ordinary radiation of heat, it is of course well known that not only do bodies radiate to cold, but also that cold bodies radiate to hot; nevertheless the general result of this simultaneous double exchange of heat always consist, as is established by experience, in an increase of the heat in the colder body at the expense of the hotter.”

        I agree with the statement above and suggest that you have not addressed it properly. You claim that CO2 allows energy to be added into the system by preventing the radiative transfer that would take place without the CO2 absorption and re-emission. This seems to make a great deal of sense but thermodynamics is not that simple and what seems common sense often isn’t.

        For one, the extra little bit of warming due to the re-emission of energy from the CO2 layer would cause outward emissions to increase and the cooling rate to increase.

        Unless you can show that the amount of energy re-radiated back to the first bar is greater than the increased net outflow of radiation caused by the claimed warming you can’t really support the claim that you just made. I believe that Clausius is implying that when you have a simultaneous double exchange of heat the colder body does not make the warmer body hotter but am willing to consider a counter argument if one is provided.

        • Clausius’ statement is not violated by my example. The second, cooler bar does indeed have its heat content raised at the expense of the hotter bar. BUT…since the hotter bar has an unlimited source of energy (albeit flowing in at a constant rate), the Clausius statement does NOT prevent the temperature of the hotter bar from increasing.

          • Vangel says:

            Actually, your have not established that example does notviolate the statement. You have yet to demonstrated that the reradiated energy will cause the temperature to go up because if it did go up the amount of energy radiated outward would increase above the levels that existed before the second bar was added. I suspect that it would not and that the amount of energy radiated outward from the two bars would be the same as it would have been without the second bar being in place.

  19. AverageIQ says:

    “How Cooler Objects Make Warmer Objects Even Hotter”???

    Shouldn’t that be: – “How HEATED Objects Make Cooler Objects Warmer” ?

    The title confuses cause and effect.

  20. Joletaxi says:

    “I agree that the concepts can get muddled with ambiguous words like “heating” and “cooling”. I’m sure I could have been more precise.”

    Let it put straight :your explanation is plenty right, but maby the “wording” is inapropriate.

    The energy budget off the warming plate,with the second plate placed against her will be disrupted.The flow off electricity rermaining constant, and receiving an small amount off photons from the second plate, the only way too maintain his energy budget is the t° too go up…. off, build up his surface(I will come back with that.)
    Secondly, you compare the atmosphere as un insulating layer.That an other way too confuse.
    Iff You intend too insulate your house, off a bottle of champagne off anithing,You will cover it with a layer who will disrupt the conductive process.. an eventualy will reduce the flow off photons going out.But the main process will be the disruptive process off conduction.Let think at the “bouteille thermos”.The iner bottle, made in the case off glass will perfectly conduct the flow off heat(in the case off an heated contend) This flow will be interrupted by the vacuum(never perfectly, because it will always be a partial vacuum) and the only outgoing energy will remain the flow off radiative energy,the inner vacuum chamber can’t stop or reduce this flow.So, with this example, We can see that de radiative budget off the thermos bottle will be… peanuts.
    Let ‘s go too the earth
    The most important place for the incoming energy off the sun too be captured is between the 30 ° North and South off the equator.And in this area, there is about 80 % off water.
    And in this area,trade wind are pushing the huge amounts off latent energy that come from evaporation toward the central equateur named in french :le pot au noir”Where continuosly,that latent energy is propelled in the upperf athmosphere(talk about that too the french pilot off Air farnce with the late airbus los)
    The water in this area can be relatively hot, but always cooler that the atmosphere surrounding, and the upper layer off water is even cooler with the evaporation process going on.Again, we can’t calculate this, but we can guess that the radiative process in the part off the earth remain… peanuts
    In the upper latitude,in the South, there mainly water, with the same, but weaker process going on.In the North, all will depend off wath is the use off the surface(Dr Pileke is here absolutely right)
    In the pole… peanuts…
    So one off the main process too aquilibrate the earth energy budget will be evaporation, convection.. and of course, the sea currents but that an other story.
    No, all will depend what is going on in the upper athmosphere, where hugue amounts off energy are propelled by the continuous process in the pot au noir,.
    Maby that an other reason they can’t find the “hot spot” in this area?
    In these surrounding, we betray in “terra incognita”
    hugue amounts off energy will flow out,depending off the contain off water vapor, and GHG,and also the t° off the surroundin stratospere.
    Iff the t° off this layer,where the chemistery off the gases is poorly know,is buiding up, we can guess that it will expand,or be mixed with the lower layer off the stratosphere?
    It also evenly probable, that this intermediate layer is not an smooth surface, but will bump out,or expand at some places depending off the events in the lower layers.
    Certainly the GHG will affect in some places the adiabatic profile off the athmosphere, but that will remain an very small effect,constrained in small areas off the earth,and the main driver will always be the evaporation convection(and the sea currents) process.

    So, The main error is too remain focused on the radiative budget, when, in the same time, the main heat tranfer process is doing the job.

    What happen in the upper layers, that would be for me, the greatest inerest iff You can explain that, or make an article

    Would You apologize for my very poor english,and I would be delighet to post my essay in French

    Best regards

    PS Your article is off the greayest intered because the battle is raging here in france, about this matter

    • As others have pointed out, yes, convective heat transport is very important. But so is infrared radiation.

      • Joletaxi says:

        I’m not conviced off that,in the lower layers off the athmosphere.
        Think about the very hugue amounts off energy that can be tranfered by evaporation,condensation process,in the upper layers.
        Think about the very tiny radiative effect off water,in contrast with the hugue evaporation potential,multiplied by the convective effect.
        And water is the main “receptor” off sunlight.
        I ‘m not aware that anything in the very dry deserts have changed.At night, it can always freeze,and in day time, ground can reach very hugue t°,GHG have not affected very much this phenomene.
        By the way, in towns,on the contrary, the effect can be predominant, and surely at night, when there is atmost no wind.
        In the upper layers, it is for sure that the only mecanism is radiative, but,as I have attempted too explain,thing are quiete difficult too figure.

  21. Dusty says:

    Hi Doc,

    I’m sorry but my brain has overheated:

    Initial Conditions:

    The power is constant

    I have one active heat source within the jar at 150 F at equilibrium

    Add an entirely passive element

    Et voila, after a period of heat exchange a new state of equilibrium is reached but now I have two heat sources, the active one of which is hotter than under the initial conditions and the input power has not changed.

    So, continuing the logic, when I add an identical second passive element at a relatively identical position (wrt the active element). At the new state of equilibrium I will end up with 3 heat sources with presumably the active source being hotter still – and I still haven’t changed my input power.

    So presumably I can continue to increase the temperature of my active source and regulate it simply by adding or subtracting passive elements and without changing my input power

    I get the impression that there is something wrong with this picture.

    • Vangel says:

      I get the impression that there is something wrong with this picture.

      There is. The ‘warmed’ body increases the amount of energy it emits and cools at a faster rate. Dr. Spencer needs to support his claim by showing that the increase of energy outflow due to warming is less than the amount of energy that is reradiated back to the warming body. I am not certain that he can do that. So while his thought experiment is interesting, I do not believe that it is supported by real world experience.

  22. eilert says:

    Dr. Spencer your thought experiment violates the First Law of Thermodynamics, which basically states that you cannot increase the temperature of system without adding some extra energy to it.
    You increase your equlibrium temperature from one heated plate at 150 degrees to 260 degrees with two plates one of which does not add extra heat to the system. The whole system (both plates together) must stay at 150 degrees for it not to violate the first law. The equilibrium temperature is determined by the input energy to the system (either one plate or both) and the rate of cooling of this system to the outer chamber.

    • Dr. Spencer your thought experiment violates the First Law of Thermodynamics, which basically states that you cannot increase the temperature of system without adding some extra energy to it.

      Eilert, please pay attention. Extra energy *IS* being pumped into the system, continuously.

      • Anonymous says:

        Dr. Spencer

        You said the energy pumped into the system only will antain
        a equilibrium at 150 degrees, which is the level it could attain with the energy pumped into. If you do not increase the energy into the system your equilibrium temperature cannot, repeat cannot, increase. What you are really doing with this experiment, you reduce the flow of energy between the plates (until it is zero), but the flow of energy from both plates to the chamber walls does not increase.
        The temperature of the first plate cannot in anyway increase
        as long as the second plate has a lower temperature than the first. It cannot absorb any of the lower energy radiation from the second plate, which is the key. It can only deflect it, without being warmed.

        • Anonymous says:

          Thinking about this a bit more:
          While the second plate warms (before it reaches the equilibrium, which you say is 100 deg), the radiation to the chamber actually reduces and only increases back to the original, once it stops warming.
          Since you have a constant supply of energy, you could actually add the 150 and 100 together and reach 250, but you cannot reach the 260 you imply from your graphic, since the radiation from the second to the first plate, does not get absorbed by the first one. Otherwise it would
          violate the First and Second Laws (see other coment further down, about the absorbtion of radiation being the key to warming an object – radiation is just energy in transit).
          Merely by radiating infrared radiation towards a surface does not neccessary warm it. The energy level actually needs to be higher, for it to be able to be absorbed again (lower energy state cannot flow to an higher state without work – the basis of the Second Law) and cause further warming.

          However comming to the anology as the second plate being the greenhouse gasses:
          It in reality would be extremely thin, since there are so few gasses of this type in the atmosphere. Thus we would need only a tiny extra amount of energy to regain equlibrium (even less if you increase the the thin plate,
          after it has gained equilibrium, with an even thiner plate -the anology to the present increase in CO2)

          A different note, not actually to do with your experiment, but related:
          The reality is, the warming is just a misnomer. It is just a dampening effect, caused by the absorbtion capabilities and other processes (the combination of which is the real greenhouse effect) of the ocean, land surface and the atmosphere. The earth schould actually be far warmer during the day time, when radiation from the sun is available and far cooler during the night when not, without atmosphere, as Apollo measurements on the moon have shown. Just by looking at the greenhouse effect of the atmosphere misses many other possible dampening processes.

          By relying solely on the Stephen Boltzman black body equations, one cannot accuratly calculate the temperature
          of a body. They do to not take account of the fact that part of absorbed radiation actually conducts inwards and
          is released at a later stage, as Apollo 14 and 17 experiments on the moon, recently rediscovered, have shown.
          These experiments show that the day side is 20 degrees celcius cooler (confirming the fact some of the radiation
          is indeed conducted inwards and not radiated), but on the night side an even greater amount of 60 degrees was found.
          One could thus say the compostion of the moons surface cause a dampening effect on the temperature derived from
          Stephen Boltzman equations.

          • Since you have a constant supply of energy, you could actually add the 150 and 100 together and reach 250, but you cannot reach the 260 you imply from your graphic, since the radiation from the second to the first plate, does not get absorbed by the first one. Otherwise it would
            violate the First and Second Laws

            You are confusing the “supply” of energy (the electricity) with temperature. Temperature is not a function of energy inflow, but of the point at which energy inflow and outflow are balanced. The heat content of the system can rise, with a constant amount of electrical resistance heating, by simply insulating the system against loss. The same thing happens when you wear a coat and gloves.

            The reality is, the warming is just a misnomer. It is just a dampening effect, caused by the absorbtion capabilities and other processes (the combination of which is the real greenhouse effect) of the ocean, land surface and the atmosphere.

            Yes, exactly.

        • wrong! temperature is NOT determined by energy input alone…it is the result of the imbalance between energy input and energy loss. Until people understand that fact, they will forever remain confused about what causes the temperature — of anything! — to change.

      • Vangel says:

        This has not been established.

        You may be right if the ‘extra energy’ that is prevented from being radiated outright is greater than the energy emitted by the body from your claimed warming but you have yet to show that.

        For the record, I do not believe that is the case when you have a simultaneous double exchange. And when it comes to the practical effects we see very quickly that other mechanisms come into play that would make a surface warming due to CO2 increases very difficult if not impossible. This last conclusion makes sense to me because if you were right we would have positive feedback and the planet would have experienced a runaway temperature increase that would have made the current temperature levels extremely low in comparison to previous periods.

  23. Demesure says:

    “If you are unconvinced of this, then imagine that the second plate completely surrounds the heated plate. Will the heated plate remain at 150 deg., and not warm at all?”

    Dr Spencer, if you surround completely the 150°F heated plate with the 100°F plate, what will the new temperature of the heated plate.
    Please, give us a clue on how to calculate it, not an schematic warming of 10°F.

    • Dr Spencer, if you surround completely the 150°F heated plate with the 100°F plate, what will the new temperature of the heated plate.
      Please, give us a clue on how to calculate it, not an schematic warming of 10°F.

      The thought experiment is meant to illustrate what causes temperature to go up or down, not to specifically calculate what the equilibrium temperature of the different components would be. Off the top of my head, that would depend upon (1) the rate of electric heating of the bar, (2) the bars total mass, (3) its thermal conductivity, (4) its infrared emissivity, (5) the shape of the bar, (6) the separation between the two bars (which alters how much of the cooled chamber walls the heated bar “sees”), and (7) how efficiently the thermal-vac chamber chiller can remove heat from the system.

      Again, these details will NOT change the basic conclusion: the introduction of a cooler mass next to a warmer mass CAN cause the temperature of the warmer mass to rise. (As others have pointed out, though, it requires the warmer mass to have a source of energy being added to it, just like the Earth does from the sun.)

      • Vangel says:

        I suggest that your surrounding plate would go up in temperature and would radiate as much energy as the previous plate did. There would be no net energy added to the system and the system would be pretty much as before.

  24. Roy Spencer says:

    OK, folks, it is obvious that some of you are not paying attention.

    Parroting things you have heard from others that sound good to you does not carry much weight, and if you want to be taken seriously with your claims, you are going to have to do a better job of backing them up.

    What I have presented does not violate either the 1st or 2nd Laws of Thermodynamics. The system I am talking about is NOT energetically closed, because neither is the Earth. I think part of your confusion arises from this fact.

    It is precisely because the surface of the Earth is continuously being heated by an external energy source (the sun), combined with the fact that it is continuously losing energy to what amounts to an infinite energy sink (outer space), that the introduction of an infrared absorber (the atmosphere) can cause the surface temperature of the Earth to rise.

    There is no magical creation of new energy. Just a change in the rate of energy loss…and when you change the balance between energy gained and energy loss by an object, you change its temperature.

    If you really believe what you say, then stop wearing coats and gloves in the winter…they obviously can’t make your skin any warmer anyway since they are colder than your skin.

  25. I hope you will respond to my comment on my blog.
    Sincerely, Claes J

  26. Tom Rowan says:

    Poppycock! Pure and simple nonsense!

    Slowing the rate of cooling does not heat something up.

    Does slowing the rate of cooling make an object warmer than it would have been otherwise? Of course. This is why we use blankets in bed.

    But to say that slowing the rate of cooling and energy loss warms, or adds heat, or heats up anything is quite frankly, loopy logic.

    The blanket keeps us warm because of a primary energy source, ie our living bodies. Putting a blanket on a corpse would not warm up anything. The heat energy must be recharged.

    So too our world. We need the sun to come up every morning to recharge the lost heat energy. It is the sun that is the primary heat source of actual heating on our planet.

    I cannot believe that this simple fact must be restated over and over again. The only warming in our atmosphere occurs due to the sun and not CO2.

    This is why the sun burns our skin. Heavy breathing of CO2 has never caused skin to burn.

    Too, water vapor and clouds make up the overwhelming fiber in the blanket of “green house gases.” And we know this to be true because a cloudy night cools more slowly than a crystal clear night. Even still, a cloudy night never warms the the world more than the sunny day does.

    If CO2 were the dominate GHG then the night’s air would cool at the same rate regardless of cloud cover. It does not. This simple “thought experiment” can be proven every night in real life. The mumbo jumbo of imagining hot plates in vacuums proves only the ridiculus extent that one must go to give any credence to an obviously false hypothesis.

    Slowing the rate of cooling does not add energy to our atmosphere, it merely slows the loss of heat energy. CO2 is an inert gas and trace gas incapable of “warming up” anything.

    Lastly, CO2 has never been demonstrated to warm anything or even slow the loss of heat in the lab or in our enormous atmosphere. It has not been proven to in the lab or any field test….ever. To say otherwise is to create a “thought experiment” outside the bounds of rational thought and factual reality.

    I have a notion that Dr Spencer is using us as lab rats and testing our common sense, logic, and scientific reason. I have no colored charts or graphs or visual aids and can not prove it.

    But I wonder if I can get a consensus on the Lab Rat Theory?

    • Tom, most of what you say I agree with.

      For instance you say:

      Does slowing the rate of cooling make an object warmer than it would have been otherwise? Of course. This is why we use blankets in bed.

      And this is exactly my point! Of course the sun is the original energy source…but once something else has absorbed some of that solar energy, IT TOO becomes an energy source. You are attributing to me a position I am not advocating!

      • Anonymous says:

        Moving the cheese again Dr Spencer??? SQUEAK!!!

        • Anonymous says:

          Tom Rowan Responds:

          Let us talk a step back from the issue of CO2 and “reflective” warming or radiative warming.

          Perhaps the most perfected example of amplification of the sun’s source heat energy is the mirror.

          Solar reflectors surround a point and direct the sun’s heat energy to super heat a point in space.

          We know this to be true because we can see this ongoing enterprise before our eyes at solar farms.

          Still, the solar reflectors still must curve the light energy and pin point it just as a magnifying glass or telescope bends light and heat energy. The original heat source must be amplified by adding reflectors and perfecting the curve of the mirrors.

          Still, in this perfect environment for reflective and radiative heat does not add to the earths heat budget. It merely redirects it and concentrates it. These mirrors do create shadows after all. When the sun goes down, these perfect reflectors do not add anymore energy nor do they concentrate any more energy. I will bet they cool off rather quickly in the desert evening. No net heat gain or new heat energy source is added.

          Now say that solar reflectors were spread evenly around the globe. Would the planet’s tempurature rise anywhere other than the pinpoints of concentrated heat energy during daylight hours. (Remember all the shadows all those mirrors would have to necessarily create.)

          Do heated homes blunt cold weather fronts? Perhaps a small “heat island” effect could be measured, but even in this preposterous scenario, it is unlikely that the earth’s entire heat budget would be increased outside of the pinpoints. If it did, my guess is that the tempurature increase would be minimal if not immeasurable. So too the “power” of CO2, man made or not, to reflect and concentrate solar energy around the world.

          First, a mole of C02 is just an inert body reflecting and deflecting any heat source that happens to bump up to it. It does not amplify or concentrate any heat source nor direct it anywhere. It is an ambiant inert gas that warms and cools along with the rest of our enormous atmsophere.

          If CO2 had ANY power to warm anything more than say, a piece of dog doo does, then wouldn’t we be constructing huge mirrors concocted of CO2 rather than glass?

          There is a point where immeasurable becomes effectively nill.

          Any measurable effect of CO2 on our atmospher is actually immeasurable and effectively nill.

          How do we know this? To my knowledge, no one has ever been able to measure the concentration of CO2 on atmospheric tempuratures. Ever.

          In accounting and tax, we measure tiny mistakes or tiny influences by their affects in real life. If these small mistakes or inflences do not really affect how much you earn or how much tax one pays, then we rationally conclude they do not account for much. These tiny mistakes are what they are.

          Immaterial.

          So too CO2.

          • That solar mirrors focus energy to increase temperature in no way invalidates other concepts involving the warming of things with radiation. I agree the direct warming effect of adding more CO2 is small…so do the IPCC scientists!! We are not discussing feedbacks here, just whether the addition of more IR absorber to the atmosphere can cause the atmospheric temperature below it to rise.

  27. Layne Blanchard says:

    No issue with your explanation. But the next question for me is: How much IR can a gas (CO2) of limited absorptive characteristics absorb and re-emit (to other gases or space)?.

    If the entire IR budget of earth which falls within those bands of absorption were capable of capture by the XXXppm of additional CO2, what percentage of the total could it add?

    Even if CO2 had the same (greater) heat capacity of water, it should offer only a miniscule increase in overall IR capture. Is not humidity orders of magnitude greater then CO2 concentration?

    A humid day should provide more readily available GH gas than a doubling, or tripling of CO2, no?

    Additional energy should simply “boil the pot” more vigorously. (understanding that some of the energy would remain within atmospheric layers to warm them)

    As day becomes night, the balance changes, and saturated vapors should condense out part of their water vapor. There are readily available processes to expend excess energy.

    • I think I agree with everything you say, Layne. Even a doubling of atmospheric CO2 is only supposed to increase surface temperature by 1 deg. C. We could probably ignore it.

      That’s why I spend most of my research time on FEEDBACKS, because that’s what the IPCC claims could cause catastrophic warming.

      • Vangel says:

        This is sensible but there still is a problem. As I have stated above, you still need to look at how the amount of reradiated energy back to the surface compares to the increase in emissions due to warming. Unless you can show that the former is bigger than the latter your thought experiment falls apart.

  28. Patrick says:

    Hi Dr Spencer,

    Thanks for writing these interesting articles, keep up the good work and also especially for taking tje time to directly answer so many of peoples responses.

    Patrick.

  29. John M. Chenosky, PE says:

    Roy I love you but you need to go back to Thermo 101.

    Heat always moves from hot to cold.

    You are thoroughly confused.

  30. Michael Snow says:

    Ahhh, a professor who is also an excellent teacher! Thank you.

  31. jae says:

    Dr. Spencer: Heat does NOT move from a cooler object to a warmer one, as you are trying to “demonstrate.” You are definitely violating the Second Law here. Please do an experiment and give us the resulting data. It will show that you are dead WRONG!

    • Anonymous says:

      In fact, you are exposing exactly what is wrong with the whole notion of backradiation “heating” or even “slowing cooling” of the atmosphere: During the day when the sun is out, a solid, say a piece of asphalt, will become hotter than the air. According to your nonsense, you would have to add the backradiation to the solar radiation to determine the maximum temperature of the asphalt. In mid-latitudes in summer, on a clear day, at noon, the solar radiation is about 1000 Wm-2. This is equivalent to a bb temp. of about 190 C–and that is about what you get if you shield the asphalt from convective currents. According to your example, I would have to add the backradiation to the solar radiation. This would lead to impossibly high temperatures, even if you added only the average K&T backradiation of 323 Wm-2.

      • Actually, the emitting temperature corresponding to 1000 Wm-2 is not 190 C, but 91 C, which is close to 190 F. The trouble with your claim is that existing heat loss mechanisms are so large (in addition to IR, convective air currents and loss to the soil below), that unless you know those half-way accurately, you have no idea of what kind of downwelling IR is happening.

    • Anonymous says:

      Dang it, the Anon. should be jae again.

    • Jae, read the article. Like John and others, you are assuming I am claiming something that I am not.

  32. jae says:

    In fact, Dr. Spencer’s example illustrates exactly what is wrong with the whole notion that backradiation from GHGs heat the earth (or even slow the rate of cooling).

    (1) The temperature of an object is a function of the total amount of thermal radiation directed upon it.
    (2) During the day, when the sun is shining, the surface of, say, a piece of asphalt is heated by the sun to a temperature that is higher than the air temperature.
    (3) The standard GHG theory says that there is backradiation, which affects the temperature of the atmosphere (preventing it from cooling off as fast as it would without GHGs). This backradiation would have to be ADDED to the solar radiation during daytime to determine the total energy being received by the asphalt. That would HAVE to lead to an even higher temperature than produced by the sun alone (see (1) above).
    (4) But that cannot happen because of the Second Law; the cooler atmosphere cannot contribute heat to the warmer surface.
    (5) Therefore, something’s wrong with the “GHG backradiation hypothesis.”

    • Anonymous says:

      Could you cite where 2nd Law is stated precisely like the way you state in 4) above

  33. Nullius in Verba says:

    It’s an excellent tutorial, and I don’t disagree with anything in it.

    I do, however, have a problem with it in applying it to the atmosphere. You allude a time or two to convection, but in treating it as a minor detail to add on later, rather than a fundamental part of how the effect works in practice, I find it causes some confusion.

    As the atmosphere is convectively coupled, the temperature gradient is approximately fixed at the moist adiabatic lapse rate. As back radiation is internal to the troposphere, passing from one part of this convectively coupled layer to another, no net ‘temperature transfer’ can occur. It’s like trying to lift oneself off the ground by your own bootstraps. There’s absolutely no doubt that your arms can apply an upward force on your boots – you can even measure it! – but there’s a logical step missing between that and concluding that it is responsible for you floating off the ground – for the increased surface temperature.

    There is also the question of the temperature of the upper troposphere, at about -54 C is 70 C colder than the surface. Thus, although everybody knows the Stefan-Boltzman calculation tells us that the greenhouse effect is about 33 C, the atmosphere is observed to apply a warming twice as large as this from bottom to top. What keeps the top of the troposphere so cold? Radiation? But the effective “temperature” of outer space (including the sun) is -20 C. Radiation to the surface perhaps? As the temperature-altitude profile is a straight line, I suggest that whatever mechanism explains the relative coldness of the upper atmosphere also has to be used to explain the relative warmness of the surface.

    Regarding the laws of thermodynamics, I use the example of a refrigerator or heat pump. This can indeed convey heat from a warm place to a cold one, by compressing and expanding a gas. As air rises and falls in the atmosphere, it also compresses and expands, and therefore can act like a refrigerator in conveying heat from one level to another. The circulation is driven by surface temperature (or rather, heat input) differences between equator and pole. Nobody doubts a heat pump can be used to warm your house, by pumping in heat from outside. You can buy one and see it for yourself.

    But is it correct, if you have such a heat pump, to say that the entire reason you are nice and toasty in your heated house is that longwave radiation from the internal walls keeps you so? (Although the walls are cooler than you are.) That you are warmer than outside because your house is insulated?

    I am a great admirer of your work. Please, please keep it up.

  34. Demesure says:

    “The thought experiment is meant to illustrate what causes temperature to go up or down, not to specifically calculate what the equilibrium temperature of the different components would be.”

    Dr Spencer,
    Thank you for your reply.
    More specifically, if I surround the 150°F plate with 2 other plates, isn’t that the new tempeture for the 2 new plates will finally set at 150°F and the initial plate will warm of some °F ?
    What I would like to know is which physical laws are used to calculate that warming so we can estimate the warming in a realistic setup (for example a 1m2 plate, heated with a 10kW constant electrical power, in a 10m3 vacuum cell with thermostatic walls at 0°F).

    BTW, with external power to maitain the walls at 0°F, can we still use the conservation law here ?

    • I think I replied to you (or someone else?) that a theoretical calculation of this sort is difficult because of at least 7 variables I thought of off the top of my head. This is why thermal vacuum chambers are built…to test the (very uncertain) calculations by engineers who must design an instrument to operate within a certain temperature range in space.

  35. Brian Strong says:

    Please perform an actual experiment with a continuously heated plate, a vacuum chiller chamber, and a second plate. Please perform all calculations. Please measure all changes in temperature. Please show your work in your lab book. Please publish your results. Calculations and measurements, remember?

    There was a time when scientists did experiments. They did not make up thought games. They did not manipulate language. They were not concerned with clip art. They were not agenda driven and did not have a template. They were not called warming alarmists. They were called and were only known as scientists. They may have been mad scientists, but they were scientists. People respected them because people believed that they were all about the truth and the scientific method. They were not driven by their computer models. They were not part of the four corners of the universe of lies. They did not go on television. They did not wear TV make-up. They did not appear on the morning shows. They were scientists.

    • Actually, I have performed an experiment that more directly addresses the question of downwelling thermal radiation from the atmosphere. When I worked for NASA, I had an instrument designed and built to measure thermal microwave radiation at 10.7, 19, 37, and 85 GHz. The Earth and atmosphere absorb and emit at thermal radiation at microwave frequencies, although at a much lower level than in the infrared, through the Planck function.

      It was calibrated by looking at a high emissivity target, and at a styrofoam container containing liquid nitrogen (77 K).

      The upward looking brightness temperatures it measure were very close to what radiative transfer theory (with downwelling radiation) predicts…as I recall, around 20 Kelvin or less at 10.7 GHz where the atmosphere is very transparent, to more like 100 Kelvin or more at 19.35 GHz (near a water vapor absorption line at 22.235 GHz), and at 85 Ghz.

      So, been there, done that.

  36. “Roy W. Spencer, Ph. D. says:
    July 23, 2010 at 3:53 PM”
    Hello, You say: “Generally speaking, the rate of absorption is not temperature dependent, but the rate of emission is very temperature dependent”. Please correct me if I am wrong, but my understanding of an absorption spectrum is that a continuous spectrum passes through a cool gas and certain wavelengths are absorbed. These wavelengths then get emitted in all directions so very few photons go in the original direction, hence the dark lines. Logically speaking, I would think that if the “cool gas” were warmer, or if more electrons were at a higher energy level, then fewer photons would be absorbed and the dark line would not be quite as dark. Is this true and would this apply to the absorption of infrared photons as well? Is it the case that the rate of absorption does not depend on temperature at all or is it just that the relative amounts we are talking about are totally insignificant? Thank you!

    • Not all of the radiation is absorbed, though. At most infrared frequencies, the atmosphere is fairly, or at least partly transparent. This is because, at higher pressures in the lower atmosphere the narrow absorption lines where absorption in very strong get broadened so a much wider range of IR wavelengths, so that at most wavelengths are party absorbed

      The warmer the air is, the more IR radiation is emitted from greenhouse gas molecules. Those molecules have an endless supply of extra energy from the surrounding non-GHG molecules (e.g. nitrogen), because they are bumping into them a gazillion times each second.

      • coturnix19 says:

        oh, here is an interesting question. We know that in the past, mean surface air pressure wasn’t constant. Oxygen content changed significantly throughout history, as well as rising of high plateaus and glaciers displace some air, changing MSLP by small but non-zero value. I wonder, how (if at all) these changes influence climate, especially on part of greenhouse effect.

      • Anonymous says:

        Yes Werner, I sympathise with your doubts about the statement ” the rate of absorption is not temperature dependent, but the rate of emission is very temperature dependent “. With respect to Roy, he is doing a sterling job trying to explain his thought experiment, which is entirely sound, and so to start bickering about details of expression seems somewhat inappropriate.
        The CO2 absorbs radiation around 15 microns, irrespective of whether the gas it is ‘cool’ or ‘hot’. Having captured the photon at this wavelength, it will almost immediately re-emit it. The rate of emission therefore depends on the rate that it absorbs, not on the gas temperature. The temperature only governs the emission of ‘blackbody’ radiation, which is emitted by any body above absolute zero. It does not govern the emission of photons from an ‘excited’ gas molecule. And it is the latter that produces the so-called greenhouse effect.

        Bomber_the_Cat

  37. Kevin says:

    Dr. Spencer,

    With respect you have produced a very nice example of a very oversimplified system that does not remove any of my objections to the “greenhouse effect” and its postulated dire effects on the future of the climate of the Earth.

    To summarize my objections to the “greenhouse effect”;

    I DO NOT disagree with the existence of backradiation, nor “back conduction”, nor” “back convection”. They do in fact all exist although they can be difficult to measure.

    I DO NOT disagree that the presence of gases that absorb and reemit infrared radiation backwards towards the surface of the Earth and thereby act to insulate the Earth and do affect the temperature of the Earth.

    I DO disagree that these gases AND ONLY these gases cause the temperature of the surface of the Earth to be 33 degrees Kelvin above what that temperature would be if the gases did not exist. I believe that fundamental calculations about the temperature of the Earth sans “greenhouse gases” are deeply flawed in that they consider perfect 2D blackbody surfaces with no depth and no thermal capacity.

    I DO disagree that increases of the magnitudes predicted in the amount of these gases can cause a permanent long term warming of the surface of the Earth. My belief is based on the fact that increases in these gases will cause more energy to flow through the system at the speed of light than at the speed of heat (thermal diffusivity). So the simple explanation about more energy flowing back towards the surface causing the temperature to rise is way oversimplified and does not consider the speed at which that energy is flowing. These simple explanations completely ignore the cyclical nature of the heat source i.e. sunrise and sunset. Given a cyclical system it is imperative to understand the speeds at which energy flows through it.

    Some observations about your example;

    The second bar in the vacuum chamber is NOT causing the temperature of the First bar to rise. The fact that you are supplying a fixed amount of energy (the electric current) into two different systems (1 bar vs 2 bars) with different heat flow (transfer, exchange) characteristics IS causing the temperature of the first bar to reach a different equilibrium temperature.

    You have chosen a nice simplified system that includes only one of three known heat transfer mechanisms and where everything is stable and at a nice equilibrium. I cannot think of an example that is further from the state of the atmosphere of the Earth. While this does help people understand some concepts of heat transfer I believe it is way too simplified and glosses over the real discussion which should be “WHAT ARE THE REAL MEASURABLE EFFECTS OF “GREENHOUSE GASES” IN THE ATMOSPHERE”.

    Your example easily dismisses the fact that the heat flow into the system cycles from a veritable deluge of photons (sunshine) to a drought of photons (moonshine and starshine). This does your example a disservice.

    Your example also nicely chooses two materials (bar #1 and bar #2) with similar thermal capacities (per unit volume). Again I cannot think of an example that is further removed from the actual Earth/Atmosphere system we are considering. It is well known that the thermal capacities of the gases in the atmosphere are less (in fact by many orders of magnitude) than the surface of the Earth.

    Thanks for your detailed response, very well done.

    But I remain a denier. Not a denier that the effects you describe exist and are having some impact on the temperatures of the surface of the Earth. But rather a denier that the effects postulated (i.e. a long term permanent warming of the Earth) are the correct ones.

    Cheers, Kevin.

  38. Mike Blackadder says:

    Dr. Spencer,

    Thanks for your post and many replies in the comments.

    You do an excellent job of explaining things while avoiding using math, but sometimes a bit of math helps clarify things. Also explaining how this particular example differs from ordinary experience can also be helpful.

    Regarding your example, I think Adriaan made a good point in his first comment when he pointed out that if the plates were touching that the first plate would not heat up and it would remain very close to 150C (which is essentially correct despite the plates being imperfect conductors of heat). The first plate’s temperature will increase only because it is surrounded by a vaccuum and by definition can only transfer heat via radiation.

    • Well, he point of the thought experiment was to explore how thermal infrared radiation operates in the context of the energy budget, and thus the temperature, of things. It was, of course, meant to be an accurate model of the way the atmosphere behaves.

      But once someone decides to mash the two plates together, the point can no longer be demonstrated.

      We know that thermal conduction is very weak compared to other energy transport mechanisms within the atmosphere, due to the large vertical distances involved, so I wanted to use a setup where conduction could be ignored.

      • Anonymous says:

        I should have mentioned in my original comment my overall impression of your post. Despite having a couple of points to add I thought you chose a pretty good model for making your point. I’ve read similar posts elsewhere where the author creates a very simplified version of an atmosphere (ie. one layer atmosphere that is transparent to solar radiation but acts like a black-body with respect to long-wave). If anything your particular chioce of a model is quite diplomatic.

        Overall I think it is unfortunate that some commenters are of the impression that your argument is false/impossible or claim that it is a violation of the laws of thermodynamics when perhaps they haven’t done the back-of-envelope calculations that illustrate how this works. However, I would say that this is the typical reaction even among scientists and mathematicians when introduced to new concepts that are difficult to reconcile intuitively (ie. anyone heard of the Monty-Hall problem?). It doesn’t make them ‘deniers’ to think that your model violates the first law of thermodynamics. Rather I think they are relying too much on their intuition and are being a bit hasty to claim you are dead wrong about a point that is easily proven.

  39. Juraj V. says:

    The back radiation theory does not work in Mars. 6,000 ppm of CO2 has no effect on its 210K average T. To get higher than black body temperature, you need bulk atmosphere, does not matter much what it is composed of. (And this is it, Mars has very thin atmosphere). The atmosphere and water (=clouds, evaporation, ice albedo) cools our day (so our daily temperature is much lower than on the Moon) and warms our nights (they keep the daily heat for some time). Night clouds also hinder convection, rising of warmer air upwards.
    In my opinion, the IR radiation is just a sign that the object (atmosphere, clouds) have their temperature above zero. Maybe there is *some* effect of IR back radiation, but it either needs the bulk atmosphere to catch it and/or it can not be alone responsible for our pleasant temperature range, considering the space as a whole.

    • As I recall, the temperature profile of the Martian atmosphere, as well as the surface temperature on Mars, HAS been explained with standard radiative absorption/emission theory and radiative transfer models.

  40. Dr. Dave says:

    Dr. Spencer,

    I ain’t had this much fun since the pigs ate Billy. I spent a pretty good chunk of yesterday reading Pielke’s explanation of the greenhouse effect and well over 200 comments on WUWT. I also plodded through the G&T article until my eyes crossed.

    This morning I came upon your post and discovered an elegantly simple explanation. I’m having a blast reading your replies. For my own benefit, I took your thought experiment and complicated it a bit. I imagined a gray sphere slowly rotating in the cold wall vacuum chamber. The sphere is heated from a single point IR source until it reaches an equilibrium temperature. Then place another (even a smaller) sphere in the chamber on the “dark side” of the heated sphere. The results are the same. You slow down the rate at which the heated sphere loses energy to the cold walls of the chamber and the heated sphere becomes warmer than the previous equilibrium temperature.

    Some who have commented cavil at the atomic level and this is absurd within the context of your excellent explanation. Others simply have not read (or understood) you post carefully. Dr. Spencer, you are one of my “climate heroes” because of your ability to describe the science. You must be a truly awesome professor.

    • Anonymous says:

      Hm, “cavil”. Honestly, I had to look that word up. I was not “raising trivial objections” to what dr Spencer was trying to explain. On the contrary – I fully agree with you – his explanation is excellent. But as this article (http://www.sciencedaily.com/releases/2009/07/090730154025.htm) shows, if you bring two objects very, very close together – at the nanoscale (without touching)! – Planck’s law breaks down and radiation heat transfer is boosted. How would you explain that if not at the atomic level? That, my good fellow, I would call “cavil”. ;-)

    • Thanks for the praise, but I don’t teach, just research. I already raised my kids…I don’t want to go through it again. :)

  41. Jere Krischel says:

    Does the surface area or mass of the plates matter? Or their distance from each other?

    Assuming some original plate mass arbitrarily dense, I would imagine increased mass given a constant surface area would serve only to slow the rate of heat loss if our energy source was removed, increasing thermal capacitance as it were. All other things held constant, the IR radiation equilibrium rate would be a function of surface area.

    So if we increased the surface area, I’m assuming that the equilibrium temperature of the first plate would go down.

    Now with the second plate, the total mass has gone up, but we’ve introduced a separation of distance. If there was no separation of distance, we’d simply have an increase of mass and surface area, and we’ve concluded earlier that this means a lower equilibrium temperature and a slower rate of heat loss if the energy source is removed.

    In the model where there is separation, we still have increased mass and increased surface area. The increased mass means that more heat energy is “stored” in the system of plates, but I’m confounded by the increase of surface area not producing a commensurate drop in temperature.

    Imagine the two plates arbitrarily close to each other, separated by a single nanometer. This configuration should be pretty close to the results of actually touching, or at least arbitrarily close.

    Now imagine the two plates arbitrarily far away from each other, separated by light years of hard vacuum. This would certainly increase the time it takes to reach equilibrium, but I’m not sure if it follows that the first plate would actually increase in temperature.

    I guess my question is this -> does the presence of Jupiter mean that the sun is hotter than it would be without it?

    When I think about the blanket analogy keeping me warm at night, things also seem to break down in some odd ways. Of course, my body is the first plate, and my metabolic processes are the energy source, and if you put one blanket on, you’ll increase my surface temperature, and if you put two blankets on you’ll increase it more…but there doesn’t seem to be a number of blankets that could drive my blood to boiling (internal temperature seems to be well regulated to 98.6)…maybe there isn’t enough time for that effect to happen, and of course, a perfect insulator might actually overwhelm the body’s compensatory mechanisms ending in death and the end of the heat source…it also doesn’t seem to matter if the blanket is touching me or not, but it does seem to matter if the “blanket” is instead a tent.

    Maybe my conclusion from this, understanding the argument being put forth, is that yes you could have an increase in temperature by adding an “insulating” body to the system described, but instead of showing 160C and 100C, it would be more like 100.000001C and 100C. Has anyone ever done a computer simulation of this with assumptions of 1g of heated iron as the first plate, a power source of 60W, and 1g of heated iron as the second plate?

    • Regarding the geometry of the setup, yes the distance between the two plates matters, because that affect the extent to which the heated plate (and the second plate) “see” the cold walls of the vacuum chamber, versus “seeing” each other.

      If you could encase yourself in 12 inches of styrofoam, and somehow recapture the heat you lose through breathing so that it doesn’t escape, then your body temperature would rise till you die…I suspect that happens somewhere around 110 F, or before. Then, you would slowly assume room temperature.

  42. commieBob says:

    A somewhat off-topic question:

    Given that the CO2 molecules in the atmosphere act as resonators, do they produce Mie scattering of the incident IR?

    Of course I am not sure whether my supposition is “not even wrong” (http://en.wikipedia.org/wiki/Wolfgang_Pauli).

  43. harrywr2 says:

    Juraj V. says:
    July 24, 2010 at 9:52 AM

    “The back radiation theory does not work in Mars. 6,000 ppm of CO2 has no effect on its 210K average T.”

    Compared to water vapor CO2 is trivial.

    IIRC A doubling of atmospehric CO2 alone causes a 1.2 degree C rise in temperature.

    So if we were to double CO2 roughly 4 times to get to Mars’s 6,000 ppm , we would get a 6 degree C rise in temperature.

    All other things remaining equal.

    The $64,000 question always have been, if we raise the temperature of the earth by 1 degree, how much more water vapor will we get, and will that water vapor be in the form of vapor, low clouds or high clouds.

    On mars, unless there are enough greenhouse gases to get the temp 272K then water vapor stays out of the picture, as it is frozen.

  44. Stephen Wilde says:

    Cooler objects don’t make warmer objects warmer still.

    They just reduce the rate at which they cool.

    Is that surprising ?

    Isn’t the title of this thread a little misleading ?

    • FINALLY! SOMEONE ASKED THE QUESTION I WAS WAITING FOR!

      And the answer is….(wait for it)….

        no less misleading than asserting that cooler objects can’t make warmer objects warmer still.

      …Especially, when one realizes that THEIR assertion does not apply to the Earth, and it is the explanation for the Earth’s greenhouse effect they are objecting to!! The Earth is continuously heated by the sun, which makes all the difference to the problem!

    • Kirk says:

      Roy, your explanation is not a case of a cold object making another object hotter.
      To simplify I think we should get rid of the vacuum bottle and replace it with outer space.
      Next we replace the plates/bars with cubes so we can think in 3d instead of flat thin 2d.

      So, let’s assume then we have a little cube floating stationary in the void of deep space. The cube has a battery and a little heater deep inside that together produce exactly 1 watt of energy. This energy will heat the cube until it reaches a temperature at which exactly 1 watt of energy is radiated from it. Let’s say that this final temperature is 120 F. We have equilibrium.\and each of the 6 sides of the little cube is the same temperature and radiating exactly the same amount .

      Along comes another cube exactly the same as the first cube but it has no battery left and no heat. It’s as cold as the space around it. It parks itself, face to face with the hot cube. It’s important that they don’t touch because that would ruin the radiation only experiment (more on that later) so lets assume there is a small gap between. The cubes begin to interact by radiate energy and will again stabilize in some new state of equilibrium.

      What happens is the cube with the heater will get a bit hotter (now121F). The cold cube from the depths of space will warm up significantly unit it hits 25F and equilibrium is again achieved.

      Why did the hot cube get hotter? Originally, it had 6 surfaces radiating into the depths of space- Now it only has 5. The 6th face is radiating directly onto a 25F face of the colder cube. The hot cube has 1 less face exposed to perfect nothingness so it can’t get rid of as much heat as before. It gets a bit hotter overall and NOW it can now dissipate the original 1 watt at equilibrium.

      We can see from this that the hot cube got hotter because it had less access to nothingness. You could say that less access to nothingness is the same as energy flowing “in”. True. You can argue that the radiant energy from the cold cube is “back flowing” and adding to the heat of the hot cube. Hmm.. Terminology? Remember the cold cube has no heater. In reality the hot cube is radiating nearly 1/6 of its energy to the cold cube. In turn, the cold cubes other exosed 5 faces are radiating that same absorbed amount back into space.

      I’m not sure if I actually have a point…
      Instead of introducing a cold cube beside the lone heater cube, you could have brought in a really hot cube instead. Something you had heated with a blow-torch for example. The outcome (equilibrium state) would be exactly the same as if the cube was cold. Original Cube with heater will end up at 121F, introduced cube will end up at 25f.

      So saying “the cold bar made the hot bar hotter” is not relevant. If you had added a hot cube or a cold cube we have the same ultimate outcome. You simply proved that “adding something to reduce a body’s radiation will reduce the body’s radiation”

      Oh yes, If the two cubes touch faces they will essential become the same temperature. They don’t need to be “perfect conductors” just better conductors than radiators.- which most things are. The new “combo-cube” will now have 10 faces radiating into space and still only a 1 watt heater. The equilibrium temperature will therefore be lower that the single cube. (equilibrium =85F)

      The numbers outlined are approximate. They have been derived using a radiant flow thermal CFD software model.

      • All I said was that the presence of a cooler object can make a warmer object warmer still. I was being deliberately ambiguous to point out that when others say this CAN’T happen, they are assuming there is no external source of energy to the hot bar, which is not what happens with the Earth’s surface being heated by the sun.

  45. Dr. Spencer,

    What a timely post by you, especially in view of our recent correspondence on white roofs, etc.

    It occurred to me that your iron bar heating metaphor might be also viewed in another mind experiment that would remove the bothersome (to some) vision of the cold bar working to make the heated bar hotter.

    Take your original model, in equilibrium, and magically slice the solo heated bar in half, and move it away from the heated bar, It seems to me easier to accept that heated bar, now of a smaller mass, will increase in temperature as the electrical heating input is unchanged. The bit sliced off will cool a bit as it is no longer being directly heated by the electric current.

    Thus there are now two bars, one is hotter (the electrically heated one), and a slightly cooler separated piece. I think that the conclusions you postulated in your original model would be the same in this, reverse engineered model.

  46. Christopher Game says:

    Dr Spencer

    You do us all a mighty service and favour by your admirable and indeed magnificent work.

    But may I offer a small correction to the above post? Specifically I refer to your diagram of radiative exchange
    http://www.drroyspencer.com/wp-content/uploads/IR-example-thermally-stratified-atmosphere.gif.

    As presently labeled, your diagram is not right. Its present labeling shows surface upward radiation absorbed into the atmosphere as much exceeding radiation from the atmosphere down into the surface; and shows no non-radiative transfer. This conceals the prime importance, as you are well familiar, of the non-radiative energy transfer from land-sea surface to atmosphere. Your diagram seems to hide this non-radiative transfer within the label of radiative transfer. As you have most expert knowledge, non-radiative energy transfer from land-sea surface to atmosphere is by conduction and evaporation, followed by convection to other places and altitudes, and is very important.

    Perhaps not so well known is the extension of the so-called “cooling to space” approximation, see for example Paltridge and Platt 1976 pages 165-172, Goody and Yung 2nd edition 1989 page 250, and Wallace and Hobbs 2nd edition page 138. Goody and Yung say that the importance of the approximation “cannot be overstated”. Wallace and Hobbs describe it as working “surprisingly well”. Yet its significance is not always fully understood. Even less widely known is that a similar finding applies also to the interface between land-sea mass and lowest atmosphere, as follows.

    When, from radiosonde data, one actually directly calculates the amount of land-sea surface radiation absorbed by the atmosphere, and compares it with the amount of radiation from the atmosphere that is emitted down into the land-sea surface, one finds for a given radiosonde sounding record, that they are mostly rather nearly equal. This calculation is not often done. Indeed, so far as I know, there is only literature source for it. If you know of others, I am keen to hear of them. The only writer that I know of who has directly compared, for a given radiosonde record, the land-sea surface radiation going up to be actually absorbed by the atmosphere, let us say Aa, with the atmospheric sourced radiation emitted down, let us call it Ed, to be absorbed by the land-sea surface, is Ferenc Miskolczi, for example at http://multi-science.metapress.com/content/nm45w65nvnj3/?p=2037ca7987b54843bc4b526ff0b6ff7c&pi=1, The stable stationary value of the earth’s global average atmospheric Planck-weighted greenhouse-gas optical thickness,’Energy and Environment’ 21(4): 243- 262, doi 10.1260/0958-305X.21.4.243 . Miskolczi himself would also like to hear of other direct calculations to check his work, but sad to say so far no one seems to have bothered. Instead of checking, they prefer to accept calculations “as a residual”, or by guesses or assumptions in terms of the static spectral definition of “the window”.

    It is important for accuracy in this work to take care to distinguish between the dynamic atmospheric window and the static spectral window. Let me quote my Wiki note on this: An atmospheric window is a dynamic property of the atmosphere, while the spectral window is a static characteristic of the electromagnetic radiative absorption spectra of many greenhouse gases, including water vapour. The atmospheric window tells what actually happens in the atmosphere, while the spectral window tells of one of the several abstract factors that potentially contribute to the actual concrete happenings in the atmosphere.

    The infrared atmospheric window is the overall dynamic property of the earth’s atmosphere, taken as a whole at each place and occasion of interest, that lets some infrared radiation from the cloud tops and land-sea surface pass directly to space without intermediate absorption and re-emission, and thus without heating the atmosphere. It cannot be defined simply as a part or set of parts of the electromagnetic spectrum, because the spectral composition of window radiation varies greatly with varying local environmental conditions, such as water vapour content and land-sea surface temperature, and because few or no parts of the spectrum are simply not absorbed at all, and because some of the diffuse radiation is passing nearly vertically upwards and some is passing obliquely and indeed some nearly horizontally.

    These complications need to be taken into account for an accurate calculation of Aa, and so far as I know Miskolczi is the only published source to do so. As above, if you know of others, please tell us.

    In summary, in the above notation, Miskolczi has shown that Aa = Ed is a fairly accurate representation of the empirically observed facts, both locally and globally.

    The physical meaning of Miskolczi’s observation is that, in effect, cooling of the land-sea surface is just by radiation direct to space through the atmospheric window, and by non-radiative transfer into the atmosphere by conduction and evaporation followed by convection; but NOT by net radiative exchange with the atmosphere. In a steady-state climate condition, the land-sea body and the atmosphere are in mutual radiative exchange equilibrium, with zero net radiative exchange. This important new fact is concealed by the present labeling of your diagram.

    Perhaps I should re-emphasize that this new fact is not yet widely known, but ought to be. When people eventually come to understand this new fact, they will likely think in fresh terms about the earth’s energy transport process.

    I suggest you alter your diagram to show this new fact. You might alter only the label, or you might alter the arrows displayed as well.

    Christopher Game
    declaration of interest: I talked with Miskolczi about the 2010 article that he was writing.

    • yes, I’ve been wading through Miskolczi’s paper. I think his paper is useful from the standpoint that it quantitatively demonstrates negative water vapor feedback from the NCEP reanalysis…although many will dispute the reality of the free-tropospheric drying in that dataset, which they will claim is a problem with early radiosonde data.

      I’m still trying to digest his other claim.

      • Anonymous says:

        Thank you, Dr Spencer, for your thoughtful and valuable reply. I look forward to reading the fruits of your digestion of Miskolczi’s claim that Aa = Ed, which I am guessing is what you mean by his other claim.

        Please excuse my clumsiness, it seems I mistakenly hit an enter key, or somesuch, before I had finished writing my reply.

        A couple of minor points:

        Miskolczi deliberately did not propose that there was climatic time-scale negative feedback by water vapour effects on the earth’s global average atmospheric true (or Planck-weighted) greenhouse-gas optical thickness. He claimed that the empirical data analysis shows that there was, over the 61 years, no significant positive feedback by water vapour effect on that optical thickness.

        The leading claim of Miskolczi’s paper is that the empirical data analysis shows that over the 61-year dataset, there was no significant increase in that optical thickness. With the premise that CO2′s putative hypothetical effect on climatic temperature is by way of its increasing that optical thickness, it is to be inferred that any increase in climatic temperature over the 61 years of the dataset cannot have been caused by increase in CO2 over that time.

        • from reading his paper, I would say that the negative water vapor feedback he measured nearly canceled out the radiative forcing from increasing CO2. That conclusion, though, depends upon whether the long-term decrease in free-tropospheric humidity measured by radiosondes can be believed. Most I have talked to don’t believe it. I haven’t seen enough to make an informed opinion.

  47. Hans says:

    So it’s clear that the colder object does’nt make the temperature of the warmer object higher. It just looks this way while the electric heating is on.

    Dr. Spencer you show us that this chamber setup looks almost as the earth system and in the same way as in the chamber the earth surface can reach a higher temperature in the presence of another radiative body.

    But you fooled us here. The problem lies with the electric heating. This device is the energy source, but is not heating the first plate by radiation. This device is the only heating source but has no direct effect in terms of radiation.

    So for the radiative budget plate 1 which represents earth is the body with the highest temperature, so that’s where things go wrong. You have cancelled out the higher radiative heating source, so it just is’nt correct.

    The source of energy is not radiation, so we can not treat this as an radiative energy question and not make relations with the 100% sun-earth radiation behavior.

    In reality we have plate Sun at 5800 K on the right of plate 2 and everything changes completely now, this is the plate with the highest temperature. This plate tells the other two how to behave.
    Plate Sun is the source of the radiation and heats plate 2 up first an after that heats up plate 1 and no colder body warms up any warmer body.

    This in analogy with your chamber, where you mix energy from electrons with photons an treat it all as 100% radiation.

    But it was a good post to get one thinking about this matter.

    • I will say it again….the thought experiment was to discuss the role of IR radiation in the heat budget, and therefore the equilibrium temperature, of the atmosphere. It was no meant to be an accurate model of how the Earth is heated by the sun. And, yes, an object heated non-radiative DOES lose energy radiatively…like a hot stove….your skin can feel the extra IR radiation it emits.

  48. Andrew S says:

    Hi Doc. In your thought experiment heating element has higher temperature then passive one. Do you agree that this is always the case? No matter how many heating elements, mirrors and lenses one uses, the heating elements can never pull temperature of heated element above their own? Then you must agree that cold atmosphere can never pull Earth surface temperature above its own.

    Either there is “backradiation” or there is second law of thermodynamics. They cannot co-exist. I choose laws of thermodynamics.

    Andrew S

  49. jae says:

    It appears all so logical and simple. So, it should be easy to provide empirical evidence (hard data) for these constructs, no? Where is that data?

  50. Kevin says:

    Ok, let’s expand Dr. Spencer’s thought experiment about the ”greenhouse effect”. Let’s replace the infinite electrical power supply feeding the 150 Watt heater with a simple battery, let’s postulate a “D” cell. Everybody should be somewhat familiar with these batteries, lots of common devices around the house use them, radios, flashlights, etc. etc.

    So, let’s propose that at the beginning of each day (i.e. sunrise) the Earth’s atmosphere is “given” a fresh (i.e. fully charged, brand new, etc.) “D” cell battery. Yes of course we need to adjust units to make things exactly equivalent to the actual situation, but please remember this is a thought experiment.

    So, what happens next in the one bar .vs. two bar example in a vacuum chamber ? Well it turns out that the single bar example draws energy from the battery at a slower rate and reaches a lower temperature, but this temperature is maintained for a longer period of time. The two bar example will indeed reach a higher temperature at Bar #1 and Bar #2, but this will only occur for a shorter time as the given “D” cell battery will discharge faster!

    In fact, if Dr. Spencer was granted an energy allocation (i.e. you can have 500,000 btu’s to perform your experiment) I suspect he would be a little bit more inquisitive about the energy that is heating adjacent elements (i.e. bar #2).

    So it still comes down to the total effect of heat flowing through a system which must consider all of the effects (i.e. more heat flow by radiation means LESS heat flow by conduction and convection). There is still no overwhelming evidence that the flow of energy by radiation versus the flow by conduction and convection causes any net (i.e. permanent) rise in the temperature of the Earth.

    Cheers, Kevin.

  51. sunsettommy says:

    I did not realize that the Earths atmosphere is a vacuum?

    Your thought experiment is devoid of many factors known to exist in the real atmosphere.

    Your VACUUM chamber leaves out Wind currents,convection,conduction,oxygen,nitrogen and so on.It is not even close to what the real world is.Why did you do it with a vacuum?

    The earths atmosphere is not a vacuum and not enclosed either.

    It would be more convincing if you did an actual experiment with a real atmosphere included and show us that the initial plate can get hotter than the energy setting that comes into it.

    Try the 1500 watt heater unit and see if you can make it exceed that upper limit,that is designed into it.Making it go to 1600 watts.

    Electrical Engineers would be very interested in your reply.

  52. Dirk says:

    Dr. Spencer,

    While it’s called the greenhouse effect, it’s not the same as an actual greenhouse, right? That is, there’s no prevention of convective flow as in a greenhouse?

    And the reason that C02 supposedly enhances this effect is it’s greater ability to absorb the energy spectra emitted by the earth/ocean than the energy spectra originally coming in from the sun, true?

    And this phenomena isn’t easily explained by a single simple equation is it? If so, why did Arrhenius adjust his sensitivity downward before he died?

    Lastly- what do you think of the CERN CLOUD experiment? Do you see the possibility that the findings to be presented at the Aerosols Conference could put a bullet through the heart of AGW if solar/cosmic ray/cloud/temperature interactions are verified and quantified?

    I can verify the cloud/temperature interaction- it definitely cools off when a cloud passes overhead.

  53. KuhnKat says:

    Apologies if this was already mentioned. Didn’t have time to read all the posts.

    I have seen this type of thought experiment and every time the same mistake is made….

    Sorry, but without bothering to read my original post and the ensuing discussion, you are making false assumptions about what I have said, or have not read my responses to the same objections others like you have made.
    -Roy

    • Anonymous says:

      1) Firstly why introduce a light bulb and then abandon it right away
      2) Energy is coming into the system constantly unless the “juice” is turned off. So if in theory you could put a perfect insulator around the plate the temperature – again in this ideal world would be run away. Of course in reality things will crash and burn.
      3) X watts per second are being radiated, but with radiation from the second plate Y watts per second are being received back. Effectively having two sources – electric and radiative, as far as the first plate is concerned. So to be at equilibrium X+Y watts per seconds need to be radiated.
      4) You are right that temp can only go up if input goes up or output is delayed. A specific photon is not delayed indefinitely but a lot (not all depending on direction of photons) would be always delayed, making the second plate behave like an insulator.
      5) Would be wonderful and add to the discussion if you can lose the apparent attitude in your post

  54. Joe Born says:

    For anyone who’s still trying to get his mind around this and would like to try running at it from different directions, I commend the following two sites to your attention: http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/#more-12889 and http://scienceofdoom.com/2010/07/17/the-amazing-case-of-back-radiation/.

  55. Scott Scarborough says:

    Hello Dr. Spencer,

    I understand that the mean free path of a photon of infrared radiation in the earth’s atmosphere (of the frequency absorbed by C02) is about 15 meters and this distance is inversely linearly related to how much CO2 is in the atmosphere. Since this distance is infinitesimal compared to the thickness of the atmosphere, most of that energy has to work its way up to the top of the atmosphere by convection. This was true before mankind existed on earth and is true today. If we double the CO2 the mean free path will be 7.5 meters – still infinitesimal compared to the thickness of the atmosphere. Is there a flaw to this type of thinking? Are the “side bands” of the CO2 absorption bands the issue that alarmist climate scientists are concerned with? That would be very very narrow frequency bands whose mean free path would be longer and could actually escape directly into space with a lower concentration of CO2 and be retained with a higher concentration of CO2. But this seems to become a vanishingly small amount of radiation. Please comment. Thanks.

    • Anonymous says:

      You are absolutely correct.The absorption side bands, between 14 and 16 microns,would absorb a bit more as CO2 concentrations increased, but the resulting heating effect would be marginal, compared to that produced by CO2 already present in the atmosphere. This is why the effect of CO2 is said to be ‘logarithmic’, or one of diminishing returns. You are also correct that Convection becomes very important in this ‘saturated’ condition.There is a nice explanation on Anthony Watt’s website at
      http://wattsupwiththat.com/2010/03/08/the-logarithmic-effect-of-carbon-dioxide/#more-17114

      Bomber_the_Cat

    • I believe you are referring to the center frequency of a CO2 absorption line. Because of pressure broadening of absorption lines in the atmosphere, CO2 absorption occurs over a much wider range of frequencies, and its the frequencies where the atmosphere is partly transparent that increasing GHG’s have their greatest effect. This effect has been validated with downward looking IR spectrometers flown in space, the latest of which is AIRS with thousands of IR channels.

  56. Alex says:

    What a funny leg-pulling excercise

  57. Baa Humbug says:

    Having read the article and all the comments thus far, the words that enter my head are “hyperthermia” and “potential”.

    A jumper, coat etc does keep us warm by restricting heat dissipation. If we stood naked at a very cold place, we would lose heat quicker than we can “manufacture” it. But standing naked at a warm location reduces heat dissipation. However, in either scenario, our body still only “manufactures” about 37degC of warmth. Those living in the tropics do not have body T’s of 40deg or more do they?

    The other word, “potential”, comes to mind because in the case of the heated bar, it does not reach it’s potential heat because it dissipates heat quickly when enclosed by much cooler surroundings. However, as soon as we introduce a warm body next to it, it’s T rises to somewhere closer to it’s potential heat. (rather like a bucket with a hole in the bottom, if at equilibrium, the level of water in the bucket doesn’t reach it’s potential. But as soon as we reduce the size of the hole, the bucket will fill to it’s potential).

    The part I have trouble with is as follows….

    Having introduced a 2nd bar, and having increased the T of the first bar, wouldn’t this create a loop? i.e. 150F goes to 160F, therefore the 2nd bar should warm to something higher than 100F, in turn warming the original higher still etc. When does it stop? I would guess at a T close to double the original, which is what the “potential” T of the heated bar would be.

    • Anonymous says:

      A) I assume you are talking about warm blooded animals that regulate their body temperature. Do not know what that has to do with the thought experiment.
      B) Perhaps you mean potential temperature if there was no heat loss due to radiation. But in a thought experiment it could be without limit. Remember no reality can intervene.
      C) The march to a new equilibrium would probably be asymptotic with respect to time. But of course it they are made of ideal materials then the march would be instantaneous.

    • as someone has previously mentioned, there is no infinite feedback loop here. As the second bar warms, the energy exchanges between the bars and the walls changes in the direction of reducing the rate of warming. Eventually a new equilibrium is reached with new temperatures.

  58. eilert says:

    The key to understanding warming through radiation is this:
    A substance needs to ABSORB radiation in order for it to be warmed.
    If it is not absorbing any radiation, transmitted towards it, it cannot warm.

    Radiation is just energy in transit.
    Radiation which is deflected, reflected, dispersed or transmitted stays energy in transit, if none of it is absorbed.

    The second plate is at a lower energy state than the first plate, thus it can intercept and absorb the radiation from the first plate and gets warmer. It than emits this radiation, some of which goes back to the first plate.
    Important however: this radiation is at a lower energy state, than that comming from the first plate.
    In order not to violate the First and Second Law of thermodynamics, the first plate cannot absorb that radiation (lower energy state cannot move to an higher energy state without work – THIS is the basis of the Second Law), but only deflects it. The first plate sences the energy state through the frequency of that radiation.

    Similar if greenhouse gasses emit radiation from a position in the atmosphere, some of it most certainly can be transmitted towards the surface (even if it is in a very much colder part of the atmosphere). But the radiation emitted from this gas is most certainly at a lower
    energy state, than when it came off the surface in the first place, thus it cannot again be absorbed by that surface. (It might find its way to another already colder surface and may be absorbed there, but the likelyhood is rare – the radiation may have cooled down, to get there)

    • Joletaxi says:

      I think You are right.

      Wat was amazing in that proposal is that:

      -You have the first plate P1 emitting the full amount off energy provided by the wired source let say 100 watts.,and for that purpose, ritching a t° T1
      -then You place a second plate,who will receive a part off the energy emited by P1,and thus warming too T2 and emiting the full amount off energy received.

      something strange happend

      with 100 watts You was emiting at T1 for the P1,and now you are emiting more energy coming from P2 emitin at P2
      - then ,some photons, off what else are hitting P1 ,and T1 increase too T1′ emiting more energy again.

      My question?
      From where come that extra energy?

    • Anonymous says:

      Please elucidate the concept of statefulness of energy and its apparent filtering effects.

      • Joletaxi says:

        My question is this:

        -Plate P1 is emiting at the rate regarding is t° let say T1,with one steady input of energy off let say 100 watt,
        let asume that the outer space off the vacuum chamber is spheric.Every place in that “ball” will receive and absorb an certain amount off energy

        -placing plate P2 .An certain amount off energy will iluminate the plate, his t° will grown,an folowing the same rules, will emit energy.

        So, we still have P1, emiting at the rate off 100 watt,an now, a second source off energy, emiting at a fraction off the 100 watt rate.The only way too solve this, is that P2 is intercepting the amount off energy flowing from P1.in front off P2.Plate P2 is shaduwing a portion off the surface off the ball.So,nothing have change except that the face P1 in front off P2, has now the surface off the full P2,and thus emiting at a rate and t° in equivalence off the P2 surface.
        But the emitting surface off P1 in front off P2 emit only towards P2,and nothing go too the ball.

        - let now P2 emiting bacwards too P1.T° is growing,and so the flow off energy,going everywhere too the ball from P1.and eventualy too P2,in the same loop.

        How could You manage that the ball will now receive a little bit more energy,that the first 100 watt input,going everywhere in the first place??

    • You are making the implicit — and false in the case of the Earth’s surface — assumption that the warmest object does not have an external source of energy.

  59. John Marshall says:

    Sorry Roy, the plate experiment is not valid. Your plate is heated internally and emits energy in balance showing a constant temperature. With the second plate some energy is reflected back to the heated plate, where heating remains so energy is still emitted and with the reflected energy will increase in temperature to a new balance. In the real atmosphere any CO2 molecule will gain energy from incoming sunlight to a level which is balanced. Remove the heating, cloud moving across the sun for instance, and this molecule will immediately start to radiate heat away. This molecule, unlike your plate, is not endowed with its own energy generator but has to rely on what is adsorbed. In reality what we experience with this cloud cover is a cooling. Trenberth’s graphic double counts energy, quite cleverly and the addition has to be done carefully. The e2nergy out from the surface is only correct IF the re-radiated energy is correct. IT IS NOT. There is NO way that a colder body can heat a hotter one. Energy can only flow in one direction like a roller skate on a hill. It can only go from high(temperature) to low.
    Another problem that reality throws into these thought experiments is that any parcel of warming air will start to ascend due to getting less dense and will then start to cool adiabatically. Another variable in this chaotic

    • Anonymous says:

      Sorry John. You can argue validity of Dr. Spencer’s conclusion about his THOUGHT experiment. But after all it is a thought experiment, not sure how it became invalid. The CO2 molecules will be simultaneously absorbing and emitting energy, not necessarily the same spectrum. They obviously will not wait for a cloud or some other cover to come by
      I did not understand the segue into cloud cover. A hot body can absorb radiation from a colder body, unless you can prove that along with the energy comes some sort of a signature that says “I am from a cold body”

    • I’m am tiring of repeating myself. The Earth’s surface DOES have an external energy source…the sun, which is necessary for what I have said to be true. It sounds to me like you have not read the article or the discussion.

  60. timheyes says:

    Thank you Dr. Spencer. I must say that I’m surprised by the number of people who argue against this interpretation of the energy flows.

    It seems some readers have either a peculiar understanding of the thermodymanics of the (dare I say it?) “model” which you propose or misunderstand the “model” to be an atmospheric model rather than a demonstration of energy flow.

    If something is above absolute zero then it by definition must radiate infra-red energy. The sensible temperature of the body is due to the BALANCE of the radiation it recieved from the surroundings or that generated interally and the amount of radiation which it emits. There is a dynamic equilibrium of energy quanta set up.

    The total input energy for the fist plate is increased because the second plate must radiate some if its energy back to the first plate. This drives the equilibrium in the direction of the energy source (1st plate) rather than the energy sink (the surroundings).

    • Andrew S says:

      Here you, Dr. Spencer and many others assume that first plate can absorb “back” radiation from second plate (which itself originated from first plate, but that irrelevant side point) and get warmer. So the argument boils down to “is this assumption correct”? Some people (me included) say it violates Laws of Thermodynamics and has no physical sense.

      • Anonymous says:

        Let’s stipulate that it does so. Can you precisely state what this law is and whence you get that definition. Make sure you cite something and not you intepretation of the law.

        • Andrew S says:

          OK. My statement that hotter object cannot absorb radiation from colder object is just my theory. However, Dr Spencer’s thought experiment cannot disprove my theory, because is assumes just the opposite. On the other hand, I may suggest other thought experiment in support of my theory with understanding that as a thought experiment it either cannot prove my theory.

          Let’s continue Dr.Spencer experiment with two plates, one is electrically heated, the other one is passively headed by the first one. Let’s assume (as Dr. Spencer does) that the first one has temp equal to 150 degrees C before adding second plate and second plate reaches 100 degrees C in presence of heated plate.

          We will however slightly modify experiment. Our heated plate will be equipped with thermometer and regulation mechanism will regulate electrical intake to keep it at 150 degrees at all tiles.

          Let’s add another heated plate that is equal in all respects to the first one. Obviously, passive plate will heat some more, however the temperature of passive plate will necessary stay below 150 degrees. Do you agree so far? Let’s as add some mirrors that will reflect heat from heated plates to passive one. It will force passive plate to heat some more, while still staying below 150. OK so far?

          We can introduce more and more object into our system, the subject of our experiment will still be below 150 as long as no added objects are hotter then 150 degrees. Can we agree on that?

          Given all the above is true, how can addition of cold GHG into the atmosphere cause Earth surface to warm above that of GHG?

          • Anonymous says:

            What you postulate in your model Andrew, is to reduce the heat input thermostatically to prevent the temperature rising above 150 deg. OK so far? But the Sun is not that condescending. It keeps blasting that radiation out whether we get warmer and warmer or not!

            Bomber_the_Cat

          • Andrew S says:

            If tone of my comment seems condescending I apologize. It wasn’t my intention. However, if there is logical fallacy in my reasoning I’d like to hear about it. Sun was neither part of original Dr. Spencer thought experiment, nor mine.

          • Once you thermostatically regulate the temperature of the heated plat you have, a priori, kept its temperature from changing, no matter how fast it can shed heat to its surroundings. In the case of adding the second bar, the tendency of the first plate is to increase in temperature as its rate of energy loss has been reduced. The thermostatic control mechanism will then reduce the rate of electrical resistance heating to keep the heated bar at a constant temperature.

      • I am not “assuming it”, and the 2nd Law is not violated because the net flow of IR is always from warmer to colder temperatures.

        • Anonymous says:

          Dr. Spencer. I do not know how much heated plate absorbs thermal radiation from passive one in your thought experiment. To work around it I put thermostatic regulation to it and shift my focus to the other plate.

          Here is my general hypothesis which may be true or false: given an object with thermal equilibrium temperature A, when another (thermostatically regulated) object with higher temperature B is put in near proximity to the (passive) object, thermal radiation from second object will heat the first one. However, temperature of the passive one will stabilize below B and at no point in time can it exceed B.

          Now, there are three possible outcomes. If my hypothesis is correct, it trivially follows that there is no such thing as “backradiation? If my hypothesis is false, it equally easy to show that Second Law of Thermodynamics is broken. The third outcome is that there is logical fallacy in my reasoning.

          I’m genuinely interested to know which one is the correct one.

          • Anonymous says:

            “Here is my general hypothesis which may be true or false: given an object with thermal equilibrium temperature A, when another (thermostatically regulated) object with higher temperature B is put in near proximity to the (passive) object, thermal radiation from second object will heat the first one. However, temperature of the passive one will stabilize below B and at no point in time can it exceed B.”

            To clarify, we have an object maintained at a temperature B (let’s call this object B). We introduce a 2nd ‘passive’ object at a lower temperature A (let’s call this object A). Will the temperature of object A ever rise above that of B (if its source of heat is only from object B) ?
            Is that the question? If so the answer is NO, so your hypothesis, as I understand it, is absolutely sound.

            Unfortunately, I cannot follow your non-sequitur, “If my hypothesis is correct, it trivially follows that there is no such thing as backradiation?”. How? Why? What did you expect to happen?

          • Here is my general hypothesis which may be true or false: given an object with thermal equilibrium temperature A, when another (thermostatically regulated) object with higher temperature B is put in near proximity to the (passive) object, thermal radiation from second object will heat the first one. However, temperature of the passive one will stabilize below B and at no point in time can it exceed B.

            The situation you present could be used (I think, not sure) to demonstrate the same concept IF you monitored the amount of electricity being used before and after putting the second bar in the vacuum chamber. Importantly, the the comparison must be done between equilibrium states, i.e., after the two set-ups have equilibrated to a constant temperature. You would find that less electricity would be used to maintain the actively heated bar at a constant temperature.

          • Andrew S says:

            “Unfortunately, I cannot follow your non-sequitur, “If my hypothesis is correct, it trivially follows that there is no such thing as backradiation?”. How? Why? What did you expect to happen?”

            Here is how. In my thought experiment I have control over temperature of object B as it is thermostatically regulated.
            I can lower temperature of object B to be as close to that of object A as I want. Now, look what’s happening. I put object B with temperature arbitrary close to that of object A side by side with that object A. Still, object A is not allowed to warm past arbitrary chosen temperature B, as you just agreed with me.

            Compare that to Dr. Spencer’s experiment where object at temperature equilibrium of 150 degrees gets warmer in the presence of colder object nearby. Do you agree that two scenarios are mutually exclusive? Which one is correct? Why?

          • Andrew S says:

            “The situation you present could be used (I think, not sure) to demonstrate the same concept IF you monitored the amount of electricity being used before and after putting the second bar in the vacuum chamber. Importantly, the the comparison must be done between equilibrium states, i.e., after the two set-ups have equilibrated to a constant temperature. You would find that less electricity would be used to maintain the actively heated bar at a constant temperature.”

            That is the essence of my disagreement with “backradiation” theory. You say warmer object absorbs “back” radiation from colder one, hence less electricity is needed to preserve it temperature. I say warmer object cannot be warmed by colder one, hence there is exactly same amount of electricity is required to keep the object’s temperature.

          • yes, I know that is what you are claiming.

          • Andrew S says:

            Dr. Spencer. I’m not sure your reply addressed to me or somebody else, so I (naturally) assume it’s me you are talking to ;-)

            After thinking about your experiment I found another line of (admittedly naive) argument why your experiment doesn’t work. Assume that it does work as you say. An object in your experiment reaches 150 degrees by using constant electrical intake. You put colder object nearby and temperature of the first one increases somewhat. The reason for increased temperature is the object now has an additional source of heat.

            That means the object increases frequency it emits photons. Frequency (energy level, “temperature”) of the emitted photons also increased. However, increase of radiation rate must be equal to the rate that object absorbs radiation from colder object (after changed system reaches equilibrium, of course). But the object absorbs “colder”, lower frequency photons from colder object. That must mean it absorbs more photons (per unit of time) from colder object than difference between radiation frequencies of the object in new state and original state. That is, in simple terms, for every 5 absorbed photons at “temperature” 100 degrees it will emit 2 photons back at “temperature” 160 degrees to keep its equilibrium state.

            I believe that precisely this mechanism handful of backradiation theory skeptics call “unscientific heat pump.”

            Any comments?

        • Andrew S says:

          Sorry, forgot to sign it.

          Andrew S.

  61. Mindert Eiting says:

    Dear Dr. Spencer,
    As a non-physicist I would call what is termed thermal equilibrium the heat capacity of a body: a plate of iron does not cool instantaneously when the current is shut off. Is that correct? Did you increase the heat capacity by adding the second plate? Are we talking about the heat capacity of our atmoshpere? Does that depend on the air pressure (number of molecules per volume)? I can understand that CO2 adds to the atmospheric heat capacity by its radiative properties but that is not because of pressure. Am I correct? I do not know the literature well enough, but apart from clouds and winds, why not an experiment in a large hall (greenhouse) with our present atmosphere. Warm it from the outside by the sun and add CO2. What will happen? I remember one article about such an experiment around 1908 or something like that (forgotten the name of the physicist).

    • yes, the heat capacity of the system was raised when the second bar was added. But what I am talking about is the temperatures after *equilibrium* has been reached. Heat capacity is indeed important to the temperature of the system while the temperature is changing, but (after thinking about it some mpore) not to the final temperature of the system when it has reached equilibrium.

  62. Derekcolman says:

    Dr. Spencer, your hypothetical experiment is very impressive, and very convincing, but I think I have spotted the flaw in it. You have a vacuum between the two plates, but in the atmosphere the gap between CO2 molecules and the ground is not a vacuum, but is filled with atmospheric gases. These other gas molecules which vastly outnumber the CO2 will be convecting heat upwards. I would contend that if the space between your two plates was filled with air which could freely move, the amount of back radiation to the warmer plate would be substantially reduced as those moving gas molecules would surely absorb some of the radiation and convect it away. I subscribe to the view that CO2 does have a greenhouse effect, but that at only 385 ppmv its influence on global temperature is far smaller than some would have us believe, and a rise of 100 or so in ppmv over a century cannot account for a 0.7 deg.C rise in temperature.

    • Anonymous says:

      The amount of radiation is predominantly dependent on the Temperature and the of course the natureof the material. Convection may or could be larger that radiation but cobections presence does not — by any known mecahnism — stop or impede radiation.

    • Anonymous says:

      At the risk of doing my Karnak act and divining Dr. Spencer’s reason for using a vacuum was precisely not to muddy the radiation argument. The presence of gas molecules is only germane if we are talking radiation absorbing molecules. Otherwise they tend to bring in convection which could be useful in an overall argument but not in a radiation only scenario. Convection might overwhelm radiation – depending on the situation – but does not switch it off.

    • Again, the vacuum chamber example was only meant to show that a temperature increase can indeed occur if the rate of radiative cooling is reduced.

      The MAGNITUDE of the CO2 absorption of radiation is a separate question, I agree with you on that.

  63. Colin Henderson says:

    The net flow of heat is only in one direction, by slowing it down you can cause a buildup of energy and therefore temperature but that increase in temperature is in no way caused by energy flowing backwards. The concept that the net flow of heat can be reversed at all, let alone by a trace gas is absurd.

    • Anonymous says:

      Actually it is caused by energy flowing backwards.

      When there was only one plate it reached an equilibrium temperature of 150C (which means the rate of energy inserted by the heater is equal to radiation emitted from plate when at 150C).

      When we add the second plate the only way that the first plate can reach a higher equilibrium temperature is if it has an additional source of energy. It receives energy from the second plate, because it is emitting radiation.

      Unless you are suggesting that the first plate stops radiating as much energy because it has knowledge of the presence of the second plate.

      I don’t see anything absurd about what Dr. Spencer is saying. If there are 1000 cars per hour driving north on 1st street and 100 cars per hour driving south are you saying that there aren’t any cars driving south? Are you suggesting that it would be irrelevant if the city removed the south bound lane?

      • Anonymous says:

        Colin, sorry I missed adding my name on that last comment.

        Mike Blackadder

      • Andrew S says:

        “When we add the second plate the only way that the first plate can reach a higher equilibrium temperature is if it has an additional source of energy. It receives energy from the second plate, because it is emitting radiation.”

        Is there any substantiation in Thermodynamics to suggestion that hotter object (first plate) can absorb thermal radiation from colder object (second plate)?

        • Anonymous says:

          Any object can absorb any radiant energy or perhaps reflect or r-emit it. Your version of Thermodynamics is unfamiliar to me, can you cite anything that the Laws of Thermodynamics say about radiation absorbtion or the absorbtion’stempature dependence.

          • Andrew S says:

            Sure, atmosphere, for example, (virtually) cannot be warmed by visible light. Polished silver ball (virtually) cannot be warmed by any radiation at all. It all depends on properties of objects receiving radiation. These properties change with temperature. As physical body warms its radiative spectrum changes, so changes spectrum of absorbed radiation.

            I am surprised that ability of colder object to warm warmer object remains as a thought experiment. It seems easy to reproduce that experiment in real life and resolve this issue once and for all.

  64. Mike Blackadder says:

    I think my last post has been lost somehow (if it reappears then I apologize for repeating myself).

    I only wanted to say that in my original comment I wasn’t clear about my overall impression of your post. I think your example is actually a pretty clear way of demonstrating the point about how radiation from a colder body will increase a warmer body’s temperature.

    I’ve seen similar posts elsewhere that use an over-simplified model of an atmosphere. If anything your choice of model was somewhat diplomatic.

    I wouldn’t accuse other commenters of being in denial or non-scientific. This is a typical reaction to arguments like yours that are difficult to reconcile intuitively (ie. anyone heard of the Monty Hall problem?). It’s unfortunate though that some commenters accuse you of being dead-wrong about a point which is easily proven. ie. Your ‘gut’ isn’t a good way to prove that Roy’s violated the laws of thermodynamics.

  65. Mike Blackadder says:

    Ok, the jury may be out regarding my understanding of physics, but I am definitely inept at posting comments. Once again, I apologize if I am repeating myself with multiple comments.

    Colin,

    Energy flowing backwards (from 2nd plate to 1st plate) is the only explanation for the increase in temperature of the first plate. When we had only one plate rate of energy from the heater was equal to the radiated energy from the plate when at 150C.

    When the temperature of the 1st plate increases its rate of radiated energy also increases. How then can it be in equilibrium at a higher temperature? Because it receives radiation from the 2nd plate, because the 2nd plate has non-zero temperature and clearly in this example has the ability to absorb and emit radiation.

    Unless you are suggesting that the first plate reduces its emission at 160C because it has knowledge of the presence of the second plate.

    Eg. 1000 cars per hour are driving north on 1st street. 100 cars per hour are driving south on 1st street. Are you arguing that there are no cars driving south? Would it be irrelevant if the city removed the south bound lane?

  66. Stephen Wilde says:

    I think I can make this very simple.

    A cooler object near a warmer object will slow the rate of cooling of the warmer object because of the exchange of energy to and fro between the two objects before that energy departs the combined system.

    If there is a constant energy input throughout then the slowing down of the rate of cooling will cause the warmer object to settle at a higher equilibrium temperature than would otherwise have been the case.

    Thus the presence of the cooler object does indeed result in the warmer object becoming warmer than it otherwise would have done.

    That is the essence of the greenhouse effect which is a concept I have always accepted despite the misleading nomenclature.

    Any sceptical viewpoints that rely on denying those simple facts must be rejected because they weaken the sceptical cause.

    • DEEBEE says:

      Nicely distilled

    • Colin Henderson says:

      I see your point, the cold body acts as a secondary heat source increasing the amount of energy going into the warm bodys and increasing the warm body temperature equilibrium. The warm body, now being at a higher temperature increases the radiation falling on the cold body, which in turn radiates more energy to the warm body, further warming it. I just can’t seem to get my head around where this loop ends, it seems like a perpetual motion machine and/or an M.C. Escher drawing.

      • Anonymous says:

        Colin,

        There is no ‘loop’.

        There is only a slowing of the rate of cooling. No energy is being added to either of the two bodies.

        With a constant source of energy from an outside source one only needs to slow the rate of cooling to achieve a higher equilibrium temperature.

        Stephen Wilde

      • You are creating confusion by thinking of the colder body as a “heat source”. It is better to think of it as a thermal insulator that operates radiatively. There is no new energy generated.

    • Andrew S says:

      Stephen says:

      “A cooler object near a warmer object will slow the rate of cooling of the warmer object because of the exchange of energy to and fro between the two objects before that energy departs the combined system.”

      So says the “backradiation” theory. Mind you it *is* a theory. Is there any validation of that theory in real life? I mean except that diagram in IPCC report.

      “Any sceptical viewpoints that rely on denying those simple facts must be rejected because they weaken the sceptical cause.”

      So, anyone how doubts the theory you support is in denial and must be ignored. Well said!

  67. DEEBEE says:

    Dr. Spencer you are a courageous soul. ScienceOfDoom tried this a few weeks ago. The similarity of flames is amazing. The detractors here only seem to be
    a)Trying to emulate you (looking for an honorary PhD?) by extending your thought experiment. Rejecting it outright because it is not real, or trying to add “reality” to it with personal views of scientific factoid they have assimilated.

    b)Propound undocumented (illegal?) personal versions of the Laws of Thermodynamics.

    c)Chastise you for not doing real science but though experiments. “Paging Stephen Hawkings find a new job”

    The biggest difference is that none of the respondents, as yet, have waxed ineloquent into Kirschoof’s Law, even though you had the “temerity” to mention it. But then again the post is still young.

  68. Dirk says:

    The idea that CO2 is a stronger greenhouse gas has something to do with the structure of two oxygen atoms coming off a carbon atom as opposed to just two oxygen atoms, and the ability of CO2 to absorb infrared radiation and then re-radiate it.

    I’m not sure how this is measured (as opposed to modeled), but I know one of the facts cited to support it is the temperature of the planet with the atmosphere vs. what it would be if not for the atmosphere.

    One question- in this calculation of the temperature of the earth vs. a theoretical blackbody- is there an accounting for the internal heat of the earth? We don’t even know the temperature at the core of the earth- it could be thousands of degrees- couldn’t this account for some of the warmer nature of the earth?

    • Just about everyone has ignored heat from from the Earth’s interior through the Earth’s surface. As I recall, it is estimated to be something like 1 W/m2. For many land regions, the temperature 5 or 10 feet underground is the same as the average air temperature at that location (annually-averaged), supporting the assumption. But at the sea bottom, who knows? I’ve never seen a detailed study of this, but it might be out there somewhere.

  69. old construction worker says:

    I’m sort of a day late on this discussion, but I don’t think I’m a dollar short.

    You have set the bar being heated at 150 degrees f and the walls of the chamber at O degrees f. You have put the second bar in the chamber which make the first bar “hotter”, 160 degrees. What happens if the second bar is moved to and touches the walls of the chambers? Would the second bar be equal the O degree like the walls and the heated bar stay at 150 degrees f or would the “back radiation” be so slight that it could not be measured?
    Same thing with the “blanket”. Lets say I’m in a tent camping out in the middle of winter and I add another blanket to to kept warm, but instead of putting the blanket over me I put the blanket over the tent. Can the blanket effect be measured?

  70. Scott Scarborough says:

    Old construction worker- Would the second bar be equal the O degree like the walls and the heated bar stay at 150 degrees f or would the “back radiation” be so slight that it could not be measured?

    The answer is NO. The bar would not be at 0 F because real materials do not have infinite thermal conductivity. As you move the colder bar further away from the warmer bar (heat source) it will have less of a warming effect on the warmer bare.

    Old Construction worker – Same thing with the “blanket”. Lets say I’m in a tent camping out in the middle of winter and I add another blanket to to kept warm, but instead of putting the blanket over me I put the blanket over the tent. Can the blanket effect be measured?

    Yes putting a blanket over the tent will have an effect. In fact, If the blanket is made thicker to make up for the fact that the tent has more surface area than you your steady state temperature would be the same as putting the standard, thinner blanket over you. But it will take longer to reach that steady state temperature.

  71. benpal says:

    Dr Spencer, thanks for trying to explain this phenomenon in simple terms. As a layman I’m somewhat surprised that the very basic question of radiation doesn’t seem to be solved in physics. After all these billions poured into climate research, shouldn’t there be a “consensus” on how radiation works based on reproducible experiments?

    I think the term “back-radiation” confusing. Each of the bars absorbs whatever radiation it gets from the surrounding. And each bar emits according to the heat energy it contains. So the second bar doesn’t radiate “back”, it simply radiates based on its energy content, not knowing where it got its energy from in the first place, and it would do so even if the first bar was removed.

  72. jae says:

    Roy, re: 7/24 at 8:56 am. where you say:

    “Actually, the emitting temperature corresponding to 1000 Wm-2 is not 190 C, but 91 C, which is close to 190 F. The trouble with your claim is that existing heat loss mechanisms are so large (in addition to IR, convective air currents and loss to the soil below), that unless you know those half-way accurately, you have no idea of what kind of downwelling IR is happening.”

    Yeah, I screwed up with the degree C/F. But the point stands. I don’t think it is possible that the heat losses can be THAT large, and you have presented absolutely no evidence to the contrary, only vague speculation (which is all too common in climate science). We will never know, unless you can you give me a reasonable figures (that means actual data) for “backradiation” at an average noon, July 21, Atlanta, FL on a clear day. Can you do this? Based on past experience with these unending debates, I doubt it. We desperately need empirical data to confirm the GHG effect, and I can’t find any that support any GH theory, despite YEARS of seeking such (something is really wrong with this picture, eh?).

    You and all the other folks who are pushing this hypothesis have absolutely no empirical evidence for it. Therefore, there is no science here, yet. Only speculation, which is no better than my own speculation. No better than Thieme, Miskolski, Siddons, etc. Consult the writings of Dr. Feynman for more info. on this problem.

    Please admit to some uncertainity here and stop being so damn dogmatic.

  73. jae says:

    BTW, Roy, you can shield the asphalt surface with an IR transparent film, like NaCl, to stop convection), like Woods did, and I will bet you $100 that you will still not see an increase in temperature. Something is definitely wrong with your hypothesis, I think. We desperately need some empirical evidence for all this GHG stuff!!!

    • Anonymous says:

      Fascinatin g to watch you accuse Roy of being fogmatic and not having any science but sputing yourdogmas. The pot callingg ……

    • there is all kinds of empirical evidence, Jae. Upward looking IR radiometers, spaceborne downward looking radiometers, lab measurements of IR absorption of CO2 at various temperatures and pressures, energy budget calculations to explain (for instance) the day-night range of temperature changes throughout the atmosphere.

      • Anonymous says:

        But why then do I find in a Wiki-table that the specific heat capacity of CO2 is less than nitrogen, oxygen or normal air? If our atmosphere consisted entirely of CO2 it would cool more rapidly than at present.

  74. Kevin says:

    It’s the speed of the effect that matters. I suggest that if you allocate just 1 kilowatt hour to Dr. Spencer’s thought experiment you will find that bar #2 may reach a higher peak temperature but it will then cooldown for a longer portion of each day. Therefore the average temperature does not increase. For the “greenhouse effect” to cause a slow increase in the temperature of the Earth, it must exhibit a speed of heat flow that is comparable to the Day/Night cycle (i.e. 24 hours), last I checked a photon travels from the surface of the Earth to the top of the Atmosphere in about 100 microseconds, so even if it is absorbed and remitted by a “greenhouse gas” a thousand times it still travels through the atmosphere of the Earth in a second. That seems a very slight fraction of a 24 hour day.

    Cleary case #2 (two bars) requires more ENERGY from the
    electric company to fill bar #1 and bar #2 with ENERGY. If you examine the “energy budget” of the Earth postulated by the climate scientists you will see that all of the units are units of power (watts/meter squared). Any real budget must be careful to employ the correct units of the effect (i.e. ENERGY) that it purports to “audit”.

    I suggest a simplified thought experiment whereby bar #1 is “filled” at sunrise with 1 megaJoule of energy. Then we can examine the temperature of bar #2 versus time after that energy is supplied. I postulate that in the two cases presented (1 bar .vs. 2 bars) the single bar case will reach a lower peak temperature but maintain that temperature for a longer period of time. In the two bar case bar #1 will reach a higher temperature, but it will cool down faster (due to the thermal conduction inside bar #2 being more effective at cooling bar #1 than the radiation only condition present with just one bar). In either of these cases the “average” temperature of bar #1 does not increase because of the substitution of heat transfer mechanisms (i.e. more radiation means less conduction/convection).

    I suggest the following thought experiment:

    1 – At the beginning of each cycle (i.e. at sunrise) a fixed amount of energy is supplied (Joules, BTU’s, Calories, you choose)

    2– Please demonstrate how you can adjust the ratio of heat transfer mechanisms (some of which, i.e. radiation which is postulated to be much faster that the cycle time of the energy input to the system) to show that the “average” temperature at the end of each cycle is higher ?

    Cheers, Kevin.

    • no matter what the time of day is, adding “radiative insulation” will cause a warmer temperature for the heated surface. The magnitude of the temperature increase might change day-vs-night, though.

      • Anonymous says:

        Dr. Spencer,

        With respect my point is not that the added insulation varies in it’s effect during the daily cycle. It probably does not vary by much.

        Care to repeat your experiment and turn the power supply off after 1 kilowatt hr has been delivered to each system. Please demonstrate that the total energy (energy in bar #1 + energy in bar #2) goes up at the end of each cycling of the power supply.

        If this is the case the temperature of the earth must ratchet up at the end of each day, everyday since the beginning of time. Surely we should be at 1 million degrees by now.

        Or perhaps as the Earth warms just a little bit it radiates FASTER and automatically cools backdown ! Seems like the speed of heat through the atmophere “self-modulates” and prevents a permanent warming trend !

        Cheers, Kevin.

      • Anonymous says:

        Interesting, I posted a refined thought experiment regarding turning off your power supply to the heater after 1 kilo watt hr was delivered to each system. Care to post that reply ?

        To refine my reply, please explain how the energy in the two systems (1 bar vs 2 bars) is greater 24 hours after the power supply is turned on then off ? (1 kWh at 150 watts is about the cycle time of daylight if I have my thought experiment math mostly correct)

        Or perhaps you wish to keep this discussion going for a little longer ?

        I respect your work, and I find some of the comments most enlighting, but really how much longer can this hoax go on ?

        Clearly if the insulative properties of “GHGs” cause the Earth to warm, it will also radiate faster to cool back down. The predominate cause of the temperature of the Earth is the thermal capacity (in totality) of the Oceans (HUGE), the Rocks (BIG) and lastly be the gases (TINY).

        Cheers, Kevin.

        • Anonymous says:

          Dr. Spencer, I apologize if my “second” post was a bit snarky, my internet access is somewhat sporadic and I did not realize that my “first” post was published in a timely manner WITHOUT censorship (thanks for that BTW).

          However I still suggest the thought experiment whereby the power supply to the heater is turned off after 1 kiloWatt hr is supplied ? Please describe how the temperature of the bars within the vacuum chamber continues to slowly rise.

          Note that the Apollo Lunar Rovers (a real feat of engineering at the time) used WAX to store and then later release the heat generated by the electronic systems in those “electric cars”. Again, I still strongly believe that the solution to this conundrum of AGW is the speed of heat through a system regardless if the system is a lunar rover, or the atmosphere of the Earth.

          Cheers, Kevin.

          • once the power is shut off, the temperature of the heated bar will start to fall. Insertion of the second, COOLER bar would then slow the rate of temperature drop, but could NOT cause the temperature of the first bar to rise.

  75. jae says:

    NO EMPIRICAL DATA = NO SCIENCE FULL STOP

    • Anonymous says:

      There goes Theoratical Physics. Do notunderstand why views have to be expressed in concrete terms. Tomy eyes that seems like you and other detractors are trying to shut the dialogue. This sort of exchange can either silence one side because it is a waste of time or degenerate into “PROVE IT, NO YOU PROVE IT”. If you have an assertion to make give some next level reasoning for the same so we can follow along. Snideness is not a subsitute.

      • Anonymous says:

        You just don’t get it, sir. It is up to Einstein to prove his theory, not me. And he did. Now, it’s time for the GHG scam artists to prove their arguments with ACTUAL DATA. And that ain’t gonna happen, because they have no real data, just bullshit words.

  76. Alex says:

    Dr. Spencer you are a cruel and devious man. My compliments. I’m curious how you are going to explain it away.

  77. Alex says:

    I am plugging in my 1000 watt heater and filling my room with inert objects. If I don’t get $10 million dollars immediately, I will plug in a second heater and fill it with inert objects as well. If I don’t get my money I will destroy the world with my supernova. If you are concerned about the vacuum, don’t worry. I will attach the vacuum hose to my head.

  78. william says:

    ‘Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Stil’

    wow, hope to see this used in thermal insulation,

    getting more heat out of the initial heat source!!

    • William, you are hurting your credibility. Are you saying that adding thermal insulation does not increase the temperature on the warm side of the insulation? You do not have to create new energy to increase temperature when energy is continuously being pumped in from an external source.

      • william says:

        i put a thermometer under my arm and measured my temp, i then put on a massive coat and measured after a while, temp was the same.

        • we’ve already discussed that a thermostatically controlled heat source is a special case…although if you could take the temperature of exposed skin, then covered skin, it would go up slightly.

  79. william says:

    i am sure if co2 actually caused warming it would be used in thermal insulation, how easy to make clear tubes, put in high levels of co2 and let the sun do the rest,

    anyone for clear co2 tubes on their roof tops.

    perhaps we should ask the guys who have greenhouses where they pump up the co2 levels to 1000ppm whether they see any difference in temps to the greenhouses without.

    • Anonymous says:

      William, perhaps you realize how close to the truth you are. The insulative power of gases.Because of the relative rarity (low density) of molecules, if you can stop convection, they make great insulators. And having fluffy fibre glass is just that.

    • the path length is too small to see such an effect. Remember, doubling CO2 through the entire depth of the atmosphere (approximately 1 mb) adds only 1 deg. C. At sea level 10 meters of air is about 1 mb, so (*very roughly*) a vertical depth of 30 feet would add only 0.001 deg. C.

  80. Baa Humbug says:

    Back here a second day and it seems we’re all still scratching our heads.

    Dr Spencer is saying that the addition of a second bar will increase the temperature of the first bar from 150F to 160F.

    Following the above thought experiment, what would happen in the following real life example?

    Take your average 1000W bar heater. This puts out about 190F of heat. Can I increase it’s heat output to say 200F just by introducing a bar alongside of it?
    Should I start designing patents for super efficient bar heaters? We can add 3 or 4 benign bars to the heater and increase it’s output to 220F, all for the same 1000w of input.

    • Anonymous says:

      To my view the numbers were just forillustration so as perhaps not get into symbols. Would you be more mollified if the Doc had used 150 and 150.00001. The magnitude, I think, was not esential to explicate the concept.

    • like others, you are confusing temperature with “heat output”.

  81. william says:

    stop convection! that’s the point, heat rises, quickly in the desert at night with little moisture.

    if you stop the movement of heat from anything be it stone concrete etc it would be wonderful.

    I always like the example of heated water in pot A on the stove, pour into pot B, put pot A back on the stove, pour water immediately back into pot A has pot A become hotter.

    of course imagine pot B is higher up than than pot A

    This is more the theory of co2 causing warming or rather not.

  82. william says:

    in fact to get the heat from pot B to pot A would i not need a heat pump like a fridge.

  83. TheLastMan says:

    William
    July 26, 2010 3:36 AM

    Take your average 1000W bar heater. This puts out about 190F of heat. Can I increase it’s heat output to say 200F just by introducing a bar alongside of it?
    Should I start designing patents for super efficient bar heaters? We can add 3 or 4 benign bars to the heater and increase it’s output to 220F, all for the same 1000w of input.

    What a terribly confused person you are.

    A 1000w bar heater does not put out “190F” of heat, it puts out “1000 watts” of heat, that is why it is called a “1000w heater”, not a “190f” heater.

    Think about it carefully, put a 1000w heater in a small metal box 4ft high x 4ft wide x 2ft deep with mineral fibre cladding located in Houston, Texas. Let’s call it an “oven”.

    Put another 1000w heater in a 500,000 sq ft warehouse with a 80 ft ceiling located on the West Antarctic peninsula. Lets call it the “woefully inadequate space heater”.

    Would the oven in Houston reach a higher temperature than Antarctic warehouse?

    If we surround a 1000w bar heater with 3 or 4 “benign bars” we end up with 3 or 4 hot bars. The total heat output is still 1000w, some of the heat is emitted by the bar heater some by the benign bars.

    If we stick 10ft thick panels of highly effective ceramic foam cladding (the sort they stick to the bottom of the Space Shuttle) immediately surrounding a 1000w bar heater such that any heat is almost completely prevented from heating. Turn on the 1000w heater. What temperature will the 1000w heater reach? Probably high enough to cause the bar heater to melt.

    To me, Dr Spencer’s piece is very well argued and perfectly clear.

    • TheLastMan says:

      Sorry that should have been “any heat is almost completely prevented from escaping”.

    • william says:

      it will only get as hot as the source!! you can not make the source heat hotter.

      funny you cpmpare buildings to extra co2, if you would like to use this analogy of causing warming,

      at 380ppm of co2 this would be like about 99.7 % of the roof missing- some roof to cause warming!!

      as we know with the desert heat, at night this is lost in a few hours, no heavy cladded roof there to compare with.

      • TheLastMan says:

        it will only get as hot as the source!! you can not make the source heat hotter.

        You do not state what “it” is. You talk previously about a “1000w” heater, so I presume that is what you are talking about. You also do not talk about which of my scenarios you are referring to, so I will assume you mean the last one, that is where the 1000w bar heater is encased in an almost perfectly insulating material.

        In which case… you are just plain wrong I am afraid. You are displaying your ignorance and I suggest you go away and take some very basic physics lessons.

        You are confusing heat output with temperature. A 1000w bar heater has a given heat output NOT a given temperature. The reason it does not melt in your living room is that the heat it emits is the same as the heat it generates i.e. 1000 watts. That is because your living room is a very effective heat sink.

        If you put something generating 1000w of heat into a perfectly insulated chamber then that chamber (and the things in it) will rise in temperature indefinitely.

        Of course nothing is a perfect insulator, and every man made heat source and/or insulator will break down above a certain temperature. In my last example the copper wires taking the current to the bar heater will melt when the temperature inside the chamber gets to about 1,085c (1,358f), which it would – eventually.

        How do you think insulation works? In winter, with an outside temperature of 35f, a well insulated room will get to 75f using just a 100w light bulb, whereas a poorly insulated room will fail to reach 50f with a 1,000w bar heater.

        As has been stated time and time again by Roy and others in these comments, the temperature of a heat source will rise until the heat emitted is in balance with the heat input. Anything that slows down the ability of the heated item to radiate that heat away (something blocking, or insulating it) will cause the heated item to rise in temperature.

        A rising proportion of CO2 in the air, by reducing the radiation of heat from the surface of the Earth, will cause the temperature of the air to rise.

        This is a fact. This is not in dispute by any serious scientist. The dispute is about “how much” the temperature will rise not “whether” it will rise. What are the feedbacks? Are they positive or negative?

        I am done with this conversation. If you still do not understand it then reading blogs like this is pointless. You need to go and get a decent basic education.

        • william says:

          as we know the atmosphere is not a well insulated room at all, we see this in the desert at night with low humidity a rapid loss of heat. from 100 degrees to minus degrees in hours.

          • william says:

            we need to stop the cpmparisons with buildings/ greenhouses where there is no movement of air.

          • William, do you realize you are helping to make my point? It cools much more rapidly in the desert than it does here in Alabama (we see about a 20 deg. F swing this time of year, closer to 30 deg F in the winter).

            The more rapid cooling is because the downwelling IR from the sky is much less when the humidity is low than when it is high.

      • “hot as the source” has no meaning here…the electric heater pumps in energy at a constant rate. If you insulate the thing being heated, its temperature will rise.

        Yes, in the desert at night it cools off, which means the surface (and lower atmosphere) are losing IR to space. This does not prove that there is no downward IR radiation.

        • william says:

          insulation, we have to think of insulation as a complete body surrounding a heat source ,lagging of waterpipes to stop freezing,

          as we know with co2 this is not the case. co2 at 380ppm or so would be the equivalent of a roof with 90% missing. does co2 store heat , if so an amazing quality, if it releases the heat this would be upwards, there is no insulation around co2. in simple terms the release of heat upwards is same amount of heat that entered the co2. as Mr spencer said this heat is being added to the whole time, the co2 is not a fuel that adds heat itself.
          .

          moisture in the desert, a green house gas, i would assume during the day it has a cooling quality, at night has the ability to stop us freezing, once again not really adding heat.

  84. TheLastMan says:

    I think the responses to your very clear and simple article demonstrate very clearly why the current level of debate on this subject has reached such a nadir.

    Some people just do not understand the science, and even rather make up their own theories (as if nobody had bothered to think about this stuff in the last 4,000 years of scientific endeavour).

    No laws of thermodynamics are violated by the theory of greenhouse gases. Dr Spencer is trying to educate you about well proven and established science. It is pointless arguing with him, he has probably forgotten more than you will ever know about this subject.

    Among scientists working on climate, the only point of contention is the extent of feedbacks, and their sign. Every properly qualified sceptical scientist understands and accepts:
    – The earth has warmed over the last century
    – Adding CO2 to the atmosphere increases the greenhouse effect.

    Now stop arguing about stuff that is already very well proven and demonstrated and start discussing the real science.

    • Colin Henderson says:

      While most people agree that:

      A) The earth has warmed over the last century
      B) Adding CO2 to the atmosphere increases the greenhouse effect.

      there is no agreement, and no data, that proves a cause and effect relationship between A and B.

      • I would say the evidence is tenuous at best.

        • Anonymous says:

          Colin Henderson (July 26, 2010 at 12:19 PM) said:

          “While most people agree that:

          A) The earth has warmed over the last century
          B) Adding CO2 to the atmosphere increases the greenhouse effect.

          there is no agreement, and no data, that proves a cause and effect relationship between A and B.”

          In reply, Roy W. Spencer, Ph. D. (July 26, 2010 at 12:34 PM) said:

          “I would say the evidence is tenuous at best.”

          In the following remarks, I follow up on Henderson’s comment and Spencer’s reply by falsifying the thesis that there is a “CO2 atmospheric greenhouse effect.”

          To begin my proof, for every “effect” there is a “cause.” That there is a cause and effect relationship implies that: a) the event of the cause preceeds the event of the effect and b) given the event of the cause, the event of the effect occurs with a probability of 1.

          Spencer’s thought experiment identifies a cause and effect relationship. The cause is introduction of the unheated plate. The effect is a rise in the temperature of the heated plate.

          By analogy, one might reason that there is a cause and effect relationship in the atmosphere wherein the cause is a rise in the CO2 concentration and the effect is a rise in temperatures at points in Earth’s surface. However, this reasoning is refuted by the numerous observations of events in which a rise in the CO2 concentration was followed by a fall in temperatures at points in Earth’s surface. These observations falsifiy the existence of the “CO2 atmospheric greenhouse effect,” for given the cause, the probability of the effect is not 1.

      • TheLastMan says:

        While most people agree that:

        A) The earth has warmed over the last century
        B) Adding CO2 to the atmosphere increases the greenhouse effect.

        there is no agreement, and no data, that proves a cause and effect relationship between A and B.

        I didn’t say there was. My point was there is nothing to be gained in arguing over well established and proven facts. The argument has moved on and people should get over it.

        The argument is over how much of the observed warming is caused by increased greenhouse gases and how much from other causes.

        My difficulty with the non-greenhouse gas argument is what else could be causing the warming?

        The sun is going through a very low output phase, yet despite this we are seeing close to record temperatures (as shown in Dr Spencer’s satellite data).

        Dr Spencer postulates that cloud cover may be a negative feedback, in which case the global average air temperature is rising despite the presence of a significant negative feedback.

        If the warming is an internal forcing, there must be some unknown multi-decadal cyclical effect trapping more of the Sun’s heat in a way not linked to rising CO2 (and other GHG) levels. El Nino / La Nina run on too short a time scale and nobody has yet shown a mechanism that would link the PDO and longer Atlantic cycles to an overall warming of the atmosphere (as opposed to a local or regional effect).

        I am not saying there isn’t one, just that as things stand we have a prime suspect (CO2), but maybe not quite enough evidence yet to convict. I suspect we are not living through an Agatha Christie novel with a surprise ending, and that the obvious suspect is likely to be found guilty in the end.

        • If the warming is an internal forcing, there must be some unknown multi-decadal cyclical effect trapping more of the Sun’s heat in a way not linked to rising CO2 (and other GHG) levels. El Nino / La Nina run on too short a time scale and nobody has yet shown a mechanism that would link the PDO and longer Atlantic cycles to an overall warming of the atmosphere (as opposed to a local or regional effect).

          Except that El Nino, which causes global-average warming, has been more frequent since the 1970s.

  85. Dave Dardinger says:

    Roy is right (99% anyways). You folk who think he isn’t don’t understand the physics correctly. It’s that simple. It’s a shame you don’t seem to be willing to learn as that makes it easier for the CAGW crowd to put down all skeptics.

    Please notice that Ray isn’t trying to produce a total skeptical position. He’s just trying to correct the common mistake of thinking that a cooler body can’t make a warmer body even warmer by, as it were, shadowing part of the cooling path of the warmer body.

    One more thing, which I’ve rarely if ever seen pointed out is that thermodynamics is a macro-explanation of the atomic theory. Trying to mix thermo and atomic explanations can result in occasional errors which are sometimes hard to catch.

    Ok, still one more thing. While excited CO2 molecules may emit IR rapidly by human standards, it’s still much slower than the mean time between collisions (at least in the lower atmosphere) This means that essentially all CO2 gives up its energy gained from absorption of IR to the surrounding atmosphere. It only emits IR on rare occasions when it happens to have picked up enough energy from a collision (and only rarely then. It might take thousands of such collisions before it happens to emit.)

  86. frederic says:

    Roy, you said
    It is precisely because the surface of the Earth is continuously being heated by an external energy source (the sun).

    Where is that place on the earth surface?

    • TheLastMan says:

      It is precisely because the surface of the Earth is continuously being heated by an external energy source (the sun).

      Where is that place on the earth surface?

      That is not what he said. Roy said “the Earth” not “every location on the Earth all at the same time”.

      At any one point in time, half of the Earth’s surface is receiving energy directly from the sun, therefore the Earth is continuously being heated.

      Or do you think somebody turns off the Sun at night just so as you can get some sleep?

      I think deliberate and wilful misunderstanding should be made a criminal offence.

    • ?? If there is a point you want to make, please make it.

  87. Ron says:

    Surely, for Dr Spencer’s thought experiment to be entirely appropriate, his initial plate(and the second incidentally) would have to be warmed by external radiation rather than an internal electric current. To be sure, clouds complicate matters but as a first instance, just assume uniform radiation which would necessarily be “seen” by both plates. You’d then probably need to cool the second plate to simulate the CO2 in the various layers of the atmosphere.

    No argument with Dr Spencer’s argument as applied to this system (or people under blankets – who are also net heat sources rather than warmed from outside) but it seems to me to be there are real differences between the model and the earth.

  88. william says:

    probably the biggest mistake is to compare cladded buildings with extra heating, the atmosphere is constantly shifting moving.

    with a heavily cladded building the heat would rise and heat from top to bottom, as we know
    from lack of hot spots in the atmosphere this would not be the case in the real world.

    • william says:

      the cladded building, see above , would have to be a very strange comparison to the real world , the building would be hotter the higher you are, in the real world the higher you go the colder it gets,

      very strange reply above!!

  89. Kirk Myers says:

    Thanks for the nice explanation of the workings of the so-called “greenhouse effect,” which, as you point out, is misnamed. It more accurately should be referred to as the “atmospheric effect,” but I guess we’re stuck with “greenhouse.”

    I notice you don’t spend much time explaining the role of conduction, convection and vaporization and its contribution to the earth’s cooling. Is it possible that those atmospheric processes overwhelm what little impact the earth receives from CO2-induced warming?

    Dr. Ferenc Miskolczi’s peer-reviewed findings indicate that there is a “greenhouse equilibrium,” probably the result of changes in atmospheric water vapor transport and heat transport, cancelling out the thermal impact of increased levels of CO2.

    As Miskolczi notes: “Our atmosphere, with its infinite degree of freedom, is able to maintain its global average infrared absorption at an optimal level. In technical terms, this “greenhouse constant” is the total infrared optical thickness of the atmosphere, and its theoretical value is 1.87. Despite the 30 per cent increase of CO2 in the last 61 years, this value has not changed. The atmosphere is not increasing its absorption power as was predicted by the IPCC.

    He continues:

    “The conventional greenhouse theory does not consider the newly discovered physical relationships involving infrared radiative fluxes. These relationships pose strong energetic constraints on an equilibrium system . . . Nobody thought that a 100-year-old theory could be wrong. The original greenhouse formula, developed by an astrophysicist, applies only to the stars, not to finite, semi-transparent planetary atmospheres. New equations had to be formulated.”

    Your thoughts?

  90. william says:

    i have a lovely high green house at home, just measured the temps, wow, a big difference in the top two feet ,very hot, to the bottom 8 feet, not so hot, phew! lucky this is not the reality of the real world,

  91. Joletaxi says:

    Sacré D. Spencer!!!
    Following Your latest answer

    I will tell You that in French:

    vous nous avez bien eu,chapeau bas.

    La présentation de votre proposition tendrait à nous faire admettre qu’un corps froid placé à côté d’un corps chaud pourrait rendre celui-ci plus chaud? Bien joué
    En réalité, la proposition devrait être:si je place un corps froid près d’une source de chaleur, cela va-t-il ralentir le refroidissement de celle-ci?
    Vous nous faites découvrir l’isolation. Merci

    I love this kind off trick. In my youth I had a professor off mechanical engineering who use that kind off trick,and we do’nt enjoyed that much in face off a white sheet off paper….

    Congratulations

    • william says:

      for fun i translated of google translate,

      you’ve been well, hats off.

      The presentation of your proposal would tend to make us accept a cold body placed next to a warm body could make it warmer? Well done
      In reality, the proposal should be: if I put a cold body near a heat source, it will he slow down the cooling of it?
      You let us discover the insulation. Thank you

      • Joletaxi says:

        Thanks William
        When You have been fouled, You do’nt like people know it,that the reason I posted that in french

        anyway thanks

  92. Phil. says:

    Join the club Roy, this is a concept that I have been trying to explain for some time, most recently on WUWT coincidentally at the same time as your post. I have tried to explain it there with reference to the shielding of thermocouples in flames to reduce radiation losses and thus read a more accurate (higher) temperature. I’ve linked to a NIST report their which does a complete thermal analysis which explains why adding a shield which is cooler than the thermocouple but hotter than the surroundings will cause the temperature of the thermocouple to go up.
    Someone on here asked about making the surrounding walls hotter which is perhaps the easiest way to deal with it.
    Consider the single heater system above the heater will reach a steady state temperature when its heat input equals its heat loss, the heat loss is given by:

    Loss=C(Th^4-Tw^4)

    So for a given input the temperature of the heater, Th, will go up if Tw is increased because loss has become less than the heat input until a new steady state has been reached.

    If instead of changing the temperature of the wall we insert an object (the second heater) which is maintained at a temperature intermediate between Th and Ts (T2) then we get:

    Loss=C(Th^4-(aTw^4+bT2^4)) (a+b= 1.0)

    under the conditions stated (aTw^4+bT2^4) is greater than Tw^4 therefore losses go down therefore Th goes up.

    Hope that helps.

  93. Phil. says:

    jae says:
    July 25, 2010 at 6:54 PM
    BTW, Roy, you can shield the asphalt surface with an IR transparent film, like NaCl, to stop convection), like Woods did, and I will bet you $100 that you will still not see an increase in temperature. Something is definitely wrong with your hypothesis, I think. We desperately need some empirical evidence for all this GHG stuff!!!

    Woods analysis was all messed up, as he himself admitted he didn’t spend much time on it.
    For those of you who made comments about commercializing this effect, you’re too late.
    Incandescent bulbs produce a lot of IR whereas what’s needed is white light, there are now bulbs made that have a coating on the glass surface which reflects the IR back but passes the white light, the returned IR heats up the source as explained above, giving ‘hotter’ light with less IR. The energy savings are about 30%.

  94. william says:

    i love this story about a vacuum,

    While your own normal blood pressure will keep your blood from boiling, the saliva in your mouth could very well begin to do so. In 1965, while performing tests at the NASA facility now known as Johnson Space Center a subject was accidentally exposed to a near vacuum (less than 1 psi) when his space suit leaked while in a vacuum chamber. He did not pass out for about 14 seconds, by which time unoxygenated blood had reached his brain. Technicians began to repressurize the chamber within 15 seconds and he regained consciousness at around the equivalent of 15,000 feet of altitude. He later said that his last conscious memory was of the water on his tongue beginning to boil.

    guess strange things happen in a vacuum!!

  95. Andrew S says:

    Here is my hypothesis: given an object with thermal equilibrium temperature A, when another (thermostatically regulated) object with higher temperature B is put in near proximity to the original (passive) object, thermal radiation from second object will heat the first one. However, temperature of the passive one will stabilize below B and at no point in time can it exceed B.

    If this hypothesis is incorrect, that would mean heat flows from colder (second) object to warmer one.

    If, on the other hand, my hypothesis is correct, than it follows that there is no such thing as “backradiation” or in other words “Cooler Objects Can *Not* Make Warmer Objects Even Warmer Still”.

    • Colin Henderson says:

      Can a blanket at room temperature make you warmer? The second cold object (due to its back radiation) is acting as an insulator and ultimately reducing the overall heat loss of the warm object, consequently raising its equilibrium temperature. In a vacuum with no absorption and re-radiation of energy this is theoretically possible.

      The way Roy Spencer has worded and presented this idea will push you to think through, and realize the absurdity of radiative forcing, because we aren’t in a vacuum, and none of this theoretical (climate model) stuff applies to the real world- quite clever of him really.

      • Anonymous says:

        Colin, blanket certainly can make my body warmer. At least in cases where root temperature is lower than my body temperature. In some warm countries people wear warm thick cotton coats to save themselves from unbearable heat. However, convection and conduction mechanisms are not the issues here. We are discussing radiative heat transfer only.

        You avoid address my statement and repeat what already being said. Can you show me in which way my hypothesis is wrong.

        Can an object with equilibrium temperature of A degrees, in the presence of another object with higher temperature B warm to temperature above B? Why?

      • yes, I am trying to fool everyone. I’ve been found out.

  96. william says:

    insulation, we have to think of insulation as a complete body surrounding a heat source ,lagging of waterpipes to stop freezing,

    as we know with co2 this is not the case. co2 at 380ppm or so would be the equivalent of a roof with 90% missing. does co2 store heat , if so an amazing quality, if it releases the heat this would be upwards, there is no insulation around co2. in simple terms the release of heat upwards is same amount of heat that entered the co2. as Mr spencer said this heat is being added to the whole time, the co2 is not a fuel that adds heat itself.
    .

  97. william says:

    moisture- cooling during the day insulation from freezing at night?

  98. old construction worker says:

    Scott Scarborough says:
    July 25, 2010 at 6:06 PM

    ‘As you move the colder bar further away from the warmer bar (heat source) it will have less of a warming effect on the warmer bare.
    So “distance” is a factor with the “back radiation” in the vacuum chamber. So if distance is between the two bars is large enough, “back radiation” from the second would flat line or to a point where it can not be measured.

  99. Wilson says:

    Roy,

    does the metal bar heat up because heat is being radiated from a cooler object to a warmer object ?

    or does it warm up because the generated heat has less avenues to dissapate ?

    Similarly, does the body heat up because the cold blanket is radiating heat back or because the body continues to generate heat and has less opportunity to cool due to the presence of a warmed blanket ?

    Thanks.

  100. Brad says:

    Dr. Spencer:
    Your thought experiment was interesting. A more accurate title would be “Yes, Virginia, Cooler Objects Can APPEAR To Make Warmer Objects Warmer Still.” The cooler bar did not actually make the heated bar warmer. You have shown the value of insulation. You have also shown the value of how semantics plays in scientific discussion. I’m sure that those of the Catastrophic AGW faith are making great use of your blog’s title. From what I have read, one of the reasons the AGW theory fails is because it requires CO2 to be a super-insulator.
    I have not read all the comments so I don’t know if anyone challenged the statement in your reply to John Marshall July 24, 2010 at 12:53 AM regarding radiative transfer of energy in the atmosphere where half the energy goes up and half down. I have heard the statement before but, it doesn’t ring true and accurate. Energy emitted up and to the sides is a far greater quantity than that going down. The higher up in the atmosphere a molecule gets, the less likely energy transfer will occur back toward the ground. I don’t have the mathematical proofs but, I think a thought experiment using a light bulb over a sphere should prove illuminating.
    Thanks for the brain exercise.

    Brad

  101. Phil R says:

    william says:
    July 26, 2010 at 12:29 PM

    Geez,

    I’m glad I’m not an engineer or a physicist, just a lowly geologist so never had to overthink thermodynamics. My very basic understanding is heat (energy) can be transferred three ways: conduction, convection and radiation. All three are at work in the atmosphere. Dr. Spencer’s model was only intended to help describe and clarify the radiative component without suggesting that the others do not occur or are not important. Your answer (and many others) seems to conflate (confuse?) radiation and convection. The missing “roof,” which would affect convective heat transfer (as in a GREENHOUSE), is not analogous with radiative heat transfer.

  102. jae says:

    I’m not reading all the comments, since I have limited time. Has anyone yet produced ANY even small amount of empirical evidence to show that these “very elementary calculations/suppositions” are true? I think the answer is NO. Even Einstein was forced by the terrible denialists of his day to demonstrate his theory of relativity by means of empirical observations (light bent by the gravity of the Sun, remember?) Where is there a similar proof of the magic GHE?

    Still just an hypothesis, folks.

    Actual observations that I have been making for the past few years shows absolutely no increase in temperatures with increases in GHGs. Compare deserts and humid areas at the same latitude/elevation: the EXACT OPPOSITE rules!

    • Anonymous says:

      Note: if the CO2 theory had any credibility, we would be seeing increasing temperatures for the last 15 years, barring some other effect, which nobody seems to be able to idendify. A rational person has to conclude that the whold GHG global warming hypothesis is just another environmentalist SCAM! Which is undoubtedly the truth.

  103. northern says:

    Allright, here’s simple question from average Joe: what if we put IR emitting source outside of such coated bulb?

    Does this kind of coating prevent IR from entering inside the bulb?

  104. Phil. says:

    Brad says:
    July 26, 2010 at 6:14 PM
    Dr. Spencer:
    ……
    I have not read all the comments so I don’t know if anyone challenged the statement in your reply to John Marshall July 24, 2010 at 12:53 AM regarding radiative transfer of energy in the atmosphere where half the energy goes up and half down. I have heard the statement before but, it doesn’t ring true and accurate. Energy emitted up and to the sides is a far greater quantity than that going down. The higher up in the atmosphere a molecule gets, the less likely energy transfer will occur back toward the ground. I don’t have the mathematical proofs but, I think a thought experiment using a light bulb over a sphere should prove illuminating.
    Thanks for the brain exercise.

    OK try that experiment out, remember to get the scale right, that would require say a 7″ radius ball (a soccer ball) and a light bulb 0.01″ above its surface. If you had access to a 7′ radius ball then you could position the light 1/8″ above the surface.

  105. Phil. says:

    Brad says:
    July 26, 2010 at 6:14 PM
    Dr. Spencer:
    Your thought experiment was interesting. A more accurate title would be “Yes, Virginia, Cooler Objects Can APPEAR To Make Warmer Objects Warmer Still.” The cooler bar did not actually make the heated bar warmer.

    Actually it did by radiating back proportional to 310^4 rather than the 273^4 of the wall it replaced (8×10^9 vs 5.5×10^9).

    • Anonymous says:

      Phil,

      and as the unpowered bar radiates back and heats the heater, it increases the temperature of the heater which then radiates even more toward the unpowered bar which increases its temp so that it radiates even more to the heater increasing its temp which radiates even more to the unpowered bar raising its temp…

      Where is your math and the theorems to support this new physical principle that MUST have a limit to prevent this unlimited feedback from melting everything??

      Additionally, when will you and others supporting this map the ENERGY flows through this system to illustrate that it is plausible?? Both Steffan-Boltzman and Poynting Vectors are methods of computing NET energy flows between objects radiating against each other that would seem to refute your position. Can you point us to new math/theorems to update them??

  106. Daniel says:

    Dr Spencer?

    I’m no scientist, rather a layman, but extremely interested in the development of man-made hysteria around climate change and warming.
    Regarding your post. I’ve some questions for you :

    In my understanding, no formal, peer reviewed to Gerlich & Tscheuschner was ever published (some were announced, at least one prepublished on Arxiv…) ; as their paper is frequently criticized or even ridiculed / discredited, how is it that nobody was in a position to issue a ‘true’ rebuttal paper ?

    Other point : I read some short papers by Heinz Thieme, the German engineer linked to the ‘Heiligenroth manifest’ (something like the US Oregon petition); he points two things about backradiation :
    - first, if there is any backradiation effect, in any case, there is a balance between downwards backradiation(from GHG in atmosphere to earth) and upwards (from earth to GHG in atmosphere) so that in the end both effects neutralize each other ( a point already stressed by other comments)
    - second , instead of backradiation, the higher temperature of air close to the earth has much to do with gravity, compressed gaz close to the earth having to be at higher temperature due to the ‘weight’ of the air column, and air high in the upper atmosphere being much less compressed and consequently colder ; of course warmer air tends to gain in altitude, which leads to the point of the atmosphere’s intrinsic instability. Is this something you have been working on ?

    many thanks in advance for your feed back

    Daniel

    • Anonymous says:

      Daniel,

      Gerlich & Tscheuschner now have a peer reviewed paper out. There is no reasonable refutation of their paper. Some have tried and failed miserably to address what ia actually in their paper. Refuting strawmen does not make a valid presentation.

      The only real disagreement is over whether there is a GLOBAL AVERAGE TEMPERATURE. I believe their stance was to the effect that since it varied so much and could not be tracked in a way to predict it, there was no basis for claiming we could compute a valid value.

      • I’m afraid I have never understood the claim that a global average temperature doesn’t exist. I DO agree that there are a myriad of different ways in which you could define it, and there are not equivalent to each other in terms of heat content. But once you pick a definition, then you stick with it and monitor it over time. We (John Christy and I) believe that one of the most useful measurements is the global average tropospheric temperature.

    • So, *IF* there is backradiation, then it just happens to balance the upwelling radiation? That’s convenient. :)

  107. Mindert Eiting says:

    Dear dr. Spencer,
    Thanks for your interesting blog. Also thanks to some of your (anonymous or pseudonymous) commenters, especially those who made it clear to me that this is about insulation.
    I have learned that a body, continuously fed by energy, arrives at a higher equilibrium temperature when insulated.
    For me it would have been easier if the example was not a vacuum. I would have seen air molecules flying around, doing their energy-transportation job (convection). I see a cold molecule zigzagging between the heated plate (1) and the insulator plate (2). It is heated by (1), transports energy to (2), cools down and bumps back to (1), is heated again, etc. This does not mean that (1) is heated by the molecule. It still is cooled but at a slower rate than without the insulator (2).
    In the vacuum we only have radiation. There I still can see photons flying around. May I translate ‘molecule’ = ‘photon’? The back radiation story should be consistent with thermodynamics. One of your commenters suggested that we have to translate heating and cooling here as becoming blue or red during the job.
    From other literature on the internet I have learned that in the troposphere radiation is not very important. At the outside of the stratosphere cooling can only be done by radiation. Do you agree with this lay-man summary?

  108. Deech56 says:

    Dr. Spencer, thanks for your clear explanation of something that can sometimes be hard to grasp. I am tempted to see your atmosphere picture as a way to explain stratospheric cooling, but every time I try to simplify it in my own mind I am reminded that the concept is more complex.

  109. Fred Staples says:

    Roy, if all you are doing is demonstrating that thermal insulation will raise the temperature of a heated object, why all this discussion? The insulation inhibits the energy transfer from the heated body, and a higher temperature is required to overcome the thermal resistance and restore the original transfer rate. As you say to Andrew, very early on, “yes exactly”.

    But Roy, you are making a different claim. You are postulating something called “radiative-insulation”, and suggesting that it will act like thermal insulation. That is much more problematical, as Joss pointed out.

    To do anything at all net radiative energy must flow from a source to a sink, in one direction only, so as to increase the entropy of the system. If (actually a large assumption) the earth surface is warmer than the lower atmosphere the earth will be cooled and the atmosphere will warm.

    Increasing the CO2 content of the atmosphere will increase the absorption and warm the atmosphere further, reducing the net radiation and the surface cooling.

    But since it is primarily the atmosphere which radiates to space, this will increase outgoing radiation and cool the whole system down, as someone else pointed out.

    Which is the dominant effect? Woods tried to find out by comparing two enclosures, radiative and non-radiative, everything else being equal. The glass (radiative) must have been warmer than the non-radiative (rock-salt). The suggestion (repeated on Page 18 of Global Warming by John Houghton) was that the glass interior would have to radiate more energy (incoming solar plus back-radiation), increasing its temperature to do so. So its interior should have been warmer.

    It was not.

    Why not? Possibly because the increased convection from the warmer glass counter-balanced the reduction in outgoing radiation. Whatever the explanation, their was no sign of a “back-radiation” effect from the (cooler) glass to the (warmer) interior.

    Science, as opposed to thought experiments, should start from there.

  110. Eli Rabett says:

    Andrew S, the photons that are absorbed by the warmer object are instantly converted into thermal energy (vibrations of the atoms in the sold object), in emitting photons from that warmer object thermal energy is converted into photons. The conversion to thermal energy is something you have not considered. Photon number and frequency/wavelength is not conserved. Photon energy is.

  111. donald penman says:

    I was a bit confused over your treatment of the result that there is a logarithmic decrease in warming with increase in co2 in the atmosphere in your recent book . if co2 is added constantly to the atmosphere then warming will slow down and eventually stop more or less,but when you say that the rate of warming caused by a forcing of co2 has a logarithmic decline as it reaches its maximum effect this seems to contradict the idea that warming decreases as co2 increases in the atmosphere,because all you have to do to increase the temperature more rapidly is add another forcing of co2 no matter how much co2 the atmosphere already contains.

    • I’m not sure why you are confused. There is a logarithmic decline in the RATE of warming as more CO2 is added. That means that the most total warming will be with the largest total CO2 increase.

  112. Massimo PORZIO says:

    Dear Dr.Spencer,
    I agree with your concept of delaying the heat by means of an object placed close to the heater, but I’ve a doubt about that simplified concept of back-radiation.
    For example, let me assume that initially the heater is the only energy source of the system (by the way, the walls of the chamber chiller must be at 0K otherwise it’s a source of power too, that because of your back-radiation concept, where colder bodies warm warmer ones).
    And let me assume that a very little first bar radiation impinges the second bar, which absorbs it and re-radiate. Then only a very little of this radiation impinges back to the first bar. So by your point of view (correct me if I’m wrong here), the sum of the initial power flux and this even small amount of back-radiated flux becomes the new outgoing flux from the first bar, which should increase its temperature.
    This system should be represented by two algorithms:

    F2(ni)=F2(ni-1)+F1(ni-1)*Kr

    And

    F1(ni)=F1(ni-1)+F2(ni)*Kbr

    Where:
    ni is the iteration number (ni = 1…infinite).
    F1(ni+x) is the power flux outgoing from the first bar at the iteration ni+x (x = 0|-1).
    F2(ni+x) is the power flux outgoing from the second bar at the iteration ni+x (x = 0|-1).
    Kr is the coefficient of radiation from the first bar to the second bar.
    Kbr is the coefficient of back-radiation from the second bar to the first bar.
    Of course F1(ni=0) is the initial power flux for the first bar and F2(ni=0) is the initial power flux for the second bar.

    If you apply this simplified math the system will never find an equilibrium, because the algorithms lead the two bars fluxes to run away to infinite (when ni = infinite), even if you choose very little Kr and Kbr.

    Since the IR radiation is an electromagnetic energy, I guess that you should use the vector math involving the Poynting theorem to get realistic results instead. But it’s a really tricky job.

    I read what Kirk said on July 24, 2010 at 2:13 PM, and I better agree that explanation, even if I’m not sure of the temperature lowering when the two cubes are joint together, because the volume doubles while the surface just became 10/6 of the original cube surface. Given a constant thermal conductivity, the average thermal path from the barycentre and the surface of the two cubes joint together should be greater than the one of the original single cube, so I guess the two cubes joint together should be warmer than the original one. Maybe I’m wrong here, because this is just a speculation based on old school remembrances of when we approached the heatsinks efficiency by their shapes.

    Anyways excuse me if I wrote a silly post here and I missed some fundamental points in your article, also because English is not my first language.
    I would like to know you opinion about.

    Massimo

    • Anonymous says:

      Ok, I did the computation, Kirk should be right and I was wrong. The two cubes joint together should be colder than the original one, but not so colder as what it was if they were arranged to have a shape of a cube having two times the volume of the original cube.

      Massimo

  113. Andrew S says:

    Eli Rabett says:
    July 27, 2010 at 6:15 AM

    “The conversion to thermal energy is something you have not considered.”

    Of course, I did. That’s why I use terms “absorb” and “emit”.
    Don’t you find it a little bit counter-intuitive that when body is bombarded with photons, its response is to emit smaller number but higher energy photons back? I certainly do. Can objects behave this way?

  114. Dr. Spencer:
    You’re high on the geothermal heat flux. It’s more like 0.1 W/m2, which is why it’s so routinely ignored. You can see a map of it at http://geophysics.ou.edu/geomechanics/notes/heatflow/global_heat_flow.htm

    The highest values are mostly on mid-ocean ridges. Given how weakly stratified the oceans are, and how slow-moving, oceanographers do pay some attention to this effect. Meteorologically, it’s still pretty small.

  115. bluegrue says:

    PLEASE DON’T POST

    Dear Dr. Spencer,
    a comment of mine seems to have gone into the spam bucket. I pointed Daniel to the reply to Gerlich and Tscheuschner by Halpern et.al. complete with links to the abstracts and free copies of the papers where available. Could you please look for it? Otherwise I’ll try to repost after 12 hours.

    Before I forget, thank you for trying to teach people about the misconceptions you’ve addressed in this post.
    TIA
    bluegrue

    • Anonymous says:

      Are you sure your comment have gone into the spam bucket?
      I say that, because I still found a comment about Gerlich and Tscheuschner made on July 27, 2010 at 6:30 AM by “Anonymous”.
      It could be yours.
      Note that this post will be assigned to an “Anonymous” poster even if I filled the Name field.
      It seems that Dr.Spencer’s blog has some problem with replies to other posts.
      More, for a strange reason my last post left here today has been placed before some other posts left yesterday.

      Anyways, I don’t believe Dr.Spencer trashed any post indeed.

      Massimo

  116. KuhnKat says:

    Basically there is a difference in view of the physics. I disagree that your experiment will have the result you claim.

    Find someone to actually PERFORM this experiment:

    Take two lasers of identical low energy type. align them
    toward each other as exactly as possible. (let’s use CO2 lasers for kicks and giggles) Set one to half the output of the other and start their firing cycle to occur together. Let’s see if the higher power laser is heated by the lower powered one!

    You should be able to instrument the lasers for temperature and electrical loads to tell if there are feedbacks. If I and the others who believe thermodynamics applies to radiation are right only the lower powered laser should be heated above normal operating levels for that output. If the AGW science is right they BOTH should be heated. (no the internal energy flow of a laser does not prove this one way or the other as there are physical components that heat from conduction inside the laser! The net energy flow is outside the laser! While they are pointed at each other there would be extra internal heating of both if AGW physics is valid.)

  117. KuhnKat says:

    Dr. Roy,

    In reference to your experiment Dr. Roy, what do you think is happening with energy flow?? While the insulator is warming you claim there is warming of the heater by the insulator? That is, what is the model of energy flow before, during, and after the insulator is introduced and finally achieves a new equilibrium?? Will it acheive equilibrium? Where does the energy come from for the heater to radiate at a 160c rate when the power supply could only drive it at a 150c rate before the insulator was introduced. The longer the heater radiates at this higher energy level the more the difference there is in what the energy supply could have provided and what is being radiated from the system in this environment. Won’t the insulator approach the temperature of the heater?

    Apparently you are claiming that energy that should normally be radiated out of the experiment is being directed back into the heater. Let’s go with that idea for a minute. The radiation flow then raises the temperature of the heater to 160c (or 151.c) and the heater then increases its radiation rate to meet its temp. Well, as its radiation increases the amount of energy the insulator receives to send back goes up. If it is going up it should raise the temperature of the heater even more. Where is the limit on how much energy can flow through your system Dr. Roy?? This is a feedback with no extra external energy that appears to have no limit?? Why can’t we increase the output of anything radiating by putting an insulator next to it??

    When is someone going to sit down and do the math of energy flow over time in one of these thought experiments to show whether they even appear plausible??

    Sorry, you may THINK that radiation from a cooler object can add energy to a warmer object, but, if that were so the earth would have already melted.

    The compromise position is that there IS some limit on the amount of energy that can flow from cool to warm. Steffan-Boltzman and Poynting Vectors both compute NET flows when 2 or more objects are radiating at each other. In other words, if there is energy added to the hotter object it must be lost somewhere else to prevent it from radiating more. That implies that it will not raise the temperature of the object as temperature determines the amount of radiation. Could you please explain the theorem and the mathematics that explains how this works without unlimited feedback, or refer us to a physicist that can?? I simply have never heard of this area of science.

    • bluegrue says:

      If it is going up it should raise the temperature of the heater even more. Where is the limit on how much energy can flow through your system Dr. Roy?? This is a feedback with no extra external energy that appears to have no limit??

      You’ve just invented a new variant of Zeno’s paradox. Your appeal to ridicule assumes, that an infinite sum of temperature rises necessarily diverges. That’s wrong. As an example for a convergent infinite series that 1 + 1/2 + 1/4 + 1/8 + … + 1/2n + …., which converges to 2. Anyway, if you are only interested in the temperatures in Dr. Spencer’s example, you can bypass the time evolution of cooling/warming and simply calculate the temperatures of the bodies in the steady state.

      bluegrue

    • bluegrue says:

      Testing HTML support of character entities for an upcoming reply

      σ

    • bluegrue says:

      Could you please explain the theorem and the mathematics that explains how this works without unlimited feedback, or refer us to a physicist that can??

      At most a Bachelor level understanding of physics is needed, high school may or may not suffice already. Due to limiations in the HTML available for comments the formating in the following will be a little ugly; carets “^” indicate power.

      Let B1 and B2 be black bodies, surrounded at a large distance by a background Bs, which is also a black body, in a vacuum. Bs is a heat sink at constant temperature Ts. Let B1 and B2 be large, thin plates, parallel to each other with a small gap in between, such that emission along the edges of the plates and out of the gap between them can be neglected. Let B1 be heated with a power per unit area j0.

      Let’s look at the steady state. B1 receives blackbody radiation from B2 on one side and Bs on the other plus additional heating by j0. B2 receives blackbody radiation from Bs on one side and B1 on the other without additional heating. For each plate incoming and outgoing fluxes must balance. Therefore
      2 σ T1^4 = σ Ts^4 + σ T2^4 + j0
      2 σ T2^4 = σ Ts^4 + σ T1^4
      where σ is the Stephan-Boltzmann constant and the T are the temperatures of the respective bodies.

      Rearranged:
      2 T1^4 – T2^4 = Ts^4 + j0/σ
      -T1^4 + 2 T2^4 = Ts^4

      Solving for T1^4 and T2^4
      T2^4 = Ts^4 + 1/3 j0/σ
      T1^4 = Ts^4 + 2/3 j0/σ

      Now remove B2, then B1 will then settle to a temperature T1′. The flux balance in this case is
      2 σ T1′^4 = 2 σ Ts^4 + j0
      or
      T1′^4 = Ts^4 + 1/2 j0/σ

      Therefore T1′ < T1, as explained by Dr. Spencer without resorting to maths.

      • Anonymous says:

        Hi Bluegrue.
        I’m not a physicist, but I guess that there is something wrong in your fluxes balance computation.
        You assume T1 and T2 constant during the computation but they don’t. Only the chiller temperature Ts is constant by definition. You sould iterate you computation along the time and doing that you get a temperature runaway.
        You solved the equations supposing that T1 and T2 were indipendent variables, but in this (supposed) system during the initial (and infinite by my point of view) transition T1 is a function of T2 and vice versa T2 is a function of T1.

        So this step of your computation:

        > Rearranged:
        > 2 T1^4 – T2^4 = Ts^4 + j0/?
        > -T1^4 + 2 T2^4 = Ts^4
        >
        > Solving for T1^4 and T2^4
        > T2^4 = Ts^4 + 1/3 j0/?
        > T1^4 = Ts^4 + 2/3 j0/?

        It’s mathematically right, but physically wrong
        The first system of equations should be write:

        T1^4 = (Ts^4 + j0/? + T2′^4)/2
        T2^4 = (Ts^4 + T1′^4)/2

        Where T1′ is the former value of T1 and T2′ is the former value of T2. Since T1 is always greater than T1′, for each successive iteration T2 will be always greater than T2′, for this no equilibrium at all.
        Here in Italy we say something like “it’s the typical cat which bites its tail”.

        Massimo

        • bluegrue says:

          Hi Massimo,

          sorry for the confusion, I had hoped I was clear in my writing. I’m not looking at an iterative that you envision. It can be done, but that approach is needlessly difficult. When I wrote “Let’s look at the steady state” I meant that I am not looking at the time evolution of the temperatures, at all. I’m just looking at the final state, no matter how it was reached.

          The reasoning is as follows. I assume that I can freely choose j0 and Ts and that there is a final steady state, where temperatures no longer change [1]. It comes down to simple bookkeeping. In the final steady state the power put into a plate and coming out of that plate need to be equal; otherwise the plate would either heat or cool, contradicting the assumption of a steady state. We have two variables and two independent equations for T1^4 and T2^4 in the case of two plates. We have one equation for T1′^4 in the case of one plate only. The equations can be solved for T1, T2 and T1′ as demonstrated in my above post. So my initial assumption turns out to be correct, there is a steady state and the choice of j0 and Ts fully determines T1, T2 and T1′.

          If I wanted to look at the dynamics, the calculations would get much more involved. I would have to include the heat capacities, thickness and thermal conductivities of the plates as well. However, to prove Dr. Spencer’s example it is enough to only look at the steady state. It is a common trick of the trade in physics. What remains to be shown is that the steady state can be reached, irrespective of the initial conditions. Here I ask you to accept my judgment as a physicist, that there are no obstacles to reaching that state, because a formal proof would require way more work.

          I hope I have explained it better this time. :)

          bluegrue

          [1] If this assumption should turn out to be wrong, I would get a mathematical contradiction somewhere down the road in the calculation or would be unable to find a solution.

          • Anonymous says:

            Hi Bluegrue,
            I must apologize for my misunderstanding, but I admit that I’ve some problem reading the English language. Since this thread has an unbelievable huge number of posts, I read it without the needed care and I missed the meaning of your statement. Re-reading your initial post you were clear from the beginning.
            I must learn to read better the post before posting a reply next time.

            Excuse me again.

            Massimo

          • bluegrue says:

            Hi Massimo,

            no problem. Misunderstandings happen.

            bluegrue

    • You might be forgetting that as the cooler bar warms, it then emits more energy to the cold vacuum chamber walls. You cannot determine a final temperature of any of the bars without taking into account the flows between all elements. This is exactly what is done in radiative transfer models with many layers of the atmosphere and the Earth’s surface. WE HAVE DONE THIS, with code we wrote in-house, and got basically the same results as everyone else.

      It is those who dispute the existence of “back radiation” (an admittedly poor term) who have yet to produce a time-dependent model of temperature that conserves energy and results in a final equilibrium temperature profile for the atmosphere.

  118. Gord says:

    I see my post dated July 26, 2010 at 8:13 PM is still awaiting moderation??

  119. Douglas W. Cooper, Ph. D. says:

    Let’s try again. Mechanisms of heat transfer are conduction, convection and radiation.

    Ignoring solar radiation to start, assume a surface temperature of To. If there is no atmosphere, the ground radiates to space (T=0 absolute) at a rate proportional to the fourth power of T.

    If there is a non-absorbing layer of gas at temperature Tg, it radiates back to the ground and to outer space, the same rate in both directions. There is also conductive heat transfer from the hotter To to the cooler Tg and conductive transfer from the hotter Tg toward space. Such conductive transfer is likely to be small in both directions because it is inversely proportional to the “distance” between the entities. Convection can also play a role, too complicated to include in this simple example.

    If there is a partially absorbing layer of gas at temperature Tg’, it will be heated by the radiation from the earth, making Tg’>Tg, and it will in turn radiate more heat in both directions and lead to a slower cooling of the earth, even though the gas Tg’ is less than Earth’s To .

    Dr. Spencer’s analysis is correct. The complications come in adding all the real – world details. At night, my blanket never reaches 98 F, but it keeps me warmer than I would be without it.

  120. Bill Hunter says:

    I have learned at least one thing from this, a new idiom. “Thought Experiment”.

    For the life of me I can’t imagine nobody has actually conducted this thought experiment as a real experiment.

  121. Kevin says:

    Dr. Spencer, funny but my last response (I did intend it to be respectful and I hope it was received with that spirit) to your thread on this topic received a reply from you, but did not include an opportunity to reply to your reply to my reply to your reply…. Probably an administrative oversight, they are so rare aren’t they.

    Have I been banned from additional replies? Is there a limit on how many replies I am allowed? I expect not, you are clearly trying to resolve this conundrum as are the rest of us.

    I suggest an expanded thought experiment that expands on your next post. Let’s built a 100 foot tall atmospheric simulator. This simulator would be made of 4” diameter pvc (low speed of heat) pipe sliced into 10 foot tall sections. Between each 10 foot section of pipe would be a thin window of synthetic diamond, it does exist and has nearly transparent optical characteristics and demonstrates very high speed of heat characteristics. Each section of pipe would have its proportions of gases adjusted to match the expected proportions in the atmosphere (scaling to make 10 feet act (optically speaking) as 10,000 feet may be necessary). We can easily control the pressure in each section.

    Next we place a surface at the bottom of the ATMSIM (Atmosphere Simulator) that allows us to adjust the ratio of water / ice / Earth surface. It also allows us to adjust albedo, and if we chose to apply a thermostatically controlled system we can simulate any temperature of the surface of the Earth.

    At the top of the ATMSIM we place a Xenon Arc Lamp, these lamps closely mimic the spectral distribution of the Sun (i.e. 6500K). With appropriate parabolic collectors and power supplies we can easily simulate the Sun’s cycles during a 24 hour period.

    Above the Arc Lamp we then place a “light trap”, these are well known in the optics field and are actually quite effective at making photons disappear. We can simulate the cold vacuum of space and make many photons effectively disappear.

    The ATMSIM has many advantages:

    We can simply drill a 100 foot deep hole then assemble the ATMSIM as we lower it into the hole.

    We can adjust the granularity of the elevation increments by simply making the pipe section lengths smaller (perhaps 3 feet) if necessary at lower elevations.

    The parts are largely standardized; we could easily make the ATMSIM in 1 or 10 foot sections if desirable.

    The time to execute actual experiments is actually quite short, we can fill it with the “optical” equivalent of 300 ppm GHG’s and make some measurements, then fill it with 500 ppm GHG’s and repeat the experiment.

    The cost is not outrageous; I believe if we can find an old ICBM silo we could do this without drilling any new bores for probably less than 50 million dollars.

    Sorry, I don’t have the time right now for a nice illustration but if this peaks your thought processes I can find the time to make a nice drawing.

    Crazy isn’t it……… But no more crazy than thinking we could travel all the way to the Moon and back…

    This is just a somewhat wacky engineer’s musings on the topic. Boy it sure would be nice to have a definitive experiment to answer the question.

    Thanks for your efforts.

    Cheers, Kevin.

  122. Gord says:

    My second attempt to get this posted.

    Re: Vacuum Chamber with plates.

    First, identify the ONLY energy source in the Vacuum Chamber with an electric heater.

    The ONLY energy source is the ELECTRIC HEATER that heats a plate with electricity to a temperature of 150 deg F or 338.56 K.

    Asuume an emissivity = 1 and a surface Area for the plate = 1 m^2

    Using the Stefan-Boltzmann Law, the Watts provided by the Electric Heater is:

    P = e*BC*A*T^4

    Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

    P = (5.67X10^-8) X 1m^2 X (338.56 K)^4 = 744.95 Watts

    (***That’s ALL the Energy Available and cannot be exceeded without CREATING ENERGY***)

    The EM field produced by the plate is 744.95 Watts/ 1 m^2 = 744.95 w/m^2
    ——–
    If another identical “non-heated” and colder plate is inserted into the Vacuum Chamber next to the heated Plate then:

    The 2nd Plate also has an emissivity = 1 and a surface Area for the 2nd plate = 1 m^2

    We can easily determine the equilibrium temperature of both plates by using the Stefan-Boltzmann Law and The Law of Conservation of Energy.

    The TOTAL amount of energy available is 744.95 Watts and both plates will have the same temperature at equilibrium, so they can be considered to be a single radiating body with double the radiating surface area.

    The area of both plates = 2 m^2 so the Radiation emitted by both plates at equilibrium = 744.95 Watts / 2 m^2 = 372.48 w/m^2

    And the equilibrium temperature for both plates will be 284.69 K or 53 deg F.
    ————-
    NOTE:

    This is simple effect is demonstrated every day and is happening in your Computer right now.

    It’s called a “Heat Sink” and is used to cool the microprocessor in your computer by increasing the radiating surface area of the microprocessor.

    ———
    When Roy Spencer said the colder 2nd Plate will heat up the warmer Plate from 150 deg F to 160 deg F, Energy was CREATED!

    PROOF:
    160 deg F = 344.11 K and that means it would require:
    P = (5.67X10^-8) X 1m^2 X (344.11 K)^4 = 795 Watts!

    795 Watts exceeds the TOTAL Energy available which is 744.95 Watts.

    An obvious violation of The Law of Conservation of Energy.
    ————–
    “Second Law of Thermodynamics: It is NOT POSSIBLE for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.”

    http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html#c3

    Whenever one violates the 2nd Law, The Law of Conservation of Energy is violated.

    • bluegrue says:

      Hi Gord,

      sorry, multiple errors.
      1) The plates in the vacuum chamber are not a closed system – where the law of conservation of energy would apply – as there’s energy going into it from the heater and leaving via radiative losses; these two fluxes need not be equal at all time, as some can go into heating / come from cooling of the plate(s).
      2) Watts is a unit of power, Energy per time.
      3) You postulate that both plates have the same temperature.

      For a proper calculation see my posts further up at
      July 27, 2010 at 2:27 PM and
      July 28, 2010 at 1:24 AM
      (sorry, no links, otherwise the post goes into moderation)

      bluegrue

      • Anonymous says:

        Hi bluegrue,

        1) The Law of Conservation of Energy applies everywhere. What Grade are you in?

        2) Joule = Watt-sec, so multiply my Watts by one second and you will have Joules. The numbers remain the same.

        3) Equilibrium requires that the temperatures be the same.

        The calculations I did are correct.

        It is exactly the same calculations used to design “Heat Sinks” for Vacuum Tubes, Transistors, Diodes, Integated Circuits, Microprocessors….for the last hundred years.

        Increase the Radiating Surface Area, the Watts/m^2 drops and Temperature Drops. (See the Stefan-Boltzmann Law)

        If you want an increase in Temperature then decrease the Radiating Surface Area.

        That’s why Edison used a thin filament as opposed to a chunk of iron, when he designed the Electric Light Bulb for your Easy-Bake Oven.

        • bluegrue says:

          Hi Gord,

          The Law of Conservation of Energy applies everywhere.
          Nope, not the way you are using it. It is a twopart observation. 1) Energy can’t be destroyed, it can only change into another form, be it chemical, kinetic, gravitational, heat, etc. 2) In a closed system, the total amount of energy is constant.
          The second follows from the first and is a matter of bookkeeping. You are using 2) in your reasoning for an open system, that’s why your calculation is wrong.
          Look up “conservation of energy” + “closed system” at books.google.com

          What grade are you in?
          Ad hom? Let’s not go there, as firstly neither of us will be able to verify any claim to education made by the other and secondly you’d lose. Let’s stick to the physics.

          Joule = Watt-sec, so multiply my Watts by one second and you will have Joules.
          But that’s the point, you don’t have constant energy, you have constant incoming energy flux! In our setup we have fixed the amount of incoming energy flux, but not the amount of outgoing energy flux. While the plate is heating up the outgoing flux will be smaller than the incoming flux, while the plate is cooling the outgoing flux will be larger than the incoming flux. It’s like the water level at a dam in a river; it’s not the amount of water running over the dam that determines the level of water at the dam, but the height of the water outlet.

          Equilibrium requires that the temperatures be the same.
          This only applies in thermal equilibrium with the bodies in thermal contact. However, the plates are only in a steady state. It’s not a thermal equilibrium because of the energy flux going through the system. If you have an energy flux you can even have a temperature gradient within the same body; take e.g. a long bar that is heated on one side only.

          Heat sinks don’t work by radiative loss, it’s energy transfer to the surrounding gas by surface contact and subsequent transport of the heated gas by convection or forced air streams.

          Edison used filaments in order to achieve the electric current densities required to heat the material to high temperature. Furthermore, light is only emitted at the surface, so heating the bulk is a waste.

          Have you looked at and understood my calculation that I pointed out to you?

          bluegrue

  123. Richard Henderson says:

    Fascinating stuff! Thank you Dr Spencer.

    Here is my layman’s quibble: energy from the sun to the surface is changing due to changes in albedo.

    “Back radiation” from CO2 cannot come from the full atmospheric column. Lower layers will absorb that radiation and include that energy in the conductive convective processes of the atmosphere. The adiabatic lapse rate of the “dry atmosphere” (below dewpoint) does not appear to be unduly effected by varying greenhouse gases.

    I accept (always have) that back radiation from the lower atmosphere will REDUCE the rate of cooling of the surface, but when these arguments are applied to water surfaces I have many doubts. With some 70% of the planet’s surface covered with water I doubt the “greenhouse warming effect”.

  124. Gord says:

    Why not measure the “Back Radiation” (from the cold atmosphere) heating of the Earth directly?

    This has been done thousands, if not millions of times.

    The device to use is called a Parabolic Mirror.

    As I’m sure you are aware, a parabola has a focal point where all Electromagnetic Fields (and even sound waves) can be concentrated at a focal point.

    This is the basis for Satellite Dish antennas, parabolic microphones and even Parabolic Mirror Solar Ovens used over the entire Earth.

    Here is a link to some measurements done by the Physics Dept. at Brigham Young University where they have used Solar Ovens to COOL water and even FREEZE water when the Solar Oven is pointed at the cold atmosphere during the Day and Night.
    ——
    Solar Cookers and Other Cooking Alternatives

    “The second area of solar cookers I looked at was their potential use for cooling. I tested to see how effective they are at cooling both at night and during the day. During both times, the solar cooker needs to be aimed away from buildings, and trees.

    These objects have thermal radiation and will reduce the cooling effects. At night the solar cooker needs to also be aimed straight up towards the cold sky. During the day the solar cooker needs to be turned so that it does not face the Sun and also points towards the sky.

    For both time periods cooling should be possible because all bodies emit thermal radiation by virtue of their temperature. So the heat should be radiated outward.

    Cooling should occur because of the second law of thermodynamics which states that heat will flow naturally from a hot object to a cold object.

    The sky and upper atmosphere will be at a lower temperature then the cooking vessel. The average high-atmosphere temperature is approximately -20 °C.
    So the heat should be radiated from the cooking vessel to the atmosphere.”

    http://solarcooking.org/research/McGuire-Jones.mht
    ——–
    This link shows that heating of the Earth’s surface cannot occur from the colder atmosphere.

    In fact, the article shows how to COOL items placed in the Solar Oven at NIGHT AND DAY!

    All you have to do is point the Oven away from the Sun during the Day and the Oven will transfer heat from the WARM object in the Oven to the COOLER atmosphere!

    It can even be used to produce ICE when the ambient air temp is +6 deg C!

    “If at night the temperature was within 6 °C or 10°F of freezing, nighttime cooling could be used to create ice. Previous tests at BYU (in the autumn and with less water) achieved ice formation by 8 a.m. when the minimum ambient night-time temperature was about 48 °F.”

    And, this also confirms the validity of 2nd Law of Thermodynamics….heat energy CANNOT flow from Cold to Warm objects.

    Further, if the Back Radiation (which the AGW’ers say is available Day and Night) actually could reach and heat the Earth, Solar Ovens would produce heating DAY and Night!

    Our energy problems would be over!

  125. John Shade says:

    Well done here, Dr Spencer! A heartwarming reaction to all the discussion triggered by your thought-experiment: you have been patient and helpful.

    We need a lot more discussion like this, of heated and unheated bodies, of stationary and rotating ones, of variations in rates of external heating and cooling by radiation; with various, including mixed,surfaces; with and without covering layers of various kinds and abilities. And of course, more experimentation.

    It is odd that we have not seen more experiments on ‘glass greenhouse effects’ since 1909. Given the huge sums of money being allocated in support of such an effect + positive feedback in the climate, it would seem reasonable to fund further exploration in the laboratory. Surely modern equipment would allow more precise temperature measurements, and more precise control of, in particular, radiative environments.

  126. John C says:

    Regarding the example of how cloulds moving in on a clear, cold night causing a temperature increase due to increased back radiation. Actually, clouds moving in means a warm front has moved in – you know, relative warm, moist air running into cold air causes clouds. Forget all that radiation crap, at least for that example.

  127. custardapd says:

    Dear Dr. Spencer, I hope you don’t consider this off topic, but what’s your opinion of Miskolczi’s theories?

  128. RobjMitchell says:

    Well here are my thoughts.
    Back radiation as a forcing does not really exist, rather the insulating properties of greenhouse gases act to alter the relationship between the force (shortwave energy) and the equilibrium temperature (effectively the pressure at which in and out balance)
    The maximum effect of a greenhouse gas is to reduce the outwards flow of photons by 50% at 100% absorption, effectively a doubling if it occurred. It also means that any increase in shortwave forcing, such a the observed 5% decrease in clouds, would be amplified by the greenhouse effect.

    some other points,
    CO2 cannot re-emit a photon of the same wavelength as it absorb because photons have momentum. To do so would require emitting the photon in exactly the same direction which is infinitely unlikely. Instead a lower energy photon is emitted and the residual energy must be dissipated by conduction or subsequent photons before the molecule can absorb another photon. Increasing temp increase the time it takes for this energy to be dissipated.

    If a perfect insulator existed, the temperature inside would be limited to the temperature of the energy being put into the system. Ie the photons in sunlight correspond the temp at the solar surface. fluro lights on the other had produce light of a much higher energy/temp yet the wattage is lower than a cooler incandescent light.

    falsifying AGW is easy.
    1/ The OLR has been increasing as the temp increased, this can only be caused by shortwave as you cant loose more energy and heat up unless you put more in.
    2/ water vapor is decreasing not increasing
    3/ observed 5% decrease in cloud cover during 1990s(=3.5W) is 20 times size of CO2 forcing during period yet only caused a 0.4C increase in temp.

  129. Ani says:

    Just a quick note to say thank you. I learned from this post. I hope others have too. Guess you can teach old dogs new tricks.

  130. Gord says:

    Hi bluegrue,

    1) The Law of Conservation of Energy applies everywhere. What Grade are you in?

    2) Joule = Watt-sec, so multiply my Watts by one second and you will have Joules. The numbers remain the same.

    3) Equilibrium requires that the temperatures be the same.

    The calculations I did are correct.

    It is exactly the same calculations used to design “Heat Sinks” for Vacuum Tubes, Transistors, Diodes, Integated Circuits, Microprocessors….for the last hundred years.

    Increase the Radiating Surface Area, the Watts/m^2 drops and Temperature Drops. (See the Stefan-Boltzmann Law)

    If you want an increase in Temperature then decrease the Radiating Surface Area.

    That’s why Edison used a thin filament as opposed to a chunk of iron, when he designed the Electric Light Bulb for your Easy-Bake Oven.

  131. Jerome Hudson says:

    I think Bluegrue is on the right track. I tackled the problem slightly differently, and also put in some numbers, which I think is a good idea.

    First, the simpler problem of a plate at some initial temp Ti in equilibrium with the surrounding heat sink. All energy by radiative transfer, except for the fact that the plate is plugged in to a wall socket, and is heated by resistance to maintain the constant heat lost to the sink. Let F stand for energy flux, in watts per square meter. Also let’s convert Dr. Spencer’s Farenheit numbers to Kelvins. Let To = temp of the sink, at 255K (0F). Ti = 339K (150F). Assume emissivity of 1.0.

    The plate receives flux Fo = sigma To^4 = 5.67E-8 (255)^4 = 240W/m^2, plus let us say W = 510W from the wall, and assume an area of 1 m^2. The plate emits flux Fi = sigma Ti^4 = 5.67E-8 (339)^4 = 750W/m^2, roughly. The flux flow balance is then:

    Fo – Fi + (W/A) = 0.

    or
    240 – 750 + 510 = 0, Watts/m^2.

    Now the fun begins. Dr. Spencer slips in a second plate, of, say, the same area 1 m^2, and close enough that it blocks about half of the radiation coming from the resistively heated plate. Let the hot plate temperature and outward flux be T1 and F1, and the cold plate T2 and F2. Balancing the energy flows into plate 1 (the energized one), we have:

    Fo/2 – F1 + F2/2 + (W/A) = 0,

    and for the second,

    Fo/2 + F1/2 – F2 = 0.

    We consider F > 0 means the plate is receiving flux. The two equations then can be set up, thus:

    Fo – 2 F1 + F2 + 2(W/A) = 0,
    Fo + F1 – 2 F2 = 0.

    This, I think agrees perfectly with Bluegrue, except a bit simpler.

    The solutions are:

    F1 = (3 Fo + 4W/A) / 3,
    F2 = (3 Fo + 2W/A) / 3.

    Converting to temperatures, T1 = (F1/sigma)^1/4, etc., giving

    T1 = 357K = 183F,
    T2 = 318K = 113F.

    These are at variance with Dr. Spencer’s numbers, but one could fiddle around with W/A to get a 10 degree F rise, rather than my 33 degrees. The point is made, I think.

    - Jerry

  132. Anonymous says:

    Bluegrue…

    The Law of Conservation of Energy says Energy cannot be created or destroyed.

    If you wish to believe otherwise, then why don’t you build a Vacuum Chamber with a multiple heating plates and a colder plate beside each one.

    Each one of these Heating Plate – Colder Plate combinations will Create energy.
    —–
    We can easily predict the power gain each one of these plate combinations using the Stefan-Boltzmann Law and computing a ratio.

    -The heated plate will rise in temperature from 150 deg F (338.56 K) to 160 deg F (344.11 K)

    Power Out/ Power In = (T hot / T cool)^4

    (Notice that this is totally independent of the surface area of the plates, as long as both plates have the same surface area and emissivity, so the plates and the Vacuum Chamber can be made very small…maybe even put on a small wafer chip like Integated Circits are!)

    Power Out/ Power In = (344.11/338.56)^4 = 1.067 or 106.7%

    106.7%….now that’s what I call GREAT EFFICIENCY!!

    Put 100 watts in and you get an EXTRA 6.7 Watts OUT!
    ———
    You will quickly become VERY RICH and FAMOUS!

    Gord

    • bluegrue says:

      Gord said:
      Bluegrue…

      The Law of Conservation of Energy says Energy cannot be created or destroyed.

      If you wish to believe otherwise, [SNIP]

      In reply to my:
      1) Energy can’t be destroyed, it can only change into another form, be it chemical, kinetic, gravitational, heat, etc. 2) In a closed system, the total amount of energy is constant.

      Gord, don’t you read the comments you pretend to reply to, don’t you understand them or is it just that you prefer to shred straw men?

      bluegrue

      • Anonymous says:

        Here is what You said about about The Law of Conservation of Energy:

        “1) The plates in the vacuum chamber are not a closed system – where the law of conservation of energy would apply – …”

        “Nope, not the way you are using it. It is a twopart observation. 1) Energy can’t be destroyed, it can only change into another form, be it chemical, kinetic, gravitational, heat, etc. 2) In a closed system, the total amount of energy is constant.
        The second follows from the first and is a matter of bookkeeping. You are using 2) in your reasoning for an open system, that’s why your calculation is wrong.”

        The Law of Conservation of Energy applies EVERYWHERE, including both Closed and Open systems!
        ——–
        I have already shown that Roy Spenser’s Vacuum Chamber with two plates not only violates the 2nd Law, it violates The Law of Conservation of Energy.

        I identified the only energy input to the Vacuum Chamber as the Heater, which provides 744.95 Watts.

        The Heated plate magically increased in energy producing 795 Watts…an OBVIOUS violation of The Law of Conservation of Energy.

        I showed that this has a Power Gain (using the Stefan-Boltzmann Law) of Power Out/ Power In = (344.11/338.56)^4 = 1.067 or 106.7%

        106.7%….now that’s what I call GREAT EFFICIENCY!!

        Put 100 watts in and you get an EXTRA 6.7 Watts OUT!

        You CANNOT get a more energy out of the Vacuum Chamber by putting a heated Plate and a colder plate into it!

        The Law of Conservation of Energy says Energy cannot be created or destroyed…EVER.
        ——————-
        Re: Your calculation on July 27, 2010 at 2:27 PM.

        It is so far from reality, it is Hilarious!

        Here is just one example of your of MANY, MANY errors,…you CREATED energy by Doubling the amount of watts/m^2 of BOTH PLATES because they have TWO sides!!

        Why not have, 4 sides and get 4 times the w/m^2?
        Want 6 times the the w/m^2…simple, just make them cubes with 6 faces.
        Want an infinite w/m^2, no problem, just make them Spheres that have an infinite number of faces.

        What a HOOT!

        What Grade are you in?

        Gord

        • bluegrue says:

          Which of the following statements do you object to?
          1) A body at constant temperature radiates uniformly across it’s whole surface.
          2) The surface area of a thin plate is approximately twice its width times its height.

          Here is just one example of your of MANY, MANY errors,…you CREATED energy by Doubling the amount of watts/m^2 of BOTH PLATES because they have TWO sides!!
          In addition to not getting the surface area of a plate right, you once again showcase your utter inability to tell apart energy from power (energy flux). All I put into my calculation is that in the steady state the incoming power (electric heating, absorbed radiative power from the chamber walls / other plate) is balanced by radiative losses from the complete surface. So once again, don’t you read, don’t you understand or do you prefer straw men?

          • Anonymous says:

            The Stefan-Boltzmann deals with the ENTIRE radiating surface area of an object!!!

            P = e*BC*A*T^4

            Where P = net radiated power (Watts), e = emissivity, BC = Stefan’s constant 5.67 X 10^-8, A = area and T = temperature of radiator in K

            A plate with one m^2 TOTAL AREA (THAT MEANS BOTH SIDES!!), emissivity = 1 and a temperature of 150 deg F (338 K) will need 744.95 Watts to produce this temperature.

            According to your silly calculation, 744.95 Watts will produce 2 X 744.95 = 1489.90 w/m^2 because it has TWO SIDES!

            PROOF:

            Here is your SILLY calculation.

            “Let’s look at the steady state. B1 receives blackbody radiation from B2 on one side and Bs on the other plus additional heating by j0. B2 receives blackbody radiation from Bs on one side and B1 on the other without additional heating. For each plate incoming and outgoing fluxes must balance. Therefore
            2 ? T1^4 = ? Ts^4 + ? T2^4 + j0″

            Take Ts (the Vacuum sphere) and T2 (the cold plate) out and you get the heated plate (T1) and the watts (j0) required to heat it.

            2 ? T1^4 = j0

            YOU HAVE DOUBLED THE WATTS PER SQUARE METER BECAUSE IT HAS TWO SIDES!
            ————–
            What Grade are you in?

          • bluegrue says:

            Dear Gord,

            we are not making any progress. The exchange is long enough for others to form an educated opinion on who is right and who is wrong. I’m content with that.

            Cheers,
            bluegrue

  133. Irwin Hasenwinkle says:

    Dear Dr. Spencer,

    In your post of July 23rd , “Yes Virginia Cooler Objects Can Make Warmer Objects Even Warmer Still”, you made the following statements: “…radiation flow involves energy flow in both directions. “ and “Can Energy Flow Uphill? In the case of radiation the answer to that question is, yes”. With all due respect, I think these comments are not correct and suggest the following for your consideration:

    The Poynting Vector characterizes the power of any electromagnetic wave, including infrared radiation. The Poynting Vector is the cross (or vector) product, ExH, where E and H are the associated electric and magnetic field vectors.

    Assume two plane-polarized electromagnetic sinusoidal waves propagating in exactly opposite directions. If each wave is considered separately, the scalar magnitudes of the Poynting Vectors for each at any given point in space is [1]

    P1=(E1^2)(cos^2(wt))/n

    And

    P2=(E2^2)(cos^2(wt))/n

    Where:

    E1, E2 are the peak magnitudes of each of the two electric fields in Volts per meter.
    n is the intrinsic impedance in ohms.
    w is the frequency in radians per sec.
    t is time in sec
    P1, P2 are the values for power in Watts per square meter

    The direction of each Poynting Vector can be shown to be in the direction of propagation.

    The questions are: If both waves occupy the same space, can one treat these two Poynting Vectors as if they are completely independent so that energy flows both directions? If so, what is the net energy flow? If these waves were truly independent, would not the net energy be the vector sum of the respective Poynting Vectors?

    But wait, the electric and magnetic field components of the two waves are also vectors. Hence, the actual value of the electric or magnetic field at any point in space is the vector addition of the individual field values. For electric fields that are exactly opposite, this value is E1 – E2. The scalar value of the Poynting Vector for this resultant field is:

    P2=((E1 – E2)^2)(cos^2(wt))/n

    This result is clearly not the same as P1 – P2, the vector addition of the individual Poynting Vectors, since E1^2 – E2^2 does not equal (E1 – E2)^2.

    Note that the direction of the latter Poynting Vector is based on cross product of E and H, the vector sums of the individual electric and magnetic field vectors. It has but one direction!

    Which is correct? In my opinion, the Poynting Vectors (i.e. power flow) calculated from individual electromagnetic waves cannot be treated independently. This is analogous to the superposition theorem in circuit theory where total current flow of a linear circuit is calculated by considering each voltage source in the circuit separately and then by summing the results. One may not, however, sum the losses produced by each of these calculated currents to find the total power dissipated in the circuit since power losses are based on the square of the current. Similarly, Poynting Vectors are based on the square of the electric field.

    Thus, in the case where a hot body and a cold body both emit electromagnetic radiation, the direction of electromagnetic energy flow will be from the hot to the cold body since the hot body will produce a stronger electromagnetic field. Energy does not flow both ways. That being the case, energy cannot “flow uphill” and the second law of thermodynamics remains intact.

    [1] Skilling, H. H., Fundamentals of Electric Waves, p135, John Wiley and Sons Inc., 1948

  134. Sten says:

    Dr. Spencer,

    I would just like to thank you for a good explanation, and for taking the time to personally respond to everyone asking the same question(s) over and over again

  135. Hank Roberts says:

    > …. the direction of electromagnetic energy flow
    > will be from the hot to the cold body since the
    > hot body will produce a stronger electromagnetic
    > field. Energy does not flow both ways.

    Hmmm. How would you test your theory?

    I have a blue LED and a red LED.
    I pointed them at each other and turned both on.
    Each of them lit the other one up.
    What am I doing wrong?

    I also tried it with a blue laser and a red laser.
    Same result.

    • Anonymous says:

      Here is something I posted on another thread that answers your questions.

      If you bothered to read the established Laws of Science and EM Physics pertaining to your questions, you could answer them yourself.

      Example: Pointing a flashlight at a more powerful Car headlight.

      If the Photons could actually move from the weaker Flashlight to the more powerful Car headlight, they would be absorbed and heat up Car headlight filament, producing more light.

      The now more powerful Car headlight would move Photons back to the Flashlight causing it to heat up and produce more light.

      The now more powerful Flashlight would move even more Photons back to the Car Headlight causing it to heat up even more and produce even more more light.

      The cycle would continue untill both the Flashlight and the Car Headlight were producing INFINITE energy!

      A perpetual motion machine in a positive feedback loop.

      ——–
      from wikipedia…

      Perpetual motion
      “The term perpetual motion, taken literally, refers to movement that goes on forever. However, the term more generally refers to any closed system that produces more energy than it

      consumes. Such a device or system would be in violation of the law of conservation of energy, which states that energy can never be created or destroyed.”

      “Perpetual motion violates either the first law of thermodynamics, the second law of thermodynamics, or both”

      “A perpetual motion machine of the first kind produces energy from nothing, giving the user unlimited ‘free’ energy. It thus violates the law of conservation of energy.”
      ————-
      Gord

  136. John Onsager says:

    The raw basis of your argument is that any type of interface reflects some of the IR energy back and acts as an insulator.

    The basis for my skepticism about the CO2 bogeyman is that it is no more of a greenhouse gas than most other atmospheric gases. Water vapor is actually a far more potent greenhouse gas. The Earth actually benefits from this characteristic making for a relatively limited range of surface temperature extremes when compared to our atmosphere-less Moon.

    Let’s look at the energy input/output for the Earth. This energy can be measured as heat units BTUs or Calories. The Sun delivers a relatively steady rate of energy input affected by sunspot activity (re Maunder Minimum), orbital variations, and axial tilt. This “Heat Load” is increased minimally by geologic heat radiating from the Earth’s core. The bulk of the Heat Load is radiated back into space if the Earth was in a steady state condition with some being absorbed (stored) by the biomass.

    The present Anthropogenic Global Warming hysteria is somewhat akin to the fable of the 3 blind men examining an elephant.
    1. Decades are too short an interval to determine actual Climatic change. It is more weather data
    2. Climate has been warmer and cooler within recorded history. Witness the Roman and Medieval Warm Periods and the Little Ice Age
    3. Geologic is rife with truly hot periods as while as numerous Ice Ages.

    Question: Is Man theoretically causing the Earth to warm?
    Answer: Yes, in much the same way as the burning of a candle can add heat to a room, a house, a town, and ultimately, the Entire Earth. The heat comes from ancient solar energy that was biochemically stored by the ancient biomass. Much of our present energy use of fossil fuels is allowing this release of ancient sunlight.

    Question: Is this warming irreversible?
    Answer: No, without this combustion, the Earth would ultimately return to its original steady state. CO2 has nothing to do with the added heat. Actually, increased plant growth would absorb more of this energy and CO2.

    Question: Won’t nuclear power eliminate CO2 and Global Warming.
    Answer: No. Heat is a tremendous waste product of all nuclear plants. Electric power generation, transmission, battery storage, and release is VERY inefficient and creates tremendous amounts of waste heat. Electric cars would increase the Earth’s heat load especially if more nuclear power is used to create the electricity.

  137. spendulus says:

    Dr. Spencer, thank you for a good and simple explanation of the subtlies of “cooler warming warmer”. I do have a question that you didn’t cover.

    Slightly more IR escapes than is trapped, due to the spherical nature of atmospheric layers. At any point’s altitude, draw a tangent line to the surface to see that this is so.

    This means that the thicker and more effective an “atmospheric blanket” is, correspondingly more heat is lost to space. It’s a fraction of a percent, but of course that adds up.

    Taking this offsetting factor into account, do you stand by the general assertion that thicker or more dense layers of IR trapping gas still can act to insulate? If so, is this conclusion limited to certain altitudes?

    Thank your for your time and effort in these endeavors.

    Sincerely,

    Spendulus

  138. Willis Eschenbach says:

    Dr. Roy is 100% correct. If you surround a constant heat source (the heated bar) with a solid shell, the temperature of the heated bar will increase. This is due to the “back-radiation” from the surrounding shell.

    For a further discussion of the concepts that Dr. Roy explains so well here, see my post “The Steel Greenhouse”.

    http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/

    Dr. Roy, many thanks for your most lucid explanation, and for your patience in answering the questions. Science at its finest.

    • Anonymous says:

      I read your hilarious “The Steel Greenhouse” fiction.

      Let’s see how many Watts are required for your first metal spherical shell to produce 235 w/m^2.

      Watts In = 225 w/m^2 X 4*Pi*R^2

      ***That’s ALL the Watts available !***

      Yet, you say putting another metal spherical shell around the first shell will increase the 235 w/m^2 of the first shell to 470 w/m^2

      The Watts are doubled! Watts = 470 w/m^2 X 4*Pi*R^2.

      Add a third metal spherical shell and the Watts are tripled! Watts = 705 w/m^2 X 4*Pi*R^2.

      Add even more shells and the Watts continue to rise! Watts = 235 w/m^2 X No. of Shells X 4*Pi*R^2!!!

      The Power Gain of this fantasy perpetual motion machine is;

      Watts Out / Watts In = No. of Shells

      Did you ever hear about The Law of Conservation of Energy?
      —————
      Further, the temperature of the first shell is constantly increasing because the Power is increasing over the SAME AREA (4*Pi*R^2).

      Stefan-Boltzmann Law

      Power/Area = Boltzmann Constant X Temp^4

      The Temperature Gain is:

      T final / T start = (No. of Shells)^0.25
      ——————-
      If you want to know what really happens:

      1) Read the Stefan-Boltzmann Law and you will see that increasing the radiating surface area by adding another shell, will drop the w/m^2 and DROP THE TEMPERATURE!

      2) Open your computer and look at what is mounted on top of the microprocessor.

      It’s called a “Heat Sink” and is used to COOL the microprocessor by INCREASING THE RADIATING SURFACE AREA OF THE MICROPROCESSOR!

      According to your delusional “science”, the Heat Sink should HEAT UP the microprocessor.

      What a HOOT!

      Gord.

  139. Adam R. says:

    Whew!

    Dr. Spencer gets the Patience of Job award for this one.

    If the “blanket on your body” analogy doesn’t turn on the light for some of these folks, I suspect nothing will.

  140. John Millett says:

    “The plate will eventually reach a constant temperature (let’s say 150 deg. F.) where the rate of energy gain by the plate from electricity equals the rate of energy loss by infrared radiation to the cooled chamber walls”

    Would the equilibrium temperature be determined by the power of the energy source and the surface areas of the chamber and plate?

    “Now, let’s put a second plate next to the first plate. The second plate will begin to warm in response to the infrared energy being emitted by the heated plate. Eventually the second plate will also reach a state of equilibrium, where its average temperature (let’s say 100 deg. F) stays constant with time”.

    Wouldn’t the two plates reach the same equilibrium temperature? And wouldn’t this temperature, determined as for the single plate case, be lower because the combined surface area is greater?

    The colder plate has cooled the warmer one by absorbing radiation from it in accordance with the 2nd law of thermodynamics.

  141. Brian says:

    Dr. Spencer, thank you for taking the trouble to present such a clear explanation of some of the greenhouse effect arguments.

    The model you present doesn’t correspond with the situation of planet Earth vis-a-vis the Sun. You appear to have a heat source where the Earth is but there are no heat sources of any significance in or on the Earth.

    If the Earth was like you describe then something like what you write would apply.

    What you describe is a heat source surounded by some material through which heat must flow to escape, presumably reflecting a little and offering some diffusive heat transport to the radiating surface that interfaces with deep space at about 3K. Now suppose you add more material then there is more ‘stuff’ between your heater and deep space, the temperature near the heater rises, of course!

    But the Earth doesn’t get its heat from a heater, it gets it from the Sun, so adding material will slow the heat coming in just as it slows the heat going out.

    If you model the Sun/Earth system properly you will find that the planets stabilise at a a temperature defined by the temperature of the Sun and the distance to the planet, it doesn’t matter one jot what are the thermal properties of the material the planet is made of.

    If you think about your model, replace the atmosphere with mutilayer aluminised mylar insulation, the radiation from the Sun will not get past it but neither will the heat from your 100W electric stove be able to escape, it will just accumulate, raising the temperature continuously (untill something melts!)

    Similarly with your corpse, sorry – body. The blanket keeps a body warm because it restricts the flow of heat out of the body, no heat (a corpse) cold corpse, just like in a morgue.

    I think other comments have pointed this out, it remains true.

    Thank you for the opportunity to make these remarks.

    • The rate of solar absorption of the Earth will indeed be matched by the emitting IR, which then corresponds to an effective emitting temperature. That does NOT tell you what the surface temperature will be.

      • Anonymous says:

        Of course not, the Earth also an atmosphere. Are you thinking that the atmosphere is a source of heat in addition to any sunlight it absorbs?

        You must distinguish between heat and temperature, the temperature in the atmosphere varies with altitude because the pressure varies with altitude. This variation with altitude is more easily seen if you account for it by change of gravitational potential energy (GPE). GPE affects the gas temperature directly because it reduces the specific energy in the gas. Thermal (kinetic energy) is exchanged with GPE but the total energy remains constant. This is clear from the lapse rate, which is uniform and linear at -6.5K/km over the whole planet, equator to poles. GPE is the only energy that changes uniformly with altitude over the entire planet.

        Further you seem to place no importance on the placing of the heat source. I do not understand how you can do your analysis with a heat source on (or in) the planet. The Earth generates about 0.1W^2 from radiactive decay but this is not really significant.

        Is it not clear from the banket analogy? If you try to heat a corpse in a blanket by putting it in sunlight the corpse will not get warmer than the exterior temperature of the blanket. Put a live person inside the blanket then that person (because a live person is a heat source of between 100W – 150W) will raise the temperature inside the blanket with or without the sunlight.

        The temperature increase for the live person will depend on the insulating quality of the blanket, the temperature of the corpse will be independent of the blanket quality.

        Regards, Brian.

      • Anonymous says:

        It’s Brian again.

        You write:-

        “That does NOT tell you what the surface temperature will be.”

        Planets without an atmosphere have a surface temperature which is a function of the solar intensity. The solar intensity is a function of the Sun’s surface temperature and the distance from the Sun.

        Solar intensity is also a function of the angle the surface makes with the incoming sunlight, that is why it is colder at the poles than the equator.

        Planets with an atmosphere and oceans are rather more complicated because the atmosphere and the oceans both transport heat from the equator to the poles.

        Regards, Brian.

  142. Brian says:

    “in the case of thermal radiation a cooler object does not check what the temperature of its surroundings is before sending out infrared energy”

    I’m afraid a photon does ‘check the temperature of the surroundings’before departing. The fundamental property of a photon is its energy ‘hv’ where h is Planck’s constant and v is the frequency. The energy of a photon is a function of its source temperature. This is the whole principle of thermal radiation as described by the laws determined by Max Planck and Wilhelm Wien, it is tha basis of all quantum theory.

    • by “surroundings” I meant objects at a distance from the source of the photon.

      Brian, why are you piling up a bunch of non-issues related to my various posts? FYI, I’ve had the opinion advanced that you and those like you are just playing with me to waste my time. I discounted the suggestion at first….

      • Anonymous says:

        Dear Dr. Spencer

        you wrote “Brian, why are you piling up a bunch of non-issues related to my various posts? FYI, I’ve had the opinion advanced that you and those like you are just playing with me to waste my time.”

        I am sorry you feel this way, it was (and is not) my intention to waste time. This is your blog and I would like to feel that I was contributing in a positive way.

        The question “where the heat enters a system” is by no means time wasting, nor is the energy of photons, both are at the heart of modern physics. If you consider these matters as ‘wasting your time’, I am afraid I have little else to say.

        I apologise for introducing the difference between a live and dead body as a source of heat but it is apt!

        With best wishes.

        Yours sincerely.

        Brian

  143. Kevin says:

    Dr. Spencer,

    Reading through all these exchanges, I’m struck by something — you are in a lonely spot. On the one hand, you are a serious scientists — you understand the physics & theory of AWG while you disagree with the consensus view of positive feedbacks. Yet you work patiently to explain to the “uninformed skeptics” the basic science that underlies the theory.

    I agree, there is debate within the scientific community, but not certainly on many of the points that the political class or amateurs seem to have siezed as their primary talking points/means of attack.

    Do you ever find it frustrating that the people who are most adamantly opposed to the idea that AGW may be a problem make such outlandish claims, presenting themselves as the “face” of skepisism in general?

  144. Anonymous says:

    10 years later he came with the 1,2 degr. for 2xCO2 figure.

    Make that +1,6 for 2xCO2 and -1,5 for 1/2xCO2