Guest Post by Willis Eschenbach
OK, a quick pop quiz. The average temperature of the planet is about 14°C (57°F). If the earth had no atmosphere, and if it were a blackbody at the same distance from the sun, how much cooler would it be than at present?
a) 33°C (59°F) cooler
b) 20°C (36°F) cooler
c) 8° C (15°F) cooler
The answer may come as a surprise. If the earth were a blackbody at its present distance from the sun, it would be only 8°C cooler than it is now. That is to say, the net gain from our entire complete system, including clouds, surface albedo, aerosols, evaporation losses, and all the rest, is only 8°C above blackbody no-atmosphere conditions.
Why is the temperature rise so small? Here’s a diagram of what is happening.
Figure 1. Global energy budget, adapted and expanded from Kiehl/Trenberth . Values are in Watts per square metre (W/m2). Note the top of atmosphere (TOA) emission of 147 W/m2. Tropopause is the altitude where temperature stops decreasing with altitude.
As you can see, the temperature doesn’t rise much because there are a variety of losses in the complete system. Some of the incoming solar radiation is absorbed by the atmosphere. Some is radiated into space through the “atmospheric window”. Some is lost through latent heat (evaporation/transpiration), and some is lost as sensible heat (conduction/convection). Finally, some of this loss is due to the surface albedo.
The surface reflects about 29 W/m2 back into space. This means that the surface albedo is about 0.15 (15% of the solar radiation hitting the ground is reflected by the surface back to space). So let’s take that into account. If the earth had no atmosphere and had an average albedo like the present earth of 0.15, it would be about 20°C cooler than it is at present.
This means that the warming due to the complete atmospheric system (greenhouse gases, clouds, aerosols, latent and sensible heat losses, and all the rest) is about 20°C over no-atmosphere earth albedo conditions.
Why is this important? Because it allows us to determine the overall net climate sensitivity of the entire system. Climate sensitivity is defined by the UN IPCC as “the climate system response to sustained radiative forcing.” It is measured as the change in temperature from a given change in TOA atmospheric forcing.
As is shown in the diagram above, the TOA radiation is about 150W/m2. This 150 W/m2 TOA radiation is responsible for the 20°C warming. So the net climate sensitivity is 20°C/150W-m2, or a temperature rise 0.13°C per W/m2. If we assume the UN IPCC canonical value of 3.7 W/m2 for a doubling of CO2, this would mean that a doubling of CO2 would lead to a temperature rise of about half a degree.
The UN IPCC Fourth Assessment Report gives a much higher value for climate sensitivity. They say it is from 2°C to 4.5°C for a CO2 doubling, or from four to nine times higher than what we see in the real climate system. Why is their number so much higher? Inter alia, the reasons are:
1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.
2. The climate models underestimate the increase in evaporation with temperature.
3. The climate models do not include the effect of thunderstorms, which act to cool the earth in a host of ways .
4. The climate models overestimate the effect of CO2. This is because they are tuned to a historical temperature record which contains a large UHI (urban heat island) component. Since the historical temperature rise is overestimated, the effect of CO2 is overestimated as well.
5. The sensitivity of the climate models depend on the assumed value of the aerosol forcing. This is not measured, but assumed. As in point 4 above, the assumed size depends on the historical record, which is contaminated by UHI. See Kiehl for a full discussion.
6. Wind increases with differential temperature. Increasing wind increases evaporation, ocean albedo, conductive/convective loss, ocean surface area, total evaporative area, and airborne dust and aerosols, all of which cool the system. But thunderstorm winds are not included in any of the models, and many models ignore one or more of the effects of wind.
Note that the climate sensitivity figure of half a degree per W/m2 is an average. It is not the equilibrium sensitivity. The equilibrium sensitivity has to be lower, since losses increase faster than TOA radiation. This is because both parasitic losses and albedo are temperature dependent, and rise faster than the increase in temperature:
a) Evaporation increases roughly exponentially with temperature, and linearly with wind speed.
b) Tropical cumulus clouds increase rapidly with increasing temperature, cutting down the incoming radiation.
c) Tropical thunderstorms also increase rapidly with increasing temperature, cooling the earth.
d) Sensible heat losses increase with the surface temperature.
e) Radiation losses increases proportional to the fourth power of temperature. This means that each additional degree of warming requires more and more input energy to achieve. To warm the earth from 13°C to 14°C requires 20% more energy than to warm it from minus 6°C (the current temperature less 20°C) to minus 5°C.
This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase. As a result, the equilibrium value of the climate sensitivity (as defined by the IPCC) is certain to be smaller, and likely to be much smaller, than the half a degree per CO2 doubling as calculated above.
Steve Goddard (18:48:19) :
“One reason Phoenix is warm at night is because of the humidity from lawns, golf courses, etc. I have seen dew points in excess of 70 degrees there.
Also, why do you think it is extremely cold on the top of Mt. Everest? Could it have anything to do with the lack of atmosphere?”
Well, forget Phoenix. and look at Daggett, California, then: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/23161.txt
You won’t find many lawns there.
I don’t understand the comment about Mt. Everest.
JAE,
The reason it is cold at high elevation is because there is less atmosphere above to keep it warm.
Folks, MY bottom line is that I have been looking for an empirical demonstration of the “atmospheric greenhouse effect” for a long time, and I still have found absolutely NO empirical demonstration of the phenomenon, despite the fact that many learned scholars have expounded upon the theory for over 100 years. I see a lot of really cool ideas, including the K&T diagram that Willis provided above. The theory looks good and offers a nice tidy way to explain some things. The problem is that it does NOT explain many other things, which I mentioned in my earlier posts. It is simply not any hotter in areas that have copious amounts of greenhouse gases than it is in areas which have very minor amounts of greenhouse gases. WHY? Ahrrenius provided a THEORY, not a FACT. And nobody to my knowledge has done better.
Re: Dave Springer (Mar 19 05:37),
You say elsewhere that the temperature on the moon is measured at -23C at a depth of 50cm.
All black body calculations have to do with skin surface temperatures. Even a few mm down from the skin, be it water or ground, infrared is blocked and only convection/contact works.
Thus one should go from the average skin temperature and calculate how long it takes to heat a body at near absolute zero up to -23C by conduction from a surface heated by the sun with an average ( day night year) temperature of “xx” which you do not quote.
This is not a simple problem, for all I know it might take a billion years, as regolith is a good insulator. I was not able to find a global skin surface temperature for the moon, only calculations.
My point is that the -23C at 50cm argument is not relevant to the discussion.
Re: Steve Goddard (Mar 19 09:37),
The amount of warming generated by the atmosphere is about 7C per kilometer or 70-80C. This is well established,
I am puzzled. Not by the temperature gradients, but of the claim it is the atmosphere that is generating the heat to produce the gradient. Link please?
“Once again, see my post, “The Steel Greenhouse”. It shows that the greenhouse effect has nothing to do with gas, or lapse rate, or any of that. If you have problems with the concepts there, let me know.”
I don’t think it would be it would be twice the temperature on surface planet as compared to the temperature of the outer shell. As you said:
“Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K).
I agree the that outer shell radiates the same amount of energy. And we will disregard the rather insignificant increase of the area of the outer shell.
Instead the surface temperature could depend upon how the energy is conducted to the outer shell- conduction, convection, and/or radiation.
It make simple; let’s assume it’s a complete vacuum- and so only allowing radiation.
In that case the surface of the planet should be the same temperature as the shell.
Now, let’s return to the Moon- which is a nearly perfect vacuum.
What temperature would the moon surface have if it was a light year or two from the Sun?
I would say the surface temperature would be close to the present temperature in it’s deep polar craters which are never exposed to sunlight. This surface temperature is well above the temperature of 4 kelvin which is said the be the universe’s temperature. I don’t want to argue about what the temperature exactly is- say somewhere around 20 to 100 kelvin and say that at least 20 Kelvin of that heat is from the Moon’s core heat.
So if we put the Moon a couple light years from the Sun and say it’s surface is somewhere around 20 to 30 K. And then we put a steel shell around it- the shell would be around 20 to 30 K and the surface would remain at 20 to 30 K. And it doesn’t matter how many shells you put around it, it will still be 20 or 30 K at the surface. And you hang a mile of fiberglass insulation from the sphere- it of course wouldn’t make any difference. Or if you made a perfect mirror inside the sphere it would not make any difference.
But if you increase the surface area on the outside of the sphere- put spikes on it, or simply have bumpy surface- and that could make a difference [you could double or triple the surface area]- so you could lower the outside temperature of the sphere. Though of course it can not get below 4 Kelvin.
Though, since a sphere can be made into a smooth exterior surface and the Moon [or any planet] would be rougher surface [and therefore more surface area], such reduction in surface area would reduce it’s ability to radiate heat and therefore increase the temperature on the lunar surface.
I would think that one possible indicator of changes in the ground-level greenhouse effect, if any, would be a year-by-year listing of the coldest near sea-level temperature recorded for each of those years in the Arctic dark zones. I believe it would be the clear cold quiet periods in this region that would be most affected by changes in the classical greenhouse effect.
Said differently make a box a meter cubed. Have it have a sandwich of vacuum and thin sheets of metal. Say have sheet 2 mm thick separated by 2 mm of vacuum- so that gives you 250 “shells”. Put this on the ground on the moon in the shade- and it will not increase the temperature of the ground it is on.
anna,
The only significant heat source in the climate system is the sun. A coat keeps you warm without generating any heat.
Why are you puzzled by my statement, but apparently not by the premise of this article?
Joel Shore (19:32:16) :
we know that all bodies radiate when they are at a nonzero temperature
I was being lectured that a white body [emissivity = 0] does not radiate at any temperature [100C, 400C or millions of degrees] …
Joel Shore (19:36:57) :
Bryan:
I would invite anyone who has any doubts on the matter to go to Arthur P Smith website and the thread “the arrogance of Physicists”.
Joel, Arthur was asked by Fred what difference would be made to the thermodynamics of the atmosphere if it was composed entirely of N2 and O2 and he replied that he thought it would be ISOTHERMAL from the surface up to a height of 30Km.
Not only does Arthur seek to contradict the 2nd law of THD but now also the Ist law of THD and the slight matter of abolishing gravity thrown in for good measure.
Yet you find this type of Physics perfectly OK!
Fred Staples and Terry Oldfield have a background in heat transfer thermodynamics.
I would suggest that before Arthur writes more nonsense he should attend some thermodynamics classes
Correction
Name above should read Terry Oldberg not Terry Oldfield
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Bill Illis (18:46:57) :
“HankHenry (12:54:15) :
Beng-
“The cold deep-water is a remnant of the previous ice-ages. ” ”
The deep ocean temperature is a reflection of the past several hundred years of the coldest, densest ocean water there is on the planet.
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That’s prb’ly right, Bill, tho I’ve read it takes a couple thousand yrs for the deep ocean to recycle. So the cold deep water isn’t exactly a remanent of the last ice-age (10000 yrs ago), but it is of the last thousand or so yrs of cold polar weather and sinking cold water there, which continues today. Some study I read said that before Antarctica was glaciated (before 35 mil yrs ago), the deep water was as warm as 60F (15C).
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HankHenry (12:54:15) :
Beng-
“The cold deep-water is a remnant of the previous ice-ages. ”
This occurred to me but I find it surprising. If heat were able to conduct through seawater at just 1 meter per year and the average ocean depth is just 3.7 km that would only take us back to a couple thousand BC, and I usually think of the last ice age as ending something like 10,000 BC. I wonder if there could be another phenomenon having to do with a pressure gradient and/or density at work “fighting” against downward heat conduction in the ocean?
If it is true that deep ocean coldness is an ice age remnant, it raises more questions. Now I wonder in maintaining the equilibrium you mention how much heat is extracted in a year and whether it is something worth putting into the models…. or is it all a subsystem that can be disregarded?
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Hank, read my response to Bill above — the cold deep-water is prb’ly no more than a thousand or so yrs old, but still is an accumulated remanent of cold polar regions & resulting sinking water. So your thermal-conduction rate is prb’ly pretty close. But more cold water is sinking & replacing it in polar regions even now in the interglacial.
As far as “fighting” conduction — no such action is necessary, just relatively simple heat-transfer laws. Thickness of the insulating layer is the key — kilometer-thick, static layers slow down conduction to a crawl. That’s the same reason the ground isn’t hot — a few kilometers of insulating crust keep us shielded from 2000C molten rock.
And as far as I know, the deep cold-water isn’t included in GCM models — even ocean-heat content (OHC) measurements only go down several hundred meters (700 meters?), so basically water below that is ignored. Not sure if this is reasonable or not.
Brian G Valentine says:
No, it is not, as I explained in my post of (19:32:16). You can’t just assert that something violates the Second Law because you just feel that it does. You have to demonstrate that it does. I have explained why it doesn’t.
Leif Svalgaard says:
Do you know of any object in nature that is an ideal white body? At any rate, I don’t think it is vital to my argument that such an ideal white body (or something very close to it) exists does not exist. I was just interested in the transfers of radiation between two bodies at different temperatures.
Bryan says:
Well, I would have to think more about whether I agree with Arthur on it being isothermal (if that is really what he said). However, the important point is that some people seem to think that one can create higher temperatures at the earth’s surface simply by virtue of the pressure, even if the atmosphere is transparent to IR radiation. And, the point is that this is wrong. The earth still has to satisfy the first law of thermodynamics, which means that in steady-state the amount it radiates away has to be equal to the amount of energy it receives and this is what would determine the temperature at the surface in an IR-transparent atmosphere where all of the radiation from the surface will escape into space.
“I suppose that someone reading the exchange who does not have the necessary background to evaluate the scientific arguments might find Fred Staples or Terry Oldberg’s arguments compelling; those of us who do, do not find them at all compelling”
The same old worn out thing, thrown at people who disagree, I hate that, damn it
Brian G Valentine:
Well, look at it from our point of view: There are some people who, by virtue of their lack of background combined with their own preconceptions based on strongly-held beliefs, one seems unable to reach by scientific arguments, no matter how hard we try. When they say, “Convince me that…” they are asking us to do the impossible, which essentially means they are just wasting our time.
In the end, many people will believe what they want to believe…and there is only so much one can do to change minds.
Re: Steve Goddard (Mar 20 00:59),
I have expressed my questions with respect with assumptions in posts up ways.
The coat keeps me warm because I am the heat source and it is an insulating body, i.e. its heat capacity and conductivity of heat and permeability to air currents is such that it delays the cooling of my body by radiation, conduction and convection (lets not enter into the biological system that keeps me at 37C playing havoc with entropy).
Generally insulation delays equilibrium with ambient temperature, but if there is no heat source, ambient wins.
The difference between moon conditions and earth conditions makes it self evident that the atmosphere plays such a function for the surface of the earth, miss named “a greenhouse effect”, and misused with CO2 runaway imaginations.
In the same way that the coat does not heat me, the atmosphere does not heat nor create the temperature gradients. It is gravity and the thermodynamic and dynamical properties of atmospheric gases (including convection precipitation evaporation etc) that define the temperature gradients that the input sun energy and radiating earth generate in the atmosphere.
Gerlich and Tscheuschner have further developed their ideas regarding atmospheric physics. March 2010
Readers will require a background in Physics to follow the article completely. Consensus advocates are now getting a bit reluctant to fully engage G&T on matters such as heat transfer and radiative balance.
arxiv4.library.cornell.edu/pdf/1003.1508
Beng – Thank you for the knowledgeable and thoughtful response. I started my comments in this thread with the remark, “Two miles down on the continents the earth is quite warm, but two miles down in the oceans it’s quite cold.” I think that is a true statement based on things I read about boreholes, the thermal gradient, and deep mines in South Africa. My comment probably doesn’t bear directly on the article at the top of the thread about radiative heat. It just strikes me as a peculiar phenomenon that nature presents us. Perhaps something of as much interest as the much vaunted and misnamed “greenhouse effect.” I also want to say there must be more at play than insulation. There seems to be some process actively removing heat from the deep ocean. I think that process is probably the “sinking and replacing” that you mention.
Joel Shore (06:58:15) :
“I was being lectured that a white body [emissivity = 0] does not radiate at any temperature [100C, 400C or millions of degrees] …”
Do you know of any object in nature that is an ideal white body?
A black hole is perhaps the closest to a perfect white body [as its emissivity is zero – quantum effects aside], although the concepts of temperature and entropy begin to lose their meaning here. There are also no ideal blackbodies.
At any rate, I don’t think it is vital to my argument
You are quite correct.
Joel
……The earth still has to satisfy the first law of thermodynamics, which means that in steady-state the amount it radiates away has to be equal to the amount of energy it receives and this is what would determine the temperature at the surface in an IR-transparent atmosphere where all of the radiation from the surface will escape into space…….
There is no such law as the conservation of radiation.
The energy that the Earth receives from the Sun can be accounted for in a number of ways for instance Chemical Potential Energy (Plants included), stored as internal energy in the oceans etc
The Planet can warm up and cool down over periods lasting several centuries and still be in harmony with the Ist law of THD if the Suns energy is stored as other energy forms on Earth.
Well, look at it from my point of view: There are some people who, by virtue of their lack of background combined with their own preconceptions based on a strongly-held need to believe in AGW, one seems unable to reach by scientific arguments, no matter how hard I try. When they say, “Convince me that…” they are asking me to do the impossible, which essentially means they are just wasting my time.
In the end, many people will believe what they want to believe…and there is only so much I can do to change minds. But I’m not going to let nonsense and paranoia about AGW cause other people to suffer if I can help it.
anna,
If there was no atmosphere, there would be no temperature gradient through the atmosphere. Without a blanket you are cold. The sun warms the earth, and the atmosphere keeps the temperature up. Both are required to keep the earth livable.
@anna v
Of course the lunar regolith temperature at 50cm is relevant. The surface temperature in full sunlight fluctuates by hundreds of degrees C from day to night. Every centimeter you drop below the surface the fluctuation decreases. By the time you get to 50cm there is no more measurable diurnal fluctuation and so what you’re measuring is predominantly the average surface temperature.
Lunar regolith is, by the way, a much better insulator than typical earth soil. You have to go much deeper into the earth before diurnal temperature swings become too small to measure.
Re: Dave Springer (Mar 20 11:50),
Relevant to what?
The moon radiates heat in the infrared according to the formula flux= CxT^4
and that T must be the temperature of the first mm of surface . Because infrared does not penetrate either way more than the skin.
The temperature at 50cm is irrelevant to the heat radiated out. It is heat that is important, not a steady temperature. It fluctuates widely on the skin surface? Too bad for the calculations, but those are the numbers that take part in the problem.
The shell at 50cm, is isolated from the skin surface because the regolith is such a good insulator.