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.
Leif Svalgaard (07:43:34)
Is my writing really that opaque? My conclusion is that the climate sensitivity is on the order of half a degree C per doubling of CO2, and the equilibrium sensitivity is much less than that.
blackswhitewash.com (07:45:35)
Oh, please. If you have something substantial to say, bring on the math. Otherwise, keep lurking and wait until we see if it is Swiss cheese or not. Are there any substantive claims that I have not disputed?
mkelly (09:08:39)
Please see my post called “The Steel Greenhouse” which answers your question.
brent (09:34:26)
I view them as a first order approximation, which is quite adequate for the purposes of my analysis.
[quote Willis Eschenbach (10:35:46) :]
Are there any substantive claims that I have not disputed?
[/quote]
Well, um, yes. Mine. Trenberth’s model uses made up numbers and ignores real life measurements. Using his work to check your results means you’re comparing your results to made up numbers.
That’s not a flaw in your line of reasoning, but it is a flaw with what you’re using for a sanity check on your results.
Which is why I was wondering if you feel up to the task of creating a more realistic energy balance. One that uses real-world numbers.
Willis Eschenbach (10:34:11) :
Is my writing really that opaque? My conclusion is that the climate sensitivity is on the order of half a degree C per doubling of CO2, and the equilibrium sensitivity is much less than that.
As far as I can see you concluded that based on the 20C figure. You never used the 8C figure for anything, which is good because it is wrong.
scienceofdoom (23:02:15)
Thanks for the response to brian W. I find your blog a wonderful resource for people get an understanding of things like back radiation.
There is enough detail there for those with technical backgrounds and your way of explaining things to those with “not so technical” backgrounds is very good. I often wonder what the blog wars would have been like if reasonable voices like yours had been in charge of real climate.
Scienceofdoom: WRT Willis contention. It seems a first good guess that
the first order effect of albedo is linear, it’s effectively a TSI knob, but one thing that bothers me about the diagram is what happens to the 29W/sq meter that get reflected from the surface. Shouldnt some portion of that be absorbed by trop and lower strat ( as is the case with that same radition when it is incoming)? And shouldnt some be relected back down by clouds and aerosals? is the 29 a net net net. Let me put it this way. Short wave reflected at the surface has to travel back through the same phyiscal medium it took to get to the earth.. probably should not read too much into the cartoon I suppose.
Willis,
There is no need for laboratory experiments. The earth itself is the laboratory, with some locations having much less atmosphere than others. Temperatures in the upper elevations of the Himalayas average 40-50C cooler than similar latitudes at lower elevations, because of the lack of atmosphere. Your claim below is simply incorrect.
WTF? Environment Canada says that this winter was 4.0°C above average?
http://www.msc.ec.gc.ca/ccrm/bulletin/national_e.cfm
The national average temperature for the winter 2009/2010 was 4.0°C above normal, based on preliminary data, which makes this the warmest winter on record since nationwide records began in 1948. The previous record was 2005/2006 which was 3.9°C above normal. At 3.2°C below normal, the winter of 1971/1972 remains the coolest. As the mean temperature departures map shows all of the country, but for a small area over the southern Prairies, was above normal, with some areas of the arctic and northern Quebec more than 6°C above normal. Southern Saskatchewan did have a cooler than normal winter, with temperatures more than 1°C below normal. This past spring was the first season in the past 5 years with temperatures below normal, as shown in the consecutive seasons graph.
Leif Svalgaard (10:20:53)
I’ve never heard of a “white blackbody”. I’m using the standard definition of a blackbody. Here’s one, from Mathworld:
Here are the only numbers of interest in my discussion above:
Temperature of a body at the distance of the earth from the sun, with an albedo equal to that of the earth: about 20°C cooler than the present earth.
TOA radiation from the present earth: about 150 W/m2
Since 150 W/m2 TOA radiation gives a warming of about 20°C, this gives a climate sensitivity of about half a degree for a doubling of CO2. This is only one-sixth of the UN IPCC canonical value of 3°C.
I don’t care if I’m off by 10%, this is a first-order analysis. My answer is way below the IPCC answer, that’s the issue, not the exact details.
I greatly regret putting up the global energy budget diagram, as people are overlooking the subject of the thread, which is climate sensitivity. I invite everyone to discuss whether this method of calculating climate sensitivity is valid and whether those numbers are correct, and leave downwelling radiation and pV=nRT and “white blackbodies” for another thread.
Leif Svalgaard (10:42:01)
Leif Svalgaard (10:20:53)
Can’t have it both ways … the 8C is right for a black blackbody. Since I was not talking about a white blackbody, the 8C figure is not wrong.
Willis Eschenbach (10:34:11) :
Is my writing really that opaque?
The initial quiz was a bit opaque. I would have agreed completely if the premises were ‘no atmosphere and totally black’. The ‘blackbody’ bit is somewhat off, because in the infrared, the Earth is an almost perfect blackbody and that is where it radiates.
Willis Eschenbach (10:59:56) :
Can’t have it both ways … the 8C is right for a black blackbody. Since I was not talking about a white blackbody, the 8C figure is not wrong.
But you were not explicitly talking about a ‘black’ blackbody either so the built-in assumption is that you were talking about a blackbody with an albedo [0.15] as that the surface actually has…
But perhaps we have generated enough heat by now without much new light. 🙂
mkelly (08:33:01) :
David L (07:50:25) :
“Are you saying that the 5.3million giga tons of atmosphere has no effect on the earth’s temperature? If I leave T as an unknown in the formula and use standard pressure for P and figure for a volume up to say 100km I get a temperature of 0 deg C. ”
Not due to PV=nRT realistically. Are you saying that because there’s pressure and gas molecules and a volume (what’s the volume of the earth’s atmosphere by the way?) then we have a temperature? Turn off the sun, shut down the geothermal heat, and see how long it takes for the atmosphere to become pools of liquid N2 on a frozen ocean. Pressure is not driving temperature here. It’s the other way around. Classic example of mixing up Causation with Correlation. Temperature and Pressure are correlated…but which causes which? If your highschool student compresses air, the compressed air temp changes….but that temperature is soon absorbed by the surroundings. Ever touch a high pressure gas cylinder? How hot is it? It’s room temperature…But it’s under high pressure? how can that be? Now put that tank in the sunlight….guess what…temp goes up AND pressure goes up.
So why would the amount of air, the earths atmospheric volume, and pressure have anything to do with global temperatures according to PV=nRT? Look up R….what are the terms? J/K mol. It’s the amount of energy per temperture per mole….it’s all about the energy folks Know the energy flow and you’ll know all the temperatures!!!!!! If you’re not doing work on the system you aren’t moving energy around!! Just having an atmosphere is not doing work!!!! You have to do something to it…like heat it up with sunlight!!!!!!
“Take a white sphere and heat it to 400C, then take a black sphere and heat it to 400C, they both radiate the same amount. If not, we would violate the thermodynamics, where we could get work out of the radiative difference [use it to power a steam engine] between two bodies at the same temperature.”
This warmist logic! The body with the lower emissivity will radiate less, that is what emissivity means: it radiates less at the same temperature.
As for breaking fundamental laws of thermodynamics, if that were true, then thermos flasks and space blankets both break your “fundamental” laws as the low emissivity material prevents irradiation of heat compared to similar “blackbody” materials at the same temperature.
magicjava (10:25:39)
The balance shown in the head post is my own. Much of it is based on actual data, just as much of Trenberth’s is based on actual data. I used my own simple radiation/convection/evaporation model available here to calculate the values.
Unlike Trenberth’s model, however, mine actually works. His cannot work, as it shows different fluxes up and down from the atmosphere, which is not physically possible.
For the reasons I lay out in The Steel Greenhouse, a model of the earth’s greenhouse system has to have two physically isolated radiating layers to concentrate enough energy to equal earth conditions. The upper of these is just above the tropopause.
These layers must be physically separated from each other, because if there is more than a very small amount of thermal leakage between them, the efficiency of the system drops below that which is required to model the earth.
These constraints limit the possible solutions to the problem. I have used my model to fit the known parameters. One of these is the temperature of the TOA, which is on the order of -50C, or on the order of 150 W/m2. So the number I am using to calculate the sensitivity is not supported by my/Trenberth’s model alone.
w.
Richard Sharpe (10:01:57) : I don’t have “smiley” faces to put at the end of sentences, but you it the nail on the head. It was a rhetorical question.
Most of the people on the earth live in areas that get above the temperature of where CO2 radiates. That’s part of the point.
If CO2 would not warm up coffee or keep gespacho luke warm how is it able to warm up an entire global. The atmosphere has conduction and convection. Gases dissipate heat. My frozen turkey will warm to room temperature but no more. No amount of CO2 in my house will heat the turkey higher than what the room temperature is.
Is there a ratio of convection/conduction by CO2 versus radiation by CO2 when dissipating heat?
If I remember my heat transfer class when two discs at different temperatures radiate at each other the lower will rise to the higher then it becomes a push/pull and nothing further happens. So the best that could happen with CO2, is it could rise to the temperature of the earth and then all stops. The pot does not warm the burner.
Willis, I think inadvertently you have stumbled upon the key wrong assumption in this whole AGW debacle. The value for Trenberth back radiation is very wrong. In the paper they show the upward radiation from the surface and then from the top of the atmosphere. At the TOA we can see the absorption peaks and how there is a reduction in the spectra. The difference in W/m2 is the difference in the integrals of these curves i.e. what does not come out of the TOA is the difference in the integrals. If those curves are integrated an eyeball estimate is about 10 to 15% difference. If we have 339 going up how can we then have 321 coming back?
Secondly using the same curves, and using the measured CO2 15 µm band, which is heavily saturated for the current ppm of CO2. Any increase in ‘the wings’ is 4 orders of magnitude different than the main peak. This puts any decrease in transmissivity of that order of magnitude. Now considering that the difference in back radiation is around 30 to 50 W, how exactly is a doubling of CO2 going to produce 3W/m2 extra forcing. The simple answer is it can’t hence the assumption that water vapour increases; an assumption that as yet has not been tested.
Just realised, when I mentioned thermos flasks, I forgot to mention I was referring to the silver coating used to reduce heat loss.
groundRe: magicjava (Mar 17 04:43),
Thanks for the link. One can see the difference between air and skin surface. in the maps.
mkelly (09:08:39)
For anyone a question. If CO2 can cause an increase in temperature based on reradiation of IR then why are thermos bottles not filled with 100% CO2 instead of using a vacuum?
By the IPCC formula of 5.35ln(C/CO) you could get up to 75 W/m2 of extra “heat” going back into your coffee. OUCH!
Let’s see if I can explain. Heat is transferred by Convection, Conduction and radiation. See the sun? It’s in a vacuum. There is a lot of nothing around it.
So the heat of the sun get here by radiation. Take a metal rod in your hand hold it in a fire. That’s conduction.
In a thermos you build a vacuum chamber. You try to get as few as particles in there as possible to prevent conduction. Filling that chamber with anything would defeat the purpose of reducing the heat transfered by CONDUCTION. How do you prevent radiation through this vacuum? Easy, you put a SILVERED LINING on the metal.
That shiny metal “blocks” or reflects the IR. It does this much better than C02 does. C02 only “blocks” certain regions of Longwave.
http://www.howstuffworks.com/thermos.htm/printable
When you look at how radiation move through a material you have to look
at all the ‘windows’ in the material. .
This may help:
http://www.crisp.nus.edu.sg/~research/tutorial/atmoseff.htm
Then go google radiative transfer equations.
Technically, if we have now reduced the greenhouse effect/atmospheric pressure impact to 20K, the TOA forcing relative to that figure is not 150 W/m2 anymore – it would only be about 95 W/m2.
The equilibrium emission temperature would now be 268K rather than 255K and 95 W/m2 gets you up to the surface temperature of 287K or 14C.
So the average over the entire 95 W/m2 would be 0.21K /W/m2. [As I said the first part of the 95 watts would have more impact than the last one which is just 0.18K /W/m2].
Technically, it also wouldn’t be the TOA forcing either but just the 3.1 km high forcing.
[quote Willis Eschenbach (10:57:20) :]
I greatly regret putting up the global energy budget diagram, as people are overlooking the subject of the thread, which is climate sensitivity.
[/quote.
My apologies for being one of those people. When I see Trenberth’s diagram, it just sets off a giant red flag for me.
There are so many topics covered in the post that it’s difficult for me to find the core substance of it all. I _think_ the point to the entire thing is:
[quote Willis Eschenbach:
This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase.
[/quote]
… but that was saved until the next-to-last sentence.
Also, you’re tossing around a lot of unsubstantiated claims. Examples include almost everything you say about the IPCC. For example, does the IPCC really calculate the effects of Evaporation incorrectly? And no one ever noticed this? With all the things coming out about the IPCC lately, this may very well be the case, but statements like this need some supporting evidence.
The use of 20°/8° is also very confusing to me. It’s not clear when one or the other applies to what you’re saying.
I’d suggest stripping out as many non-essential points as possible, so folks like me don’t get tripped up over Trenbreth, and folks like Lief don’t nit-pick you to death on black blackbody and white blackbody stuff.
If your main point is the increase in temperature due to increase in W/m-2 is logarithmic, start with that and make it clear.
Willis Eschenbach (10:35:46) :
Are there any substantive claims that I have not disputed?
Derek says – Well, yes actually, for instance see the below.
AND my own post elsewhere.
http://www.globalwarmingskeptics.info/forums/thread-609.html
Terry Oldberg (09:44:22) :
In the language of thermodynamics, there is no such thing as an “energy flow.” The only energy that “flows” is heat. In a Kiehl-Trenberth diagram, some of the “flows” are heat. Others are radiation intensities. It is clear that the “back radiation” is not heat for it “flows” from cold to hot matter; if it were heat, it could not flow in this manner, under the second law of thermodynamics. However, while we have a conservation principle for heat flows, we do not have one for radiation intensities. Thus, the proposition that the K-T diagram portrays some kind of “balance” is false.
John A wrote (21:46:51):
How of the incoming energy from the Sun is converted into atmospheric and oceanic convection? None, according to Kielh and Trenberth. Which is amazing. Climate physics is a fascinating subject.
What causes atmospheric and oceanic convection to slow down (the only reason solar energy would be needed to keep them going)? Friction. Where does the energy consumed in overcoming friction appear? As heat. Physics is a fascinating subject.