I’ve been in a number of discussions over at Climate Audit (www.climateaudit.org) regarding the basic thermodynamics of the greenhouse effect. Not the physics of absorption spectra, just the basic thermodynamics.
That’s not a discussion that Steve McIntyre (the host at Climate Audit) wants us to have. Fair enough.
The most recent discussion (with Gunnar) was quite long, and eventually came down to this:
Gunnar asserted that a small mass could not cause substantial warming in a much larger mass, and I disagreed (wholeheartedly).
Since the discussion keeps coming up, I thought I’d provide a forum for it. And here it is…
Michael Smith | 03-Dec-07 at 11:28 am | Permalink
I would never assert that “a small mass cannot cause a substantial warming in a much larger mass” — not without stipulating a time interval. So let me begin by asking: over what length of time is the alleged heating to take place? If we could instantaneously increase the amount of CO2 in the atmosphere by 100 ppmv, how long would it take that CO2 to heat the rest of the atmosphere?
John V | 03-Dec-07 at 12:57 pm | Permalink
To me, the argument about whether a small mass can significantly heat a large mass comes down to basic thermodynamics. I wrote a post with equations over at Climate Audit, but it was deleted. Luckily, I thought that might happen so I saved the post. Here it is (with some minor editing where comments did not make sense without context):
=====
Gunnar, let’s work out the details with your model of CO2 purely absorbing energy. Think of the CO2 molecules as tiny black rubber balls distributed in the atmosphere. Each ball gets warmed continuously by incoming radiation reflected from the earth’s surface. This causes each ball to become warmer than the air around it. Necessarily, the warm ball will heat the surrounding air.
That’s the summary. Now for the details:
First, consider the incoming radiation only. The radiation power is Pr (J/s) and the heat capacity of each ball is H (J/degC). The rate of warming for the ball is:
dT/dt = Pr/H
Second, consider the heat transfer from the ball to the atmosphere. Use deltaT for the temperature difference between the ball and the air. The rate of heat transfer is linear with deltaT for conduction and cubic (?) for radiation. However, for small temperature differentials it’s reasonable to linearize the radiative heat transfer. The rate of heat transfer from the ball to the atmosphere is:
Pa = a*deltaT, where a is a constant
And the rate of cooling for the ball is:
dT/dt = a*deltaT/H
At equilibrium, the rate of warming must match the rate of cooling:
Pr/H = a*deltaT/H
Pr = a*deltaT
That is, at equilibrium the heat transfer from the ball to the air (a*deltaT) exactly matches the warming from the incoming radiation. The deltaT between the ball and the air depends on the heat transfer propeties of the ball/air interface, and not on the heat capacity of the ball. In fact, the heat capacity of the ball drops out of all the pertinent equations.
Whew. That was a long one. I hope it makes sense.
John V | 03-Dec-07 at 1:55 pm | Permalink
#1 Michael Smith:
I’m glad that you would never make that assertion. It’s plainly wrong, but that was Gunnar’s assertion over at CA. His opinion is that CO2 does not have enough mass or heat capacity to warm the atmosphere and the ocean.
Of course, heat capacity affects the rate of warming for a given forcing, but not the total warming possible.
Instead of dealing with a 100ppm increase in CO2, let’s talk about instantaneously doubling CO2. The immediate effect is an energy imbalance of 3.7W/m2. This extra energy has to go somewhere:
- warming the atmosphere
- warming the land
- warming the ocean
- evaporating water
- melting ice
- etc
This is where it gets complicated. Warming the ocean causes more CO2 to be released because of the reduced solubility of CO2 in warm ocean water. Evaporating water increase cloud cover which decreases warming, and increases water vapour which increases warming. There are many other complications.
Now I will commence to wave my arms and think while typing. There are likely some mistakes in the following, but I did my best:
To keep it very simple, let’s avoid the feedback mechanisms and pretend all of the energy imbalance goes to heating. This will give the instantaneous rate of heating right after the CO2 is magically doubled. Since the ocean is the dominant heat sink, I’ll focus on the ocean alone:
- M, mass of the ocean: ~1.4e21 kg
- h, specific heat capacity of sea water: ~4.2 kJ/kg/degC
- A, surface area of earth: 510e12 m2
- F, flux imbalance: 3.7 W/m2
The warming rate, R, is
R = AF / Mh
R = (510e12 * 3.7) / (1.4e21 * 4.2)
R = 3.2e-7 degC/s
R = 0.028 degC/day
R = 10.0 degC/year
Yikes — that’s way too high to be true for any significant length of time. As the earth warms the outbound IR radiation increases (by definition) until a steady state is reached. The next question is therefore determining the increase in the earth’s temperature to reach steady state.
Radiation varies with the
cubefourth power of temperature. Again using ballpark figures, the average temperature of the earth’s surface is about 15 degC, or 288K. The total outgoing longwave radiation is ~240 W/m2 (see link below). To achieve energy balance, the longwave radiation must increase to 243.7 W/m2. The steady state temperature is therefore something like:Ts = 288K * cuberoot(243.7 / 240) = ~289.5KTs = 288K * quadroot(243.7 / 240) = ~289.1K
http://stephenschneider.stanford.edu/Climate/Climate_Science/EarthsEnergyBalance.html
That is, the total warming (without any feedbacks) should be about
1.51.1 degC.I did a quick simulation using only the increased radiation from a warming earth and found that the warming would take less than 6 months if the oceans and atmosphere warmed together. The difference comes from the fact that the atmosphere and land surface heat quickly, thereby increasing outbound radiation to reach energy balance. The oceans are then warmed by convection and conduction from the atmosphere, which is much slower.
=====
I found a good reference for the effect of doubling CO2 from the pre-political days of AGW:
http://www.cs.ntu.edu.au/homepages/jmitroy/sid101/uncc/fs016.html
Michael Smith | 03-Dec-07 at 4:00 pm | Permalink
Where does the value for F, 3.7 W/m2, come from?
John V | 03-Dec-07 at 4:24 pm | Permalink
I can’t remember my source for that (it’s been a while).
Try Googling on “co2 double forcing 3.7″.
John V | 03-Dec-07 at 11:24 pm | Permalink
There’s clearly something wrong with my analysis is #3, but I don’t know what. Some ideas:
- arithmetic error
- latent heat of evaporation
- latent heat of fusion (melting)
- increased heat loss due to convection and other processes
Michael Smith | 04-Dec-07 at 5:06 am | Permalink
John, I don’t see anything wrong math-wise, except this. You wrote:
- M, mass of the ocean: ~1.4e21 kg
- h, specific heat capacity of sea water: ~4.2 kJ/kg/degC
- A, surface area of earth: 510e12 m2
- F, flux imbalance: 3.7 W/m2
The warming rate, R, is
R = AF / Mh
R = (510e12 * 3.7) / (1.4e21 * 4.2)
R = 3.2e-7 degC/s
Where does the unit of time, the “s” in this last equation, come from? I guess it must be part of “F”. I don’t see it in any of the other equations.
Also, I’ll try, as you suggest, “Googling” the expression “co2 double forcing 3.7″, but I believe that this is the very question Steve McIntyre has been asking for some time now without getting an answer.
In any event, proving that “a small amount of CO2 can warm the earth” reduces to the issue of proving the value of “F”. Everything after that is just math, unless I’m missing something.
John V | 04-Dec-07 at 8:37 am | Permalink
Michael Smith:
The unit of time “s” does indeed come from the forcing term. F is in W/m2 which equals J/s/m2. That is, 1 Watt = 1 Joule per second.
The radiative forcing for doubling CO2 in isolation (3.7W/m2) is pretty well established and accepted. Steve McIntyre’s question was how this radiative forcing can lead to a 2.5 degC temperature increase. A simple radiation balance shows a temperature increase of only ~1.1C. The remaining warming is caused by feedbacks which are much more difficult to quantify without a model. Steve McIntyre knows this but continues to ask for an answer without the use of a model. As some have said at CA, it’s like asking for the total lift of a Boeing 747 from first principles (no modelling allowed).
In #2 I tried to show that the warming power of the atmospheric CO2 in isolation is unrelated to the mass or heat capacity of the CO2. It comes down to the heat transfer properties only. The key questions are how much a given amount of CO2 affects the radiation balance (F = 3.7W/m2) and the effect of feedback mechanisms. As far as I can tell, the former is well accepted but the latter is not. The feedbacks are very complicated — here’s a small sample:
- increased temperature leads to more water vapour in the atmosphere (water vapour in isolation causes warming);
- increased water vapour leads to more clouds (clouds probably cause cooling);
- increased temperature leads to more CO2 from the oceans due to decreased solubility at higher temperatures;
- increased temperature leads to more plant growth, which decreases CO2;
- increased temperature leads to less ice (less ice causes more warming);
- etc
Michael Smith | 05-Dec-07 at 10:40 am | Permalink
How was the 3.7W/m2 value established?
John V | 06-Dec-07 at 2:29 pm | Permalink
#9 Michael Smith:
As far as I know, 3.7W/m2 was established from numerical simulations. Wikipedia has a list of radiative transfer codes:
http://en.wikipedia.org/wiki/List_of_atmospheric_radiative_transfer_codes
MODTRAN has a web interface here:
http://geosci.uchicago.edu/~archer/cgimodels/radiation.html
jae | 08-Dec-07 at 8:07 am | Permalink
John V: I think you forgot something in #8. Increased temperature does cause increased evaporation, but at an “expense” of 0.69 watt-hr/g/m^3. This is the latent heat of evaporation, and it is a negative feedback, since you don’t get most of that energy back anywhere near the surface (except dew). See http://www.esnips.com/web/climate
Also, all that solar energy is spent in heating the air column and the surface. All those molecules are “trading” radiation, and at some elevation (maybe 5km?) there is more going to space than back and forth. But there’s no “cascading” of radiation toward the ground, as is commonly thought. That would be adding energy. If the radiation hypothesis is true, tell me why it is hotter in July (average and maximum temperatures) in Dagget, CA than anywhere in the tropics? After all there is no shortage of “greenhouse gases” (water vapor) in the tropics. I have been asking this question for two years now, and nobody even attempts to answer it…
John V | 08-Dec-07 at 9:41 am | Permalink
jae, I did mention the latent heat of evaporation in #6. Using common SI units, the value is 2258 kJ/kg. I’m not sure what your units are — is the g for grams and m for metres? If so, your units don’t work out. Converting your value to more standard units I get:
0.69 watt*hr/g/m^3
=0.69 watt*(3600s)/g/m^3
= 0.63 J/g/m^3
= 0.63 kJ/kg/m^3 –> what’s the m^3 doing there?
Anyways, putting the units aside and focussing on your point, here’s my opinion:
The *surface* temperature is higher in deserts because of albedo and evaporation. But don’t forget that the latent heat of evaporation is recovered when the water vapour condenses into clouds. Do you have data for atmospheric temperatures at a couple of thousand feet in these areas?
ClaytonB | 09-Dec-07 at 11:01 am | Permalink
“As some have said at CA, it’s like asking for the total lift of a Boeing 747 from first principles (no modelling allowed).”
Yes, but I’m quite certain that they performed wind tunnel tests…
John V | 09-Dec-07 at 4:11 pm | Permalink
ClaytonB, they have indeed performed wind tunnel tests and/or flight tests. But that’s not the point. The point is that Steve McIntyre’s request for the derivation of the climate sensitivity from first principles is basically impossible. There are two ways to determine the value:
1. Empirically using CO2, temperature, and solar reconstructions;
2. Numerically using computer models;
Another problem with determining climate sensitivity is that it has different values depending on the timescale. There are slow feedbacks (eg. glaciers melting) and fast feedbacks (eg. water vapour).
jae | 09-Dec-07 at 7:56 pm | Permalink
12: “higher in deserts because of albedo and evaporation.”
Now this is really an odd response. Both of these things tend to LOWER temperatures. And, sorry, if I put the m^3 in there, it was a mistake. The latent heat of evaporation is about 0.69 watt-hr/gram of water, and that is a NEGATIVE feedback. That energy is spent at the surface. Of course you get the energy back, but too high in the atmosphere to affect your thermometer. And please tell me why the radiation models dwell on CO2 and completely ignore the most important GHG–water vapor. It falls out of the sky, unlike CO2, but at any given time, it’s there! I have no qualms with the radiation demonstrations that show the average temperature of the Earth. But when you start creating this bogus “layer in the sky” that decides how much IR goes to Earth, and how much goes to space, I get very skeptical.
Someone, please tell me why it’s hotter in Phoenix than in Guam, if the so-called GHGs have ANYTHING to do with temperature! It’s all about heat storage, not radiation. I hope I can find a way to prove this.
John V | 09-Dec-07 at 9:04 pm | Permalink
jae:
Obviously I meant the temperature is higher because of lower albedo and evaporation, but thanks for the correction.
You seem to have found that water vapour has a negative feedback on *surface* temperature. I merely asked if the same pattern existed at the level of clouds. Based on your response, I assume you don’t have an answer. It’s ok to say that.
Alan D. McIntire | 12-Dec-07 at 6:28 pm | Permalink
Water vapor does indeed have a negative feedback. The dry adiabetic lapse rate for the atmosphere would be about 9.8 C per 1000 meters. For moist air, that drops to roughly 5 C, depending on the original temperature of the air. The moral of the story is that dry air cools more rapidly with height than moist are. The converse is that moist air is warmer at higher altitudes, and this radiation of moist air from higher altitudes is a negative feedback-
John V | 12-Dec-07 at 9:40 pm | Permalink
Alan:
I don’t believe your conclusion is correct. The moist air lapse rate only applies when the air is 100% saturated, such that additional expansion (cooling) causes condensation (warming). If the relative humidity is 98%, the dry adiabatic lapse rate still applies.
I had thought that RH would go up with altitude, but Paolo M over at CA has taught me otherwise.
For the record, I made a similar mistake about moist vs dry air but came to a different conclusion.
You are also neglecting the greenhouse gas properties of water vapour, which is the primary reason it’s considered a positive feedback. Other water vapour properties such as reflective clouds and surface evaporation are negative feedbacks.
sue | 14-Dec-07 at 11:28 am | Permalink
I’m not a scientist (duh), but I have direct observational experience from working in commercial greenhouses, that show the impact of vegetation/moisture on temperature when all other conditions are identical. Yoder Bros. a large commercial grower, had a California plant that produced chrysanthemum cuttings. The plant consisted of more than a dozen identical clear glass and wood frame greenhouses. Each greenhouse had eight long parrallel, beds about 18″ off the ground, and four giant fans at the back of the green house, at blew air accross large fiber mats that water trickled over.
Many times during my tenure at this operation I had occasion to work in the same day, (even within the same hour) in greenhouses that were completely full of plants and those that were completely empty of growing plants (removing dried remains and preparing beds for new planting). The only differences between the comparison greenhouses were the presence or absence of plants and the presence or absence of the water vapor resulting from the twice a day watering of the growing plants. It was entirely typical for the growing house to have a temperature of 86 degrees F while the bare house (right next door at the same time of day) had a temperature of 120 degrees F. Thermometers were located every few feet in all greenhouses and easy to read. Thirty to forty degree differences between houses were commonly observed through out the summer.