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Webster’s: : spout (transitive verb), definition 2a: to speak or utter readily, volubly, and at length

Perhaps you would prefer “blurting” or “reciting”?

“To start the history contest, Jack recited the first sentence of the Declaration of Independence. In response, John spouted back the entire Declaration Of Independence from memory, earning huge applause and an “A” for his effort.”

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My point it that this paper is full of hedges, assumptions, over and under estimations. If you want to quantify the energy absorbed by GHG, you need to break it up into 2 pieces. The fraction where the surface is not covered by clouds and the fraction covered by clouds. Where the surface is cloudless, the Plank distribution of surface energy is filtered by the absorption spectrum between the surface and space. Where there are clouds, the Plank distribution of cloud energy is filtered by the atmospheric absorption spectrum between the clouds and space. The difference is mostly the amount of water vapor, which is almost absent from clouds tops to space.

GHG absorption in the atmosphere between clouds and the surface has almost no net effect. Clouds are already mostly opaque to IR radiation. Even a very thin cloud cover will absorb nearly 80% of surface IR. The concentration of GHG between the clouds and the surface makes no difference to the amount of absorbed energy, as it’s all being absorbed by the atmosphere anyway (although in this case, by clouds). Of the energy absorbed by the atmosphere half is directed up and half down. This must be true for the atmosphere with or without clouds. You should already understand steradains and how they apply to BB radiators, perfect or otherwise.

The numbers I presented in the last post are from a properly performed analysis which even includes accounting for surface energy passing through clouds. This analysis matches the energy balance required to match satellite data within a few percent. This analysis also predicts that doubling CO2 from 280ppm to 560ppm increases the amount absorbed by CO2 by 3.6 W/m^2. This is consistent with the consensus claim of 3.7 W/m^2. The consensus value is from older absorption line data, while mine is based on the latest 2008 HITRAN data.

In any event, absorption between the surface and clouds has zero impact on the Earth’s energy balance.

George

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Robin,

Did you read the paper you cited?

Not all of it.

In the same paragraph where 333 W/m^2 was mentioned, it stated:

” …note that considerable uncertainties exist, and especially that there were problems in accurate simulation of thermal emission from a cold, dry, cloud-free atmosphere, and a dependence on water vapor content. The latter may relate to the formulation of the water vapor continuum.”

Sure. It’s still a lot more than 80W.

It went on to say that there is some evidence that this is overestimated.

Is that your reading of the following paragraph:

“It has been argued that downward LW radiation is more likely to be underestimated owing to the
view from satellites which will miss underlying low clouds and make the cloud base too high. Wild
and Roeckner (2006) have argued that the longwave fluxes should typically be rather higher than
lower in climate models, which, in turn, are higher than the best estimate given here. Nevertheless,
as they discuss, uncertainties are substantial. Zhang et al (2006) found that the surface LW flux was
very sensitive to assumptions about tropospheric water vapor and temperatures, but did not analyze
the dependence on clouds. Yet the characteristics of clouds on which the back radiation is most
dependent, such as cloud base, are not well determined from space-based measurements (Gupta et
al. 1999), and hence the need for missions such as CloudSat (e.g., Stephens et al. 2002; Haynes and
Stephens 2007). There are also sources of error in how cloud overlap is treated and there is no
unique way to treat the effects of overlap on the downward flux, which introduces uncertainties.
For mid and upper level clouds the cloud emissivity assumptions will also affect the estimated
downward flux. Another source of error is the amount of water vapor between the surface and the
cloud base. In the tropics, the effect of continuum absorption strongly affects the impact of cloud
emission on surface longwave fluxes.”

Because they seem to me to be talking more about a likely underestimation than an overestimation.

In any case 333 is their best estimate as at last year. And that it a lot more than 80W. (And of that 80W, not all is lost to space … it looks like half to two thirds is back-radiated; So the figure for heat lost to the atmosphere/earth system by convection is nearer 30W).

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Did you read the paper you cited? In the same paragraph where 333 W/m^2 was mentioned, it stated:

” …note that considerable uncertainties exist, and especially that there were problems in accurate simulation of thermal emission from a cold, dry, cloud-free atmosphere, and a dependence on water vapor content. The latter may relate to the formulation of the water vapor continuum.”

It went on to say that there is some evidence that this is overestimated.

George

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Please explain what you mean by “spouting”, and give some examples of how the word is used in this meaning without trying to offend.

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Okay, I didn’t do any numbers after I spotted your slip.

The numbers are:
Heat transfer to the atmosphere by evapotranspiration: 80 Watts/m²
Greenhouse gas back radiation: 333 Watts/m²
Earth’s global energy budget.

It’s measurable, but it doesn’t dominate. And it is understood and included in climate analysis including modelling.

And water vapour feedback is strongly postitive.

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Nope. It’s intending to communicate the truth, with some grace.

I can’t admire your people management skills then.

Do you have any friends?

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That’s not intending to offend?

Nope. It’s intending to communicate the truth, with some grace.

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Yes, you’re right, the heat of vaporizing water is different than the energy to boil water, it’s actually higher. If you think about it, you must add a lot more energy to a pot of boiling water to evaporate all of the water. A gram calorie is equal to about 4.18 Joules, which is the conversion factor I used. I said Kcal in the previous post, but I meant gram-cal. There are 4180 joules per Kcal.

At 18 grams of H20 per mole, it takes 18*100 gram-calories to raise one mole of water from 0C to 100C. 18*100*4.19 = 7542 Joules/mole, as compared to the latent heat of evaporation of 44000 Joules/mole. So in fact, starting from OC, it takes 5.8 times more energy to evaporate water than it does to boil it.

To clarify again, it’s not so much the rain that cools the planet, but the complete cycle of evaporation and condensation. Rain amplifies the effect on land. As you pointed out earlier, rain falling on a hot surface quickly evaporates to start the cycle over again.

To clarify yet again, oceans always cool as water evaporates from them. Oceans will cool when rain falls only if the temperature of the rain is lower than the temperature of the ocean, which is often the case. Boiling water effectively cools as steam evaporates from it, which is why the temperature of boiling water never exceeds 100C at 1 ATM. The rate of heat going in is exactly equal to the energy required to produce the steam boiling off.

A final point is that what I presented is not a hypothesis, but a test. There’s a big difference. You also need to understand that AGW as a substantial driver of climate change is not even a hypothesis, but simply a speculation reinforced by the Tinkerbell effect.

George

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Are you going to tell me that falling snow does not cool the ground? That is hard to believe.

If it melts, then it takes heat from the ground to overcome the latent heat of fusion of the water.

If it doesn’t melt and the ground is warmer than the snow, the surface of the ground will change temperature towards than of the temperature of the snow and vice-versa by conduction.

In the latter case heat is not lost. All the heat in the ground and all the heat in the snow are still there, just redistributed between the ground and snow.

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You are correct that falling rain doesn’t destroy energy, but the rain that falls is far cooler than the water that originally evaporated. This results in a net loss of heat from the planet.

A drop in temperature is not a loss in heat. Rainfall does not destroy heat, nor any other form of energy. And neither does it remove it from the planet.

Suppose snow was falling (which is only water after all). Are you going to tell me that falling snow does not cool the ground? That is hard to believe.

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To raise 31.7mg of water from 16C (average surface temperature) up to 100C (equivalent temperature of energy to vaporize) requires 2.66 Kcal/sec

You don’t need to boil water to get it to evaporate. The latent heat of vaporisation of water at 16°C is a little less that 44,000 J/mol. (source)

44,000 J/mol / (18 g/mol) * 31.7mg in kilocalories is only 0.0185 kilocalories. Not 2.66.

You are correct that falling rain doesn’t destroy energy, but the rain that falls is far cooler than the water that originally evaporated. This results in a net loss of heat from the planet.

A drop in temperature is not a loss in heat. Rainfall does not destroy heat, nor any other form of energy. And neither does it remove it from the planet.

BTW, as water evaporates from the ocean, the temperature of the ocean drops.

My point was that this is not faster after (or during) rain, because the ocean is already wet on the surface.

The flaw in the paper you pointed me to was the selection of climate sensitivity estimates subject to the analysis.

Then you need to show what was wrong with their selection methodology, because you certainly haven’t shown that their selection methodology was flawed yet.

Remember, it only takes one experiment to invalidate a hypothesis and mine is not the only one that shows a far lower climate sensitivity.

Under that logic your hypothesis was invalidated by every paper analysed by the paper. And these are not the only ones that show a far higher climate sensitivity.

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Consider that the world wide yearly water flux in and out of the atmosphere is about 1m (average annual rainfall). This requires evaporating 31.7nm of water per second, which is about .0317 cc/sec over a square meter, which is about 31.7 mg per second per m^2. To raise 31.7mg of water from 16C (average surface temperature) up to 100C (equivalent temperature of energy to vaporize) requires 2.66 Kcal/sec, which is 11.12 J/sec per m^2 -> 11.12 W/m^2. In other words, it takes 3% of the incident solar energy to transport 11.12 W/m^2 into the atmosphere to drive the weather. The peak rate is twice this since evaporation primarily occurs during the day. The energy to drive the weather comes from the heat which is extracted from evaporated water as it condenses into clouds and rain (look at how hurricanes work). The net result is a loss of 11.12 W/m^2 from the surface, which is otherwise not contributing to increasing surface temperatures.

You are correct that falling rain doesn’t destroy energy, but the rain that falls is far cooler than the water that originally evaporated. This results in a net loss of heat from the planet. BTW, as water evaporates from the ocean, the temperature of the ocean drops.

The flaw in the paper you pointed me to was the selection of climate sensitivity estimates subject to the analysis. Remember, it only takes one experiment to invalidate a hypothesis and mine is not the only one that shows a far lower climate sensitivity.

George

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Do you really think you’ve found a flaw in the climate sensitivity estimates of “Using multiple observationally-based constraints to estimate climate sensitivity”, with the claim that your personal estimates were not included in the analysis?

Can you see why I don’t think that that is a reasonable flaw?

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