SOD:

“..Basic Science is Accepted – This blog accepts the standard field of physics as proven. Arguments which depend on overturning standard physics, e.g. disproving quantum mechanics, are not interesting until such time as a significant part of the physics world has accepted that there is some merit to them.”

I understand that you don´t want discussions about QM being wrong. But why do you use quantum physics to explain bulk properties like temperature? Since when do we count photons to calculate heat transfer?

And why do you base your argument on contradicting the whole theory of emission, temperature and heat if you care so much about arguments being accepted physics?

]]>Stefan-Boltzmann law with inverse square law.

You need more?

“The moderator reserves the right to just capriciously delete comments which use as their premise that standard textbook physics is plain wrong.”

I expect this to happen. Someone is wrong and my posts will be deleted.

]]>Lasse,

1360W-390W/4=242.5W=effective temperature

Textbook and formula please.

Check the Etiquette:

*..Basic Science is Accepted – This blog accepts the standard field of physics as proven. Arguments which depend on overturning standard physics, e.g. disproving quantum mechanics, are not interesting until such time as a significant part of the physics world has accepted that there is some merit to them.*

*The moderator reserves the right to just capriciously delete comments which use as their premise that standard textbook physics is plain wrong..*

If you wanted to ask I could have given you the textbook equation but you have arrived with huge confidence that you understand textbook heat transfer.

Ok, so provide the formula.

[That doesn’t mean – randomly put numbers on a page. It means cite a formula under any relevant conditions and note what each element of the formula represents.

It also doesn’t mean – 5 paragraphs of words.]

]]>1360W-390W/4=242.5W=effective temperature

The heat transferred to the surface is what it takes to heat the atmosphere at the same instant. It´s logical that the atmosphere cools the surface and heat is transferred in the same rate it emits.

“I’ll give you a clue. The solar radiation is still solar radiation, with a peak around 0.5μm, and a total flux of around 1360 W/m2”

Always the bad tone with you people.

I´ll give you a clue. The emitted energy depends on temperature only. Is the atmosphere with it´s increasing amount of dry ice a part of the surface internal state?

“In fact, the reflectivity of snow and ice at solar wavelengths is quite high. That is the snow and ice reflects 30-80% of solar radiation. On the other hand in the longwave radiation range it acts almost as a blackbody.”

Wake up, we are talking temperature, not spectral composition. It is only your eyes that makes a difference in visual and infrared radiation. Do you think your eyes has anything to do with the relationship of emission and temperature?

Do you accept the fact that a place with snow, ice and the necessary cold temperatures is a place that receives heat from all locations of higher temperatures in the surroundings?

]]>If temperature decrease the absorption increase. This is an article about heat transfer, and you forget to include heat transfer?

Please provide your **formula** for absorption of solar radiation.

The solar constant acts as a radiating body at ~395 Kelvin.

You are just making up random assertions. Please provide a formula. And a reference or a textbook.

I’ll give you a clue. The solar radiation is still solar radiation, with a peak around 0.5μm, and a total flux of around 1360 W/m^{2}.

With 51% IR in solar radiation the ice and snow is like a blackbody spreading her legs.

What has the % of an arbitrary construct known as IR got to do with anything. Provide a formula.

In fact, the reflectivity of snow and ice at solar wavelengths is quite high. That is the snow and ice reflects 30-80% of solar radiation. On the other hand in the longwave radiation range it acts almost as a blackbody.

]]>If temperature decrease the absorption increase. This is an article about heat transfer, and you forget to include heat transfer?

The solar constant acts as a radiating body at ~395 Kelvin. Heat transfers according to difference in temperature to the surface.

“And of course, alert readers will have noticed that it all changes when the surface freezes as the heat conductivity through ice will be different, and the albedo of the earth will change..”

Yes, when the surface gets cold a lot more heat is transferred from the solar constant and other warm locations in the surroundings. With 51% IR in solar radiation the ice and snow is like a blackbody spreading her legs.

]]>Chris G

In both calculations K&T assumed the Earth is a perfect black body emitter(e=1)

Even Wikipedia do not go as far as that:

Emissivity is…..

0.96 to 0.99[5][6] (except for some small desert areas which may be as low as 0.7). Clouds, however, which cover about half of the earth’s surface, have an average emissivity of about 0.5[7]“…….

Now you might think that K&T did a rough and ready calculation and just picked the nearest whole number.

However they give the surface radiation as 396W/m2 that is to three figures of accuracy.

Now I suspect that in every well run university physics department the students are told that if they give a result to three significant then all data used to calculate that result should be accurate to three figures or better.

What conclusion can we come to?

K&T don’t know any better and just muddle along?

Or is it a deliberate exaggeration so as to inflate figures to make the so called “Greenhouse effect” appear to have some reality!

Brian,

(I suspect this is a dead subject, but anyway…)

It looks to me that you are missing the point that the later K&T contains 396 where the prior has 390 is not a result of rounding up, but a result of fixing a problem in the math used on the prior paper.

It is nice that they fixed it, but it’s not a huge problem any more than there is a huge difference between 390 and 396.

]]>You’re welcome.

]]>