Tropospheric Hot Spot- Why it does not exist...

[Wuwei]

WTF are you talking about? Heat is energy, it is measured in Joules. If it isn`t energy then what is it ?
And then "Furthermore radiatitive heat transfer by EM energy ". The adjective for radiate is radiative not "radiatitive".
Typical gibberish which clearly indicates that you are pretentious and have no idea what any of the terms you re-defined here actually mean.
Like this nonsense:"Entropy says nothing about the flow of energy one way or another"
Really?
That shows that you have no clue whatsoever.
Entropy is about the transfer of energy, as opposed to enthalpy which is the energy content of a system and not about the transfer of energy
Typical indicators that you are one of the science impostors grand standing here ! I bet anything that the moment your WiFi gets cut your "expertise" in physics goes poof in an instant.

Do you also believe that thermal radiation cannot emit from a colder object to a hotter one?

 

[Billy_Bob]

the whole point was that the AGW hypothesis is based on radiation from CO2...and it is built entirely on radiation and assumes that CO2 radiates all the energy it absorbs...it doesn't take convection or conduction even into consideration.....and the amount of energy that CO2 actually radiates as opposed to losing to conduction is about one billionth the amount of energy that the AGW hypothesis assumes that CO2 is radiating in all directions.

Water vapor is also triatomic and has the same type of vibration states as CO2. Do you use the same "logic" for water too? If it weren't for water vapor the entire earth would be near freezing.

CO2's primary absorption/emission band is 16-21um. Water Vapor is 6um to 65um. When we consider that water vapor dwarfs CO2 in concentration by a factor of 2,400 it becomes painfully obvious we have a gorilla in the room that can stop all of CO2's activity and not even breath hard.

 

[Wuwei]

You mean IR..which is exactly what radiative heat is..which is measured in joules and exposure to it is expressed in terms of joules per square meter?

Read it again. The diagram from the hyperphysics site I posted does not refer to IR. It refers to energy transfer using Freon (or whatever they use nowadays.)

 

[Wuwei]

You are a f'ing idiot.....did you not read what polar bear posted for you? Heat is energy...it is measured in joules...radiant exposure...which is, in fact, exposure to IR radiation is expressed in terms of joules per square meter....energy is energy you doofus...and all energy is governed by the second law of thermodynamics which says that it won't move spontaneously from cooler objects to warmer objects...

even when it is given to you in explicit terms, your mind won't allow you to read and understand the words.

Very good! You and polarbear can look up the units of energy. But you should cool down and don't blow a gasket. The units of energy are well known, but what you aren't considering is there are many forms of energy, not just heat. The subject is that thermal radiation (photon or EM energy) from any object is not impeded from striking any other object at any temperature.

You and polarbear disagree with that. Read about thermal equilibrium. I gave you references a while ago.

 

[polarbear]

WTF are you talking about? Heat is energy, it is measured in Joules. If it isn`t energy then what is it ?
And then "Furthermore radiatitive heat transfer by EM energy ". The adjective for radiate is radiative not "radiatitive".
Typical gibberish which clearly indicates that you are pretentious and have no idea what any of the terms you re-defined here actually mean.
Like this nonsense:"Entropy says nothing about the flow of energy one way or another"
Really?
That shows that you have no clue whatsoever.
Entropy is about the transfer of energy, as opposed to enthalpy which is the energy content of a system and not about the transfer of energy
Typical indicators that you are one of the science impostors grand standing here ! I bet anything that the moment your WiFi gets cut your "expertise" in physics goes poof in an instant.

Do you also believe that thermal radiation cannot emit from a colder object to a hotter one?

What makes you think I believe that?
Sure it does, but the net transfer of energy is from the hotter to the colder one because the hotter one radiates more energy than the colder one. In the process the hotter one will cool off and the colder one will warm up with the radiation it absorbed from the hotter one.
There is no net gain of energy. For that to happen you need another heat source.
What`s the big mystery here?
If you got this other heat source then the temperature will increase if the colder body obscures the warmer one from an even colder back ground..so what ?
The problem is not if it does, but by how much !
If you want to see how much CO2 does then you have to measure it and that was done by Heinz Hug:
The Climate Catastrophe - A Spectroscopic Artifact
The radiative forcing for doubling can be calculated by using this figure. If we allocate an absorption of 32 W/m^2 [14] over 180º steradiant to the total integral (area) of the n3 band as observed from satellite measurements (Hanel et al., 1971) and applied to a standard atmosphere, and take an increment of 0.17%, the absorption is 0.054 W/m^2 - and not 4.3 W/m^2.This is roughly 80 times less than IPCC's radiative forcing.

It`s also a joke to say that 1/2 of the absorbed energy goes down and is re-absorbed.
Why? Because it ignores the angle of incidence of the down dwelling radiation.
7/10 th of the earth surface is water and for that part the reflectance increases with the a.o.i. for all light, IR included.

 

[polarbear]

You are a f'ing idiot.....did you not read what polar bear posted for you? Heat is energy...it is measured in joules...radiant exposure...which is, in fact, exposure to IR radiation is expressed in terms of joules per square meter....energy is energy you doofus...and all energy is governed by the second law of thermodynamics which says that it won't move spontaneously from cooler objects to warmer objects...

even when it is given to you in explicit terms, your mind won't allow you to read and understand the words.

Very good! You and polarbear can look up the units of energy. But you should cool down and don't blow a gasket. The units of energy are well known, but what you aren't considering is there are many forms of energy, not just heat. The subject is that thermal radiation (photon or EM energy) from any object is not impeded from striking any other object at any temperature.

You and polarbear disagree with that. Read about thermal equilibrium. I gave you references a while ago.

I don`t have to look up what energy is, but you should.
Obviously you should else if you knew you would never say:
The units of energy are well known, but what you aren't considering is there are many forms of energy, not just heat. The subject is that thermal radiation (photon or EM energy) from any object is not impeded from striking any other object at any temperature.
That`s not just one error but a whole cluster fuck of errors.
I`ll pick apart the worst one:
What do you get after you factor this equation with the 4th power of Temperature [in deg K]?
σ = 5.67×10^−8 [W m^−2 K^−4]
you get watts per m^2 right?
And if you knew the area and use that factor you get watts right ?
And what`s a watt?
It`s power, not energy right?
And if you knew the time during which you applied the number of watts then and only then do you know how much ENERGY that was.

 

[Crick]

Do you think you're the only one to have thought of that [Angle of incidence]? And where does the reflected light go? It still has the entire atmosphere, at a very shallow angle, to penetrate.

 

[SSDD]

You mean IR..which is exactly what radiative heat is..which is measured in joules and exposure to it is expressed in terms of joules per square meter?

Read it again. The diagram from the hyperphysics site I posted does not refer to IR. It refers to energy transfer using Freon (or whatever they use nowadays.)

Sorry guy....it doesn't say that at all...in fact, it is written at a 6th grade level and you still don't seem to be able to understand it...you feel the need to try and interpret it to mesh with your fouled up notions....try just reading what it says rather than trying to read something into it that it doesn't say.

 

[Crick]

WTF are you talking about? Heat is energy, it is measured in Joules. If it isn`t energy then what is it ?
And then "Furthermore radiatitive heat transfer by EM energy ". The adjective for radiate is radiative not "radiatitive".
Typical gibberish which clearly indicates that you are pretentious and have no idea what any of the terms you re-defined here actually mean.
Like this nonsense:"Entropy says nothing about the flow of energy one way or another"
Really?
That shows that you have no clue whatsoever.
Entropy is about the transfer of energy, as opposed to enthalpy which is the energy content of a system and not about the transfer of energy
Typical indicators that you are one of the science impostors grand standing here ! I bet anything that the moment your WiFi gets cut your "expertise" in physics goes poof in an instant.

Do you also believe that thermal radiation cannot emit from a colder object to a hotter one?

What makes you think I believe that?
Sure it does, but the net transfer of energy is from the hotter to the colder one because the hotter one radiates more energy than the colder one. In the process the hotter one will cool off and the colder one will warm up with the radiation it absorbed from the hotter one.
There is no net gain of energy. For that to happen you need another heat source.
What`s the big mystery here?
If you got this other heat source then the temperature will increase if the colder body obscures the warmer one from an even colder back ground..so what ?
The problem is not if it does, but by how much !
If you want to see how much CO2 does then you have to measure it and that was done by Heinz Hug:
The Climate Catastrophe - A Spectroscopic Artifact
The radiative forcing for doubling can be calculated by using this figure. If we allocate an absorption of 32 W/m^2 [14] over 180º steradiant to the total integral (area) of the n3 band as observed from satellite measurements (Hanel et al., 1971) and applied to a standard atmosphere, and take an increment of 0.17%, the absorption is 0.054 W/m^2 - and not 4.3 W/m^2.This is roughly 80 times less than IPCC's radiative forcing.

It`s also a joke to say that 1/2 of the absorbed energy goes down and is re-absorbed.
Why? Because it ignores the angle of incidence of the down dwelling radiation.
7/10 th of the earth surface is water and for that part the reflectance increases with the a.o.i. for all light, IR included.

Critical angles are determined by each material's index of refraction. Index of refraction is frequency dependent.

Wikipedia

Snell's law states that the ratio of the sines of the angles of incidence and refraction is equivalent to the ratio of phase velocities in the two media, or equivalent to the reciprocal of the ratio of the indices of refraction:

with each

as the angle measured from the normal of the boundary,

as the velocity of light in the respective medium (SI units are meters per second, or m/s),

as the wavelength of light in the respective medium and

as the refractive index (which is unitless) of the respective medium.

 

[Wuwei]

the net transfer of energy is from the hotter to the colder one because the hotter one radiates more energy than the colder one. In the process the hotter one will cool off and the colder one will warm up with the radiation it absorbed from the hotter one.

Great. Now try to tell SSDD that. He sincerely believes the colder object can't radiate to the hotter one because he doesn't understand the second law or the S-B equation. Have you ever argued those points with him?

I don`t have to look up what energy is, but you should.
Obviously you should else if you knew you would never say:
The units of energy are well known, but what you aren't considering is there are many forms of energy, not just heat. The subject is that thermal radiation (photon or EM energy) from any object is not impeded from striking any other object at any temperature.
That`s not just one error but a whole cluster fuck of errors.
I`ll pick apart the worst one:
What do you get after you factor this equation with the 4th power of Temperature [in deg K]?
σ = 5.67×10^−8 [W m^−2 K^−4]
you get watts per m^2 right?
And if you knew the area and use that factor you get watts right ?
And what`s a watt?
It`s power, not energy right?
And if you knew the time during which you applied the number of watts then and only then do you know how much ENERGY that was.

Very good. Now try to get SSDD to believe that. He doesn't.

 

[SSDD]

the net transfer of energy is from the hotter to the colder one because the hotter one radiates more energy than the colder one. In the process the hotter one will cool off and the colder one will warm up with the radiation it absorbed from the hotter one.

Great. Now try to tell SSDD that. He sincerely believes the colder object can't radiate to the hotter one because he doesn't understand the second law or the S-B equation. Have you ever argued those points with him?

I don`t have to look up what energy is, but you should.
Obviously you should else if you knew you would never say:
The units of energy are well known, but what you aren't considering is there are many forms of energy, not just heat. The subject is that thermal radiation (photon or EM energy) from any object is not impeded from striking any other object at any temperature.
That`s not just one error but a whole cluster fuck of errors.
I`ll pick apart the worst one:
What do you get after you factor this equation with the 4th power of Temperature [in deg K]?
σ = 5.67×10^−8 [W m^−2 K^−4]
you get watts per m^2 right?
And if you knew the area and use that factor you get watts right ?
And what`s a watt?
It`s power, not energy right?
And if you knew the time during which you applied the number of watts then and only then do you know how much ENERGY that was.

Very good. Now try to get SSDD to believe that. He doesn't.

Like I said goober...you be sure to tell me when they rewrite the second law of thermodynamics to relate energy transfer in terms of net as opposed to gross energy movement....it will be big news...but let me know anyway.

 

[Wuwei]

Sorry guy....it doesn't say that at all...in fact, it is written at a 6th grade level and you still don't seem to be able to understand it...you feel the need to try and interpret it to mesh with your fouled up notions....try just reading what it says rather than trying to read something into it that it doesn't say.

Reread the caption of the drawing on the right. You didn't look at all of the words:

Spontaneous flow of heat from a cold area to a hot area would constitute a perfect refrigerator, forbidden by the second law.

They are talking about heat, not any sort of energy
Furthermore they label the flow of energy as Q. That symbol means heat energy in thermodynamics. So you are incorrect. And, as you say, it's written at 6th grade level.

 

[Wuwei]

Like I said goober...you be sure to tell me when they rewrite the second law of thermodynamics to relate energy transfer in terms of net as opposed to gross energy movement....it will be big news...but let me know anyway.

It is no longer big news. It was rewritten over 100 years ago in terms of entropy. Tell me how the second law written in terms of entropy disallows radiation from a colder body to a warmer body. (Note that the warmer body radiates more to the colder body so the net energy is from the warmer to the colder.)

 

[SSDD]

Sorry guy....it doesn't say that at all...in fact, it is written at a 6th grade level and you still don't seem to be able to understand it...you feel the need to try and interpret it to mesh with your fouled up notions....try just reading what it says rather than trying to read something into it that it doesn't say.

Reread the caption of the drawing on the right. You didn't look at all of the words:

Spontaneous flow of heat from a cold area to a hot area would constitute a perfect refrigerator, forbidden by the second law.

They are talking about heat, not any sort of energy
Furthermore they label the flow of energy as Q. That symbol means heat energy in thermodynamics. So you are incorrect. And, as you say, it's written at 6th grade level.

Alright...you have proven beyond any doubt that you are an idiot and can't read and comprehend words written at a 6th grade level....I believe you are mostly illiterate..you can stop demonstrating now...and for the last time you f'ing idiot...heat is energy...it is measured in units like all other forms of energy...and the second law applies to all energy in any possible form.....

 

[SSDD]

Like I said goober...you be sure to tell me when they rewrite the second law of thermodynamics to relate energy transfer in terms of net as opposed to gross energy movement....it will be big news...but let me know anyway.

It is no longer big news. It was rewritten over 100 years ago in terms of entropy. Tell me how the second law written in terms of entropy disallows radiation from a colder body to a warmer body. (Note that the warmer body radiates more to the colder body so the net energy is from the warmer to the colder.)

Lying doesn't make it true...if it were as you say, surely as respected a physics department as that of Georgia State would be aware and write it properly...

 

[Wuwei]

Alright...you have proven beyond any doubt that you are an idiot and can't read and comprehend words written at a 6th grade level....I believe you are mostly illiterate..you can stop demonstrating now...and for the last time you f'ing idiot...heat is energy...it is measured in units like all other forms of energy...and the second law applies to all energy in any possible form.....

My gosh you get angry. Try to cool down; it will let you think better. When you lash out at me you are lashing out at all science and scientists. Yes,it is absolutely true that the second law applies to all energy. In the case of radiation it applies to net energy. Want me cite the dozen sources that say that radiation between objects at any temperatures is two way radiation exchange?

 

[Wuwei]

Lying doesn't make it true...if it were as you say, surely as respected a physics department as that of Georgia State would be aware and write it properly...

They did write it properly. Look at the the diagram on the right where it talks about heat and uses the heat symbol Q. Look at the definition of Clausius at other university sites and tell me if you disagree with them.

The did write it with the understanding that the energy they were referring to is heat.

Edit: you will find that many word it the same as the hyperphysics site. Some have the entire paragraph identical to that site. But it should not be confusing to the students who understand the subject.

 

[Wuwei]

Do you believe that the second law of thermodynamics written in terms of entropy is valid?

 

[polarbear]

Do you think you're the only one to have thought of that [Angle of incidence]? And where does the reflected light go? It still has the entire atmosphere, at a very shallow angle, to penetrate.

Gee I almost never noticed your post under all the other clutter that buried it.
Your next post only proves that you can copy&paste wikipedia.
I guess you had to go there to look up what angle of incidence means.So you came across Snell`s law.
After that ou came up with your own theory how AGW is supposed to happen.
At least the IPCC factors in how much light has been reflected (albedo)and does not warm up the surface. Your`s uses light that has been absorbed, radiated down, warming the surface, which then re-radiates it up, to be absorbed again, going down, heating the surface even more, which radiates it up again and so on and on, till you get 400 watts out of 4 watts.
No wonder people like you think that conventional power plants are obsolete
And why did you copy that:...???
Snell's law states that the ratio of the sines of the angles of incidence and refraction is equivalent to the ratio of phase velocities in the two media, or equivalent to the reciprocal of the ratio of the indices of refraction:
Obviously because you had no idea what I was talking about but wanted to appear as somebody who knows physics inside out....as long as your WiFi works that is!

 

[mamooth]

.and for the last time you f'ing idiot...heat is energy..

But heat has no meaning for a single atom or photon. It is a macroscopic quality only, a statistical description of a system of matter and photons.

Hence, the second law, being it only talks about heat, doesn't apply to single atoms or photons.

All non-tards understand that. Hence, you don't. Hence, everyone laughs at you hard.

Now, there's nothing wrong with being way below average in intelligence, SSDD. I'm sure you're a nice boy, and a good boy, and maybe even handsome, so you don't need brains. You just need to understand how it is a problem when stupid people get belligerent with their stupidity, as you do.