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Utopia Talk / Politics / Seb science
Sam Adams
Member
Sat Apr 07 09:00:50
Yes seb, that is the definition of work. But you correctly said backscatter doesnt do work. Increased hurricanes is work. Increased energy flow from one temperature to another is work. So what is doing that work, if not backscatter?

Magic?
Seb
Member
Sat Apr 07 09:13:44
Sam:

"So what is doing that work, if not backscatter?"
I have answered your question many, many times.

So lets start again.

Must work be done to raise an objects temperature? Yes or no?
Sam Adams
Member
Sat Apr 07 09:36:28
"I have answered your question many, many times. "

No. You have tried and failed to answer my question many times. Like now, where you avoid the question. You know its not backscatter... you just said so. You know its not the sun... thats not changing. Something must be causing your fabled work increase. So... what?
Sam Adams
Member
Sat Apr 07 09:49:55
"Must work be done to raise an objects temperature? Yes or no? "

Ahhh i see why you are confused. You think something in our atmosphere is increasing work to warm the surface when in reality it is the opposite. The temperature is going up because our atmosphere is doing less work on the rest of the universe. Not only is this not an increase in hurricanes, it is not a temporary abberation that will not last.
Sam Adams
Member
Sat Apr 07 09:51:00
It is a temporary.
Seb
Member
Sat Apr 07 10:23:36
Sam:
It absolutely is the sun.

Recall work is the energy flowing between systems without accompanying entropy.

What's changing is the internal structure of the earth as a system.

So, I ask again:

Does work have to be done to raise an object's temperature?

Do not speculate, just answer the question.


Seb
Member
Sat Apr 07 10:28:37
Btw: the argument you present violates your assumption that for the gradient to increase requires work to be done.

You end up with a hotter atmosphere radiating to space of the same temperature - i.e. increased gradient, while arguing the atmosphere is doing less work on space.

Also, evaporation requires a phase change of water and lofting it to higher altitudes.

Is this work? And if adding CO2 cannot create temperature gradients because it would constitute work, how is it able to result in higher humidity?




Seb
Member
Sat Apr 07 10:28:55
You really don't understand this concept.

Seb
Member
Sat Apr 07 10:42:06
Also Sam, you have nearly understood the concept in your last post.

As we showed in the last thread, the change in outgoing power spectrum means the maximum work of the outgoing power spectrum is reduced.

A greater proportion of insolation is being disipated as heat. The mechanism being...
Seb
Member
Sat Apr 07 10:42:26
Only it's not temporary.
Sam Adams
Member
Sat Apr 07 13:31:59

"What's changing is the internal structure of the earth as a system. "

Right, that would be backscatter, doing the work. Something you know is impossible. The sun is not doing any more or less work, i assure you.

"You end up with a hotter atmosphere radiating to space of the same temperature"

Wrong. The concept of the effective radiating level is beyond you. The radiating temperature is actually reduced, then returns to previous.

"Only it's not temporary."

The warming of the atmosphere is temporary. It soon reaches a new equilibrium and stops warming.

You have yet again failed to grasp the vertical profile of the layered column model, an especially egregious mental error since it is the only axis. You are yet again unable to apply one law without violating another. You are unable to think through to the final equilibrium state.

Given your difficulties thinking through the physics on your own, i suggest you actually use a layered column model, a one dimensional radiative convective model. Input earths various parameters and let it settle on a final equilibrium state. Ask yourself why. Tweak something and see what it does.

There is almost no chance you can create code like that on your own, so i suggest you download someone elses. Also the number of calculations is large. Dont use a newb language like matlab or even worse a slugish shell script. I suggest c.
Seb
Member
Sat Apr 07 14:06:56
Sam:

Recall the definition of work, and think about how you've defined your thermodynamic system. Define it clearly. Is it closed?

So, back to the thought experiment:

Does raising the temperature of an object require work to be done?
jergul
large member
Sat Apr 07 14:34:03
Seb
Are you reminding yourself not to teach? Its not this painful in real life. At least not at grammar school levels and above (and you would be doing it at uni levels).
Sam Adams
Member
Sun Apr 08 05:48:20
"Does raising the temperature of an object require work to be done? "

Yes. What does that work? And is that source changing?

The sun, obviously not changing. Neglecting trivial albedo changes, its doing the same work on earth that it was.

So why is the temperature rising?

Because the earth is doing less work on the rest of the universe.

A blanket doesnt do work, it actually prevents work.

The current thought path you are going down actually has the opposite sign of what you were hoping.

Lol.

Sam Adams
Member
Sun Apr 08 05:55:43
I should say, more specifically, why is the temp rising in this case?

And answer with the reduced earth universe work while sun earth work remains constant.

So that the easily confused seb@jergul are maybe a little less confused (but probably not)
Seb
Member
Sun Apr 08 06:19:48
jergul:

I earn way more than I would get as a teacher.
Seb
Member
Sun Apr 08 06:24:49
Sam:

Temperature is a measure of the distribution of energy in a system.

If an objects temperature rises, then it's entropy is decreased.

Explain to me how this occurs without work being done.

Also, when you say "the earth is doing less work on the rest of the universe", what does that mean? Are you suggesting the entropy content of the outbound radiation is less?

I'm sorry, you are not defnining work correctly. You are using it to mean "power flow", which is not the same thing at all.

For example, consider a power station providing a flow of superheated steam. Initially, it is venting that steam directly into an isothermal heat sink - all the energy is thermalised and no work is done. Then, the steam is diverted into a turbine, and the residual steam out of the turbine dumped into an isothermal heat sink.

In the first case, no work is done.

In the second case, work is done.

In both cases the power outflow is the same.

Work is the flow of energy between systems without accompanying entropy.
Seb
Member
Sun Apr 08 06:25:20
I'll ask again:

Can you raise the temperature of an object out of thermal equilibrium without doing work on it?
jergul
large member
Sun Apr 08 09:02:15
Seb
*woosh*
Seb
Member
Sun Apr 08 10:38:28
Jergul:

My determination never to teach again is no joking matter!
Sam Adams
Member
Sun Apr 08 11:30:31

"Explain to me how this occurs without work being done."

That is literally the opposite thing i said in my last post. Not just my last post, the very first word in my last post.
you cant even read english anymore? Yes means no? No wonder you struggle in those consent threads.


And if you cant figure out what "yes" means, no wonder you cant reason through atmospheric physics either.
Seb
Member
Sun Apr 08 11:53:53
Sam:

"That is literally the opposite thing i said in my last post."

So you agree that for the temperature to be raised, work must be being done on the object.

Right.

Now consider a black body powered by some internal energy source delivering a constant 5W.

That black body, when in power ballance will have some temperature and radiate a constant 5W black body spectrum.

Now surround that black body with a partially reflective surface (as this is a question of principle, lets say one that is equally reflective at all spectrums) that transmits 50% of what is incident and reflects the remaining 50% back to the object.

What will happen to the objects temperature?
Sam Adams
Member
Sun Apr 08 12:25:29
It warms. Duh. But the 5w source is the same, doing the same work. Less energy is travelling up and out. The object is now doing less work on its surroundings. Which is a temporary change until the new equilibrium is reached. Then its back to normal flows. Backscatter can prevent work from being done, but cant do work itself. Like i have tried to teach you many times, it is a delaying mechanism only. So not only is the effect the opposite of what you are looking for, it is temporary, you couldnt understand the word "yes", and jergul is on your side. Yikes. Rough day for seb.
Seb
Member
Sun Apr 08 13:58:27
Sam:

You answer is garbled. Answer in steady state, I.e. after the object is in equilibrium.
Seb
Member
Sun Apr 08 13:59:08
*power balance. Let's not confuse with formal thermodynamic equilibrium.
Sam Adams
Member
Mon Apr 09 09:44:52
After equilibrium is reached, the work being done settles back on its original value.

This is fundamental.

Lets think through this in a slightly different way to try to clear up some of your confusion.

A packet of energy leaves the sun at 6000 degrees K. It is then absorbed by the earth's surface in a pre-industrial world and is eventually reradiated to space from the effective radiating level at a temperature of 220K.

In scenario 2 we have everything the same- except it is now a post industrial world where the internal structure of the atmosphere has changed but settled on a new equlibrium. The packet still leaves the sun at 6000K and still leaves the earth at 220K from an effective radiating level that is higher altitude but the same temp as before.

The work done in both cases is the same.
Seb
Member
Mon Apr 09 10:45:08
The objects temperature is higher.

So s=Q/T.

Has the entropy contet of the object increases?

Seb
Member
Mon Apr 09 11:00:15
Sam:

It doesn't leave at 220k. That's an abstraction.

220k is the temp of a black body that emits the total power incident on the earth.

The whole point of the greenhouse effect is the consequences of spectroscopic effects on particular wave bands deviating from black body.

As we discussed, and you eventually agreed the impact of the changes to the outbound power spectrum is the equivalent of a reduced temperature (a higher entropy content).

Sam Adams
Member
Mon Apr 09 12:35:16
"
As we discussed, and you eventually agreed the impact of the changes to the outbound power spectrum is the equivalent of a reduced temperature (a higher entropy content). "

Incorrect. The atmosphere settles into a
new equilibrium such that the work done is identical... if extra work was being done the temperature would still change! This is the very definition of equilibrium. Out=in.

You seem to be confused about the concept of an effective radiating level/temperature. I might have confused you once by using the word average instead of effective(even though in this case the difference is quite minor)... however the effective radiating temp is absolutely unchanged. It could not be otherwise once the new equilibrium is reached.
Seb
Member
Mon Apr 09 14:56:49
Sam:

1. It's a dynamic system. It requires work to be done to maintain the Earths temperature out of thermal equilibrium of it's surroundings (which is 3k).

2. "This is the very definition of equilibrium. Out=in." No, the definition of thermal equlibrium is that two objects have the same temperature. Power in = power out is powe ballance.

3. I understand effective radiating temperature well. Effective in physics implies a parameterisation of some time. The formal definition of effective radiating temperature is simply "the black body temperature corresponding to the power radiated by the planet". As we discused previously, if you examine the emitted radiation power spectrum, the distribution of power in the spectrum will be different in the final state than the pre-industrial state. It's entropy content will be greater, corresponding to a lower temperature. The addition of CO2 takes the atmosphere further away from black body like characteristic.


Now, stop waffling and answer the question:

Our black body with a spherical mirror around it.

The temperature increases.

Recalling that S=Q/T, has the entropy of our object decreased, increased or remained the same?







Sam Adams
Member
Mon Apr 09 20:19:00
2. Different types of equilibrium. Equilbrium is simply a balanced state.

3. "the distribution of power in the spectrum will be different in the final state than the pre-industrial state."

True.

"It's entropy content will be greater"

False.

"has the entropy of our object decreased, increased or remained the same? "

The entropy of our object is mostly irrelevent. We are talking about the flows, not the stocks. the entropy of the effective radiating layer does matter- but that is unchanged.

"The addition of CO2 takes the atmosphere further away from black body like characteristic. "

Irrelevent. The planet will settle onto a new state where the outbound power and entropy are the same as before. It cannot be otherwise. Its radiative equilibrium state is such that out=in. And since in is unchanged... neither is out. If out was changed, temperature would change so that out becomes equal to in.

Sam Adams
Member
Mon Apr 09 20:22:16
I think its important to note there is a difference between effective or integrated temperature(wrt power) and average temperature.

You can change power distribution while keeping entropy the same so long as average temperature also changes to match.
Seb
Member
Tue Apr 10 04:41:31
Sam:

You are applying things that hold only in a static equilibrium - formal thermal equilibrium - which is a closed system (no work can be done), to a dynamic equilibrium - power balance - in an open system.

Hence your errors.

"False"
It is demonstrably true.

Think about it.

S=Q/T right?

So consider the outbound radiation just after you add co2.

At this point the spectrum immediately changes. Q - the heat content, I.e. total power - is reduced due to backscatteter.

So the entropy content of the radiation must reduce. This would imply that somehow for the radiation that is emitted, despite a greater proportion going through more scattering events, less is being thermalised.

The effect of increasing co2 on the outbound power spectrum has to be the equivalent of a reduction in temperature.

So when back in power balance, S=Q/T will be higher than previously.

As to the idea that in power balance the sun does no work on the earth, lol. Seriously? Where do you think the stored energy in coal comes from, for example?
What drives hurricanes?





Seb
Member
Tue Apr 10 07:09:50
"I think its important to note there is a difference between effective or integrated temperature(wrt power) and average temperature."

Describe what you no think the relationship is between the effective temperature and power. Preferably as an equation.

When you say average temperature, averaged over what?

"You can change power distribution while keeping entropy the same so long as average temperature"

Temperature averaged over what? Temperature is *defined* as a power distribution. You cannot change the power distribution without changing the temperature.
Sam Adams
Member
Tue Apr 10 09:42:14
"The effect of increasing co2 on the outbound power spectrum has to be the equivalent of a reduction in temperature. "

Until the planet warms up such that it reaches its new radiative equilibrium.

"So when back in power balance, S=Q/T will be higher than previously. "

No.

Its actually s=integral(q/t),wrt to height,temp,or power. Or effective radiating temp, for short. Which is the same as the prior equilibrium state.

At least you see how the radiated power is the same, so you are halfway there.

Now imagine a case where an object was not in radiative equilibrium. It would continue to change temp until radiative equilibrium is reached, no?

Seb
Member
Tue Apr 10 09:50:23
Sam:

"Until the planet warms up such that it reaches its new radiative equilibrium."

Are you saying that when the atmosphere warms, the dip in the LW band disappears?

" actually s=integral(q/t),wrt to height,temp,or power"

Mathematical and physical nonsense.

This is S of the radiation field.

The Q of the radiation field is defined as integral Power (lambda) over all lambda.

Q is not a function of T so cannot be integrated.

Neither Q nor T of the outbound radiation are functions of altitude. By definition the outbound radiation field is beyond the TOA.




Sam Adams
Member
Tue Apr 10 09:57:42
"Are you saying that when the atmosphere warms, the dip in the LW band disappears? "

Of course. Your basic SB law.


Q is not a function of T "

Seb
Member
Tue Apr 10 10:01:46
Sam:

Stephan Boltman law applies to a black body spectrum. The outbound radiation is not a black body spectrum.
Sam Adams
Member
Tue Apr 10 10:02:08
"
Q is not a function of T "

Very wrong.

"Neither Q nor T of the outbound radiation are functions of altitude. "

Very wrong.


"By definition the outbound radiation field is beyond the TOA. "

Well then in that case the integral has already been done and the effective radiating temp can be used, unchanged.

There is no way for you to wiggle an entropy,work,or power change out of this. It is impossible.

"
Now imagine a case where an object was not in radiative equilibrium. It would continue to change temp until radiative equilibrium is reached, no? "

Now answer that question.

Sam Adams
Member
Tue Apr 10 10:11:42
"Stephan Boltman law applies to a black body spectrum"

In its simplest form yes but you add emissivity for real cases, duh. Meaningless comment.
Seb
Member
Tue Apr 10 10:15:25
Sam Adams:

"Very wrong."

Ok. Express Q, the total power of tge radiation field emitted by the planet, as a function of T, the temperature of the radiation.


"Very wrong."

Ok. Express Q and T of the outbound radiation field in the spherical shell 300 m thick at 100,000 and at 1,000,000 km from earth.


"There is no way for you to wiggle an entropy,work,or power change out of this. It is impossible."

Not only is it possible, it is necessary.

And I will show you if you would answer the questions I pose.

Let me restate:

Our black body object with an internal energy source editing 5w at some temp T_0 has a 50% reflective shell placed around it.

After a new power balance is reached, the temperature of the object has risen.

Recalling S=Q/T, has the entropy of the object increased?

You may assume the object has a heat capacity C.


Sam Adams
Member
Tue Apr 10 10:40:41
"After a new power balance is reached, the temperature of the object has risen. "

The surface temp increases. The temperature of the shell, the radiating layer, has not risen at all. Man, that vertical axis is hard for you. You should probably try to understand this effective radiating level thing. Without that key bit of understanding, you will remain lost.
Seb
Member
Tue Apr 10 13:32:12
Sam:

I asked about the object.

RTFQ.

The shell is a mirror, not a black body. 50% reflected, 50% transmitted. 0% absorbed so no radiation. That would be a fundamentally different scenario (instead of a mirror, a black body). We will get to that later.


The temperature of the object has increased.

The object is in power balance at a new temperature.

What has happened to the objects entropy?

You may assume heat capacity C.
jergul
large member
Tue Apr 10 13:42:02
Seb attempt to lure sammy into using actual maths. A worthy venture!
Sam Adams
Member
Tue Apr 10 15:09:24
"I asked about the object"

Irrelevant. The shell is the effective radiating surface.


"The shell is a mirror, not a black body. 50% reflected, 50% transmitted. 0% absorbed so no radiation."

Not how the atmosphere works. But even in your 0 absorption case... that is merely an emissivity change. Effective blackbody temp = the same.


"What has happened to the objects entropy?"

Dont care. Not relevent to the radiations entropy.

Why are you avoiding the question seb? If the atmosphere was doing more work on the universe than our share of the suns work, the planet would cool, no.
Sam Adams
Member
Tue Apr 10 17:30:43
Ok time for the big guns. In a trusted and well verified radiative convective equilibrium column model, run for both an unwarmed and warmed world(delta 2 degrees C), the integrated radiation entropies are identical... within 0.1%... within the base truncation and rounding errors of the tool in question.

Muhahahahaha.
Sam Adams
Member
Tue Apr 10 17:48:08
Oh i should be clear thats top of atmosphere outbound radiation.
Seb
Member
Wed Apr 11 01:57:13
Sam:

The shell isn't the effective radiating surface. In this hypothetical scenario, it's not radiating.

This is what I mean by you not having a good understanding of science. You are attempting to apply a model that is fundamentally different (and you will see why when we replace the mirror with a black body later in this thought experiment).

"Not how the atmosphere works"
Duh.

But the point of this experiment is to understand how entropy, work and radiation work. Which presently I do not think you do. If we work through this simplified problem, we should end up with a shared understanding of how those work. Either you will be right or I will be right, but we should agree or it should be very clear where and why we disagree and it will be possible to resolve that one way or another.

"Dont care. Not relevent to the radiations entropy."

It is however relevant to the work being done in the system.

Does the entropy of the object rise, or fall.

"If the atmosphere was doing more work on the universe than our share of the suns work, the planet would cool, no."

Not clear by "Our share of the suns work" means.

See my example of the steam exhaust from a power plant. The amount of work it does is entirely depenent on the system that it is going in to. It is quite possible that the earths reradiated energy does no work on the universe whatsoever, travelling infinitely through empty space until the expansion of the universe redshifts it to that of the cosmic background radiation. Equally, I could construct a half spherical sterling engine shell around the night side of the earth that exploits the temperature difference between deep space and the earths radiation to harness that energy and do work.

Work is defined as the movement of energy between systems without the acompanying flow of entropy, and the radiation is carrying entropy with it.

A better thing to look at would be exergy. But we will come to that.

Sam:

"In a trusted and well verified radiative convective equilibrium column model, run for both an unwarmed and warmed world(delta 2 degrees C), the integrated radiation entropies are identical... within 0.1%... within the base truncation and rounding errors of the tool in question."

Without working there is no way to verify how you obtained that result or critique it. I suspect your defintions are wrong.

For example, in the following two papers we see the opposite:

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2871903/

In this paper a simple 1-d model based on BB spectra demonstrates the increase in optical depth results in greater production of entropy in the atmosphere.

"It is noteworthy that the results obtained from this study reveal that the atmospheric net radiation entropy flux at all altitudes is intrinsically connected with the overall atmospheric LW optical depth, which further implies the sensitivity of the atmospheric net entropy flux (or production rate) to greenhouse gases (i.e. increased overall atmospheric LW optical depth)."

Meanwhile, here is a formal analytical method for calculating entropy from power spectra:

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5432030/

In figure 5 its applied to a line-by-line radiative transfer code and demonstrates the impact of spectrocopic effects of atmospheric chemistry (it creates and it's deviation from BB radiation, increasing overall entropy).

If we then were to compare an atmosphere with CO2 vs the sun to obtain a delta S, and then a black body curve vs the sun, we would see different delta S. The same is true if you add more CO2 to obtain a different line spectrum.

Anyway, this is second order stuff.

We first need to go back to the basic principles and understand work in the context of an open system.

So, if you would be so good:

Does the entropy of our hypothetical object increase or decrease?

Sam Adams
Member
Wed Apr 11 09:54:19
I have seen those papers. They neglect convection. Ooops.

Furthermore, they find a very minor entropy change for a truly massive emissivity change. I dont know their error bars for other variables, so i cannot say whether that is all error or not. I cant say if the temp changes are so large they drive minor but noticeable sw albedo changes... which certainly would change the radiative equilibrium and entropy of the flows.

Given the shortcomings above, i ran the integrals myself.

Perhaps you should get or create some code and do that as well. I suspect that capability is beyond you, but it has higher odds of teaching you something than you just banging your head into a wall confusing stocks and flows.
Seb
Member
Wed Apr 11 11:44:23
Sam:

"I have seen those papers. They neglect convection."

Convection isn't a radiative process, and we are seeking to understand the entropy content of the radiation emitted at the top of the atmosphere.

Your contention is the entropy change is and the exergy of the outbound radiation is identical in steady state despite changes in CO2.

Convection doesn't come in to it.

"Furthermore, they find a very minor entropy change"

Your argument is one of principle. There should be zero.

Saying something is minor is a relative statement that needs to be qualified. The energy content of all weather integrated over any given period is tiny compared to the energy flows. So yeah, if your baseline is the power flows, it will look like a rounding error, but will be significant in terms of weather.

"I cant say if the temp changes"
There is nothing in the paper about temp changes, you are making an argument from thermodynamic principle - you have said that there will be no change in work as a consequence of changes in line ratios in the outbound power spectrum; as a matter of thermodynamic principle.

These papers show that is not the case.

"are so large they drive minor but noticeable sw albedo changes... which certainly would change the radiative equilibrium and entropy of the flows."

What you are essentially arguing here though is that spectroscopic changes *can* in fact drive persistent changes in the exergy of outbound radiation and how energy is distributed in the climate *but* that feedback effects limit it.

That's a completely different argument. If you were pursuing the line you were arguing in your previous post, you would have no need to look at that because as a matter of principle entropy content remains fixed in power balance.

You are still avoiding the question though.

If you are willing to work through a few simple scenarios we can see what is the case and what is not the case.

Either you are incapable of doing so, or you don't want to do so because you know it will demonstrate you are incorrect.

There is simply no reason not to.



Sam Adams
Member
Wed Apr 11 13:21:32
"There should be zero. "

After a trillion floating point ops, it very well might not be exactly zero in the sims.


"There is nothing in the paper about temp changes"

Lol. A radiative equilibrium model with different outcomes has no temp changes? You are completely confused.

"is that spectroscopic changes *can* in fact drive persistent changes in the exergy of outbound radiation"

Lol comparing sw and lw changes as if they were the same.

You are confusing stocks with flows, you are confusing sw with lw, and im not sure you even understand what radiative equilibrium is in a vertically stratified environment, much less radiative convective or how to model one.

You are completely lost.
Seb
Member
Wed Apr 11 16:08:26
Sam:

If you think it's a numerical error, then say that. If it's a numerical error, why are you invoking potential feedback mechanisms such as albedo changes in response which implies any change is a physically valid outcome.

Complete incoherence.

"A radiative equilibrium model with different outcomes has no temp changes"

Yeah, because both are only looking at upwelling radiation in isolation.

"
Lol comparing sw and lw changes as if they were the same"

Nowhere is that being done

"You are confusing stocks with flows"

Really? Where?

Sam, you are just babbling now.

Rather than babble, why not work through the example I set.
Sam Adams
Member
Wed Apr 11 16:59:48
"Yeah, because both are only looking at upwelling radiation in isolation. "

Completely wrong.

You are completely confused by the concept of an atmosphere in radiative equilibrium.
Sam Adams
Member
Wed Apr 11 17:33:03
You should try to use some of the code. You might weasel some understanding out of it
Seb
Member
Thu Apr 12 03:28:59
Sam:

The subject of both papers is methods for determining entropy content of upwelling radiation in an atmosphere. Both explicitly state that application to understanding climate is subject for future work.

Both are taking fluxes *in power balance* scenarios as calculating entropy content.

One paper compares different power balance.

One compares the real spectra to bb again in power balance.

In both cases, the power balance is an input to the analysis.

Your bizare claim appears to be that by simply calculating entropy the paper should be modelling feedback effects. But nothing's changing. Simply calculating entropy doesn't have any effect on planetary albedo.

It seems apparent you haven't read the paper and correctly understood what analysis is being performed.
Seb
Member
Thu Apr 12 03:29:44
So, does entropy of the object increase? Or does it decrease?
Sam Adams
Member
Thu Apr 12 09:44:33
Are you still babbling about how temperature change is not part of radiative equilibrium?
Seb
Member
Thu Apr 12 10:42:45
Sam:

Are you saying that the global temperature will change WHEN the planet is in power balance?

That seems spectacularly... unlikely.

However no, I am asking a very simple question.

Our black body object with an internal energy source editing 5w at some temp T_0 has a 50% reflective shell placed around it.

After a new power balance is reached, the temperature of the object has risen.

Recalling S=Q/T, has the entropy of the object increased?

You may assume the object has a heat capacity C.


Sam Adams
Member
Thu Apr 12 11:35:58
"Are you saying that the global temperature will change WHEN the planet is in power balance?"

Lol seb you are completely confused. The temperature changes with height and time as an atmosphere settles into radiative equilibrium. The temperature then changes with only height after the equilibrium state has been reached. Obviously two different equilibrium states have temperature differences.

Its gonna be very hard for you to understand integrated atmospheric entropy if you cant even grasp the basic structure of an atmosphere.
Seb
Member
Thu Apr 12 13:39:00
Sam:

The radiation release at TOA has a fixed temperature and entropy Sam. I've invited you to show any difference.

Both of these papers describe entropy changes in eqm so temperature profile is fixed and invariant in time. Spatial temperature profile is fixed in each scenario and modelled.

Did you not understand the paper?

So why would modelling any change in albedo be relevant to simply looking at entropy of the radiation?


You simply are not making any sense.

Seb
Member
Thu Apr 12 13:39:33
And clearly can't answer simple thermo problems.
Sam Adams
Member
Thu Apr 12 14:22:40
Oh so there are indeed temperature changes with both height and between the different scenarios, contrary to your prior posts. Lol seb=oops.

You do realize the outgoing radiation depends on the temp, right?

"So why would modelling any change in albedo be relevant to simply looking at entropy of the radiation? "

Because in huge global warming scenarios sw albedo changes could indeed start to show up in entopy fields, but that is completely different from our discussion. But you confuse sw and lw changes, so this will probably confuse you too.
Seb
Member
Thu Apr 12 14:29:41
"Oh so there are indeed temperature changes with both height and between the different scenarios"

I don't think it was ever claimed otherwise.

What I said was that simply calculating the entropy shouldn't require consideration of any changes to albedo because, well, it's a static scenario and nothing is changing.

"You do realize the outgoing radiation depends on the temp, right?"

Your entire argument is that it *doesn't*, and I'd also say your causal argument is back to front. It is the line radiation (determined by atmospheric chemistry) which sets the temperature of the planet. This is why changing the chemistry changes the temperature.

"Because in huge global warming scenarios sw albedo changes could indeed start to show up in entopy fields"

Indeed it could, but none of that is under study in those papers - rather they were exploring how to claculate the entropy changes in the radiation field as a consequence of changing chemistry - and my point in citing these studies is because you claimed that entropy content of outbound radiation was invarient.


"In a trusted and well verified radiative convective equilibrium column model, run for both an unwarmed and warmed world(delta 2 degrees C), the integrated radiation entropies are identical... within 0.1%... within the base truncation and rounding errors of the tool in question."

Now you are saying that albedo changes can change the entropy content, which they can do of course.

Can you make your mind up please?
Sam Adams
Member
Thu Apr 12 16:18:31
"I don't think it was ever claimed otherwise. "

Seb
Member Wed Apr 11 16:08:26

"A radiative equilibrium model with different outcomes has no temp changes"

Yeah


Sam Adams
Member
Thu Apr 12 16:24:51
"and my point in citing these studies is because you claimed that entropy content of outbound radiation was invarient."

I specifically stated such neglected sw albedo changes. You are confusing sw and lw radiation again. If you make a sw albedo change there will most certainly be an energy budget and storm strength change. Unfortunately for your argument, sw albedo is not changing in any significant way.

Sam Adams
Member
Thu Apr 12 16:28:58
"Your entire argument is that it *doesn't*,"

You are confusing something here. Perhaps temperature with effective radiating temperature, perhaps radiation spectrum with integrated radiation spectrum. You are so confused i cant really tell what you are confused about with this one.
Sam Adams
Member
Thu Apr 12 16:38:14
"well, it's a static scenario and nothing is changing. "

First, its 2 scenarios in each paper with clear differences between them. Second, the equilibrium models made nearly infinite changes to themselves while settling into their equilibrium states. The radiation analyzed at the end depends partly on the integrated result of those changes.

So ya, plenty of change there.
Seb
Member
Thu Apr 12 17:20:42
Sam:

Your post 16:18:31 makes no sense. You appear to be quoting me, quoting you. The phrase

"A radiative equilibrium model with different outcomes has no temp changes"

first appears in this thread in your post April 11 13:21:32 - albeit as one of your "hilarious" summaries which don't reflect anything I have said.

Like I said, you should stop doing this because you can't keep it straight in your own head who actually said what.

"I specifically stated such neglected sw albedo changes."

The paper is looking soley at the entropy content of radiation as it progresses through the atmosphere in equlibrium (in one case) and in the other the generalised means to apportion entropy by spectral line - with a worked example of the specrtum produced at TOA.

In neither case is it relevant to look at changes to albedo - because while that would also change the entropy further, the simply point here is to look at how atmospheric chemistry alone changes the entropy content of the radiation.

And the sole point I made in this was to point out how wrong you were to conclude entropy content would not change with changes in the power spectrum.

A point so comprehensively made that you are now making an entirely different argument.

"You are confusing sw and lw radiation again."
Not at all. In the second paper, it's the entropy content of the full spectrum. In the fist, the entropy is specifically of the SW.

"If you make a sw albedo change there will most certainly be an energy budget and storm strength change."
Neither of these papers are predicated on changes of anything other than the chemistry and the impact *that* has on radiative transfer and the entropy that introduces to the radiation field.

"Unfortunately fBeither or your argument, sw albedo is not changing in any significant way."

Nothing in either paper is predicated on albedo being constant or not constant. It's simply irrelevant.

The point stands: radiative transfer introduces entropy to the radiation field as it propogates up through the atmosphere; and this entropy content will change if the atmospheric chemistry changes.

Previously, you argued that this was not the case: that the entropy content of the radiation was fixed.

"First, its 2 scenarios in each paper with clear differences between them."

Dear lord... do you understand what a static scenario is? Obviously the two scenarios are different, but both are steady state scenarios. Albedo changes aren't relevant here. It's simply looking at the change in entropy of the radiation field. If the planetary albedo changes in a full coupled model, that just means the initial spectrum at the ground will be different. But the impact of atmospheric chemistry on the radiation field and the entropy it introduces isn't going to change just because the albedo changes.


Seb
Member
Thu Apr 12 17:23:17
i.e. a molecule of CO2 doesn't know or care that the SW photon it's scattering came from a surface with one albedo or another.

The impact that has on the overall increase in entropy of the radiation field with each scattering is still going to add entropy to the radiation field, irrespective as to whether the initial radiation field is different in each scenario.
Seb
Member
Thu Apr 12 17:30:07
P.s. note that in the paper this argument applies to, the initial entropy at 0 altitude is different (i.e. at the very least ground temperature is being modeled).

Adding albedo effect changes due to temperature is a very low order effect.

And doesn't alter the principle. Atmospheric chemistry changes change the overall entropy content of the outbound radiation. This can in principle be calculated. It is therefore wrong to claim that in steady state the entropy difference between inbound and outbound radiation is always the same.

It therefore follows by looking at that, that the exergy of the outbound radiation is reduced and your argument regarding "work" (it's not work, it just dissipation of energy inputted from the sun) is incorrect.
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