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

[IanC]

What about the 343 K walls in my home? Did they spontaneously achieve that temperature?

Let me guess..you believe your HVAC generates spontaneous energy movement. And I suspect you should learn to convert temperatures...you just claimed that the walls in your home are heated to 157 degrees F.

Let me guess..you believe your HVAC generates spontaneous energy movement.

Let me guess, you don't know how my walls reached 70 F?

And I suspect you should learn to convert temperatures...

You got me, I added 70 F to 273 K.
294 K.......better?

Now, back to your claim that energy can move from cooler matter to warmer matter if work is done........

It's easy to just own up to simple calculation mistakes, at least for some people.

 

[IanC]

the second law says no such thing...you say that..and your references say that...but the second law doesn't say that...an unobservable, unmeasurable, untestable mathematical model says that...but not the second law...

The second law is very general. Consider a cold object and a warm object. The entropy of the system will be given by some value S0. After a while the warm object will be colder and the cold object will be warmer. The entropy is a larger value S1. There is no place in the entropy representation that says the two objects cannot have a two way exchange of energy, with, of course, the end result that more energy went from the warmer to the colder object.

Again...that isn't what the second law says....I suppose you are arguing that the second law of thermodynamics is a "living" law....meaning that you can change it when it isn't convenient...face it guy...neither heat nor energy move spontaneously from cool objects to warm objects...

and again...is energy moving from the warmer surface of the earth to the cooler atmosphere a natural process? easy question...yes or no answer is all that is needed.

The SLoT is defined by changes in entropy. You imagined your version. And your imagination is wrong. It violates entropy.

Energy moving up hill..moving from a more disorganized state (cooler) to a more organized state (warmer) violates entropy which states that energy always moves towards a more disorganized state and that all natural processes are irreversible precisely because energy is always moving towards a more disorganized state?

Your description of what is allowed or prohibited is at odds with reality, therefore it is either wrong or incomplete.

SSDD says the transfer of energy can only go downhill. No exceptions.

Let's make up a simple example and see what hIfappens.

A small packet of 99 molecules of gas has a range of speeds from one unit to 99 units, one molecule for each possible speed. Let's say that only one unit of speed can be transferred by any single collision.

After every collision the standard deviation decreases as slower molecules speed up and the fast molecules slow down, all heading for the average speed of 50 units. As more and more molecules reach 50, there are fewer SSDD allowed collisions and more and more SSDD prohibited collisions that cannot happen. At some point in time all speeds are 50, and no further collisions are allowed.

This isokinetic condition is never found in reality. Any real set of molecules is found to have a range of speeds centred around the average, in a Bell Curve shaped distribution.

That means some molecules are always getting bumped to a higher energy level than the average. Contrary to SSDD'S vision of the second law of thermodynamics.

 

[Wuwei]

Efficiency is a big part of the real world. The most efficient way to enclose area with a defined circumference is a circle. The radius is equal in all directions. A square is not as good but still better than a rectangle.

The moon receives as much solar energy as the Earth so why isn't it as warm? The main reason is the length of a lunar day/night cycle. It is very hot in the day and very cold at night. The energy inputs and outputs are extreme, less efficient. Like how a circle has more area per circumference than a flat rectangle. Only more so, because area is a ^2 relationship and temperature is a ^4 relationship. If the lunar 'day' was half as long the average temperature would be warmer. If you halved it again, warmer still. And so on, ignoring the effects of spinning the mass.

The predicted temperature of the Earth is -18C, from the average amount of sunlight hitting it. Now hold on a minute. We were talking about efficiencies. An Earth day is shorter than a lunar day but it is still 24 hours. Shouldn't the hypothetical max (circle) be greater than the actual Earth (square) which is greater than the Moon (rectangle)?

Obviously the atmosphere has something to do with it. But how does the atmosphere raise the average temperature ABOVE the theoretical maximum derived from the solar input?

I'll talk more about this if anyone is interested. Or we can just leave it as another unanswered question.

Hey Wuwei, you like theoretical problems.

Don't you find it interesting that the GHE is large enough to more than compensate for the inefficiencies caused by unequal input and output energies?

While it is just a variation on the theme of why the surface is warmer than the Sun's input, don't you find it amazing that entropy gained by converting sunlight to IR can 'power' such a large effect?

I realize the actual energy comes from radiation not lost to space but it seems to me that entropy gained in the total area causes entropy loss by the storage mechanism in a subset area.

I could be wrong. What do you think?

I'm not ignoring you. I'm working on it. Your question in quotes is easier. I once wrote an internal company paper on how to find the rotation angle of an image using entropy. The problem was that nothing was known about the image except that it most likely had a few straight edges at right angles with a bit of clutter thrown in.

The math of the solution is rather similar to your discussion of the lunar day length and how it affects entropy.

 

[Wuwei]

SSDD says the transfer of energy can only go downhill. No exceptions.

Let's make up a simple example and see what hIfappens.... etc.

This is surprisingly similar to the previous problem you posed. More on that later.

 

[SSDD]

This law-

And the law that energy cannot be created or destroyed.

You can only control the amount of radiation produced by an object by adding or subtracting to the amount of energy available to be converted to radiation.

Any object above absolute zero will attempt to cool down by radiating. It always radiates but it only cools if it emits more energy than it receives.

That law states no such thing...that law states that a perfect black body radiating in a vacuum devoid of any other matter radiates according to its temperature to the 4th power...an ideal, perfect black body in a perfectly empty universe...a fantasy.

Bring matter into the equation and j starts changing in accordance with the difference between the temperature of the radiator and its cooler surroundings. You keep forgetting the basics and jumping to knee jerk conclusions...think ian, I know that you are capable of it.

There is no law that says that the rate of energy emitted by an object can not change if its environment changes...quite the contrary..the application of the SB law relating to radiators that are not perfect black bodies radiating into perfectly empty space says exactly the opposite...P keeps changing as the difference between the temperature of the radiator and the temperature of its surroundings change.

This equation is obviously not for a perfect Blackbody because it has a term for emmisivity.

You should also note that it has no term for area. It is an intensive property.

If you divide the j by area then you get an actual amount of power, an extensive property. But you then have to do the calculations for every angle that is in line-of-sight of the area you defined.

Your confusion seems to stem from confusing the intensive property of j with the extensive property of P.

The object exists and radiates. The environment that surrounds the object does not change the radiation, which is always proportional to the temperature of the object.

We can turn the intensive property of radiation into the extensive property of power by defining the radiating surface area of the object. We can turn the intensive property of temperature into the extensive property of total heat content by measuring the mass and the specific heat for its constituents.

With the two extensive properties of power and heat content we can calculate the rate of cooling as time passes. Quickly at first and then progressively slower.

If we add a second object nearby, it also radiates in the same fashion as the first. Some of the radiation from the first will be intercepted by the second, and vice versa. The amount will be determined by the line-of-sight connection. The amount of radiation energy absorbed will add to the total energy content of the two objects, thus slowing the rate of their cooling. But both objects will still be cooling overall.

I will leave it there for now.

You can appeal to as much complexity as you like ian, but that law does not state that an object can not radiate less when it is in the presence of other matter..

The whole topic is moot as IR does not warm the air.

 

[SSDD]

Efficiency is a big part of the real world. The most efficient way to enclose area with a defined circumference is a circle. The radius is equal in all directions. A square is not as good but still better than a rectangle.

The moon receives as much solar energy as the Earth so why isn't it as warm? The main reason is the length of a lunar day/night cycle. It is very hot in the day and very cold at night. The energy inputs and outputs are extreme, less efficient. Like how a circle has more area per circumference than a flat rectangle. Only more so, because area is a ^2 relationship and temperature is a ^4 relationship. If the lunar 'day' was half as long the average temperature would be warmer. If you halved it again, warmer still. And so on, ignoring the effects of spinning the mass.

The predicted temperature of the Earth is -18C, from the average amount of sunlight hitting it. Now hold on a minute. We were talking about efficiencies. An Earth day is shorter than a lunar day but it is still 24 hours. Shouldn't the hypothetical max (circle) be greater than the actual Earth (square) which is greater than the Moon (rectangle)?

Obviously the atmosphere has something to do with it. But how does the atmosphere raise the average temperature ABOVE the theoretical maximum derived from the solar input?

I'll talk more about this if anyone is interested. Or we can just leave it as another unanswered question.

Hey Wuwei, you like theoretical problems.

Don't you find it interesting that the GHE is large enough to more than compensate for the inefficiencies caused by unequal input and output energies?

While it is just a variation on the theme of why the surface is warmer than the Sun's input, don't you find it amazing that entropy gained by converting sunlight to IR can 'power' such a large effect?

I realize the actual energy comes from radiation not lost to space but it seems to me that entropy gained in the total area causes entropy loss by the storage mechanism in a subset area.

I could be wrong. What do you think?

There is no greenhouse effect...infrared radiation does not, can not, and never will warm the air.

 

[SSDD]

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

 

[Toddsterpatriot]

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater..

Excellent!

 

[SSDD]

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater..

Excellent!

So...if your infrared heater does not warm the air in your home...how do you suppose that infrared radiation warms the air outside your home? Is there some magic at work that makes air inside your home (which has a far higher CO2 concentration by the way) invisible to IR but has exactly the opposite effect on air outside the home?

How can there be a radiative greenhouse effect in the atmosphere when infrared radiation does not warm the atmosphere?

 

[Toddsterpatriot]

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater..

Excellent!

So...if your infrared heater does not warm the air in your home...how do you suppose that infrared radiation warms the air outside your home? Is there some magic at work that makes air inside your home (which has a far higher CO2 concentration by the way) invisible to IR but has exactly the opposite effect on air outside the home?

How can there be a radiative greenhouse effect in the atmosphere when infrared radiation does not warm the atmosphere?

So...if your infrared heater does not warm the air in your home

If I turn off the infrared heater, will the air in my home remain at the current 70 F?

 

[SSDD]

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater..

Excellent!

So...if your infrared heater does not warm the air in your home...how do you suppose that infrared radiation warms the air outside your home? Is there some magic at work that makes air inside your home (which has a far higher CO2 concentration by the way) invisible to IR but has exactly the opposite effect on air outside the home?

How can there be a radiative greenhouse effect in the atmosphere when infrared radiation does not warm the atmosphere?

So...if your infrared heater does not warm the air in your home

If I turn off the infrared heater, will the air in my home remain at the current 70 F?

Of course not...all the solid objects in your room will begin to cool and in turn, not convect as much heat. which solid objects in the atmosphere do you believe the infrared radiation emitting from the surface of the earth is warming? How does a radiative greenhouse effect work if infrared radiation does not warm the air?

 

[Toddsterpatriot]

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater..

Excellent!

So...if your infrared heater does not warm the air in your home...how do you suppose that infrared radiation warms the air outside your home? Is there some magic at work that makes air inside your home (which has a far higher CO2 concentration by the way) invisible to IR but has exactly the opposite effect on air outside the home?

How can there be a radiative greenhouse effect in the atmosphere when infrared radiation does not warm the atmosphere?

So...if your infrared heater does not warm the air in your home

If I turn off the infrared heater, will the air in my home remain at the current 70 F?

Of course not...all the solid objects in your room will begin to cool and in turn, not convect as much heat. which solid objects in the atmosphere do you believe the infrared radiation emitting from the surface of the earth is warming? How does a radiative greenhouse effect work if infrared radiation does not warm the air?

Of course not...all the solid objects in your room will begin to cool and in turn, not convect as much heat.

Will they radiate as much heat?

 

[SSDD]

[

Let me guess, you don't know how my walls reached 70 F?

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater..

Excellent!

So...if your infrared heater does not warm the air in your home...how do you suppose that infrared radiation warms the air outside your home? Is there some magic at work that makes air inside your home (which has a far higher CO2 concentration by the way) invisible to IR but has exactly the opposite effect on air outside the home?

How can there be a radiative greenhouse effect in the atmosphere when infrared radiation does not warm the atmosphere?

So...if your infrared heater does not warm the air in your home

If I turn off the infrared heater, will the air in my home remain at the current 70 F?

Of course not...all the solid objects in your room will begin to cool and in turn, not convect as much heat. which solid objects in the atmosphere do you believe the infrared radiation emitting from the surface of the earth is warming? How does a radiative greenhouse effect work if infrared radiation does not warm the air?

Of course not...all the solid objects in your room will begin to cool and in turn, not convect as much heat.

Will they radiate as much heat?

They will radiate according to the difference between their own temperature and the temperature of their surroundings...but the radiation from the walls and objects will not warm the air either.


What's the matter toddster...afraid to answer the question?...how does a radiative greenhouse effect work if infrared radiation does not warm the air?

 

[Toddsterpatriot]

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater...they did not receive any energy from the air as the IR radiation coming from your heater does not warm the air.

One of two ways...either they absorbed the radiation from your radiative heater, or they were warmed via conduction from the air being blown across the element of your radiative heater..

Excellent!

So...if your infrared heater does not warm the air in your home...how do you suppose that infrared radiation warms the air outside your home? Is there some magic at work that makes air inside your home (which has a far higher CO2 concentration by the way) invisible to IR but has exactly the opposite effect on air outside the home?

How can there be a radiative greenhouse effect in the atmosphere when infrared radiation does not warm the atmosphere?

So...if your infrared heater does not warm the air in your home

If I turn off the infrared heater, will the air in my home remain at the current 70 F?

Of course not...all the solid objects in your room will begin to cool and in turn, not convect as much heat. which solid objects in the atmosphere do you believe the infrared radiation emitting from the surface of the earth is warming? How does a radiative greenhouse effect work if infrared radiation does not warm the air?

Of course not...all the solid objects in your room will begin to cool and in turn, not convect as much heat.

Will they radiate as much heat?

They will radiate according to the difference between their own temperature and the temperature of their surroundings...but the radiation from the walls and objects will not warm the air either.


What's the matter toddster...afraid to answer the question?...how does a radiative greenhouse effect work if infrared radiation does not warm the air?

They will radiate according to the difference between their own temperature and the temperature of their surroundings

They won't radiate if the air is warmer?
They won't radiate at me if I'm warmer?

how does a radiative greenhouse effect work if infrared radiation does not warm the air?

How does radiation absorbed by the air not warm the air? Of course it does.

 

[SSDD]

They won't radiate if the air is warmer?
They won't radiate at me if I'm warmer?

Energy moves from warm to cool...not the other direction.

how does a radiative greenhouse effect work if infrared radiation does not warm the air?

How does radiation absorbed by the air not warm the air? Of course it does.[/QUOTE]

The radiation is immediately emitted on towards cooler pastures...millions of hours of repeatable, measurable, observable evidence toddster...infrared does not warm the air...

so again...how does a radiative greenhouse effect work if the atmosphere is not warmed by infrared radiation?

 

[Toddsterpatriot]

They won't radiate if the air is warmer?
They won't radiate at me if I'm warmer?

Energy moves from warm to cool...not the other direction.

how does a radiative greenhouse effect work if infrared radiation does not warm the air?

How does radiation absorbed by the air not warm the air? Of course it does.

The radiation is immediately emitted on towards cooler pastures...millions of hours of repeatable, measurable, observable evidence toddster...infrared does not warm the air...

so again...how does a radiative greenhouse effect work if the atmosphere is not warmed by infrared radiation?[/QUOTE]

Energy moves from warm to cool...not the other direction.

Unless work is done.

The radiation is immediately emitted on towards cooler pastures.

Or toward the ground. Or transferred to other molecules in the air by collision or emission.

so again...how does a radiative greenhouse effect work if the atmosphere is not warmed by infrared radiation?

You just admitted, however briefly, the IR is absorbed. Matter that absorbs radiation is warmed.

 

[IanC]

This law-

And the law that energy cannot be created or destroyed.

You can only control the amount of radiation produced by an object by adding or subtracting to the amount of energy available to be converted to radiation.

Any object above absolute zero will attempt to cool down by radiating. It always radiates but it only cools if it emits more energy than it receives.

That law states no such thing...that law states that a perfect black body radiating in a vacuum devoid of any other matter radiates according to its temperature to the 4th power...an ideal, perfect black body in a perfectly empty universe...a fantasy.

Bring matter into the equation and j starts changing in accordance with the difference between the temperature of the radiator and its cooler surroundings. You keep forgetting the basics and jumping to knee jerk conclusions...think ian, I know that you are capable of it.

There is no law that says that the rate of energy emitted by an object can not change if its environment changes...quite the contrary..the application of the SB law relating to radiators that are not perfect black bodies radiating into perfectly empty space says exactly the opposite...P keeps changing as the difference between the temperature of the radiator and the temperature of its surroundings change.

This equation is obviously not for a perfect Blackbody because it has a term for emmisivity.

You should also note that it has no term for area. It is an intensive property.

If you divide the j by area then you get an actual amount of power, an extensive property. But you then have to do the calculations for every angle that is in line-of-sight of the area you defined.

Your confusion seems to stem from confusing the intensive property of j with the extensive property of P.

The object exists and radiates. The environment that surrounds the object does not change the radiation, which is always proportional to the temperature of the object.

We can turn the intensive property of radiation into the extensive property of power by defining the radiating surface area of the object. We can turn the intensive property of temperature into the extensive property of total heat content by measuring the mass and the specific heat for its constituents.

With the two extensive properties of power and heat content we can calculate the rate of cooling as time passes. Quickly at first and then progressively slower.

If we add a second object nearby, it also radiates in the same fashion as the first. Some of the radiation from the first will be intercepted by the second, and vice versa. The amount will be determined by the line-of-sight connection. The amount of radiation energy absorbed will add to the total energy content of the two objects, thus slowing the rate of their cooling. But both objects will still be cooling overall.

I will leave it there for now.

You can appeal to as much complexity as you like ian, but that law does not state that an object can not radiate less when it is in the presence of other matter..

The whole topic is moot as IR does not warm the air.

Actually I am appealing to simplicity.

j = sigmaT^4 is the most simple of the S-B equations. It makes the assumption of a Blackbody and gives the amount of radiation for any temperature. Regardless of the environment that it is imbedded in, because it doesn't have a term for the environment. The radiation is always that much. The only way you can change the radiation is to change the temperature.

The first, but still reasonably simple step up in complexity is to add the emmisivity term. Emission and absorption are still exactly equal but are less than the unity value of a blackbody. Emmisivity can be further defined by individual wavelengths. Eg, CO2 has near perfect absorption at 15 microns near perfect non-absorption at 10 microns.

The next step up is to include the environment. Any of the equations that include some variation of the term (T^4 - Tc^4). This is a huge complication, because we are now dealing with radiation both going out and coming in. We are dealing with at least two objects, and imbedding them in three dimensional space. We now have to add the area term with complex line-of-sight calculations. And emerging temperature gradients. Etc.

 

[SSDD]

Actually I am appealing to simplicity.

j = sigmaT^4 is the most simple of the S-B equations. It makes the assumption of a Blackbody and gives the amount of radiation for any temperature. Regardless of the environment that it is imbedded in, because it doesn't have a term for the environment. The radiation is always that much. The only way you can change the radiation is to change the temperature.

But it places the object in a space where there is no other matter...when you bring other matter into the picture, the equation changes and J starts to change with the difference in temperature between the radiator and its surroundings.

And again...moot. IR does not warm the air.

 

[IanC]

But it places the object in a space where there is no other matter...when you bring other matter into the picture, the equation changes and J starts to change with the difference in temperature between the radiator and its surroundings.

That is incorrect. The radiation j is always there, always sigmaT^4.

The original cavity experiment produced the data from which this relationship was derived. It was a clever use of topology to have both a radiating surface AND a space to radiate into, without any loss of the radiation produced or the temperature drop from it.

I say the cavity surface is emitting at sigmaT^4 at all times, with the radiation crossing the cavity to be absorbed on the other 'side', with no net change because the object is absorbing EXACTLY as much as it emits.

You say no radiation is allowed to be emitted into the cavity. That radiation can only be produced if it has a lower temperature absorber to accept it.

Would you accept that radiation was being produced if a probe was inserted into the cavity that produced a photoelectric effect? With the threshold energy being in the top half of the range for the Planck spectrum?

I really don't understand how the sigmaT^4 relationship was found if no radiation is being produced so that it can be sampled. Perhaps you could expand on that. We would love to hear your explanation.

 

[IanC]

The radiation is immediately emitted on towards cooler pastures...

You keep saying this but you never explain what you actually think is going on.

I think we all agree that the surface emits 15 micron energy. I think we all agree that this radiation is absorbed to extinction within just a few metres of the atmosphere.

Once absorbed, where does it go? SSDD has two mutually exclusive statements on this.

Sometimes he says it is immediately re-emitted. Sometimes he says it is transferred to nearby molecules by collision, 99.9999% of the time. So which is it?

What does 'towards cooler pastures' mean? SSDD claims that energy in all forms, but specifically radiation, can only 'spontaneously' from warm to cool. Therefore radiation is prohibited from being emitted toward the surface (presumably by some unknown mechanism). Even though radiation can be consistently measured coming from the atmosphere in the direction of the surface. Apparently we are being 'fooled by instrumentation'.