2nd law and energy flows question.

[weza]

Hello all.

As I'm a newbie here please excuse any misconceptions in my following question concerning the 2nd law of thermodynamics and heat/energy flows.

Although energy will not spontaneously flow from a colder to a warmer object - the colder object still emits IR photons toward the warmer object.
As I understand it this lower energy/freq photons are not absorbed by the warmer object (2nd law) - but what actually happens to them.

My understanding (most likely wrong) is that the lower energy photons are effectively reflected away by the warmer object?
Or are they absorbed and re-emitted immediately without transferring any energy?

Am I missing something here or is my view simply wrong?

Thanks in advance for any guidance on this.

 

[Simon Bridge]

You are talking about radiation transfer of heat.
Everything is absorbing and emitting radiation.
Some things emit radiation at a greater rate than others.
In your example and all other things remaining equal, you will get more radiation emitted by the hot object than is absorbed by it so, on balance, it loses energy. Reverse for the cool object.

The result is that Heat appears to flow from the hot object to the cold one.
Note: IR photons are not the same as heat - it's just the range more readily absorbed by matter.

The situation is a tad more complicated than that but I believe that answers your question.

 

[weza]

Thanks Simon.

I still have a little trouble with this.

You say that even though the warmer object is radiating more energy than the cooler object - the cooler object still passes energy to the hotter object?

How does the energy radiated by the cooler object effect the warmer object?
Obviously it cannot add heat to it as this would violate the 2nd law.
The only effect I can think of is a slowing of the rate of transfer.

Actually...(thinking a little more about it) if the above is correct...
As the 2 objects approach equilibrium, shouldn't the cooler object radiate more and more energy back to the warmer object? And wouldn't this slow the transference of energy down even more?

 

[DrewD]

The laws of thermodynamics are true on average. When you talk about IR photons, you are talking about microscopic fluctuations. If you only take one or two fluctuations into account, the laws of thermodynamics would not necessarily be true.

If we only consider conduction, the same thing happens. At any given moment, if we focused on the contact point between the two bodies you proposed (I know they weren't touching, but let's consider that now), there is a possibility that you will see some energy transfer from the cooler object to the hotter object. But, this will happen very rarely in comparison to energy transfer in the other direction.

The 2nd law of thermodynamics is not violated by either situation because, on average, the heat (or radiation) is going from the hotter body to the colder body. Even on quite short time scales and very small areas this is true, but if you start trying to track individual photons, things must be considered differently.

If the two bodies are radiating energy and absorbing energy, the cooler one will heat up and the hotter one will cool since the hotter one emits more energy through radiation and absorbs less and the opposite is true for the cooler object. Yes, as they approach equilibrium, the time derivative of the temperature difference should decrease.

 

[Darwin123]

Hello all.

As I'm a newbie here please excuse any misconceptions in my following question concerning the 2nd law of thermodynamics and heat/energy flows.

Although energy will not spontaneously flow from a colder to a warmer object - the colder object still emits IR photons toward the warmer object.
As I understand it this lower energy/freq photons are not absorbed by the warmer object (2nd law) - but what actually happens to them.

My understanding (most likely wrong) is that the lower energy photons are effectively reflected away by the warmer object?
Or are they absorbed and re-emitted immediately without transferring any energy?

Am I missing something here or is my view simply wrong?

Thanks in advance for any guidance on this.

>As I'm a newbie here please excuse any misconceptions in my following question >concerning the 2nd law of thermodynamics and heat/energy flows.
As long as you don’t get angry at me for correcting those mistakes and misconceptions.
>Although energy will not spontaneously flow from a colder to a warmer object – the
Not true. Energy in the form of heat can not spontaneously flow from a colder to warmer object. Energy in the form of work can flow from any object to any other object.
A better expression for the second law is that “entropy will not spontaneously flow from a colder object to a warmer object.” Or “energy can not be carried by entropy from a colder to warmer object.”
>colder object still emits IR photons toward the warmer object.
>As I understand it this lower energy/freq photons are not absorbed by the warmer object >(2nd law) - but what actually happens to them.
The low energy photons can be absorbed by the warmer object. However, the absorption of energy is balanced by a flow of photons from the warmer object to the colder object. When there is a thermal equilibrium, the flow of energy of the black body photons will be equal.
It would be useful to avoid thinking what happens to the "photons". Instead, ask yourself what happens to the "entropy".
>My understanding (most likely wrong) is that the lower energy photons are effectively >reflected away by the warmer object?
No. The photons absorbed by the warmer object are destroyed. The warm object both destroys and creates photons. The energy absorbed by the warmer object is remitted as new photons. Some of the new photons have a higher energy than the photons that were destroyed.
>Or are they absorbed and re-emitted immediately without transferring any energy?
It doesn’t have to be immediately. There could be some sort of delay on a microscopic level. The equilibrium hypothesis in thermodynamics is that all the measurements are being performed very slowly, so that the rapidly completed processes are “invisible”.
>Am I missing something here or is my view simply wrong?
It helps to look at the history of science. When thermodynamics was first being developed, entropy was thought of as an indestructible gas. The word “caloric” was used instead of “entropy”, but mathematically it is the same thing. Carnot analyzed his engine in terms of this indestructible gas called caloric (i.e., entropy). Lavoisier listed caloric (i.e., entropy) as an element in his version of the periodic table. If you don't understand that heat=entropy to these gentlemen, then understanding the science of what they said can very difficult.
The definition of entropy was developed as an analog to the ideal law of gases. Temperature was the pressure that the entropy was under.
The jargon used in thermodynamics is an atavistic holdover in the idea that entropy is a gas. There are articles on this, but they haven't become popular. I only have a paper copy of this article, which is very old. However, it has helped me a lot. Try to look for a copy of:
“Entropy in the teaching of introductory physics” by Hans U. Fuchs. American Journal of Physics 55(3), 215-219 (March 1987).
The article suggests that students should be warned about the ambiguity in the word “heat”, often propagated in textbooks that are otherwise very good. It suggests that the concept of entropy should be used more often in introductory physics. With this idea in mind, let me say the following.
The picture of absorption and remission is closer to the truth than the reflection picture. What may be confusing you is that photons are sometimes “work” and sometimes “heat”. In your example, all the photons are “heat”. However, your picture is slipping back and forth between photons as “heat” and photons as “work”. On an atomic level, what distinguishes work from heat is statistics. A short cut that avoids statistical analysis is the heuristic idea that entropy is a substance. For example:
You have associated all low energy photons with heat. However, this is not the case. The problem is that some photons may not be part of the “heat energy”. For example, if a high frequency laser beam shines on an object, the photons of the laser have to be counted as work rather than heat. So the laser beam can heat the warm object without limit.
On an atomic level, there is no difference between heat and work. On a macroscopic level, the only difference between heat and work is that heat energy is carried by entropy. On a macroscopic level, entropy is a concrete concept.
One of the things that you may be missing is that the word “heat” is often used ambiguously. Sometimes, the word “heat” refers to energy. However, the word “heat” can also refer to entropy. In many statements referring to thermodynamics, the word “heat” should be replaced with the work entropy.
Try replacing the word “heat” in a few places of your textbook with the word “entropy”. Try replacing the word "heat" in the Laws of Thermodynamics with the word "entropy". Some of the laws will make more sense if the word heat is used to refer to entropy rather than to energy.

 

[Studiot]

Obviously it cannot add heat to it as this would violate the 2nd law.

Why not? it gives off energy, not heat. This energy may be gathered by another body where it will increase the total internal energy of that body regardless of the temperature difference.
As others have said, this is a two way process so the hotter body gives off more energy than the colder so the colder body gains internal energy and the hotter body looses it as a net result.

Since you are asking about the second law I assume you are familiar with the first.

IR photons are not 'heat energy' in the sense of thermodynamics laws.
Sure, they carry energy but they are not heat and the energy they carry is not heat.

If the quantity was sufficient we would consider it as an additional term in the first law.

dU = q + w + e_r + e_e + e_m...

where e_r is radiation energy
e_r is electrical energy
e_m is magnetic energy

and so on.

The second law refers specifically to q, not to the q plus any other energy.

 

[Simon Bridge]

:) Notice how everyone is insisting that the photons are not the same as heat?
I suspect that this is key to your understanding here. "Heat" may spontaneously move only from hot objects to cooler ones, but energy can go either way quite happily. You can as easily kick a hot ball as a cold one ;) so a photon may as easily be absorbed by a hot body as a cooler one. After all, how does the body know where the photon came from?

It probably helps to describe the 2nd law in terms of entropy rather than heat flow - which is an historically observed consequence.
Confining ourselves to one process: a body A releases a photon, which is absorbed by another body B, what happens to the overall entropy if T_A < T_B and what happens forT_A > T_B? If the total entropy decreases for either, then the 2nd law has been violated right?

You'll find it hasn't.
The description you have [in terms of heat flow] is quite limited and, as you have seen, it fails to model nature in the fine details. You can modify it without loss by noting that it is the net heat flow that is from hot to cold... it's all you'll see with tools like thermometers anyway. Fortunately it is only part of the 2nd law which, if you look it up, you'll see is properly, and more usually, described in terms of entropy.

Take heart: it's all good. At the early stages in your education you will be handed information in easily digested lumps .. well, easy for some. These have to be incomplete and flawed if only because you won't yet have the math/language for the bigger pictures. As you get used to questioning them, testing the limits of what you have been told, you'll discover that others have been there before you and come up with something to sort out the problems. [After a while this stops happening and you find those areas of nature where the science is very much still a work in progress. That's where all this practice testing and questioning comes in real handy.] Then you'll learn more and move on to tougher problems.