2
$\begingroup$

I need help to interpret the following paragraph from Kittel and Kroemer, Thermal Physics 2ed, page 228-229:

Work can be completely converted into heat, but the inverse is not true: heat cannot be completely converted into work. Entropy enters the system with the heat, but does not leave the system with the work. A device that generates work from heat must necessarily strip the entropy from the heat that has been converted to work. The entropy removed from the converted input heat cannot be permitted to pile up inside the device indefinitely; this entropy must ultimately be removed from the device. The only way to do this is to provide more input heat than the amount converted to work, and to eject the excess input heat as waste heat, at a temperature lower than that of the input heat (Figure 8.1). Because $\tfrac{\delta Q}{\sigma}=\tau$, the reversible heat transfer accompanying one unit of entropy is given by the temperature at which the heat is transferred. It follows that only part of the input heat need be ejected at the lower temperature to carry away all the entropy of the input heat. Only the difference between input and output heat can be converted to work. To prevent the accumulation of entropy there must be some output heat; therefore it is impossible to convert all the input heat to work!

enter image description here

Especially, two sentences are unclear to me:

(1) "A device that generates work from heat must necessarily strip the entropy from the heat that has been converted to work."

What does it mean to "strip the entropy from the heat"?

(2) "The entropy removed from the converted input heat cannot be permitted to pile up inside the device indefinitely"?

Why the "The entropy removed from the converted input heat cannot be permitted to pile up inside the device indefinitely"

$\endgroup$
3
  • $\begingroup$ Please provide more context regarding the quote $\endgroup$
    – Bob D
    Jul 12, 2023 at 22:13
  • 1
    $\begingroup$ Kittel and Kroemer have a strange way of phrasing the issue. $\endgroup$ Jul 15, 2023 at 0:35
  • $\begingroup$ @DavidWhite knowing what excellent physicist Kittel was and Kroemer is, the word strange is the correct adjective but given the still controversial (!) nature of the subject even 130 years after Clausius it was understandable that they wished to stay "politically correct" and deviate the least from orthodoxy. Just remember the hostility Truesdell had to face about the same time that the K&K book was written. Even today, 40+ years later, is still considered controversial... $\endgroup$
    – hyportnex
    Jul 15, 2023 at 2:30

3 Answers 3

4
$\begingroup$

What Kittel & Kroemer are quietly saying is that "heat" does not convert to work, only work converts to work, if one uses a conventional meaning of the English word "convert". The reason for this is that the total entropy in all, yes all, thermodynamic transformation can never decrease. In other words, entropy is indestructible the same way as gravitating mass or electric charge are but with the difference that if it does not stay the same then it increases.

If you assume that heat converts to work then it should follow that the entropy being carried with thermal energy representing "heat" somehow disappears for work has no entropy. Since that is not allowed, Clausius introduced the concept of simultaneous heat compensation, a separate irreversible process to restore that disappearing entropy in the working fluid before rejected at a lower temperature.

What Kittel & Kroemer are quietly saying with "A device that generates work from heat must necessarily strip the entropy from the heat that has been converted to work" is that Clausius is wrong about that "heat compensation", there is no compensation, instead the entropy in the thermal energy is, by some magic, removed from it and it is the same entropy that is to be rejected at the lower temperature. Kittel & Kroemer do not explain that if that is indeed the case then what is being converted in the first place, and how does the converted heat become work.

There is one consistent explanation to all that is to view the entropy and not heat as being the agent of the work performed. No conversion, just entropy transport from a higher to a lower temperature, the same way as a gravitating mass dropped from a higher gravitational potential to a lower one performs work.

And to your question why entropy cannot pile up, is because you are interested in a cycle, and you want to return the engine to its starting state, and whatever entropy was absorbed at the higher temperature it is rejected at the lower one along with any irreversibly generated excess entropy (say, from friction or conduction).

$\endgroup$
1
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Aug 16, 2023 at 19:28
2
$\begingroup$

Heating shifts entropy; doing work via an impermeable barrier (e.g., expansion work) doesn't.

Therefore, energy that participates in the former and then the latter is automatically separated from the associated entropy. This is the "strip[ping]" being referred to.

Why can't we have entropy pile up indefinitely?

We do; it piles up in the cold reservoir or in the engine. For a cyclic heat engine that must return to its original state every cycle, only the former remains as a possible entropy depository.

$\endgroup$
1
$\begingroup$

Why is this the case? Why can't we have entropy pile up indefinitely?

What Kittel & Kroemer are referring to is entropy cannot "pile up" over a complete heat engine cycle. The reason is entropy is a system property and has only one value at any equilibrium state. By definition, a cycle returns a system to its original equilibrium state and thus the change in entropy of the system after completing a cycle is necessarily zero. Thus any entropy a system acquires during some process in the cycle must be removed (discarded to a lower temperature reservoir) in the form of heat, to return the system to its original state.

For example, in the Carnot Cycle for an ideal gas the system acquires entropy during the reversible isothermal expansion when heat is taken in and work is performed equal to the heat taken in. The entropy acquired during the process is temporarily kept in the system. In order to return the system entropy to its original value at the start of the cycle, a reversible isothermal compression is performed to transfer heat (and the acquired entropy) out of the system and to a low temperature reservoir. That "waste heat" is not available for work.

Hope this helps.

$\endgroup$
6
  • $\begingroup$ You wrote "in the Carnot Cycle the system acquires entropy during the reversible isothermal expansion when heat is taken in and work is performed equal to the heat taken in". This equality of incoming heat with performed work is true only when an ideal gas is the working fluid or more generally in systems whose internal energy depends on temperature alone. It is not true for a real gas or some other systems, such as a simple electric capacitor whose dielectric volume is temperature dependent. $\endgroup$
    – hyportnex
    Jul 14, 2023 at 14:41
  • $\begingroup$ @hyportnex I meant it for an ideal gas. Have clarified. $\endgroup$
    – Bob D
    Jul 14, 2023 at 18:20
  • $\begingroup$ the clarification is good but it just begs the question: what happens in the Carnot cycle, (reversible cycle whose 4 stages are (1) isothermal $\to$ (2)adiabatic $\to$ (3)isothermal $\to$ (4)adiabatic), when the working fluid is not an ideal gas? $\endgroup$
    – hyportnex
    Jul 14, 2023 at 18:26
  • $\begingroup$ @hyportnex I'm not sure what this has to do with the question of entropy "piling up". The ideal gas Carnot Cycle was only an example of a cycle which receives entropy but must reject the same amount of entropy to complete the cycle. What does it matter whether or not it is a real gas or an ideal gas. The same principle holds, does it not? $\endgroup$
    – Bob D
    Jul 14, 2023 at 18:31
  • $\begingroup$ if you reread the original post you will notice that it has two questions the 1st being 'What does it mean to "strip the entropy from the heat" '. Your 2nd paragraph seems to have been answering that for the case of ideal gas, hence my question. Also it is being "piled up" or not is just that entropy that has been "stripped", so they are related. $\endgroup$
    – hyportnex
    Jul 14, 2023 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.