1
$\begingroup$

Everywhere in the literature, it is mentioned that "Heat cannot be converted into work with 100% efficiency in a cyclic process because, in a cyclic process, some portion of heat received from the source will end up in the sink, so all the heat will not convert into work". I understand and agree with this statement.

My questions are the following:

  1. If heat cannot be converted into work in a cyclic process with 100% efficiency, then can it be converted into work with 100% efficiency in a non-cyclic / single-step process?

  2. Can heat at all, in any case, be converted into work with 100% efficiency in any kind of "Practically Possible" process(es)? If yes, then isn't it a violation of the second law of thermodynamics? If no, then why not?

  3. If heat cannot be converted into work with 100% efficiency in any kind of practically possible process then can we say that "Isothermal" processes are not practically possible because they claim to convert heat into work with 100% efficiency?

  4. If the heat cannot be converted into work in any possible practical way, then why is it the case? What restricts this conversion? Can somebody help me understand it in detail, with some intuitive reasoning?

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ BTW I saw your other migrated post on background of Carnot work. Suggest you get a copy of Carnot's 1824 paper "Reflections on the Motive Power of Fire". You can download pdf from the internet. I found it fascinating reading from the historical perspective. $\endgroup$
    – Bob D
    Commented Dec 9, 2019 at 18:01
  • $\begingroup$ @Bob: Thanks a lot for your response. I sometimes wish that I could ask some of my simple questions to which simple answers are required, to somebody of yours staturr, who is well versed with the subject. is it possible that I can ask you some simple questions once in a while on chat or on any other platform possible for you? $\endgroup$
    – Devansh Mittal
    Commented Dec 9, 2019 at 18:43
  • 1
    $\begingroup$ Of course. I only participate on this exchange but Chat would be fine $\endgroup$
    – Bob D
    Commented Dec 9, 2019 at 18:53
  • $\begingroup$ Thanks a lot Bob. On which forum We can talk? Email? WhatsApp? Facebook? Or any other forum? $\endgroup$
    – Devansh Mittal
    Commented Dec 9, 2019 at 18:56
  • 1
    $\begingroup$ I'm an old guy from the old school that values privacy. I do not participate in any of the social media formats. But I would be happy to talk with you in Chat on this site. $\endgroup$
    – Bob D
    Commented Dec 9, 2019 at 20:21

2 Answers 2

Reset to default
2
$\begingroup$
  1. If heat cannot be converted into work in a cyclic process with 100% efficiency, then can it be converted into work with 100% efficiency in a non-cyclic / single-step process?

Yes, if it is a reversible isothermal process, such as the reversible isothermal expansion of an ideal gas in the the Carnot cycle.

  1. Can heat at all, in any case, be converted into work with 100% efficiency in any kind of "Practically Possible" process(es)? If yes, then isn't it a violation of the second law of thermodynamics? If no, then why not?

As already indicated, it can if it is a reversible isothermal process invoolving an ideal gas. However no real heat transfer process is reversible (or as you say, "practically possible") since heat transfer by definition requires a temperature difference (thermal disequilibrium). Another impracticality, that @David White pointed out in his comment below, the smaller the temperature difference the slower the heat transfer rate and the process will take an infinite amount of time. That severely limits the practicality of the cycle in terms of power output (work per unit time). In a reversible process the temperature difference is infinitesimally small so that we can say, in the limit, the process is 100% efficient and would not violate the second law, but the process would go extremely slowly. See further considerations discussed below regarding your Q4.

  1. If heat cannot be converted into work with 100% efficiency in any kind of practically possible process then can we say that "Isothermal" processes are not practically possible because they claim to convert heat into work with 100% efficiency?

A process can be irreversible yet still be isothermal. Isothermal refers to the system temperature being constant which is possible if the system can be considered a thermal reservoir, that is, its heat capacity is sufficiently large that the amount of heat transferred into or out of the system does not change its temperature.

The difference between a reversible and irreversible isothermal process is in the former the system temperature is in equilibrium with the temperature of the surroundings (temperature difference being infinitesimally small) at all times, whereas in the case of the latter there is a finite temperature difference. The ramifications of this are discussed below an response to question 4.

  1. If the heat cannot be converted into work in any possible practical way, then why is it the case? What restricts this conversion? Can somebody help me understand it in detail, with some intuitive reasoning?

An irreversible process restricts the conversion. An irreversible process generates entropy which, for a complete cycle, must be dumped back into the surroundings in order to return all system properties (including entropy) to their original state. The only way to transfer entropy to the surround is by transferring some of the heat the system got during the expansion back to the surroundings during the isothermal compression process. That heat is thus unavailable to produce net work in the cycle. To further explain:

For a reversible isothermal heat transfer (where the system and surroundings temperature is the same) the change in entropy of the system is

$$\Delta S_{sys}=+\frac{Q}{T}$$

And the change in entropy of the surroundings is

$$\Delta S_{surr}=-\frac{Q}{T}$$

For a total entropy change (system + surroundings) of

$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{surr}=+\frac{Q}{T}-\frac{Q}{T}=0$$

This satisfies the equality $\Delta S_{tot}=$ 0 for a reversible cycle.

For an irreversible isothermal process the temperature of the surroundings is greater than the system. So the change in entropy of the surroundings is

$$\Delta S_{surr}=-\frac{Q}{T_{surr}}$$

And the system

$$\Delta S_{sys}=+\frac{Q}{T_{surr}-\Delta T}$$

The total entropy change then being

$$\Delta S_{tot}=+\frac{Q}{T_{surr}-\Delta T}-\frac{Q}{T_{surr}}$$

Note that for all positive values of $\Delta T$ the first term will be greater than the second and therefore, for an irreversible heat transfer we have

$$\Delta S_{tot}>0$$

Since all real processes are irreversible, the inequality holds for all real processes.

The difference between the change in entropy of the system and the surroundings is call the entropy generated in the system. To complete a cycle and return all properties to their original values this entropy generated must be returned to the surroundings which can only occur in the form of heat. That heat is thus not available to do work in the cycle. So the net work done in an irreversible cycle will always be less than that for the same cycle carried out reversibly.

In short, for an irreversible isothermal expansion not all of the heat transferred to the system produces net work in the cycle because of the entropy generated. Some has to be returned to the surroundings.

Hope this helps.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Very informative, thanks! $\endgroup$
    – Dale
    Commented Dec 9, 2019 at 20:08
  • $\begingroup$ There is also the additional problem of power requirements. Heat transfer through an infinitesimal temperature difference requires an infinite amount of time, and no human is willing to wait that long. $\endgroup$
    – David White
    Commented Dec 9, 2019 at 20:30
  • $\begingroup$ @DavidWhite Absolutely another check mark in the impracticality column. Someone once said if you put a Carnot engine in your car you would get phenomenal fuel economy, but pedestrians would be passing you by! $\endgroup$
    – Bob D
    Commented Dec 9, 2019 at 20:33
  • $\begingroup$ @DavidWhite I added your point to my answer. $\endgroup$
    – Bob D
    Commented Dec 9, 2019 at 20:41
  • $\begingroup$ @AndrewSteane Correct. I am so used to the Carnot cycle with an ideal gas I take it for granted. Will correct. Thanks. $\endgroup$
    – Bob D
    Commented Dec 10, 2019 at 11:50
0
$\begingroup$

If heat cannot be converted into work in a cyclic process with 100% efficiency, then can it be converted into work with 100% efficiency in a non-cyclic / single-step process?

A compressed gas spring that is expanding and doing mechanical work is an example of heat being converted into work with 100% efficiency.

Another example is a power plant in the moon. Such power plant that converts moon ice to steam, then lets the steam expand to the vacuum through a turbine.

The turbine should be infinitely large, so that it can extract energy from infinitely thin gas.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ "Lowering a bucket full of hot water in a gravity field is an example of converting heat into work with 100% efficiency" is not correct. First, no heat is transferred in the process. Second, the same work would be done if the bucket was full of the same amount of cold water. And third any net work done converting gravitational potential energy into kinetic energy of the bucket is done by gravity, not by heat. $\endgroup$
    – Bob D
    Commented Dec 9, 2019 at 20:18
  • $\begingroup$ @BobD OK. I deleted that part. $\endgroup$
    – stuffu
    Commented Dec 10, 2019 at 11:20

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.