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Entropy change question

  1. Apr 6, 2015 #1
    One hundred 1:00 kg bricks are removed from a ring kiln, which operates at a temperature of 500 C; and are allowed to cool in the atmosphere at 20:0 C: If the process irreversibility is 18:8 MJ determine the entropy change of the Universe and the specic heat capacity of the material from which the bricks are constructed.
     
  2. jcsd
  3. Apr 6, 2015 #2

    MarcusAgrippa

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    Clue: what is the definition of entropy?
     
  4. Apr 6, 2015 #3
    is a measure of the number of specific ways in which a thermodynamic system may be arranged, commonly understood as a measure of disorde
     
  5. Apr 6, 2015 #4

    MarcusAgrippa

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    This is a question about classical thermodynamics, not about statistical thermodynamics or statistical physics.
     
  6. Apr 6, 2015 #5
    ds=dw/t
     
  7. Apr 6, 2015 #6

    MarcusAgrippa

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    dS = dQ/T
     
  8. Apr 6, 2015 #7
    That's only for a reversible process.

    Chet
     
  9. Apr 6, 2015 #8

    MarcusAgrippa

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    True. But most elementary exercises assume that dS is calculated in this way. Anyhow, I was correcting Piston's formula, not responding to your comments - which are valid. The question is not optimally worded. But this is a common fault in elementary courses. Perhaps Piston should check that he has transcribed the question correctly.
     
    Last edited: Apr 6, 2015
  10. Apr 6, 2015 #9
    Says who? And, do you really think that cooling a pile of bricks from 500 C to 20 C in air is a reversible process?

    Chet
     
  11. Apr 6, 2015 #10

    MarcusAgrippa

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    No. Of course it is not. Heat is transferred across a temperature gradient so it is not reversible. But it is not useful to Piston for you and me to quibble, so I shall remain silent from here on.
     
  12. Apr 6, 2015 #11
    i assume dw is the lost work which is an irrevesible process
     
  13. Apr 6, 2015 #12
    If you knew the heat capacity of the bricks C, could you write down an equation for the change in entropy of the bricks (in terms of C, the mass of the bricks (100 kg), the original brick temperature 773K, and the final brick temperature 293K)?

    Chet
     
  14. Apr 6, 2015 #13
    Piston: The way I interpret this problem statement, you are currently learning about the concept of lost work in your course. My interpretation of the phrase "process irreversibility" in the problem statement is the lost work WL. In your course, you should have seen an equation relating the lost work to the change in entropy of the system, the temperature of the surroundings (in this case 293K), and the heat received by the system Q (in this case, Q is negative). Can you write out that equation for us?

    Chet
     
  15. Apr 8, 2015 #14
    The OP has not responded to this thread in a few days, and it looks like the thread is fading into oblivion. I find this problem statement very interesting, and the solution involves some powerful and useful concepts. Is there anyone else out there who is interested in continuing this solution with me? If not, I will give it another day or two, and then lock the thread.

    Chet
     
  16. Apr 8, 2015 #15

    MarcusAgrippa

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    I would like to see what you make of it. I followed the naive route, since I could not make much sense of the problem, and assumed that "the process irreversibility" is the energy radiated by the bricks into the surroundings. With that assumption, one can calculate the specific heat of the bricks. If we then naively use dS = dQ/T, we might get the entropy change in the universe (or at least, a lower bound to it). What would you do?
     
  17. Apr 8, 2015 #16
    It seems pretty clear to me, especially from the OP's allusions to lost work, that the process irreversibility is the same thing as the lost work of this irreversible process. This is the amount of work that could have been done if the cooling of the bricks had been carried out reversibly (using, for example, carnot engines), with the rejection of any heat done reversibly at the temperature of the surroundings (i.e., the air), minus the work that was actually done in this process, which, in this brick cooling process, was zero. So the lost work LW is equal to the ideal reversible work in this situation.

    Smith and Van Ness (Introduction to Chemical Engineering Thermodynamics, Chapter 16) have presented a precise and detailed analysis of how to determine the Lost Work, and have derived the following equation for the lost work:
    $$LW=T_0ΔS-Q$$
    where ΔS is the change in entropy of the system (in this case the bricks) between the initial and final equilibrium states, T0 is the temperature of the surroundings (in this case, the air), and Q is the heat transferred from the surroundings to the system in the actual irreversible process.

    In the case of our process,

    ##ΔS=100C\ln (\frac{293}{773})=-97.01C## (kJ/K), where C is the heat capacity of the bricks.

    and

    ##Q=100C(293-773)=-48000C## (kJ)

    So, ##LW = -28424C+48000C=19576C## kJ##=19.6C## MJ

    So, ##19.6C = 18.8##

    So, ##C = 0.96 kJ/kgK##

    and, ##ΔS = -93.2 (kJ/K)##

    Hope this makes sense.

    Chet
     
  18. Apr 8, 2015 #17
    I guess that's only the entropy change of the bricks, and not the entropy change of the universe. For the entropy change of the universe, we need to add the entropy change of the air. The entropy change for the air would be -Q/293 = 157.3 (kJ/K). So the entropy change for the universe would be 64.1 kJ/K.

    Chet
     
  19. Apr 9, 2015 #18

    MarcusAgrippa

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    If one accepts that "process irreversibility" = lost work, and that the formula LW=T_0 ΔS − Q is valid, then your solution seems inevitable. So my first question is how do you get "process irreversibility" = lost work? Is this standard engineering jargon? I have not come across this term before.

    My second question is on the validity of the formula quoted. I would have calculated the lost work by first calculating how much heat is conducted from the bricks to the surroundings. I would then ask, how much work might have been gotten out of this heat had a Carnot engine been driven by the heat from the bricks being exhausted to the surroundings. Of course, this required the hotter reservoir (the bricks) to have a continually changing temperature, so the calculation is not straightforward.

    I am trying to locate Smith and van Ness to see how they prove their formula. None of the numerous physics books that I have consulted address this problem, except perhaps for Zemansky who gives a formula that I am not able to reconcile with the one you have quoted.

    This topic is more interesting than I first imagined. This issue of lost work may address a gap in my knowledge that has bothered me for some time: what is the difference in content of 1st and 2nd laws in the case of irreversible processes. The mismatch between T dS and dQ appears to be nicely illustrated by the formula you quote. It also emphasises the fact that the "T" that appears in the Clausius theorem is not the system temperature, but the temperature of the reservoir that is used to dump the rejected heat.
     
    Last edited: Apr 9, 2015
  20. Apr 9, 2015 #19
    Me neither.

    Yes. This is exactly what they addressed in Smith and van Ness. They also provided an example to give a better idea how the continuously varying temperature is taken into account. It was very interesting.
    Yes. Same here. To work this problem, I had to read up on the subject in Smith and van Ness, and I found the development very intriguing. The best relationship that I have found up to now for expressing the 2nd law mathematically is the Clausius inequality (which still is a very powerful relationship). You can see the close relationship between the lost work equation and the Clausius inequality. But this turns it into an actual equality. The development is, for some reason, presented in one of the latter chapters of Smith and Van Ness.

    This is something that I have been trying to hammer home for a long time in Physics Forums, but the textbooks out there fail to emphasize this point. As a result, student after student is profoundly confused by the Clausius inequality. In the descriptions that I write, I use TI to represent the temperature at the interface between the system and surroundings, and say that ΔS for the system is ≥ the integral of dQ/TI. My TI is what you are calling "the temperature of the reservoir that is used to dump the rejected heat." I have a brief write up on the first and second laws that I have prepared to help those poor students on PF that are struggling with all this. Any interest in taking a look?

    Chet
     
  21. Apr 9, 2015 #20

    MarcusAgrippa

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    Very interested. Do you need an email address, or can you post it here? Perhaps a dropbox link?

    I would also appreciate the page number in Smith and van Ness where they develop that formula. I now have the book and am reading it with interest. Converting the inequality into an useful equality is something that I have searched for in vain in standard texts.

    The idea of the temperature of an interface seems a bit murky.
     
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