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Homework Help: Confused about entropy.

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data
    In the entropy equation 1S2 + ∫δQ/T = S2 - S1, what is the difference between 1S2 and ∫δQ/T?

    So for example if we have two reservoirs, one hot and one cold and heat is transferred in between we have:

    Shot = Q/Thot
    Scold = Q/Tcold

    Where do the Shot and Scold terms go in that equation?
  2. jcsd
  3. Mar 13, 2013 #2

    rude man

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    A reservoir has a constant temperature by definition. So what does that enable you to do in evaluating ∫δQ/T ?

    And BTW ∫δQ/T = 1S2 only if evaluated over a reversible path. Otherwise the integral is < entropy change.
  4. Mar 13, 2013 #3
    Sorry I meant ∫δQ/T = 1S2,generated.

    But if ∫δQ/T = 1S2,gen, then that means S2 - S1 = 2∫δQ/T
  5. Mar 13, 2013 #4

    rude man

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    No it does not. S2 - S1 = ∫δQ/T over a reversible path. S2 - S1 is the entropy change in ONE of the reservoirs. Since you have two reservoirs at two different temperatures T1 and T2 , each has its own S2 - S1.

    Have you figured out how that integral is simplified if you're dealing with entropy changes of reservoirs?
  6. Mar 13, 2013 #5
    Yeah I know how to do that. Thanks!
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