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Entropy change of a reservoir after heating something up

  1. May 2, 2015 #1
    1. The problem statement, all variables and given/known data
    1kg of silver is heated by a large heat reservoir at 373 K from 273K. Calculate the change of entropy in:
    a) the silver
    b) the reservoir
    c) the universe.

    2. Relevant equations

    ΔS = ∫dQ/T

    3. The attempt at a solution

    calculating the change in the silver first

    ΔS = ∫dQ/T
    = C∫dT/T
    = Cln(T2/T1)
    = 0.312

    This is fine, I understand it. Now I need to work out the entropy change in the large reservoir. Since the change happens spontaneously, it isn't reversible and the entropy of the universe increases. This means that I can't just take the negative of the change in the silver.

    Since the temperature of the reservoir doesn't change, is there no effective change in the entropy? This obviously shouldn't be the case because it would violate the second law. I'd appreciate any insight that can be offered.
     
  2. jcsd
  3. May 2, 2015 #2
    The temperature of the reservoir doesn't change appreciably, but it does lose the same amount of heat that the silver absorbs. As far as the reservoir is concerned, because the heat transfer took place at constant temperature, it thinks it experienced a reversible change (actually, it really did experience a reversible change). On the other hand, the silver experienced an irreversible change (since there were substantial temperature gradients in the silver during the transient heating). The combined change in both the silver and the reservoir must be positive because of the irreversibility in the silver. So you have enough information now to determine the entropy change of the reservoir, and the overall entropy increase for the combination (i.e., the universe).

    Chet
     
  4. May 3, 2015 #3
    Great, thanks for your help. So doing the entropy integral for a fixed temperature, I get:

    ΔS = ∫dQ/T = (1/T) * C(ΔTsilver) = (1/373)*C*100
    = -61.66
    This number is massive compared to the integral I did above for the silver with a varying temperature. Could you perhaps tell me which number is wrong and why?
     
  5. May 3, 2015 #4
    You got the first part correct : 0.312C
    But, it looks like you made an arithmetic error (or something) in the second part. I get -0.268C.

    Chet
     
    Last edited: May 3, 2015
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