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Entropy change in an isolated system

  1. Apr 24, 2010 #1
    hi im having a few conception problems with entropy

    problem:
    a solid with constant heat capacity and initial temp To, is brought in contact with hotter reservoir temp T1 and allowed to come into equilibrium. This is all isolated.

    a) is the change in the entropy of the universe negative zero or positive thanks to the systems process. explain why
    b) calculate entropy change to solid
    c) calculate entropy change to reservoir

    a) i would have thought that the entropy change of the universe is zero because entropy change is dQ/T and the change in Q for both systems is the same (for the solid its +ve and reservoir -ve)

    b) dS = dQ/To
    c) dS = -dQ/T1

    is this right?
     
  2. jcsd
  3. Apr 24, 2010 #2

    Andrew Mason

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    The heat flow out of the reservoir is at the higher reservoir temperature.

    The total heat flow out of the reservoir (negative) has to be equal in magnitude to the (positive) heat flow into the solid. But since the temperature of the solid is always less than that of the reservoir, the change in entropy of the reservoir (dS = dQ/T), which is negative, will have a smaller magnitude than the positive change in entropy of the solid. So the total change in entropy will be greater than 0.

    That is the initial differential change in entropy. The total change is:

    [tex]\Delta S = \Delta S_{res} + \Delta S_{sol}[/tex]

    where:

    (1) [tex]\Delta S_{res} = \int_{T_{ir}}^{T_{fr}} m_{res}C_{res}\frac{dT}{T} = m_{res}C_{res}\ln{\frac{T_{fr}}{T_{ir}}[/tex]

    and

    (2) [tex]\Delta S_{sol} = \int_{T_{is}}^{T_{fs}} m_{sol}C_{sol}\frac{dT}{T} = m_{sol}C_{sol}\ln{\frac{T_{fs}}{T_{is}}[/tex]


    AM
     
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