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Overall entropy

  1. May 10, 2006 #1
    I'm not sure about this qustion:

    A piece of metal at 80 degrees C is placed int 1.22 litres of water at 72 degrees C. This thermally isolated system reaches a final temperature of 76 degrees C. Estimate the overall change of entropy for this system.

    Can I actually calculate anything for this question since I don't know the weight or the type of metal??

    Thanks
     
    Last edited: May 10, 2006
  2. jcsd
  3. May 10, 2006 #2

    Chi Meson

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    Key word: "estimate." Since the heat that flows (Q) and the temperatures are what matter for finding the changes in entropy, and since all the heat required to change the temperature of 1.22 L of water came from the metal, you have enough for an estimated answer. start by finding Q.
     
  4. May 10, 2006 #3
    Ok so I from the density of water I found that 1.2L is 1.2 kg of H20

    So Q=mcdT
    Q=(1.2kg)(1.0 kcal/kgC)(76-72C)
    Q=4.8 kcal

    So dSw = 4.8 kcal/347K
    dSw = 0.014 kcal/K

    So now that I know the change in entropy for the water how do I figure it out for the metal?
     
  5. May 11, 2006 #4

    Chi Meson

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    Q_out = -Q_in

    Use the average temperatures of each.
     
  6. May 11, 2006 #5
    So then dS for the metal would be Q/(Avg temp of the metal)

    dSm = -4.8 kcal/351K
    dSm = -0.014 kcal/K

    So adding these together would mean that the overall change in Entropy is 0? Can this be right??
     
  7. May 11, 2006 #6

    Chi Meson

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    drag your results to one more digit and it would be right. As far as appropriate sig figs go, there is "no significant change in total entropy." but there is always a net increase in entropy.
     
  8. May 11, 2006 #7

    Andrew Mason

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    This is a good illustration of when (mis)use of significant figures at intermediate stages can bury an answer.

    The important point here is that the loss of heat of the metal is exactly the gain in heat of the water. So you should express the difference between the two entropy changes in two significant figures ie:

    [tex]\Delta S[/tex] = 4.8(1/347 - 1/351) = 4.8( 3.3 x 10^-5) = 1.6 x 10^-4 J/Kg.

    AM
     
  9. May 11, 2006 #8
    Ok so if I increase the number of digits

    dSw = 0.01383
    dSm = -0.01368

    Therefore the overall change in Entropy is S = 0.01383+(-0.01368)
    S = 0.00015 kcal/K
    Does this look about right?
     
  10. May 11, 2006 #9
    I guess its right - must have been writing the last messages at the same time.

    Thanks for the help
     
  11. May 11, 2006 #10

    Andrew Mason

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    It should be 1.6 x 10^-4. (1/347 - 1/351) = 4/(347*351) = 3.3 x 10^-5.

    AM
     
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