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Change in entropy question

  1. Apr 6, 2015 #1
    1. The problem statement, all variables and given/known data
    5 kg of water at 60 degrees are put in contact with 1 kg of ice at 0 degrees and are thermally isolated from everything else. The latent heat of ice is 3.3x105 J/kg

    What is the change of entropy of the universe when 100J of energy are transferred from the water to the ice

    2. Relevant equations

    Q=+/- mL
    Q=mcΔT
    ΔS=|Q|/Tc-|Q|/TH

    3. The attempt at a solution
    mcΔT+mL+mcΔT=0
    (5)(4200)(Tf-60)+(1)(3.3×105)+2000(1)Tf=0

    23000Tf=96000
    Tf=41.74 C

    ΔS=100/(41.74+273)-100/(273)=-0.0486

    I'm not getting the right answer for this one. Help would be greatly appreciated!
     
  2. jcsd
  3. Apr 6, 2015 #2

    vela

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    You're assuming that all the ice is going to melt.
    What does the last term represent? I figure it has something to do with the ice, but I'm not sure.

    This expression also assumes that all of the ice melts. Is this possible if only 100 J of energy is transferred from the warm water to the ice? Can you describe in words what's going to happen?

     
  4. Apr 6, 2015 #3
    Firstly you can not use the relation $$\Delta S= \Delta Q/T$$ to find out entropy change for water, because here T is changing. Instead you should use the general relation $$\Delta S= \int dQ/T$$ to find out the change in entropy. This process should work out.
     
  5. Apr 6, 2015 #4
    There is a bit of ambiguity regarding this problem statement. It is not clear whether the water and ice are in intimate contact, or whether they are separated by a diathermal barrier, and that, once the 100 J are transferred, the diathermal barrier is removed and replaced by an insulation barrier. The version with the barriers is simpler to solve. If 100 J is removed from the 5 kg of water, how much does its temperature drop?

    Chet
     
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