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Find the changes in entropy for the water and for the ice

  1. Jan 29, 2005 #1
    In a well-insulated calorimeter, 1.0 kg of water at 20 C is mixed with 1.0 g (.001 kg) of ice at 0 C. What is the net change in entropy of the system by the moment the ice completely melts? The heat of fusion of ice is 3.34 x 10^5 J/kg.


    I must find the changes in entropy for the water and for the ice, then find the net change.

    Net change = /_\S1 + /_\S2

    This is what I did so far:
    S_ice = (mL)/T, where Q = mL

    S_ice = [(3.34 x 10^5 J/kg)*.001 kg]/[273 K] = 1.223 J/K ???

    How do I find the change in entropy for H2O??

    Thanks.
     
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Jan 29, 2005 #2

    Andrew Mason

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    You have found the [itex]\Delta S[/itex] for the ice. The [itex]\Delta Q[/itex] for the water is just the loss of heat to the ice. It takes place at 293K (we can see that there is not a significant temperature reduction - about .1 degree). So:

    [tex]\Delta S_{H_2O} = -334/293 = -1.14 J/K[/tex]

    AM
     
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