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Entropy ice-water

  1. Jan 21, 2009 #1
    1. The problem statement, all variables and given/known data
    A 1.30×10-2 Kg cube of ice at an initial temperature of -12.0 C is placed in 0.470 Kg of water at 50.0 C in an insulated container of negligible mass.

    mi=1.3x10-2 Kg
    mw=0.470 Kg
    Ti=-12 C
    Tw=50 C

    2. Relevant equations

    3. The attempt at a solution

    okay so first I tried to find Tf

    for entropy I tried to add up all the stages where ice->0 C + ice->water + ice water->Tf + hot water->Tf

    S=[miciln(T2/Ti)]+[miLf/T]+[micw ln(Tf/T)]+mwcw ln(Tf/Tw)
    S=[1.3x10-2*2100ln(273.15/261.15)]+[(1.3x10-2*3.34x105/273.15)]+[1.3x10-2*4190 ln(46.98+273.15/273.15)]+[0.470*4190 ln(46.98+273.15/50+273.15)]

    I still get the wrong answer when using this method so if someone could point me in the right direction that would be much appreciated.
  2. jcsd
  3. Jan 21, 2009 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    But once the ice melts, the H2O (at 0 C) then warms up to the final temperature as liquid water, not ice. So you have to use the specific heat of water for that stage.

  4. Jan 22, 2009 #3
    oh I get it now, thanks for the help.
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