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Find the entropy for the process

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Supercooled water is water that is liquid and yet BENEATH the freezing point.
    a) A sample of 131 g of supercooled liquid water freezes to solid ice at a temperature of -8.00 ° C. Using the following,
    Cp,ice = 38.09 J/molK
    Cp,liquid = 74.539 J/molK
    fusH° (at T=0 ° C)=6.01 kJ/mol,
    calculate S° for the process. Hint: you must set this up in steps! Report your answer to 3 significant figures.

    2. Relevant equations

    delta s = CpLn(T2/T1)
    delta Sfus = delta Hfus/T

    3. The attempt at a solution

    I converted the fusH 6.01kj/mol(131g/18g/mol)=.043 joules
    I know Delta S has to be an addition of the steps of freezing the water but am not sure how to get there.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 9, 2012 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    To determine the change in entropy, you have to find a reversible path between the initial and final states and then calculate the integral of dQ/T for that path. Supercooled water freezing at -8C is not reversible. Water freezing at 0C is reversible. So you have to get the supercooled water from -8C to 0C reversibly, let it freeze reversibly, and then cool the ice back to -8C. The change in entropy is the intergral of dQ/T along that path.

    AM
     
  4. Oct 9, 2012 #3
    Thank you so much for the clarification. It makes more sense now to view it in that process.
     
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