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Thermodynamic Entropy Clarification

  1. Dec 17, 2008 #1
    1. The problem statement, all variables and given/known data
    We mix 0.6 kg of water at a temperature of 25°C with 0.8 kg of alcohol at 30.5°C. They
    come to equilibrium. (Here's some useful data: The specific heat capacity for water is 1
    kcal/kg°C = 4186 J/kg°C, and for alcohol it is 0.58 kcal/kg°C = 2430 J/kg°C

    What is the change in entropy (in J/K) of the system when the two are mixed?

    2. Relevant equations
    ∆S = ∆Sw + ∆Sa = Mw*Cw*ln(Tf /TW) + Ma*Ca*ln(Tf /Ta)


    3. The attempt at a solution
    I've calculated the final temperature to be 300.55 K or 27.4 C

    so now the equation looks like:
    .6*4186*Ln(Tf/Tw) + .6*2430*Ln(Tf/Ta) = dS

    The problem is, do I use Kelvin or Celsius for the entropy calculation? I get two totally different values if I do either one... .0794 with K or 21.87 with C.
     
  2. jcsd
  3. Dec 17, 2008 #2

    rl.bhat

    User Avatar
    Homework Helper

    You have to use Kelvin.
     
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