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Homework Help: Thermodynamics - Entropy

  1. Jun 21, 2010 #1
    1. The problem statement, all variables and given/known data
    You make tea with 0.250 kg of 89.0 °C water and let it cool to room temperature (20.0 °C) before drinking it.

    1)Calculate the entropy change of the water while it cools.
    2)The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all the heat lost by the water goes into the air.


    2. Relevant equations
    Q = m*L_f (L_f = 3.34*10^5 J/kg)
    Q = m*c*(delta)T


    3. The attempt at a solution

    This question was posted before and this is the link
    https://www.physicsforums.com/showthread.php?t=200594

    My problem is when I tried the suggested method by Andrew, what I got was
    delta(S) = (0.25)(4,186)(342)/293
    = 1221.5J
    whereas if I used the edited solution by danni I get 244J which is given by the book.

    If danni's way is correct, then could you please explain why we have to use the latent heat equation when the question doesn't involve a change in the state of the object.

    Thank You
     
  2. jcsd
  3. Jun 21, 2010 #2
    One first thing: The dimension of S is J/K, not J! Be careful, it would cost you a lot of points :cry:
    Danni was incorrect, Andrew was the right one. The formula was right, the only thing wrong is that you (and Danny) calculated the wrong [tex]\Delta T_{tea}[/tex].
     
  4. Jun 21, 2010 #3
    I get it! (delta)T is 69K right?
     
  5. Jun 21, 2010 #4
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