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Simple Entropy

  1. Nov 20, 2008 #1

    pmg

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    1. The problem statement, all variables and given/known data
    Suppose that you want to freeze 0.7 kg of water for a party, and only have 12 minutes to do it. The temperature inside the refrigeration unit is 273 K, and the temperature outside is Thot = 309 K. Since you are a brilliant UW student, assume that your refrigerator has the maximum possible efficiency. Assume also that the initial temperature of the water is 273 K. The Latent Heat of Fusion of water is 3.33×105 J/kg.

    What is the change in entropy in the water per second in J/K s?

    2. Relevant equations
    change in entropy (S) = energy absorbed or expelled/constant temperature

    3. The attempt at a solution

    I have found that the Qhot/time =366.44J/sec
    Qcold/time =323.75J/sec

    By my understanding the delta S should = .7kg * 3.33E5 J/kg / (273K*720sec)
    But 1.18 J/K*sec is not an accepted answer.
    What am I doing wrong?
     
  2. jcsd
  3. Nov 20, 2008 #2
    Try restating your relevant equation; entropy is not energy. Do this and the problem is straightforward.
     
  4. Nov 20, 2008 #3
    Try showing us your intermediate steps. How did you get your Qs.

    Also, your relevant equation looks okay to me (unless I am missing something). Since, [itex]S_2-S_1=\int_1^2\frac{\delta Q}{T}[/itex]
    Assuming constant T gives [itex]\Delta S=Q_{12}/T[/itex]
     
  5. Nov 20, 2008 #4

    Andrew Mason

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    Science Advisor
    Homework Helper

    The entropy calculation does not involve mass. The mass determines the amount of heat that must be removed in order to freeze the water. You have correctly determined the rate of heat flow out of the water. Since temperature is constant it is a simple calculation to determine the change in entropy of the water per second.

    AM
     
  6. Nov 20, 2008 #5

    pmg

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    Thanks fellas,
    By taking the Latent heat of fusion times the weight of water divided by the absolute temp and the time (3.33x10^5*.7kg/(273K*720sec) I was finding the right answer. I just needed to remember that by removing heat from the water, the entropy of the system was actually decreasing--and exothermic process. Making the anwser -1.18J/K*sec.
    Thanks again.
     
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