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Calculate the change in entropy of the system

  1. Jun 3, 2005 #1
    An aluminum can, with negligible heat capacity, is filled with 485 g of water at 0°C and then is brought into thermal contact with a similar can filled with 517 g of water at 48.5°C. Calculate the change in entropy of the system if no heat is exchanged with the surroundings.

    1) I find avg temp, which turns out to be: ~298
    2) I calc change in temp for both sides, and find Q. (deltaT)(4.184)(mass)
    3) Divide each Q (they're the same, with one being negative) by each starting temperature
    4) I come up with .163599 J/K, which is wrong

    Where am I messing up? I have 4-5 problems identical to this but can't figure any of them out!!!
     
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Jun 3, 2005 #2
    Can you show your calculations?
     
  4. Jun 3, 2005 #3

    Andrew Mason

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    Since the temperature changes as the heat is absorbed/lost, you have to integrate:

    [tex]\Delta S = \int_{T_i}^{T_f} dQ/T = cm\int_{T_i}^{T_f} dT/T = cm* ln(\frac{T_f}{T_i})[/tex]

    AM
     
  5. Jun 3, 2005 #4
    Thanks so much, I've been very frustrated with how to do those and couldn't find the answer ANYWHERE. Big thanks!!
     
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