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Heat and Entropy Change HELPPPP

  1. Jan 16, 2009 #1
    Heat and Entropy Change HELPPPP!!

    Calculate change in entropy of the universe for:
    1] A can of coke, at initial temperature of 5 degrees C, with heat capacity 2500JK-1, is allowed to reach thermal equilibrium with its surroundings at 40 degrees C

    2] Two Cans of coke at 5 degrees C and 40 degrees C are placed in thermal contact with each other inside a cool bag where they are thermally insulated from their surroundings and allowed to equilibrate

    3]One of the above coke cans at 40 degrees C weighing 250g is dropped from a height of 50m into a shallow pond of negligible depth which is also at 40 degrees C


    Attempt

    1] using delta S = Cpln(Tf/Ti)
    giving 2500xln(40/5)
    =5198.6JK-1

    ii]
    using Tf=(C1T1+C2T2)/C1C2

    [2500x40]+[2500x5]/2500+2500
    =22.5 degrees C

    therefore delta S = C1ln(Tf/T1)+C2ln(Tf/T2)


    =2500ln(22.5/40)+2500ln(22.5/5)
    =2321.8

    so far so gd?

    for iii] do i use delta S = P.E / temp of surrounding ?
     
    Last edited: Jan 17, 2009
  2. jcsd
  3. Jan 18, 2009 #2
    Re: Heat and Entropy Change HELPPPP!!

    For 1), you need to change the tempratures to kelvin units.

    For 3), if I understand the question correctly, you use the connection between energy and heat transfer, Q=U+W, U=mgh, the work that being done by the bodies, W, not sure how to get it.
    Anyway, dQ=SdT+TdS, there isn't a change in the tempratures, and the change is entropy is what you're after, so dQ=TdS, dS=dQ/T, find the change in the heat transfer and I believe your'e done, not sure though.
     
  4. Jan 18, 2009 #3
    Re: Heat and Entropy Change HELPPPP!!

    cheers :)
     
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