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Homework Help: I need help with an entropy integral!

  1. Nov 6, 2005 #1
    I know the final and initial temperatures of .259kg of aluminum. Specific heat of almuninum is 900 J/(kg*K).

    The aluminum is cooled from 373K to 333.45K. I need to find the entropy change. This is what I did:

    let c = specific heat, m = mass, T = final temperature

    Q = cm(T - 373) <=> Q = Tcm - 373cm <=> T = Q/cm + 373

    Now that I had T in terms of Q, I subsituted it into the entropy integral:

    delta S = [integral from initial heat to final heat] (Q/cm + 373) dQ

    Initial heat lost is cm(373K-373K) = 0, final amount of heat lost is cm(333.45K-373K) = -9219 J

    My integration yielded -3.256 x 10^6 J/K. The right answer should be only around -20 j/k or so. I am WAY off, but my approach seems like it should work.

    What did I do wrong?
     
  2. jcsd
  3. Nov 7, 2005 #2

    lightgrav

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    Homework Helper

    if you integrate dQ/T , instead of T dQ , you might get closer to 25 J/K
     
  4. Nov 7, 2005 #3
    :surprised

    :rofl:
     
  5. Nov 7, 2005 #4
    The only problem is that it gives a horrible integral (partial fraction decomposition???? :cry: )

    This is much better:

    dQ =cmdT

    so integrate

    cm [integral from initial temp to final temp] dT/T

    Ahhhhhhh... How much nicer. :biggrin:
     
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