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Thermodynamics - Entropy/Temperature question

  1. Sep 25, 2012 #1
    http://imageshack.us/a/img402/3349/asdasdamx.jpg [Broken]

    The first part of the question was actually to find the entropy, but it wasn't so hard so I put the answer at the bottom of the picture.

    I'm having trouble understanding what part b) is asking / what to do for it.

    dU= dQ - dW = dQ -m dB

    Adiabatically dQ = 0

    dU = -m dB

    So doesn't that mean that the temperature will stay the same if B is decreased slowly?

    Thanks - greg
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 25, 2012 #2
    I just realised that the temperature could change due to the work, but how can you find out how it changes?

    The work term is dW = - m dB = dU since dQ =0

    Integrating it gives ...T/.. log(cosh(..B/T..) where the ...'s are just the other constants,

    But how can you know how the temperature is actually changing?

    Is it correct to do this

    U = -W

    mB = -...T/.. log(cosh(..B/T..)

    ...tanh(..B/T..) B= -..T/.. log(cosh(..B/T..) so B is almost directly proportional to T

    So T decreases as B decreases?
     
    Last edited: Sep 25, 2012
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