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Thermodynamics - Entropy/Temperature question

  • Thread starter Greger
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  • #1
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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
 
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Answers and Replies

  • #2
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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?
 
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