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Dissipation and Entropy

  1. Oct 4, 2012 #1
    Hello there,

    I have a question on the dissipation and entropy.

    Let us consider a newtonian dampener with viscosity coefficient η, pulled at a fixed rate e', immersed in an infinite bath at temperature T.
    The mechanical work input in time dt is then dW = ηe'*e'dt, and is all dissipated into heat.
    The bath will see an increase of its entropy of -dW/T.
    And the dampener? Some might argue that it will exchange heat with the bath, but its entropy gain cannot be opposite and equal the entropy gain of the bath, as the second law prescribes the entropy for the overall dissipative sistem has to increase.
    So how do I compute the total change in entropy? Where is the "additional" entropy coming from, to satisfy the second law?
    I am so puzzled, hop somebody can relieve me!

    Thanks
     
  2. jcsd
  3. Oct 4, 2012 #2

    DrDu

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    Science Advisor

    As long as the dampener's temperature does not increase, it's entropy remains constant. The overall increase of entropy is entirely due to the increase of entropy of the bath.
     
  4. Oct 4, 2012 #3
    Please note there is no universal "conservation of entropy law" as there is with energy.

    Entropy can increase because processes rearrange the distribution of energy, although the energy remains constant.
     
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