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Thermodynamics - heat flow of a rubber band

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a rubber band for which the tension, [itex]f[/itex], as a function of temperature [itex]T[/itex] and length [itex]L[/itex] is [itex]f = \kappa T (L+\gamma L^2)[/itex], where [itex]\kappa[/itex] and [itex]\gamma[/itex] are positive constants. Determine the heat flow between it and its surroundings when the rubber band is stretched reversibly and isothermically from length [itex]L_1[/itex] to length [itex]L_2[/itex].

    2. Relevant equations
    [tex]dE = TdS + fdL + \mu dN[/tex]

    3. The attempt at a solution
    I think the question is asking to find [itex]\left( \frac{\partial E}{\partial L} \right)_{T,N}[/itex] and integrate that. I'm not sure how to get this quantity, though, since I don't know what entropy is. Can I use the Helmholtz free energy when calculating heat transfer? I don't think so, since they are not equal...

    Thank you for your help!
  2. jcsd
  3. Sep 19, 2010 #2


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    It's not [itex]\left(\frac{\partial E}{\partial L}\right)_{T,N}[/itex], since some of the energy change dE comes from strain energy, and you wouldn't include that in the heat flow. So you only want to integrate part of dE; which part is it? (Take a look at your equation.)

    (Then I expect you'll get into the fun world of Maxwell relations.)
  4. Sep 19, 2010 #3
    As I understand it, then, the heat flow should be [itex]\int TdS[/itex], since this is the heat part of the equation. So then should I be integrating:

    [tex]T\int^{L_2}_{L_1} \left( \frac{\partial S}{\partial L}} \right)_{T,N} dL[/tex]

    Using a Maxwell relation as you said, I see that [itex]-\left( \frac{\partial S}{\partial L} \right)_{N,T} = \left( \frac{\partial f}{\partial T} \right)_{N,L}[/itex], and then the integral is straightforward.

    I am just unsure if that integral above is correct, and if it is, what is the basis on getting that from [itex]\int TdS[/itex] in this context.

    Thanks! I appreciate your help.
  5. Sep 19, 2010 #4


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    Looks good to me!
  6. Sep 19, 2010 #5
    I still am a little confused how to get to that integral. If S is a function of E, L, and N, doesn't that mean then that:

    [tex]\left( dS \right)_{T,N} = \left( \frac{\partial S}{\partial E} \right)_{L,N} dE + \left( \frac{\partial S}{\partial L} \right)_{E,N} dL[/tex]

    But instead I wrote [itex]\left( \frac{\partial S}{\partial L} \right)_{T,N} dL[/itex]. How does this match with the above equation?

    Thanks again. I'm probably missing something pretty fundamental :)
  7. Sep 19, 2010 #6


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    Let's drop N, since it's constant here. S is a function of two independent variables. We could write

    [tex]dS=\left(\frac{\partial S}{\partial E}\right)_LdE+\left(\frac{\partial S}{\partial L}\right)_EdL[/tex]


    [tex]dS=\left(\frac{\partial S}{\partial T}\right)_LdT+\left(\frac{\partial S}{\partial L}\right)_TdL[/tex]

    The first isn't of much use; the second simplifies into the integrand above. Does this make sense?
  8. Sep 19, 2010 #7
    I think I get it now, thank you. I didn't realize that one of the independent variables could be the conjugate (T).

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