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Capacitor heat problem (thermo), direction?

  1. Sep 22, 2008 #1
    Suppose the following system for a capacitor.

    Given: [tex]dE = TdS +\phi dq[/tex]
    Where: [tex]\phi[/tex] is the potential and = [tex]\frac{q}{C(T)}[/tex]

    Find the heat absorbed in increasing charge from 0 to q. (Capacitance is a function of temp).

    So, I get [tex] dF = \phi dq [/tex] and then [tex] F = \frac{q^2}{2C} [/tex]

    Next, I tried some Maxwell relations. and got [tex] T\frac{dS}{dq} = T\frac{d\phi}{dT} [/tex].

    That didn't get me anywhere because of the C(T) included in phi. So we are still left with a term C'(T) after taking that derivative. Any help out there?
  2. jcsd
  3. Sep 22, 2008 #2
    Oh, important note. The change in charge occurs at constant T.
  4. Sep 23, 2008 #3


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    Isn't it

    [tex]T\left(\frac{\partial S}{\partial q}\right)_{T}=-T\left(\frac{\partial \phi}{\partial T}\right)_{q}\,\mathrm{?}[/tex]

    And what would be the problem with having the temperature coefficient of capacitance in the answer?
  5. Sep 23, 2008 #4
    Mapes - I think that negative sign came out of it, but you might be right, I'll check back my work. Nothing would be wrong with it, except that it was not given in the problem, I thought my answer might have been complete except for the fact that we don't actually know C(T), just that it is a function of time.
  6. Sep 23, 2008 #5
    Yes, that negative sign should certainly be there.
  7. Sep 23, 2008 #6


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    Maybe I could share my Maxwell relation approach; you might find it useful. From


    we know that

    [tex]-S=\left(\frac{\partial F}{\partial T}\right)_q\qquad\mathrm{and}\qquad \phi=\left(\frac{\partial F}{\partial q}\right)_T[/tex]


    [tex]\frac{\partial}{\partial q}\left(\frac{\partial F}{\partial T}\right)_q=\left(\frac{\partial^2 F}{\partial T\,\partial q}\right)=\left(\frac{\partial^2 F}{\partial q\,\partial T}\right)=\frac{\partial}{\partial T}\left(\frac{\partial F}{\partial q}\right)_T\mathrm{,}[/tex]

    we have

    \left(\frac{\partial (-S)}{\partial q}\right)_{T}=\left(\frac{\partial \phi}{\partial T}\right)_{q}[/tex]

    In short, I define my potential function so that the differential variables are the ones in the denominator in the Maxwell relation. If the signs are different in the potential function, then there's a minus sign in the Maxwell relation. If they're the same, there's no minus sign.
  8. Sep 23, 2008 #7
    Yes, that's the same as I had, just to save time, I drew by analogy; q ~ V , [tex]-\phi[/tex] ~ P. I just got confused with the - sign, but when I worked it out, I did what you did above and got the same relation. I get the [tex]F=\frac{q^2}{2C}[/tex] because this operation is carried out at constant temp, so I set dT=0 in your dF equation. Perhaps that is the source of error?

    This is my first time with thermo, we pretty much skipped straight to stat-mech in undergrad, so I could be missing a very basic concept.
  9. Sep 23, 2008 #8
    For anyone following this thread, mapes was right, the answer should contain [tex]\frac{dC}{dT}[/tex]
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