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Thermodynamics: Lossy Capacitor

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    A capacitor is charged to Q0 and then isolated from its surroundings thermally, electrically, and mechanically. Over time the charge leaks across the dielectric filling and is neutralized. The Free Energy is given

    F=Q2/2C0 *T/T0 + κT(1-lnT/T0)

    κ, T0, and C0 are constants.

    What is the final temperature T2 in terms of the inital temp. T1? What is the change in entropy?

    2. Relevant equations

    F=U-TS
    S=-(dF/dT)Q
    1st and 2nd Law of Thermodynamics, etc.

    3. The attempt at a solution

    Calculate S=-(dF/dT)Q
    S=-Q2/(2C * T0) + κ ln(T/T0)

    Observe that U (internal energy) is constant in the initial and final states. Equate U1=F1+T1S1 and U2=F2+T2S2

    U1=Q2/2C *T1/T0 + κT1(1-lnT1/T0) + T1 [-Q2/(2C * T0) + κ ln(T/T0)]=κT1

    U2=κT2(1-lnT2/T0) + T2 [κ ln(T/T0)]=κT2

    U1=U2 -> κT1=κT2 -> T1 = T2

    Basically this implies that there is no energy at all stored in the capacitor, and that when the charge dissipates it does not increase the temperature of the dielectric. This doesn't make sense. Where have I gone wrong? I would appreciate your help greatly.

    PS. I calculated the entropy using this method but it seems pointless to post it since this answer is clearly wrong.
     
  2. jcsd
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