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Change in entropy of a capacitor

  1. Nov 16, 2015 #1
    1. The problem statement, all variables and given/known data

    What is the entropy change of the universe as a result of the following processes:

    (a) a 1μF capacitor is connected to a 100V electrochemical cell at 0°C,

    (b) the same capacitor after being charged to 100V is discharged through a resistor kept at 0°C?

    2. Relevant equations

    dS = δQ/T (for reversible processes)
    dS > δQ/T (for irreversible processes)

    Capacitor energy = ½*C*V^2

    3. The attempt at a solution

    The main thing I'm stuck on is what exactly is happening at (a). I'm not quite sure where the entropy change could come from, or what could cause it. It seems to me that there isn't enough information - are they asking for the entropy change at the instant the capacitor is connected, in which case there's been no current flow, and so has anything actually happened? Are we meant to consider the work done in physically connecting the capacitor? (I would doubt so.) If it is at the instant when they are connected, before any current has flowed, then either nothing has happened, or it's fully reversible (both of which would give ΔS = 0), or we have no idea what the entropy change is because it's to do with the work done getting it there.

    As for (b), if we say the capacitor energy is equal to the heat, then ΔS =(C*V^2)/(2*T) = 1.83x10^(-5), but I'm not sure if this is for just the capacitor or if this is the change in entropy of the universe? I'm not sure exactly what angle to approach this from, but that's my main guess at how I'd go forward, but it might be wrong.

    I'd really appreciate any thoughts on part (a) of the question - if I've missed anything especially, and also maybe some hints on how to proceed on part (b) (and if I'm going in the right direction or not). Can't seem to find examples of similar questions online either! (If you know of any, please link me!)

    Thanks!
     
  2. jcsd
  3. Nov 21, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. May 29, 2016 #3
    For part (b) the energy change through the resistor as your capacitor reduced its voltage from ##V## to ##V-dV## is:

    $$ dQ_{rev} = \frac{1}{2}CV^2 - \frac{1}{2}C(V-dV)^2 $$
    $$ dQ_{rev} = \frac{1}{2}CV^2 - \frac{1}{2}CV^2 - CVdV - \frac{1}{2}(dV)^2$$

    To first order in ##dV##:

    $$dQ_{rev} = -CVdV$$

    Using the second law for a reversible process:

    $$\Delta S_{tot} = \int \frac{dQ_{rev}}{T}$$
    $$\Delta S_{tot} = \int_{V_0}^{0} \frac{-CVdV}{T} = \frac{C}{2T}(V_0^2)$$

    Plug the numbers in to evaluate.
     
  5. Jun 16, 2016 #4

    rude man

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    I think also that there's not enough info to do this numerically. I think you need the heat capacity of the battery CZ at constant charge, then T dS = CZ dT IF the battery emf is not a function of temperature. To get CZ I think you need to know the valence number of the cell and the Faraday constant (96,500 coulombs). Bit rich for me I'm afraid.
    Since this is happening at constant temperature, total heat trasferred = initial capacitor stored energy and so ΔS = ΔQ/T.
     
  6. Jun 20, 2016 #5

    David Lewis

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    I assume they are asking for entropy change after the cap voltage has reached 100V. I believe half the energy extracted from the cell ends up stored in the electric field of the capacitor. The other half is dissipated. You won't get it back when you discharge the cap through the resistor.
     
  7. Jun 20, 2016 #6

    rude man

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    I would agree.
     
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