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Capacitor Question

  1. Feb 6, 2007 #1
    1. The problem statement, all variables and given/known data

    The energy stored in a 55.0 µF capacitor is used to melt a 7.00 mg sample of lead. To what voltage must the capacitor be initially charged, assuming that the initial temperature of the lead is 20.0°C? Lead has a specific heat of 128 J/kg°C, a melting point of 327.3°C, and a latent heat of fusion of 24.5 kJ/kg.

    2. Relevant equations

    Energy stored = (1/2)C(V)^2

    3. The attempt at a solution

    Basically I found the heat energy require to melt the lead.
    Q = mCT + mL = (0.007)(128)(327.3-20.0) + (0.007)(24500)
    Q = 446.84 J

    Then I used energy stored = (1/2)C(V)^2
    446.84 = (1/2)(55e-6)(V)^2
    sq.root(893.68/(55e-6)) = V

    V = 4030.97 V

    What am I doing wrong?Help
    Last edited: Feb 6, 2007
  2. jcsd
  3. Feb 7, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    I think you've just become confused with units. The question states 7 milligrams of lead and you have used 7 grams in your calculations. Since your latent heat and specific heat are given to be used with kilograms you have to be careful.
    Last edited: Feb 7, 2007
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