Capacitor Question

  • Thread starter Touchme
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  • #1
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Homework Statement



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.


Homework Equations



Energy stored = (1/2)C(V)^2
Q=mCT
Q=mL

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:

Answers and Replies

  • #2
Kurdt
Staff Emeritus
Science Advisor
Gold Member
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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:

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