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