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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

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