What Voltage Is Required to Melt Lead with a Capacitor?

[Touchme]

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

 

[Kurdt]

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.

 

[m2003]

Your solution appears to be correct. However, it is important to note that this is an idealized scenario and does not take into account any losses or inefficiencies in the system. In a real-life situation, the voltage required may be slightly higher due to these factors. Additionally, it would be important to consider the capacitance and voltage ratings of the actual capacitor being used to ensure it is capable of handling the calculated voltage.