help me physics?

Question

The energy stored in a 51.0 µF capacitor is used to melt a 4.00 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.

Answer 1

The energy stored in a capacitor is given as:
E = 1/2 * C * V^2
Where E is the energy, C is the capacitance, and V is the voltage on the capacitor.

The energy needed to melt something is given as:
Q = mc(delta T) + mL
Where m is the mass, c is the specific heat, (delta T) is the change in temperature the solid experiences, and L is the latent heat of fusion.

We are given that the mass of the Lead is: 4.00 mg = 4 E-6 kg
We are told that the specific heat of Lead is: 128 J/kg°C and its latent heat of fusion is 24.5 kJ/kg = 24500 J/kg.
The change in temperature the Lead experiences is the difference in its melting point and its initial temperature = 327.3°C - 20.0°C = 307.3°C.
From this information, we can now find the heat energy (Q) required to melt it.

Q = (4 E-6 kg)*(128 J/kg°C)(307.3°C) + (4 E-6 kg)*(24500 J/kg)
Q = .1573 Joules + .098 Joules
Q = .2553 Joules

The energy required to melt the Lead is the minimum energy required stored inside the capacitor.

We know the capacitance of the capacitor = 51.0 µF = 51 E-6 F.
We just found the minimum energy needed to be stored inside.

E = 1/2 C * V^2
(.2553 Joules) = 1/2 * (51 E-6 F) * V^2
Solving for V,
V^2 = 2(.2553 Joules) / (51 E-6 F)
V^2 = 10012 volts^2
V = sqrt(10012 volts^2)
V = 100 volts