Question

How much heat energy (in megajoules) must be applied to melt 50 kg of lead at 20°C? (Hint: The melting point of lead is 327°C, and the specific heat of lead is 0.031.)

Answer 1

We can solve the problem splitting in two steps.

1) First, we need to raise the temperature of the block of lead from
`20^(\circ)C` to its melting temperature (
`327 ^(\circ)`). The quantity of heat needed for this process is

`Q_1 = m C_s \Delta T`
where
`m=50 kg` is the mass,
`C_s = 0.031 kcal /(kg K) = 0.13 kJ/(kg K)` is the specific heat of lead, and
`\Delta T=(327-20)^(\circ)C=307 ^(\circ)C=307 K` is the variation of temperature.
Calculating:

`Q_1 = (50 kg)(0.13 kJ/(kgK))(307 K)=1995 kJ=2.00 MJ`

2) Once the block of lead has reached its melting temperature, we must add other heat to cause the complete fusion of the block. The amount of heat need in this process is

`Q_2 = m L`
where
`L=22.4 kJ/kg` is the latent heat of fusion of lead. Therefore we have

`Q_2 = (50 kg)(22.4 kJ/kg)=1120 kJ=1.12 MJ`

3) And so, the total amount of heat needed for the entire process is

`Q=Q_1+Q_2 = 2.00MJ+1.12 MJ=3.12 MJ`