Question

You have a 2.00 kg block of lead. Lead melts at 327.5°C. Cpb = 130.0 J/kg*C and ?Hj for lead is 2.04x10^4 J/kg . Say you start at room temperature (25.0°C). How much heat must you transfer to melt all the lead?

Answer 1

Q = 2.00kg x 130 J/kgC x 302.5C

Q = 7.87 x 10^4j

Q = 2.00kg x 2.04 x 20^4

Q= 4.08 x 10^4

Q = 7.87 x 10^4j + 4.08 x 10^4 = 1.2 x 10^5j

Step-by-step explanation:

Answer 2

`1.2\cdot 10^5 J`

Step-by-step explanation:

1) First of all, we need to calculate the heat needed to raise the temperature of the block of lead from the initial temperature (25.0 C) to the melting temperature (327.5 C). This heat is given by the equation

`Q_1 = m C_(pb) \Delta T`

where

m = 2.00 kg is the mass of the block

Cpb = 130.0 J/kg*C is the specific heat capacity of lead

`\Delta T=327.5 C-25.0 C=302.5 C` is the temperature difference

Substituting,

`Q_1 = (2.00 kg) (130 J/kgC) (302.5 C)=7.87\cdot 10^4 J`

2) Now we have to calculate the amount of heat needed to completely melt the lead, which is given by:

`Q_2 = m Hj`

where

Hj = 2.04x10^4 J/kg is the latent heat of fusion of lead

Substituting,

`Q_2 = (2.00 kg)(2.04\cdot 10^4 J/kg)=4.08\cdot 10^4 J`

3) Therefore, the total heat needed is

`Q=Q_1+Q_2=7.87\cdot 10^4 J+4.08\cdot 10^4 J=1.2 \cdot 10^5 J`