1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20 C to 40 C. Calculate the specific heat capacity of lead.

Calorimetry Numerical Question: 5

1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20^oC to 40^oC. Calculate the specific heat capacity of lead.

Solution:

Given

Heat energy supplied = 1300 J

Mass of lead = 0.5 kg

Change in temperature = (40 – 20)⁰ C = 20⁰ C = 20 K

Specific heat capacity of lead

C = △Q / [m△T]

C = 1300 /[0.5 × 20]

C = 130 J kg-1 K-1