Calorimetry Numerical **Question: 5**

**1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20 ^{o}C to 40^{o}C. 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 = 20^{0} C = 20 K

Specific heat capacity of lead

C = △Q / [m△T]

C = 1300 /[0.5 × 20]

C = 130 J kg-1 K-1