Question

A 111.6 g sample of iron was heated from 0°C to 20°C. It absorbed 1004 J of energy. What is the specific heat capacity of iron?

Answer 1

Formula

H = m*c*ΔT

Givens

H = 1004 J

m = 111.6

ΔT = 20 - 0

ΔT = 20

c = ?

Solution

1004 = 111.6 * c * 20 Multiply the right.

1004 = 2232*c Divide to get c

1004/2232 = c Do the division

0.4498 = c

Specific heat of iron = 0.4498