Question

800 calories of heat is required to raise the temperature of 0.080 kg of a liquid from 10°C to 100°C. Find its specific heat capacity (a) in calories (b) in joules.

Answer 1

(a) Rise in temperature θ_R = (100 − 10) = 90°C

∴ Heat absorbed = mcθ_R

800 cal = 0.080 kg × c × 90°С

\[c =\frac{800}{0.080 \times 90}\]

= 111.1 calkg⁻¹°C⁻¹

(b) Specific heat in joules = 111.1 × 4.2

= 466.67 Jkg⁻¹°C⁻¹