Question

0.08 kg of a substance is heated from 30°C to 130°C when 2000 calories of energy are supplied to it. Calculate the specific heat capacity of the substance in (a) calories (b) joules.

Answer 1

Given: m = 0.08 kg, Q = 2000 cal, initial temperature = 30°C, final temperature = 130°C.

Rise in temperature: ΔT = 130 − 30 = 100°C

(a) Using the heat equation:

Q = mcΔT

$$Q = mc\Delta T$$

$$c = \frac{Q}{m\Delta T}$$

$$= \frac{2000}{0.08 \times 100}$$

= 250 cal kg⁻¹°C⁻¹

Specific heat capacity in calories = 250 cal kg⁻¹°C⁻¹

(b) Since 1 cal = 4.2 J (approx):

c = 250 × 4.2

= 1050 J kg⁻¹°C⁻¹

Specific heat capacity in joules = 1050 J kg⁻¹°C⁻¹