Question

A liquid of mass 100 g loses heat at a rate of 200 Js⁻¹ for 1 minute. If the temperature of the liquid drops by 100°C, calculate the specific heat capacity of the liquid.

Answer 1

Given: Mass (m) = 100g

Rate of heat loss = 200 Js⁻¹

Time (t) = 1 minute = 60 s

Drop in temperature (ΔT) = 100°C

Heat lost (Q) = Rate × Time

= 200 Js⁻¹ × 60 s

= 12000 J

Using Q = mcΔT

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

$$= \frac{12000}{100 \times 100}$$

= 1.2 Jg⁻¹°C⁻¹