Question

Heat energy is given to 100 g of water, such that its temperature rises by 10 K. When the same heat energy is given to a liquid L of mass 50 g its temperature rises by 50 K. Calculate (a) heat energy given to water (b) the specific heat capacity of liquid L.

[Take sp. heat capacity of water = 4200 JKg⁻¹K⁻¹]

Answer 1

(a) For water, convert mass to kg:

m = 100 g = 0.1 kg

Using H = mcΔT

= 0.1 × 4200 × 10

= 4200 J

(b) The same heat energy is given to liquid L, so H = 4200 J.

m = 50 g = 0.05 kg

ΔT = 50 K

Using $$c = \dfrac{H}{m\Delta T}$$

$$= \frac{4200}{0.05 \times 50}$$

$$= \frac{4200}{2.5}$$

= 1680 Jkg⁻¹K⁻¹