Question

Heat energy is given to 80 g of alcohol (sp. heat capacity 2200 Jkg⁻¹K⁻¹) when its temperature rises by 20 K. If the same heat energy is given to 200 g of mercury of specific heat capacity 140 Jkg⁻¹K⁻¹, what is the rise in temperature?

Answer 1

For alcohol:

m = 80 g = 0.08 kg

c = 2200 Jkg⁻¹K⁻¹

ΔT = 20 K

Heat energy (H) = mcΔT

= 0.08 × 2200 × 20

= 3520 J

For mercury:

Heat energy (H) = 3520 J

m = 200 g = 0.2 kg

c = 140 Jkg⁻¹K⁻¹

Rise in temperature, $$\Delta T = \dfrac{H}{mc}$$

$$= \frac{3520}{0.2 \times 140}$$

$$= \frac{3520}{28}$$

= 125.7 K