Question

A liquid of mass 100 g at 120°C is poured in water at 20°C, when the final temperature recorded is 40°C. If the specific heat capacity of the liquid is 0.8 Jg⁻¹°C⁻¹, calculate the initial mass of water.

Answer 1

Let the mass of water be m g.

For the liquid:

Mass = 100 g

Specific heat capacity = 0.8 Jg⁻¹°C⁻¹

Fall in temperature = 120 − 40 = 80°C

Heat lost by the liquid = 100 × 0.8 × 80 = 6400 J

For water:

Rise in temperature = 40 − 20 = 20°C

Specific heat capacity = 4.2 Jg⁻¹°C⁻¹

Heat gained by water = m × 4.2 × 20 = 84 m J

By the principle of the method of mixtures:

84m = 6400

$$m = \frac{6400}{84}$$

= 76.19 g

Hence, the initial mass of water is 76.19 g.