Question

A copper vessel contains 200 g of water at 24°C. When 112 g of water at 42°C is added, the resultant temperature of the water is 30°C. Calculate the thermal capacity of the calorimeter.

Answer 1

Let the thermal capacity of the copper vessel be C' J °C⁻¹.

Heat lost by hot water:

Q_lost = 112 × 4.2 × (42 − 30)

= 5644.8 J

Heat gained by cold water:

$$Q_{\text{gained by water}} = 200 \times 4.2 \times (30 - 24)$$

= 5040 J

Heat gained by the vessel:

$$Q_{\text{gained by vessel}} = C' \times (30 - 24)$$

= 6C'

From the principle of calorimetry:

5040 + 6C' = 5644.8

6C' = 604.8

C' = 100.8 J °C⁻¹

Therefore, the vessel's thermal capacity is 100.8 J °C⁻¹.