Question

A calorimeter contains a mass m1 = 250g of water. The initial temperature of the system

(calorimeter+water) is T1 = 18 ° C. A mass m2 = 300 g of water is added at the temperature

T2= 80 ° C.

1. What would be the thermal equilibrium temperature Te of the system if the heat capacity of

the calorimeter and its accessories was negligible?

2. in fact a temperature of thermal equilibrium Te = 50 ° C is measured. Determine the heat

capacity C of the calorimeter and its accessories.

Data: Specific heat of liquid water cw = 4185 J.kg-1

.K-1

; Density of water: ρ = 1000 kg.m-3.
​

Answer 1

To find the thermal equilibrium temperature Te of the system, we can use the principle of conservation of energy, which states that the heat lost by the hot water (m2) is equal to the heat gained by the cold water (m1) and the calorimeter. Mathematically,

m2 * cw * (T2 - Te) = (m1 + m2) * cw * (Te - T1)

where cw is the specific heat of liquid water.

Simplifying and solving for Te, we get:

Te = (m2 * T2 + m1 * T1) / (m1 + m2) = (0.3 kg * 80 °C + 0.25 kg * 18 °C) / (0.3 kg + 0.25 kg) ≈ 51.5 °C

Therefore, the thermal equilibrium temperature Te of the system would be approximately 51.5 °C if the heat capacity of the calorimeter and its accessories was negligible.

If the measured thermal equilibrium temperature is Te = 50 °C, we can use the same principle of conservation of energy to find the heat capacity C of the calorimeter and its accessories. Mathematically,

(m1 + m2) * cw * (Te - T1) = C * (Te - T1)

Simplifying and solving for C, we get:

C = (m1 + m2) * cw / (Te - T1) = (0.25 kg + 0.3 kg) * 4185 J/kg.K / (50 °C - 18 °C) ≈ 3195 J/K

Therefore, the heat capacity C of the calorimeter and its accessories would be approximately 3195 J/K.