A calorimeter contains 0.5 kg of water at 30° C. When 0.3 kg of water at 60° C is added to it, the resulting temperature is found to be 40° C. The water equivalent of the calorimeter is

Question

A calorimeter contains 0.5 kg of water at 30° C. When 0.3 kg of water at 60° C is added to it, the resulting temperature is found to be 40° C. The water equivalent of the calorimeter is (A) 0.25 kg (B) 0.1 kg (C) 0.2 kg (D) 0.25 kg

Tardigrade Admin · Accepted Answer

Temperature of cold water in calorimeter,

T_{1} = 3 0^{\circ} C, m_{1} = 0.5 k g = 500 g

Temperature of hot water,

T_{2} = 6 0^{\circ} C

m_{2} = 0.3 k g = 300 g

Resulting temperature,

T_{3} = 4 0^{\circ} C

Let water equivalent of the calorimeter is

W_{g r am}

According to principle of calorimetry.
Heat lost by warm water = Heat gained by cold water + Heat gained by
the calorimeter

\Rightarrow m_{2} (T_{2} - T_{3}) = m_{1} (T_{3} T_{1}) + W (T_{3} - T_{1})

\Rightarrow 300 (60 - 40) = 500 (40 - 30) + W (40 - 30)

\Rightarrow 300 \times 20 = 500 \times 10 + 10 W

\Rightarrow 6000 = 5000 + 10 W

\Rightarrow 10 W = 6000 - 5000

\Rightarrow 10 W = 1000

\Rightarrow W = 100 g = \frac{100}{1000} = 0.1 k g

Q. A calorimeter contains 0.5kg of water at 30∘C. When 0.3kg of water at 60∘C is added to it, the resulting temperature is found to be 40∘C. The water equivalent of the calorimeter is

0.25 kg

0.1 kg

0.2 kg

0.25 kg

Q. A calorimeter contains $0.5 k g$ of water at $3 0^{\circ} C$ . When $0.3 k g$ of water at $6 0^{\circ} C$ is added to it, the resulting temperature is found to be $4 0^{\circ} C$ . The water equivalent of the calorimeter is