Question

A calorimeter contains $0.2 kg$ of water at $30{ }^{\circ} C$, in which $0.1 kg$ of water at $60{ }^{\circ} C$ is added. The mixture is well stirred and the resulting temperature is found to be $35{ }^{\circ} C$. The thermal capacity of the calorimeter is

• A. $6300JK^{- 1}$
• B. $1260JK^{- 1}$
• C. $4200JK^{- 1}$
• D. $3200JK^{- 1}$

Answer 1

Answer: (B)

Let $X$ be the thermal capacity of the calorimeter and specific heat of water = $4200Jkg^{- 1}K^{- 1}$ .
Heat lost by $0.1kg$ of water = Heat gained by water in calorimeter + Heat gained by calorimeter
$\Rightarrow $ $0.1\times 4200\times \left(\right.60-35\left.\right)$
$=0.2\times 4200\times \left(\right.35-30\left.\right)+X\left(\right.35-30\left.\right)$
$10500=4200+5X$ $\Rightarrow $ $X=1260JK^{- 1}$