Question

A calorimeter contains $0.2kg$ of water at ${30}^{\circ }C$. $0.1kg$ of water at ${60}^{\circ }C$ is added to it, the mixture is well stirred and the resulting temperature is found to be ${35}^{\circ }C$. The thermal capacity of the calorimeter is ( specific heat of water $=4200Jk{g}^{-1}{K}^{-1}$

• A. $6300J{K}^{-1}$
• B. $1260J{K}^{-1}$
• C. $4200J{K}^{-1}$
• D. $6000J{K}^{-1}$

Answer 1

The correct option is B $1260J{K}^{-1}$

Let $X$ be the thermal capacity of calorimeter.
Specific heat of water $c=4200Jk{g}^{-1}{K}^{-1}$
Heat gained or heat lost by a body $=mc\Delta T$
where, $m$ is the mass of the body, $c$ is the specific heat of body and $\Delta T$ is the change in temperature.

Heat lost by $0.1kg$ of water $=$ Heat gained by water in calorimeter $+$ Heat gained by calorimeter
$(mc\Delta T{)}_{\text{hot water}}=(mc\Delta T{)}_{\text{cold water}}+(mc\Delta T{)}_{\text{beaker}}$
where, $m$ is the mass of the body, $c$ is the specific heat of body and $\Delta T$ is the change in temperature of body.

or, $0.1\times 4200\times (60-35)=0.2\times 4200\times (35-30)+X(35-30)$

or, $10500=4200+5X$
or, $X=1260J{K}^{-1}$