Question

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

• A. $6300\text{J/K}$
• B. $1260\text{J/K}$
• C. $4200\text{J/K}$
• D. None of these

Answer 1

The correct option is B $1260\text{J/K}$

Let $x\text{J/K}$ be the thermal capacity of calorimeter.
The specific heat of water $=4200\text{J/kg-K}$.

$\Rightarrow$Heat gained by calorimeter = Thermal capacity $\times$ rise in temperature
$\therefore$Heat gained by calorimeter$=x(35-30)=5x$
$\Rightarrow \text{Heat gained by water}=mc\Delta T$
$\therefore \text{Heat gained by water}=0.2\times 4200\times (35-30)=4200\text{J}$

$\Rightarrow \text{Total heat gained}=5x+4200\text{J}$
Heat lost by $0.1\text{kg}$ of hot water when added to calorimeter
$\Rightarrow Q=mc\Delta T=0.1\times 4200\times (60-35)$
$\therefore Q=10500\text{J}$
According to principle of calorimetry,
$\text{Total heat gained}=\text{Total heat lost}$
$\Rightarrow 5x+4200=10500$
$\therefore x=1260\text{J/K}$