Question

A calorimeter contains $0.5\, kg$ of water at $30^{\circ} C$. When $0.3\, kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is

• A. 0.25 kg
• B. 0.1 kg
• C. 0.2 kg
• D. 0.25 kg

Answer 1

Answer: (B)

Temperature of cold water in calorimeter,
$T_{1}=30^{\circ} C,\, m_{1}=0.5\, kg =500\, g$
Temperature of hot water, $T_{2}=60^{\circ} C$
$m_{2}=0.3\, kg =300\, g$
Resulting temperature, $T_{3}=40^{\circ} C$
Let water equivalent of the calorimeter is $W_{ gram }$
According to principle of calorimetry.
Heat lost by warm water = Heat gained by cold water + Heat gained by
the calorimeter
$\Rightarrow m_{2}\left(T_{2}-T_{3}\right)=m_{1}\left(T_{3} T_{1}\right)+W\left(T_{3}-T_{1}\right)$
$\Rightarrow 300(60-40)=500(40-30)+W(40-30)$
$\Rightarrow 300 \times 20=500 \times 10+10\, W$
$\Rightarrow 6000=5000+10\, W$
$\Rightarrow 10\, W =6000-5000$
$\Rightarrow 10\, W =1000$
$\Rightarrow W=100\, g =\frac{100}{1000}=0.1\, kg$