A calorimeter contains \(0.5 \mathrm{~kg}\) of water at \(30^{\circ} \mathrm{C}\). When \(0.3 \mathrm{~kg}\)…
A calorimeter contains \(0.5 \mathrm{~kg}\) of water at \(30^{\circ} \mathrm{C}\). When \(0.3 \mathrm{~kg}\) of water at \(60^{\circ} \mathrm{C}\) is added to it, the resulting temperature is found to be \(40^{\circ} \mathrm{C}\). The water equivalent of the calorimeter is
\(0.25 \mathrm{~kg}\)
\(0.1 \mathrm{~kg}\)
\(0.2 \mathrm{~kg}\)
\(0.25 \mathrm{~kg}\)
Solution
Temperature of cold water in calorimeter,
\(T_1=30^{\circ} \mathrm{C}, m_1=0.5 \mathrm{~kg}=500 \mathrm{~g}\)
Temperature of hot water, \(T_2=60^{\circ} \mathrm{C}\),
\(m_2=0.3 \mathrm{~kg}=300 \mathrm{~g}\)
Resulting temperature, \(T_3=40^{\circ} \mathrm{C}\)
Let water equivalent of the calorimeter is \(W_{\text {gram }}\).
According to principle of calorimetry.
Heat lost by warm water \(=\) Heat gained by cold water + Heat gained by the calorimeter
\(\begin{aligned}
& \Rightarrow \quad m_2\left(T_2-T_3\right)=m_1\left(T_3-T_1\right)+W\left(T_3-T_1\right) \\
& \Rightarrow \quad 300(60-40)=500(40-30)+W(40-30) \\
& \Rightarrow \quad 300 \times 20=500 \times 10+10 W \\
& \Rightarrow \quad 6000=5000+10 W \\
& \Rightarrow \quad 10 W=6000-5000 \Rightarrow 10 W=1000 \\
& \Rightarrow \quad W=100 \mathrm{~g}=\frac{100}{1000}=0.1 \mathrm{~kg}
\end{aligned}\)