In an insulated vessel, \(0.05 \mathrm{~kg}\) steam at \(373 \mathrm{~K}\) and \(0.45 \mathrm{~kg}\) of ice at \(253 \mathrm{~K}\) are mixed. Find the final temperature of the mixture (in kelvin).
\( \text { Given, }L_{\text {fusion }} =80 \mathrm{cal} / \mathrm{g}=336 \mathrm{~J} / \mathrm{g},\) \( L_{\text {vaporization }}=540 \mathrm{cal} / \mathrm{g}=2268 \mathrm{~J} / \mathrm{g}, \)
\( S_{\text {ice }} =2100 \mathrm{~J} / \mathrm{kg}, K=0.5 \mathrm{cal} / \mathrm{gK} \text { and } S_{\text {water }}=\) \(4200 \mathrm{~J} / \mathrm{kg}, K=1 \mathrm{cal} / \mathrm{gK} .\)

\(0.05 \mathrm{~kg}\) steam at \(373 \mathrm{~K} \stackrel{Q_1}{\longrightarrow} 0.05 \mathrm{~kg}\) water at \(373 \mathrm{~K}\)
\(0.05 \mathrm{~kg}\) water at \(373 \mathrm{~K} \stackrel{Q_2}{\longrightarrow} 0.05\) kg water at \(273 \mathrm{~K}\)
\(0.45 \mathrm{~kg}\) ice at \(253 \mathrm{~K} \stackrel{Q_3}{\longrightarrow} 0.45 \mathrm{~kg}\) ice at \(273 \mathrm{~K}\)
\(0.45 \mathrm{~kg}\) ice at \(273 \mathrm{~K} \stackrel{Q_4}{\longrightarrow} 0.45 \mathrm{~kg}\) water at \(273 \mathrm{~K}\)
\(
\begin{aligned}
& Q_1=(50)(540)=27,000 \mathrm{cal}=27 \mathrm{kcal} \\
& Q_2=(50)(1)(100)=5000 \mathrm{cal}=5 \mathrm{kcal} \\
& Q_3=(450)(0.5)(20)=4500 \mathrm{cal}=4.5 \mathrm{kcal} \\
& Q_4=(450)(80)=36000 \mathrm{cal}=36 \mathrm{kcal}
\end{aligned}
\)
Now since \(Q_1+Q_2>Q_3\) but \(Q_1+Q_2 < Q_3+Q_4\) ice will come to \(273 \mathrm{~K}\) from \(253 \mathrm{~K}\), but whole ice will not melt. Therefore temperature of the mixture is \(273 \mathrm{~K}\).