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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).
$$
\text { Given, } \begin{aligned}
L_{\text {fusion }} & =80 \mathrm{cal} / \mathrm{g}=336 \mathrm{~J} / \mathrm{g}, L_{\text {vaporization }}=540 \mathrm{cal} / \mathrm{g}=2268 \mathrm{~J} / \mathrm{g}, \\
S_{\text {ice }} & =2100 \mathrm{~J} / \mathrm{kg}, K=0.5 \mathrm{cal} / \mathrm{gK} \text { and } S_{\text {water }}=4200 \mathrm{~J} / \mathrm{kg}, K=1 \mathrm{cal} / \mathrm{gK} .
\end{aligned}
$$
$$
\text { Given, } \begin{aligned}
L_{\text {fusion }} & =80 \mathrm{cal} / \mathrm{g}=336 \mathrm{~J} / \mathrm{g}, L_{\text {vaporization }}=540 \mathrm{cal} / \mathrm{g}=2268 \mathrm{~J} / \mathrm{g}, \\
S_{\text {ice }} & =2100 \mathrm{~J} / \mathrm{kg}, K=0.5 \mathrm{cal} / \mathrm{gK} \text { and } S_{\text {water }}=4200 \mathrm{~J} / \mathrm{kg}, K=1 \mathrm{cal} / \mathrm{gK} .
\end{aligned}
$$
Solution:
2119 Upvotes
Verified Answer
The correct answer is:
273
$0.05 \mathrm{~kg}$ 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 $273 \mathrm{~K}$.
$0.05 \mathrm{~kg}$ 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 $273 \mathrm{~K}$.
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