Heat And Thermodynamics Ques 30
- In an insulated vessel, $0.05 kg$ steam at $373 K$ and $0.45 kg$ of ice at $253 K$ are mixed. Find the final temperature of the mixture (in kelvin).
$(2006,6 M)$
$$ \text { Given, } \begin{aligned} L _{\text {fusion }} & =80 cal / g=336 J / g, \\ L _{\text {vaporisation }} & =540 cal / g=2268 J / g, \\ S _{\text {ice }} & =2100 J / kg, K=0.5 cal / g-K \\ \text { and } \quad S _{\text {water }} & =4200 J / kg, K=1 cal / g-K . \end{aligned} $$
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Solution:
Formula:
- $0.05 kg$ steam at $373 K \xrightarrow{Q _1} 0.05 kg$ water at $373 K$ $0.05 kg$ water at $373 K \xrightarrow{Q _2} 0.05 kg$ water at $273 K$ $0.45 kg$ ice at $253 K \xrightarrow{Q _3} 0.45 kg$ ice at $273 K$
$0.45 kg$ ice at $273 K \xrightarrow{Q _4} 0.45 kg$ water at $273 K$
$Q _1=(50)(540)=27000 cal=27 kcal$
$Q _2=(50)(1)(100)=5000 cal=5 kcal$
$Q _3=(450)(0.5)(20)=4500 cal=4.5 kcal$
$Q _4=(450)(80)=36000 cal=36 kcal$
Now, since $Q _1+Q _2>Q _3$ but $Q _1+Q _2<Q _3+Q _4$ ice will come to $273 K$ from $253 K$, but whole ice will not melt. Therefore, temperature of the mixture is $273 K$.
BETA
