In an insulated vessel, 0.05 ~kg steam at 373 ~K and 0.45 ~kg of...

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}

Correct answer

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 \gt Q_3

but

Q_1+Q_2 \lt 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}$.