$$ 0.05 \mathrm{~kg} \text { steam at } 373 \mathrm{~K} \xrightarrow{\mathrm{Q}_1}0.05 \mathrm{~kg} \text { water at } 373 \mathrm{~K} \text {. } $$

$$ 0.05 \mathrm{~kg} \text { water at } 373 \mathrm{~K} \xrightarrow{\mathrm{Q}_2} 0.05 \mathrm{~kg} \text { water at } 273 \mathrm{~K}$$

$$ 0.45 \mathrm{~kg} \text { ice at } 253 \mathrm{~K} \xrightarrow{\mathrm{Q}_3} 0.45 \mathrm{~kg} \text { ice at } 273 \mathrm{~K} $$

$$ 0.45 \mathrm{~kg} \text { ice at } 273 \mathrm{~K} \xrightarrow{\mathrm{Q}_4} 0.45 \mathrm{~kg} \text { water at } 273 \mathrm{~K}$$

$$ \begin{aligned} &\begin{aligned} & \Rightarrow \mathrm{Q}_1=m \mathrm{~L}=50(540)=27000 \mathrm{cal}=27 \mathrm{kcal} (\mathrm{Q}=m \mathrm{~L}) \\ & \mathrm{Q}_2=m s \Delta t=(50)(1)(100)=500 \mathrm{cal}=5 \mathrm{kcal} \\ & \mathrm{Q}_3=m s \Delta t=(4.50)(0.5)(20)=4500 \mathrm{cal}=4.5 \mathrm{kcal} \\ & \mathrm{Q}_4=(m \mathrm{~L})=450 \times 580=36000 \mathrm{cal}=36 \mathrm{kcal} \,\, [fusion]\end{aligned}\\ &\text { } \end{aligned} $$

Since, $Q_1+Q_2>Q_3$ but $Q_1+Q_2

Therefore temperature of mixture $=273 \mathrm{~K}$

(or $0^{\circ} \mathrm{C}$ )

Method II

Heat lost by steam to covert into $0^{\circ} \mathrm{C}$ water

$$ \begin{gathered} \mathrm{H}_{\mathrm{L}}=0.05 \times 540+0.05 \times 10 \times 1 \\ {\left[\mathrm{Q}=\mathrm{H}_{\mathrm{L}}=m \mathrm{~L}+m s \Delta t\right]} \\ \mathrm{H}_{\mathrm{L}}=27+5=32 \mathrm{kcal} \end{gathered} $$

Heat required by ice to change into $0^{\circ} \mathrm{C}$ water.

$$ \begin{aligned} \mathrm{H}_{\mathrm{g}} & =450(20) \frac{1}{2}+450 \times 80 \\ & =4500+36000=40500 \mathrm{cal} \\ & =40.5 \mathrm{kcal} \end{aligned} $$

$\Rightarrow$ To convert ice, we have $450 \times \frac{1}{2}=225 \mathrm{cal}$.

But we need 32000 cal.

Therefore, ice will remain as ice. Hence final temperature is $0^{\circ} \mathrm{C}$.

JEE Advanced Subjects