1 gm of ice at 0^o C is mixed with 1 gm of | vedclass

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(A) Step $1$: Calculate heat required to melt $1 \ gm$ of ice at $0^o C$ to water at $0^o C$: $Q_1 = m \cdot L_f = 1 \ gm 	imes 80 \ cal/gm = 80 \ ca

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(A) Step $1$: Calculate heat required to melt $1 \ gm$ of ice at $0^o C$ to water at $0^o C$: $Q_1 = m \cdot L_f = 1 \ gm 	imes 80 \ cal/gm = 80 \ cal$.Step $2$: Calculate heat released by $1 \ gm$ of steam at $100^o C$ condensing to water at $100^o C$: $Q_2 = m \cdot L_v = 1 \ gm 	imes 540 \ cal/gm = 540 \ cal$.Step $3$: Since $Q_2 > Q_1$,the ice melts completely and the resulting water at $0^o C$ is heated by the remaining heat from the steam.Step $4$: Heat available to raise the temperature of the water: $Q_{rem} = Q_2 - Q_1 = 540 - 80 = 460 \ cal$.Step $5$: Total mass of water is $2 \ gm$. Let the final temperature be $T$. The heat absorbed by $2 \ gm$ of water to reach $T$ is $Q = m \cdot c \cdot \Delta T = 2 \ gm 	imes 1 \ cal/gm^o C 	imes (T - 0) = 2T$.Step $6$: Equating heat gained and lost: $2T = 460$,which gives $T = 230^o C$. However,since the steam is at $100^o C$,the final temperature cannot exceed $100^o C$. This implies all steam condenses and the mixture reaches equilibrium at $100^o C$ with some steam remaining.

$1 \ gm$ of ice at $0^o C$ is mixed with $1 \ gm$ of steam at $100^o C$. What will be the final temperature $(^o C)$?

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