1 ~g of steam is sent into 1 ~g of ice. At thermal equilibrium, the...

Heat required to melt $1 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ to water at $0^{\circ} \mathrm{C}=1 \times 80 \mathrm{cal}$. Heat required to raise temperature of $1 \mathrm{~g}$ of water from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}=1 \times 1 \times 100=100 \mathrm{cal}$. Total heat required for maximum temperature of $100^{\circ} \mathrm{C}=80+100=180 \mathrm{cal}$. As one gram of steam gives 540 cal of heat when it is converted to water at $100^{\circ} \mathrm{C}$, therefore, temperature of the mixture $=100^{\circ} \mathrm{C}$.