One gram of ice mixed with one gram of steam. After thermal equilibrium, the temperature of the mixt

Question

One gram of ice mixed with one gram of steam. After thermal equilibrium, the temperature of the mixture is\

Professor PC Thomas Classes · Answer

Heat given out when 1 gram steam condenses into water at\ $100^{\circ} \mathrm{C}=\mathrm{mL}=1 \mathrm{~g} 	imes 540 \frac{\mathrm{cal}}{\mathrm{g}}$\ $=540 \mathrm{cal}$\ Heat required to raise 1 g ice at $0^{\circ} \mathrm{C}$ to water $100^{\circ} \mathrm{C}=\mathrm{mL}+\mathrm{mc} \Delta 	heta$\ $=1 	imes 80+1 	imes 1 	imes 100$\ $=180 \mathrm{cal}$\ Balance heat $=540-180=360 \mathrm{cal}$\ Thus we have 360 cal of heat and 2 g water at $100^{\circ} \mathrm{C}$.\ With 360 cal of heat, the quantity of water that can be converted into steam\ $=\frac{360}{540}=\frac{2}{3} \mathrm{~g}$\ $1 \frac{1}{3} \mathrm{~g}$ water remains at $100^{\circ} \mathrm{C}$. Therefore the temperature of the mixture will be $100^{\circ} \mathrm{C}$.\