An amount of ice of mass $10^{-3} \mathrm{~kg}$ and temperature $-10^{\circ} \mathrm{C}$ is transfor

Question

An amount of ice of mass $10^{-3} \mathrm{~kg}$ and temperature $-10^{\circ} \mathrm{C}$ is transformed to vapour of temperature $110^{\circ} \mathrm{C}$ by applying heat. The total amount of work required for this conversion is,\ (Take, specific heat of ice $=2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$, specific heat of water $=4180 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$, specific heat of steam $=1920 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$, Latent heat of ice $=3.35 	imes 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}$ and Latent heat of steam $=2.25 	imes 10^{6} \mathrm{~J} \mathrm{~kg}^{-1}$ )\

Professor PC Thomas Classes · Answer

(a) : Required amount of work, $Q$ is\ $=m_{i} s_{i} \Delta t+m_{i} L_{i}+m_{w} L_{w} \Delta t+m_{s} L_{s}+m_{s} s_{s} \Delta t$\ As, $m_{i}=m_{w}=m_{s}=m_{s}$,\ $Q=10^{-3} 	imes 10 	imes 2100+10^{-3} 	imes 3.35 	imes 10^{5}+$\ $10^{-3} 	imes 4180 	imes 100+10^{-3} 	imes 2.25 	imes 10^{6}+10^{-3}$\ $	imes 1920 	imes 10$\ $\Rightarrow \mathrm{Q}=(21+335+418+2250+19.2)$\ $\Rightarrow Q=3043.2 \mathrm{~J}$