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 \(=2100 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\), specific heat of water \(=4180 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\), specific heat of steam \(=1920 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}\), Latent heat of ice \(=3.35 \times 10^5 \mathrm{Jkg}^{-1}\) and Latent heat of steam \(=2.25 \times 10^6\) \(\mathrm{Jkg}^{-1}\) )

\(\begin{aligned} & \Delta \mathrm{Q}_1=\mathrm{m} \times \mathrm{S}_1 \times \Delta \mathrm{T}=10^{-3} \times 2100 \times 10=21 \mathrm{~J} \\ & \Delta \mathrm{Q}_2=\mathrm{m} \times \mathrm{L}_{\mathrm{f}}=10^{-3} \times 3.35 \times 10^5=335 \mathrm{~J} \\ & \Delta \mathrm{Q}_3=\mathrm{m} \times \mathrm{S}_{\mathrm{w}} \times \Delta \mathrm{T}=10^{-3} \times 4180 \times 100=418 \mathrm{~J} \\ & \Delta \mathrm{Q}_4=\mathrm{m} \times \mathrm{L}_{\mathrm{v}}=10^{-3} \times 2.25 \times 10^6=2250 \mathrm{~J} \\ & \Delta \mathrm{Q}_5=\mathrm{m} \times \mathrm{S}_{\mathrm{v}} \times \Delta \mathrm{T}=10^{-3} \times 1920 \times 10=19.2 \mathrm{~J} \\ & \Delta \mathrm{Q}_{\mathrm{net}}=3043.2 \mathrm{~J}\end{aligned}\)