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 $=2.25 \times 10^6 \mathrm{Jkg}^{-1}$)