An amount of ice of mass 10^-3 kg and temperature -10^° C is transformed to vapour of temperature 110^° C by applying heat. The total amount of work…

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