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