An amount of ice of mass \({10}^{-3}\mathrm{kg}\) and temperature \(-10^\circ \mathrm{C}\) is transformed to vapour of t…

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

[JEE Main 2025, 22 Jan (Shift 1)]