Physics Question from JEE Main 2025 · 22 January, Shift 1

An amount of ice of mass $10^{-3}$ kg and temperature $-10^\circ C$ is transformed to vapor of temperature $110^\circ C$ by applying heat. The total amount of work required for this conversion is,

(Take, specific heat of ice = $2100 \, \text{J kg}^{-1} \text{K}^{-1}$, specific heat of water = $4180 \, \text{J kg}^{-1} \text{K}^{-1}$, specific heat of steam = $1920 \, \text{J kg}^{-1} \text{K}^{-1}$, Latent heat of ice = $3.35 \times 10^5 \, \text{J kg}^{-1}$, Latent heat of steam = $2.25 \times 10^6 \, \text{J kg}^{-1}$)