JEE Main 22 Jan 2025 Shift 1 Paper

An amount of ice of mass 10−3‌kg and temperature −10∘C is transformed to vapour of temperature 110∘C by applying heat. The total amount of work required for this conversion is,
(Take, specific heat of ice =2100Jkg−1K−1, specific heat of water 4180Jkg−1K−1, specific heat of steam = 1920Jkg−1K−1, Latent heat of ice =3.35×105Jkg−1 and Latent heat of steam =2.25×106Jkg−1 )