JEE Main 2025 Physics question
An amount of ice of mass 10⁻³ kg and temperature –10°C is transformed to vapour of temperature 110° by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = 2100 Jkg⁻¹K⁻¹, specific heat of water = 4180 Jkg⁻¹K⁻¹, specific heat of steam = 1920 Jkg⁻¹K⁻¹, Latent heat of ice = 3.35 × 10⁵ Jkg⁻¹ and Latent heat of steam = 2.25 × 10⁶ Jkg⁻¹)
Answer: (2)
Solution
Sum of all heat stages: 21+335+418+2250+19.2 = 3043.2 J.