JEE Main 2026 Physics question
The heat extracted out of x gram of water initially at 50°C to cool it down to 0°C is sufficient to evaporate (1000−x) gram of water also initially at 50°C. The value of x (closest integer) is . (Take latent heat of water 2256 kJ/kg.K, specific heat capacity of water 4200 J/kg.K)
Answer: 922
Solution
210x = (1000−x)(210+2256) [heating to 100°C plus vaporising], giving x≈921.5≈922.