The heat extracted from x grams of water initially at 50°C to cool it to 0°C is exactly sufficient to convert (1000-x) grams of water, also initially at 50°C, completely into steam at 100°C. Take the specific heat capacity of water as 4200 J·kg⁻¹·K⁻¹ and the latent heat of vaporisation as 2256 kJ·kg⁻¹. Find x to the nearest integer in grams.
Physics · Thermodyanmics · heat and principle of calorimeter
JEE Mains 2026 5 April Evening Shift

Question

The heat extracted from x grams of water initially at 50°C to cool it to 0°C is exactly sufficient to convert (1000-x) grams of water, also initially at 50°C, completely into steam at 100°C. Take the specific heat capacity of water as 4200 J·kg⁻¹·K⁻¹ and the latent heat of vaporisation as 2256 kJ·kg⁻¹. Find x to the nearest integer in grams.

Physics · Thermodyanmics · heat and principle of calorimeter

JEE Mains 2026 5 April Evening Shift

G1 Physics · Accepted Answer

Use J·g⁻¹ units for convenience: c = 4200 J·kg⁻¹·K⁻¹ = 4.2 J·g⁻¹·K⁻¹, L = 2256 kJ·kg⁻¹ = 2256 J·g⁻¹. Heat released by cooling x grams from 50°C to 0°C: Q(released) = x(4.2)(50) = 210x J. For each gram of the remaining water, the heat required to raise its temperature from 50°C to 100°C and then vaporise it is q = 4.2(50)+2256 = 210+2256 = 2466 J·g⁻¹. Therefore, 210x = 2466(1000-x). Hence, 2676x = 2,466,000, x ≈ 921.52 g. To the nearest integer, x = 922 g. Therefore the answer is 922.