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 , specific heat capacity of water 4200 J/kg.K )

Physics · Thermal Properties of Matter

JEE Main 2026 — 5 April, Evening Shift — Question 21

The heat extracted out of $x$ gram of water initially at $50^{\circ}\mathrm{C}$ to cool it down to $0^{\circ}\mathrm{C}$ is sufficient to evaporate $(1000 - x)$ gram of water also initially at $50^{\circ}\mathrm{C}$ . The value of $x$ (closest integer) is. (Take latent heat of water $2256\mathrm{kJ/kg}$ , specific heat capacity of water $4200\mathrm{J/kg.K}$ )

Answer: 922

Numerical answer — enter this value.

Step-by-step solution

Heat extracted from x g: $x \times 4200 \times 50$ . Heat required to cool and evaporate (1000-x) g: $(1000-x)\times4200\times50 + (1000-x)\times2256\times10^3$ . Equating and solving gives $x \approx 922$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Physics
Chapter: Thermal Properties of Matter
Topic: Thermometry and Calorimetry