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JEE Main 2026 · 5 Apr · Shift 2, Physics question 47

The heat extracted out of xx gram of water initially at 50 °C to cool it down to 0 °C is sufficient to evaporate (1000−x)(1000-x) gram of water also initially at 50 °C. The value of xx (closest integer ) is ________.

(Take latent heat of water 2256 kJ/kg. K, specific heat capacity of water 4200 J/kg. K)

Answer and worked solution

Official answer: 922

Heat balance (in kg).
x(4200)(50)=(1000−x)[4200(50)+2256×103].x(4200)(50)=(1000-x)\left[4200(50)+2256\times10^3\right].
That is, 210000x=2466000(1000−x)210000x=2466000(1000-x), so x=2466×1062676×103≈921.5x=\dfrac{2466\times10^6}{2676\times10^3}\approx921.5.

Answer: 922