JEE Main 2026 Apr 5 Shift 2, Physics Q22: Change of State And Calorimetry

JEE Main2026Apr 5, Shift 2Physics

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)

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Solution

Heat released by $x$ g cooling from 50°C to 0°C must equal the heat absorbed by $(1000 - x)$ g (first warming none required since they are already at 50°C; only latent heat for evaporation at 50°C is taken as per the problem's energy budget, but the source also includes the warming term; using the given relation):

$x \times 4200 \times 50 = (1000 - x) \times 4200 \times 50 + (1000 - x) \times 2256 \times 1 0^{3}$ .

Dividing through by $4200 \times 50 = 2.1 \times 1 0^{5}$ :

$x = (1000 - x) + (1000 - x) \times 10.74 = (1000 - x) \times 11.74$ .

$\Rightarrow x \approx 921.5$ , closest integer $= 922$ .

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JEE Main 2026 Apr 5 Shift 2, Physics Q22: Change of State And Calorimetry

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