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) Numerical answer type. Work it out, then open the answer.

Answer: 922. Heat from x g cooling 50 → 0°C: x × 4.2 × 50 = 210x J. Heat for (1000 - x) g: warm to 100°C and evaporate: (1000 - x)(210 + 2256) = 2466(1000 - x) J. 210x = 2466(1000 - x) ⇒ 2676x = 2466000 x ≈ 921.5 ≈ 922

The heat extracted out of x gram of water initially at 50°C to cool… | JEE Main 2026 (5 Apr, Shift 2)

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Calorimetry, Latent Heat & Mechanical Equivalent of Heat

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

Numerical answer type. Work it out, then open the answer.

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Answer: 922

Heat from $x$ g cooling $50 \to 0^{\circ}\text{C}$ : $x \times 4.2 \times 50 = 210x\text{ J}$ .
Heat for $(1000 - x)$ g: warm to $100^{\circ}\text{C}$ and evaporate: $(1000 - x)(210 + 2256) = 2466(1000 - x)\text{ J}$ .
$210x = 2466(1000 - x) \Rightarrow 2676x = 2466000$
$x \approx 921.5 \approx 922$

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