Q 11-10-047JEE MainJEE Main 2026 (5 Apr, Shift 2)Medium
The heat extracted out of $x$ gram of water initially at $50^\circ$C to cool it down to $0^\circ$C is sufficient to evaporate $(1000 - x)$ gram of water also initially at $50^\circ$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)
Numerical value type. Enter your answer.
Answer: 922
Heat given out by $x$ g cooling $50^\circ$C → $0^\circ$C: $x \times 4.2 \times 50 = 210x$ J.
Heat needed by $(1000 - x)$ g to warm to $100^\circ$C and then evaporate (latent heat $2256$ J/g): $(1000 - x)(4.2 \times 50 + 2256) = 2466(1000 - x)$ J.
$210x = 2466(1000 - x) \Rightarrow x = \dfrac{2466000}{2676} \approx 922$.
Solution by Sreeraj P, M.Sc Physics