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
Concept. Heat given out by $x$ g cooling from $50°$C to $0°$C equals the heat needed to bring $(1000 - x)$ g from $50°$C to $100°$C and evaporate it.
Step 1. Heat released. $x \times 4.2 \times 50 = 210x$ J.
Step 2. Heat needed. Per gram: $4.2 \times 50 + 2256 = 2466$ J, so $2466(1000 - x)$ J.
Step 3. Solve.
$$210x = 2466(1000 - x) \;\Rightarrow\; x = \frac{2466000}{2676} \approx 922$$
Answer. $922$.
Solution by Sreeraj P, M.Sc Physics
