Q 11-10-077JEE MainJEE Main 2022 (25 Jun, Shift 1)Easy
A steam engine intakes $50\ \text{g}$ of steam at $100^\circ\text{C}$ per minute and cools it down to $20^\circ\text{C}$. If latent heat of vaporization of steam is $540\ \text{cal g}^{-1}$, then the heat rejected by the steam engine per minute is ______ $\times10^3\ \text{cal}$. (Given: specific heat capacity of water: $1\ \text{cal g}^{-1}\,^\circ\text{C}^{-1}$)
Numerical value type. Enter your answer.
Answer: 31
Heat released on condensing and then cooling from $100^\circ\text{C}$ to $20^\circ\text{C}$:
$$Q = 50\times540 + 50\times1\times80 = 27000 + 4000 = 31\times10^3\ \text{cal}$$
Solution by Sreeraj P, M.Sc Physics