Q 11-10-105JEE MainJEE Main 2020 (3 Sep, Shift 2)Medium
A calorimeter of water equivalent $20\ \text{g}$ contains $180\ \text{g}$ of water at $25^\circ\text{C}$. $m$ grams of steam at $100^\circ\text{C}$ is mixed in it till the temperature of the mixture is $31^\circ\text{C}$. The value of $m$ is close to (Latent heat of water $= 540\ \text{cal g}^{-1}$, specific heat of water $= 1\ \text{cal g}^{-1}\,^\circ\text{C}^{-1}$)
Answer: (A) $2$
Heat gained by water and calorimeter: $(180 + 20)\times1\times(31 - 25) = 1200$ cal.
Heat given by steam: $m(540 + 100 - 31) = 609m$.
$$609m = 1200 \Rightarrow m \approx 2\ \text{g}$$
Solution by Sreeraj P, M.Sc Physics