Q 11-10-106JEE MainJEE Main 2020 (4 Sep, Shift 1)Medium
The specific heat of water $= 4200\ \text{J kg}^{-1}\text{K}^{-1}$ and the latent heat of ice $= 3.4\times10^{5}\ \text{J kg}^{-1}$. $100$ grams of ice at $0^\circ\text{C}$ is placed in $200\ \text{g}$ of water at $25^\circ\text{C}$. The amount of ice that will melt as the temperature of water reaches $0^\circ\text{C}$ is close to (in grams)
Answer: (A) $61.7$
Heat released by the water cooling to $0^\circ\text{C}$: $0.2\times4200\times25 = 21000$ J.
$$m = \frac{21000}{3.4\times10^{5}} \approx 0.0617\ \text{kg} = 61.7\ \text{g}$$
Solution by Sreeraj P, M.Sc Physics