Q 11-10-115JEE MainJEE Main 2019 (11 Jan, Shift 1)Medium
Ice at $-20^\circ$C is added to $50$ g of water at $40^\circ$C. When the temperature of the mixture reaches $0^\circ$C, it is found that $20$ g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water $= 4.2\ \text{J/g}^\circ\text{C}$, specific heat of ice $= 2.1\ \text{J/g}^\circ\text{C}$, heat of fusion of water at $0^\circ$C $= 334$ J/g)
Answer: (D) $40$ g
Heat given by the water cooling to $0^\circ$C:
$$50\times4.2\times40 = 8400\ \text{J}$$
If $m$ grams of ice were added, all of it warms to $0^\circ$C and $(m - 20)$ g melts:
$$m(2.1)(20) + (m - 20)(334) = 8400$$
$$376m = 8400 + 6680 = 15080 \Rightarrow m \approx 40\ \text{g}$$
Solution by Sreeraj P, M.Sc Physics