MHTCET 19 Apr 2026 Shift 2 Paper

Concept:Heat lost by the hot water and vessel equals heat gained by the ice to melt and then warm up as water.Explanation:The vessel's water equivalent is already included in the given effective water mass of 60 g.First, check whether all the ice melts by cooling to $0^\circ\mathrm{C}$.Heat released by the water and vessel: $Q = 60 \times 1 \times 40 = 2400\ \mathrm{cal}$.Heat required to melt 15 g of ice: $Q_{\text{melt}} = 15 \times 80 = 1200\ \mathrm{cal}$.Since $2400 > 1200$, all ice melts and the final temperature is above $0^\circ\mathrm{C}$.Let the final temperature be $T$.Heat lost by the water and vessel: $60(40 - T)\ \mathrm{cal}$.Heat gained by the ice: $15 \times 80 + 15 \times 1 \times T = 1200 + 15T\ \mathrm{cal}$.Using heat lost = heat gained: $60(40 - T) = 1200 + 15T$.Solving, $2400 - 60T = 1200 + 15T$, so $1200 = 75T$.Thus, $T = 16^\circ\mathrm{C}$.Answer:The final temperature of the mixture is $16^\circ\mathrm{C}$, which is Option C.