540 gram of ice at 0° C is mixed with 540 gram of water at 90° C. The final temperature of the mixture will be [Latent heat of fusion of water = 80 cal/g] (a) 10° C (b) 15° C (c) 5° C (d) 30° C

Answer: (c). Heat from water cooling to 0°C: 540 × 90 = 48600 cal; to melt ice: 540 × 80 = 43200 cal. All ice melts; extra 5400 cal heats 1080 g of water. T = (5400)/(1080) = 5°C So the answer is (c) 5°C.

540 gram of ice at 0° C is mixed with 540 gram of water at 90° C. The… | MHT CET 2026 (20 May, Shift 1)

Q6MHT CET 2026 · 20 MAY · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

$540$ gram of ice at $0^{\circ}$ C is mixed with $540$ gram of water at $90^{\circ}$ C. The final temperature of the mixture will be [Latent heat of fusion of water = 80 cal/g]

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Answer: (c)

Heat from water cooling to $0^{\circ}\text{C}$ : $540 \times 90 = 48600\text{ cal}$ ; to melt ice: $540 \times 80 = 43200\text{ cal}$ .
All ice melts; extra $5400\text{ cal}$ heats $1080\text{ g}$ of water.
$T = \dfrac{5400}{1080} = 5^{\circ}\text{C}$
So the answer is (c) $5^{\circ}\text{C}$ .

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