Ice at –20oC is added to 50 g of water at 40oC. When the temperature… | JEE Main 2019 (11 Jan, Shift 1)

Q41JEE Main 2019 · 11 JAN · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Ice at –20oC is added to 50 g of water at 40oC. When the temperature of the mixture reaches 0oC, 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 = 4.2J/g/oC Specific heat of Ice = 2.1J/g/oC Heat of fusion of water at 0oC= 334J/g)

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Answer: (d)

Heat lost by water $= 50 \times 4.2 \times 40 = 8400\text{ J}$
Ice of mass $m$ : warming $m \times 2.1 \times 20 = 42m$ ; melting $(m - 20) \times 334$ .
$42m + 334(m - 20) = 8400 \Rightarrow 376m = 15080 \Rightarrow m \approx 40\text{ g}$
So the answer is (d) 40 g.

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