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JEE Main 2019 (12 Apr, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2019 (12 Apr, Shift 1)

Q38JEE Main 2019 · 12 APR · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

When M1 gram of ice at –10oC (specific heat = 0.5 cal g–1oC–1) is added to M2 gram of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g–1 is:

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Answer: (a)

Heat gained by ice: $M_1 \times 0.5 \times 10 + M_1L$
Heat lost by water: $M_2 \times 1 \times 50$
$5M_1 + M_1L = 50M_2 \Rightarrow L = \dfrac{50M_2}{M_1} - 5$
So the answer is (a).

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