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JEE Main 2016 (9 Apr, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2016 (9 Apr, Shift 1)

Q46JEE Main 2016 · 9 APR · SHIFT 1Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

200 g water is heated from 40oC to 60oC. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K):

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Answer: (c)

No work done, so $\Delta U = Q = mc\Delta T$ .
$\Delta U = 0.2 \times 4184 \times 20 = 16736$ J
So the answer is (c) 16.7 kJ.

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