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JEE Main 2020 (4 Sep, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2020 (4 Sep, Shift 1)

Q32JEE Main 2020 · 4 SEP · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

The specific heat of water

= 4200 J kg-1K-1 and the latent heat of

ice = 3.4 ×\times 105 J kg–1. 100 grams of ice at

0oC is placed in 200 g of water at 25oC. The

amount of ice that will melt as the temperature

of water reaches 0oC is close to (in grams):

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Answer: (b)

  1. Heat from water cooling to 0∘C0^{\circ}\text{C}: 0.2×4200×25=21000 J0.2 \times 4200 \times 25 = 21000\text{ J}
  2. Ice melted =210003.4×105=0.0618 kg≈61.7 g= \dfrac{21000}{3.4 \times 10^5} = 0.0618\text{ kg} \approx 61.7\text{ g}
  3. So the answer is (b) 61.7.
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