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JEE Main 2014 (9 Apr, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2014 (9 Apr, Shift 1)

Q48JEE Main 2014 · 9 APR · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160J/s. In how much time will the temperature of water rise from 27oC to 77oC? (Specific heat of water is 4.2 kJ/kg K and that of the container is negligible).

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Answer: (a)

  1. Heat needed =2×4200×50=4.2×105 J= 2 \times 4200 \times 50 = 4.2 \times 10^5\text{ J}
  2. Net power =1000−160=840 W= 1000 - 160 = 840\text{ W}
  3. t=4.2×105840=500 s=8 min 20 st = \dfrac{4.2 \times 10^5}{840} = 500\text{ s} = 8\text{ min } 20\text{ s}
  4. So the answer is (a).
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