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JEE Main 2025 (8 Apr, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2025 (8 Apr, Shift 2)

Q11JEE Main 2025 · 8 APR · SHIFT 2Easy

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool.

(Take g = 10 m/s2, specific heat of water = 4200 J/(kg K))

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Answer: (c)

  1. All potential energy becomes heat: mgh=mcΔTmgh = mc\Delta T.
  2. ΔT=ghc=10×2004200≈0.48\Delta T = \dfrac{gh}{c} = \dfrac{10 \times 200}{4200} \approx 0.48 K
  3. So the answer is (c).
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