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JEE Main 2019 (10 Jan, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2019 (10 Jan, Shift 2)

Q42JEE Main 2019 · 10 JAN · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

An unknown metal of mass 192 g heated to a temperature of 100oC was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4oC. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5oC. (Specific heat of brass is 394 J kg–1 K–1)

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Answer: (d)

  1. Heat lost by metal = heat gained by water + brass calorimeter.
  2. 0.192 s (100−21.5)=(0.128×394+0.240×4200)(21.5−8.4)0.192\,s\,(100-21.5) = (0.128 \times 394 + 0.240 \times 4200)(21.5-8.4)
  3. 15.072 s=1058.4×13.1≈1386515.072\,s = 1058.4 \times 13.1 \approx 13865
  4. s≈920 J kg−1K−1s \approx 920\ \text{J kg}^{-1}\text{K}^{-1}, closest option.
  5. So the answer is (d) 916 J kg−1K−1916\ \text{J kg}^{-1}\text{K}^{-1}.
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