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MHT CET 2026 (20 Apr, Shift 2) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from MHT CET 2026 (20 Apr, Shift 2)

Q7MHT CET 2026 · 20 APR · SHIFT 2Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A piece of metal of 850 K is dropped into 1 kg of water at 300 K. If equilibrium temperature of water is 350 K then the heat capacity of the metal expressed in JK\frac{J}{K} is (\Bigg(Sp. heat of water =1 calg ∘C)= 1\ \frac{\text{cal}}{\text{g}\,{}^{\circ}\text{C}}\Bigg)

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Answer: (a)

  1. Heat gained by water =1000×1×50=50000 cal=2.1×105 J= 1000 \times 1 \times 50 = 50000\text{ cal} = 2.1 \times 10^5\text{ J}
  2. Metal falls by 850−350=500 K850 - 350 = 500\text{ K}: C×500=2.1×105C \times 500 = 2.1 \times 10^5
  3. C=420 J/KC = 420\text{ J/K}
  4. So the answer is (a) 420.
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