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JEE Main 2020 (5 Sep, Shift 1) · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2020 (5 Sep, Shift 1)

Q31JEE Main 2020 · 5 SEP · SHIFT 1Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A bullet of mass 5 g, travelling with a speed of 210 m/s, strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 cal/(g –oC) (1 cal = 4.2 × 107 ergs) close to:

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Answer: (a)

  1. KE =12(0.005)(210)2=110.25 J= \tfrac{1}{2}(0.005)(210)^2 = 110.25\text{ J}; heat in bullet =55.125 J= 55.125\text{ J}.
  2. c=0.030 cal/g∘C=0.126 J/g∘Cc = 0.030\text{ cal/g}^{\circ}\text{C} = 0.126\text{ J/g}^{\circ}\text{C}, so mc=5×0.126=0.63 J/∘Cmc = 5 \times 0.126 = 0.63\text{ J/}^{\circ}\text{C}.
  3. ΔT=55.1250.63=87.5∘C\Delta T = \dfrac{55.125}{0.63} = 87.5^{\circ}\text{C}
  4. So the answer is (a) 87.5∘C87.5^{\circ}\text{C}.
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