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JEE Main 2017 · Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Calorimetry, Latent Heat & Mechanical Equivalent of Heat question from JEE Main 2017

Q45JEE Main 2017Medium

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75oC. T is given by: (Given: room temperature = 30oC, specific heat of copper = 0.1 cal/gmoC)

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Answer: (c)

  1. Heat lost by ball = heat gained by calorimeter + water.
  2. 100(0.1)(T−75)=[100(0.1)+170(1)](75−30)100(0.1)(T-75) = [100(0.1) + 170(1)](75-30)
  3. 10(T−75)=180×45=8100⇒T−75=81010(T-75) = 180 \times 45 = 8100 \Rightarrow T - 75 = 810
  4. So the answer is (c) T=885∘CT = 885^{\circ}\text{C}.
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