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JEE Main 2022 (25 Jun, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2022 (25 Jun, Shift 1)

Q40JEE Main 2022 · 25 JUN · SHIFT 1Medium

Heat Engines, Refrigerators & Second Law

A steam engine intakes 50 g of steam at 100∘^\circC per minute and cools it down to 20 ∘^\circC. If latent heat of vaporization of steam is 540 cal g−- 1, then the heat rejected by the steam engine per minute is __________ ×\times 103 cal.

(Given: specific heat capacity of water: 1 cal g−1{}^{-1} ∘C−{}^\circ\text{C}^{-}1)

Numerical answer type. Work it out, then open the answer.

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Answer: 31

  1. Heat rejected = latent heat on condensing + heat lost cooling water from 100∘C100^\circ\text{C} to 20∘C20^\circ\text{C}.
  2. Q=mL+mcΔT=50×540+50×1×80Q = mL + mc\Delta T = 50 \times 540 + 50 \times 1 \times 80
  3. Q=27000+4000=31000 cal=31×103 calQ = 27000 + 4000 = 31000\text{ cal} = 31 \times 10^3\text{ cal}
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