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JEE Main 2016 (10 Apr, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2016 (10 Apr, Shift 1)

Q68JEE Main 2016 · 10 APR · SHIFT 1Difficult

Heat Engines, Refrigerators & Second Law

A Carnot freezer takes heat from water at 0oC inside it and rejects it to the room at a temperature of 27oC. The latent heat of ice is 336×103 J kg−1. If 5 kg of water at 0oC is converted into ice at 0oC by the freezer, then the energy consumed by the freezer is close to:

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Answer: (a)

  1. Heat removed: Q2=mL=5×336×103=1.68×106Q_2 = mL = 5 \times 336 \times 10^3 = 1.68 \times 10^6 J.
  2. Carnot freezer: WQ2=T1−T2T2=300−273273=27273\frac{W}{Q_2} = \frac{T_1 - T_2}{T_2} = \frac{300 - 273}{273} = \frac{27}{273}
  3. W=1.68×106×27273≈1.67×105W = 1.68 \times 10^6 \times \frac{27}{273} \approx 1.67 \times 10^5 J
  4. So the answer is (a) 1.67×1051.67 \times 10^5 J.
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