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A Carnot engine efficiency is equal to 1 7 . If the temperature of the sink is reduced by 65 K, the efficiency becomes 1 4 . The temperature of the source and the sink in the first case are respectively

Options

  1. A620 K , 520 K
  2. B520 K , 606 . 67 K
  3. C606 . 67 K , 520 K
  4. D520 K , 610 K

Correct answer

C. 606 . 67 K , 520 K

Step-by-step solution

∵ 1 - T 2 T 1 = 1 7 (in case 1 ) ∴ T 2 T 1 = 6 7 ... ( i ) 1 - T 2 - 65 T 1 = 1 4 (in case 2 ) ∴ T 2 - 65 T 1 = 3 4 ...( ii ) From Eqs. ( i ) and ( ii ), T 2 T 1 T 2 - 65 T 1 = 6 7 3 4 = 8 7 ⇒ T 2 T 2 - 65 = 8 7 ⇒ 7 T 2 = 8 T 2 - 8 × 65 ⇒ T 2 = 520 K ∴ T 2 T 1 = 6 7 ∴ T 1 = T 2 × 7 6 = 520 × 7 6 = 606 .67 K

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