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Question: Answered & Verified by Expert
A Carnot engine efficiency is equal to 17. If the temperature of the sink is reduced by 65 K, the efficiency becomes 14. The temperature of the source and the sink in the first case are respectively
PhysicsThermodynamicsNEET
Options:
  • A 620 K520 K
  • B 520 K606.67 K
  • C 606.67 K520 K
  • D 520 K610 K
Solution:
2681 Upvotes Verified Answer
The correct answer is: 606.67 K520 K
1-T2T1=17 (in case 1)

T2T1=67 ... (i)

1-T2-65T1=14 (in case 2)

T2-65T1=34 ...(ii)

From Eqs. (i) and (ii),

T2T1T2-65T1=6734=87

T2T2-65=87

7T2=8T2-8×65

T2=520 K

T2T1=67

T1=T2×76=520×76

=606.67 K

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