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For a heat engine operating on the Carnot cycle, the work output is ¼ th of the heat transferred to the sink. The efficiency of the engine is  
  • a)
    20 %
  • b)
    33.3 %
  • c)
    40 %
  • d)
    50 %
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
For a heat engine operating on the Carnot cycle, the work output is &#...
Q1->heat transferred from source to heat engine
Q2->heat transferred from heat engine to sink
So, q1 = w + q2           --(1)
Given data
w =q2 /4        - (2)
Substitute 2 in 1 you will get
q1 = q2 /4 + q2 * q1 = (5*q2) /4     --(3)

efficiency for heat engine =1-(q2 /q1)
Now substitute 3 in above equation efficiency = 1-(4*q1/(5*q1)) = 1-(4/5)  =1/5 = 0.2*10 =20%
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Most Upvoted Answer
For a heat engine operating on the Carnot cycle, the work output is &#...
Q1->heat transferred from source to heat engineq2->heat transferred from heat engine to sinkso q1 = w + q2 -(1)
Given data w =q2/4 - (2) substitute 2 in 1 you will get
q1 = q2/4 + q2 q1 =5*q2/4 -(3)
efficiency for heat engine =1-(q2/q1) Now substitute 3 in above equation efficiency = 1-(4*q1/(5*q1)) = 1-(4/5) =1/5 =0.2*100 =20%
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For a heat engine operating on the Carnot cycle, the work output is &#...
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