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KCET2014PhysicsRotational Motion

What is the source temperature of the Carnot engine required to get ( 70 % ) efficiency ? Given sink temperature ( =27^ C )

Options

  1. A( 1000^ C )
  2. B( 90^ C )
  3. C( 270^ C )
  4. D( 727^ C )

Correct answer

D. ( 727^ C )

Step-by-step solution

Given, efficiency, =70 % sink temperature, T₂=27^ C =273+27=300 ~K Efficiency, = (1- T₂ T₁ ) 100 % 70= (1- 300 T₁ ) 100 70 100 =1- 300 T₁ 0.7=1- 300 T₁ 300 T₁ =1-0.7 300 T₁ =0.3T₁= 300 0.3 =1000 ~K or T₁=(1000-273)^ C =727^ C Therefore, source temperature =727^ C

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