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A Carnot engine, whose efficiency is 20% receives heat at 450 K. If its efficiency is increased to 30%, then the intake temperature for the same exhaust temperature is
  • a)
    514 K
  • b)
    534 K
  • c)
    554 K
  • d)
    574 K
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A Carnot engine, whose efficiency is 20% receives heat at 450 K. If it...
Given: Efficiency of Carnot engine, η1 = 20%, Temperature of heat intake, T1 = 450 K

To find: Temperature of heat intake, T2 for efficiency, η2 = 30%

Formula used: Efficiency of Carnot engine, η = (T1 - T2)/T1

Calculation:

Let's assume the temperature of heat rejection to be constant, T2 = 300 K.

From the formula, we can write:

η1 = (T1 - T2)/T1

0.2 = (T1 - 300)/T1

Solving for T1, we get T1 = 375 K

Now, we need to find the temperature of heat intake, T2 for η2 = 30%

η2 = (T1 - T2)/T1

0.3 = (375 - T2)/375

Solving for T2, we get T2 = 514 K

Therefore, the temperature of heat intake, T2 for η2 = 30% is 514 K.

Hence, option (a) is the correct answer.
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A Carnot engine, whose efficiency is 20% receives heat at 450 K. If its efficiency is increased to 30%, then the intake temperature for the same exhaust temperature isa)514 Kb)534 Kc)554 Kd)574 KCorrect answer is option 'A'. Can you explain this answer?
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