Thermodynamics Question 164

Question: Efficiency of a Carnot engine is 50% when temperature of outlet is 500 K. In order to increase efficiency up to 60% keeping temperature of intake the same what is temperature of outlet

[CBSE PMT 2002]

Options:

A) 200 K

B) 400 K

C) 600 K

D) 800 K

Show Answer

Answer:

Correct Answer: B

Solution:

$ \eta =1-\frac{T _{2}}{T _{1}} $

Therefore $ \frac{1}{2}=1-\frac{500}{T _{1}} $

Therefore $ \frac{500}{T _{1}}=\frac{1}{2} $ -..(i) $ \frac{60}{100}=1-\frac{T _{2}’}{T _{1}} $

Therefore $ \frac{T _{2}’}{T _{1}}=\frac{2}{5} $ -..(ii) Dividing equation (i) by (ii), $ \frac{500}{T _{2}’}=\frac{5}{4} $

Therefore $ T _{2}=400K $



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