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 $
BETA
