Chemical Thermodynamics Question 438
Question: For a carnot engine, the source is at $ 500K $ and the sink at $ 300K $ . What is efficiency of this engine
Options:
A) 0.2
B) 0.4
C) 0.6
D) 0.3
Show Answer
Answer:
Correct Answer: B
Solution:
- Given that $ T_1=500K,T_2=300K $
By using $ \eta =\frac{T_1-T_2}{T_1} $
$ =\frac{500-300}{500}=\frac{200}{500}=0.4 $ .
