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 $ .



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