Thermodynamics Question 158

Question: In a Carnot engine, when $ T _{2}=0^{o}C $ and $ T _{1}=200^{o}C, $ its efficiency is $ {\eta _{1}} $ and when $ T _{1}=0{{}^{o}}C $ and $ T _{2}=-200{{}^{o}}C $ , Its efficiency is $ {\eta _{2}} $ , then what is $ {\eta _{1}}/{\eta _{2}} $

[DCE 2004]

Options:

A) 0.577

B) 0.733

C) 0.638

D) Cannot be calculated

Show Answer

Answer:

Correct Answer: A

Solution:

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

$ \Rightarrow {\eta _{1}}=\frac{(473-273)}{473}=\frac{200}{473} $

and $ {\eta _{2}}=\frac{273-73}{273}=\frac{200}{273} $

So required ratio $ \frac{{\eta _{1}}}{{\eta _{2}}}=\frac{273}{473}=0.577 $



NCERT Chapter Video Solution

Dual Pane