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 $
BETA
