Thermodynamics Question 163
Question: A Carnot’s engine is made to work between 200°C and 0°C first and then between 0°C and ‘200°C. The ratio of efficiencies of the engine in the two cases is
[KCET 2002]
Options:
A) 1.73 : 1
B) 1 : 1.73
C) 1 : 1
D) 1 : 2
Show Answer
Answer:
Correct Answer: B
Solution:
In first case $ {\eta _{1}}=1-\frac{T _{2}}{T _{1}}=1-\frac{(273+0)}{(273+200)}=\frac{200}{473} $
In second case $ {\eta _{2}}=1-\frac{(273-200)}{(273+0)}=\frac{200}{273} $
Therefore $ \frac{{\eta _{1}}}{{\eta _{2}}}=\frac{1}{( \frac{473}{273} )}=1:1.73 $
