Thermodynamics Question 156
Question: The temperature of sink of Carnot engine is $ 27^{o}C $ . Efficiency of engine is 25%. Then temperature of source is
[DCE 2002; CPMT 2002]
Options:
A) $ 227^{o}C $
B) $ 327^{o}C $
C) $ 127^{o}C $
D) $ 27^{o}C $
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Answer:
Correct Answer: C
Solution:
$ \eta =1-\frac{T _{2}}{T _{1}} $
Therefore $ \frac{25}{100}=1-\frac{300}{T _{1}} $
Therefore $ \frac{1}{4}=1-\frac{300}{T _{1}} $
$ T _{1}=400K=127{}^\circ C $
BETA
