Thermodynamics Question 307
Question: A carnots reversible engine converts $ \frac{1}{6}th $ of heat input into work. When the temperature of sink is reduced 62 K, the efficiency of carnotss cycle, becomes $ \frac{1}{3}. $ Calculate temperature of source and sink
Options:
A) 372K, 310K
B) 772K, 312K
C) 672K, 610K
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \eta =1-\frac{T _{2}}{T _{1}} $
where $ T _{1}\And T _{2} $
are source & sink temperatures respectively. $ \frac{1}{6}=1-\frac{T _{2}}{T _{1}} $ ..(1)
Modified case,
$ \frac{1}{3}=1-\frac{T _{2}-62}{T _{1}} $ …(2)
Solving (1) and (2), we get
$ T _{1}=372K,T _{2}=310K $
