Thermodynamics Question 305
Question: In Carnot engine efficiency is 40% at hot reservoir temperature T. For efficiency 50% what will be temperature of hot reservoir?
Options:
A) $ \frac{T}{5} $
B) $ \frac{2T}{5} $
C) $ 6T $
D) $ \frac{6T}{5} $
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Answer:
Correct Answer: D
Solution:
[d] $ \eta =1-\frac{T _{2}}{T _{1}} $
$ T _{1}=T $ (Temperature of hot reservoir) For $ \eta =40% $
$ \frac{40}{100}=1-\frac{T _{2}}{T _{1}}\Rightarrow \frac{T _{2}}{T}=\frac{3}{5}\Rightarrow T _{2}=\frac{3}{5}T $ For $ \eta =50% $
$ \frac{500}{1000}=1-\frac{\frac{3}{5}T}{T _{1}}\Rightarrow T _{1}=\frac{6T}{5} $
