Question

A Carnot engine whose low temperature reservoir is at $ 7{}^\circ C $ has an efficiency of 50%. If the efficiency of heat engine is to be increased more by 10% then the temperature of high temperature reservoir should be increased by

• A. 107 K
• B. 500 K
• C. 140 K
• D. 200 K

Answer 1

Answer: (C)

Given, $ \eta =50$%$=\frac{1}{2};{{T}_{2}}=273+7=280\,K $
now, $ \eta =1-\frac{{{T}_{2}}}{{{T}_{1}}} $
or $ \frac{1}{2}=1-\frac{280}{{{T}_{1}}} $
$ \Rightarrow $ $ {{T}_{1}}=280\times 2=560\,K $
Now, $ \eta =50+10=60$%$=\frac{3}{5};{{T}_{2}}'=280K $
$ \therefore $ $ \eta =1-\frac{{{T}_{2}}'}{{{T}_{1}}'} $
$ \Rightarrow $ $ \frac{3}{5}=1-\frac{280}{{{T}_{1}}'} $
Or $ {{T}_{1}}'=\frac{5}{2}\times 280=700\,K $
$ \therefore $ Increase in temperature of hot reservoir,
$ =700-560=140\text{ }K $