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, $\eta =50$%$=\frac{1}{2};T_2=273+7=280K$
now, $\eta =1-\frac{T_2}{T_1}$
or $\frac{1}{2}=1-\frac{280}{T_1}$
$\Rightarrow$ $T_1=280\times 2=560K$
Now, $\eta =50+10=60$%$=\frac{3}{5};{T_2}^{\prime }=280K$
$\therefore$ $\eta =1-\frac{{T_2}^{\prime }}{{T_1}^{\prime }}$
$\Rightarrow$ $\frac{3}{5}=1-\frac{280}{{T_1}^{\prime }}$
Or ${T_1}^{\prime }=\frac{5}{2}\times 280=700K$
$\therefore$ Increase in temperature of hot reservoir,
$=700-560=140K$