Question

A Carnot engine works between ice point and steam point. It is desired to increase efficiency by $20\%$, then the temperature of sink must be changed to -

• A. $253\text{K}$
• B. $293\text{K}$
• C. $303\text{K}$
• D. $243\text{K}$

Answer 1

The correct option is A $253\text{K}$

Temperature of source $T_1=100+273=373K$.
Temperature of sink $T_2=0+273=273K$.
Efficiency of carnot engine
$\eta =1-\frac{T_2}{T_1}=1-\frac{273}{373}$
$\eta =\frac{100}{373}$

To increase efficency by $20\%$
${\eta }^{\prime }=\frac{100}{373}+\frac{100}{373}\times \frac{1}{5}=\frac{120}{373}$
${\eta }^{\prime }=1-\frac{T_2}{373}=\frac{120}{373}$

$T_2=\frac{373-120}{373}\times 373=253K$
New sink temperature is $=253\text{K}$