Question

An ideal Carnot engine having efficiency $\eta$ works between temperatures $127°C$ and $-73°C$. ${\eta }_1$ denotes the efficiency when the temperature of the source increases by $50°C$ and ${\eta }_2$ denotes efficiency when temperature of the sink decreases by $50°C$. Then the correct order is

• A. $\eta >{\eta }_1>{\eta }_2$
• B. ${\eta }_1>\eta >{\eta }_2$
• C. ${\eta }_2>{\eta }_1>\eta$
• D. $\eta ={\eta }_1={\eta }_2$

Answer 1

The correct option is C ${\eta }_2>{\eta }_1>\eta$


$T_1={127}^o=400K{T}_2=-{73}^o=200K$
effeciency ($\eta$)$=1-\frac{T_2}{T_1}$
$\eta =1-\frac{200}{400}=1-\frac{1}{2}$
=$50$%

Case 1: $T_1=450K{T}_2=200K$
${\eta }_1=1-\frac{T_2}{T_1}=1-\frac{200}{450}$
$=55$%

Case 2: $T_1=400K{T}_2=150K$
${\eta }_2=1-\frac{T_2}{T_1}=1-\frac{150}{400}$
$=62.5$%

From above observation, ${\eta }_2>{\eta }_1>\eta$