Question

In a Carnot engine, when $T_2=0℃$ and $T_1=200℃,$ its efficiency is ${\eta }_1$ and when $T_1=0℃$ and $T_2=-200℃$ its efficiency is ${\eta }_2$ , then the ratio of efficiencies is

• A. $0.577$
• B. $0.733$
• C. $0.638$
• D. Cannot be calculated

Answer 1

Answer: (A)

The efficiency $\eta$ of Carnot engine is defined as the amount of work divided by the heat transferred between the system and the hot reservoir.
$\eta =\frac{\Delta W}{\Delta Q_H}=1-\frac{Tc}{T_H}$
Where, $T_{C}and{T}_H$ are temperatures of cold and hot junctions respectively.
Ist case $T_2=0℃=0+273=273K$
$T_1=200℃=200+273=473K$
$\therefore {\eta }_1=1-\frac{273}{473}=\frac{200}{473}=0.4228\approx 0.423$ ...(i)
IInd case
$T_2=-200℃=-200+273=73K$
$T_1=0℃=0+273=273K$
${\eta }_2=1-\frac{T_2}{T_1}=1-\frac{73}{273}=\frac{200}{273}=0.732$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{{\eta }_1}{{\eta }_2}=\frac{0.423}{0.732}\approx 0.577$