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{T c \, }{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=473 \, K$
$\therefore \, \, \, \eta_{1}=1-\frac{273}{473}=\frac{200}{473}=0.4228\approx0.423$ ...(i)
IInd case
$T_{2}=-200℃=-200+273=73K$
$T_{1}=0℃=0+273=273 \, K$
$\eta_{2}=1-\frac{T_{2}}{T_{1}}=1-\frac{73}{273}=\frac{200}{273}=0.732$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{\eta_{1}}{\eta_{2}}=\frac{0.423}{0.732}\approx0.577$