Search any question & find its solution
Question:
Answered & Verified by Expert
An ideal heat engine has an efficiency \(\eta\). The coefficient of performance of the engine when driven backward will be
Options:
Solution:
1179 Upvotes
Verified Answer
The correct answer is:
\(\left(\frac{1}{\eta}\right)-1\)
For ideal heat engine, efficiency is given as
\(\begin{array}{ll}
& \eta=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1} \\
\Rightarrow & \eta=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}
\end{array}\)
\(\begin{array}{ll}
\Rightarrow & \frac{T_2}{T_1}=1-\eta \\
\Rightarrow & \frac{T_1}{T_2}=\frac{1}{1-\eta} \quad \ldots (i)
\end{array}\)
When heat engine operated in backward direction, then coefficient of performance is given as
\(\begin{aligned}
\beta & =\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2} \\
\Rightarrow \quad \beta & =\frac{1}{\frac{T_1}{T_2}-1}=\frac{1}{\frac{1}{1-\eta}-1} \quad \text { [from Eq. (i)] } \\
& =\frac{1-\eta}{1-1+\eta}=\frac{1-\eta}{\eta}=\frac{1}{\eta}-1
\end{aligned}\)
\(\begin{array}{ll}
& \eta=\frac{W}{Q_1}=\frac{Q_1-Q_2}{Q_1} \\
\Rightarrow & \eta=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1}
\end{array}\)
\(\begin{array}{ll}
\Rightarrow & \frac{T_2}{T_1}=1-\eta \\
\Rightarrow & \frac{T_1}{T_2}=\frac{1}{1-\eta} \quad \ldots (i)
\end{array}\)
When heat engine operated in backward direction, then coefficient of performance is given as
\(\begin{aligned}
\beta & =\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2} \\
\Rightarrow \quad \beta & =\frac{1}{\frac{T_1}{T_2}-1}=\frac{1}{\frac{1}{1-\eta}-1} \quad \text { [from Eq. (i)] } \\
& =\frac{1-\eta}{1-1+\eta}=\frac{1-\eta}{\eta}=\frac{1}{\eta}-1
\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.