An ideal heat engine has an efficiency \(\eta\). The coefficient of performance of the engine when driven…
An ideal heat engine has an efficiency \(\eta\). The coefficient of performance of the engine when driven backward will be
- \(1-\left(\frac{1}{\eta}\right)\)
- \(\eta-\left(\frac{1}{\eta}\right)\)
- \(\left(\frac{1}{\eta}\right)-1\)
- \(\frac{1}{1-\eta}\)
Solution
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
\(\begin{aligned}
\beta & =\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2} \\
\Rightarrow \quad \beta & =\frac{1}{\frac{T_1}{T_2}-1}=\frac{1}{\frac{1}{1-\eta}-1} \quad \text { [from Eq. (i)] } \\
& =\frac{1-\eta}{1-1+\eta}=\frac{1-\eta}{\eta}=\frac{1}{\eta}-1
\end{aligned}\)
Asked in: AP EAMCET 2020 (21 Sep Shift 1)
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