Question

The relation between efficiency $\eta$ of a heat engine and the co-efficient of performance $\alpha$ of a refrigerator is

• A. $\eta =\frac{1}{1-\alpha }$
• B. $\eta =\frac{1}{1+\alpha }$
• C. $\eta =1+\alpha$
• D. $\eta =1-\alpha$

Answer 1

Answer: (B)

$\eta =\frac{dw}{dQ}$ but $\alpha =\frac{d{Q}_2}{dw}=\frac{Q_2}{Q_1-Q_2}\Rightarrow \frac{1}{\alpha }=\frac{Q_1}{Q_2}-1\Rightarrow \frac{1}{\alpha }+1=\frac{Q_1}{Q_2}\Rightarrow \frac{\alpha +1}{\alpha }=\frac{Q_1}{Q_2}\Rightarrow \frac{Q_2}{Q_1}=\frac{\alpha }{\alpha +1}$
$\eta =\frac{Q_1-Q_2}{Q_1}=1-\frac{Q_2}{Q_1}=1-\frac{\alpha }{1+\alpha }=\frac{1+\alpha -\alpha }{1+\alpha }$
$\eta =\frac{1}{1+\alpha }$