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)

$\eta=\frac{d w}{d Q}$ but $\alpha =\frac{d Q_{2}}{d w}=\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} \, $
$\eta=\frac{Q_{1} - Q_{2}}{Q_{1}}=1-\frac{Q_{2}}{Q_{1}}=1-\frac{\alpha }{1 + \alpha }=\frac{1 + \alpha - \alpha }{1 + \alpha }$
$\eta=\frac{1}{1 + \alpha }$