The efficiency of a heat engine is ' $\eta$ ' and the coefficient of performance of a refrigerator is '…

The efficiency of a heat engine is ' $\eta$ ' and the coefficient of performance of a refrigerator is ' $\beta$ '. Then
  1. $\eta=\frac{1}{\beta}$
  2. $\eta=\frac{1}{\beta+1}$
  3. $\eta \beta=\frac{1}{2}$
  4. $\eta=\frac{1}{\beta-1}$

Solution

$\begin{aligned} \beta & =\frac{T_2}{T_1-T_2} \\ \eta & =1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1} \\ \therefore \quad \eta & =\frac{1}{1+\frac{T_2}{T_1-T_2}} \\ \therefore \quad \eta & =\frac{1}{1+\beta}\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 1)

Practice more Thermodynamics questions on Aicharya