MCQ
Efficiency $(\eta)$ of the heat engine in thermodynamics can be defined as:
- A$\eta=\frac{\text{W}}{\text{Q}_2}$
- B$\eta=1-\frac{\text{Q}_1}{\text{Q}_2}$
- ✓$\eta=\frac{\text{W}}{\text{Q}_1}$
- DNone of these.
✓
Answer
Correct option: C.
$\eta=\frac{\text{W}}{\text{Q}_1}$
For heat engine, $Q_1 = W + Q_2$
$\Rightarrow W = Q_1 - Q_2$
$Q_2$ is heat rejected to the environment.
So, $\eta=\frac{\text{Q}_1-\text{Q}_2}{\text{Q}_1}=\frac{\text{W}}{\text{Q}_1}$
$\Rightarrow W = Q_1 - Q_2$
$Q_2$ is heat rejected to the environment.
So, $\eta=\frac{\text{Q}_1-\text{Q}_2}{\text{Q}_1}=\frac{\text{W}}{\text{Q}_1}$
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