Question

A heat engine receives $50\text{kcal}$ of heat from the source per cycle, and operates with an efficiency of $20\%$. Find the heat rejected to the sink per cycle.

• A. $50\text{kcal}$
• B. $10\text{kcal}$
• C. $42\text{kcal}$
• D. $40\text{kcal}$

Answer 1

The correct option is D $40\text{kcal}$

Heat supplied, $Q_1=50\text{kcal}$
Efficiency, $\eta =20\%$
Using formula for efficiency,
$\eta =\frac{W}{Q_1}$
$\Rightarrow W=\eta \times Q_1$
i.e., work obtained per cycle $W=20\%\times 50\text{kcal}=10\text{kcal}$
Or, $W=10\text{kcal}$
From energy conservation,
$\Rightarrow Q_1=Q_2+W$
Or, $Q_2=Q_1-W...\left(i\right)$
From Eq.$\left(i\right)$ heat rejected to the sink per cycle is given by,
Heat rejected, $Q_2=50\text{kcal}-10\text{kcal}$
$\therefore Q_2=40\text{kcal}$