Question

A heat engine receives $50kcal$ of heat from the source per cycle, and operates with an efficiency of $20\%$. The heat (in $kcal$) rejected by engine to the sink per cycle is

Answer 1

Formula used: $\eta =\frac{W}{Q_1}$

Given $Q_1=50kcal$ and $\eta =20\%$

Efficiency of Carnot engine, $\eta =\frac{W}{Q_1}$ ,

So, work output, $W=\eta \times Q_1$

i..e, work obtained per cycle

$W=20\%\times 50kcal$

$=10kcal=42kJ$

Since, $Q_1=Q_2+W$

so, $Q_2=Q_1-W$

i.e. heat rejected to the sink per cycle $Q_2$

$Q_2=50kcal-10kcal=40kcal$

Final answer: $40$