Question

A cycle heat engine does $50$ kJ of work per cycle. If the efficiency of the heat engine is $75$%. The heat rejected per cycle is

• A. $33\frac{1}{3}kJ$
• B. $37\frac{1}{2}kJ$
• C. $16\frac{2}{3}kJ$
• D. $66\frac{2}{2}kJ$

Answer 1

The correct option is C $16\frac{2}{3}kJ$

$\eta =\frac{W}{Q}$
$0.75=\frac{50}{Q}$
$Q=66.67$
$\therefore$ $Q_R=66.67-50=16.67kJ$
$=16\frac{2}{3}kJ$