Answer:
The heat absorbed in each cycle is 9,234.286 J
Explanation:
Given;
power output, P = 5 kW = 5,000 W
efficiency of the engine, e = 30 % = 0.3
thermal heat expelled, [tex]Q_c[/tex] = 6464 J
let the heat absorbed = [tex]Q_h[/tex]
The efficiency of the engine is given as;
[tex]e = \frac{W}{Q_h} = \frac{Q_h-Q_c}{Q_h} = \frac{Q_h}{Q_h} - \frac{Q_c}{Q_h} = 1-\frac{Q_c}{Q_h}\\\\e = 1-\frac{Q_c}{Q_h}\\\\0.3 = 1-\frac{Q_c}{Q_h}\\\\\frac{Q_c}{Q_h} = 1-0.3\\\\\frac{Q_c}{Q_h} = 0.7\\\\Q_h = \frac{Q_c}{0.7} \\\\Q_h = \frac{6464}{0.7} = 9,234.286 \ J.[/tex]
Therefore, the heat absorbed in each cycle is 9,234.286 J.
