Question

A particular engine has a power output of 5 kW and an efficiency of 30%. If the engine expels 6464 J of thermal energy in each cycle, find the heat absorbed in each cycle. Answer in units of J.

Answer 1

The heat absorbed in each cycle is 9,234.286 J

Step-by-step explanation:

Given;

power output, P = 5 kW = 5,000 W

efficiency of the engine, e = 30 % = 0.3

thermal heat expelled,
`Q_c` = 6464 J

let the heat absorbed =
`Q_h`

The efficiency of the engine is given as;

`e = (W)/(Q_h) = (Q_h-Q_c)/(Q_h) = (Q_h)/(Q_h) - (Q_c)/(Q_h) = 1-(Q_c)/(Q_h)\\\\e = 1-(Q_c)/(Q_h)\\\\0.3 = 1-(Q_c)/(Q_h)\\\\(Q_c)/(Q_h) = 1-0.3\\\\(Q_c)/(Q_h) = 0.7\\\\Q_h = (Q_c)/(0.7) \\\\Q_h = (6464)/(0.7) = 9,234.286 \ J.`

Therefore, the heat absorbed in each cycle is 9,234.286 J.