A particular engine has a power output of 2 kW and an efficiency of 27%. If the engine expels 9085 J of thermal energy in each cycle, find the heat absorbed in each cycle. Answer in units of J.

12445 J

Explanation:

Given that

Power output, P = 5 kW

efficiency of the engine, e = 27% = 0.27

Thermal energy expelled, Q(c) = 9085 J

Heat absorbed, Q(h) = ?

Using the formula

e = W/Q(h)

e = [Q(h) - Q(c)] / Q(h)

e = 1 - Q(c)/Q(h)

Now, substituting the values into the formula, we have

0.27 = 1 - 9085/Q(h)

9085/Q(h) = 1 - 0.27

9085/Q(h) = 0.73

Q(h) = 9085 / 0.73

Q(h) = 12445 J

Thus, the heat absorbed is 12445 J

answer: option c.

explanation: