Question

A lawnmower engine with an efficiency of 0.26 rejects 4700 J of heat every second. What is the magnitude of the work that the engine does in one second?

Answer 1

Efficiency of the lawnmower engine = $\eta$ = 0.26

Heat rejected by the lawnmower engine per second = Q_c = 4700 J

Work done by the engine in one second = W

Heat input to the lawnmower engine per second = Q_h

Q_h = W + Q_c

$\eta = \frac{W}{Q_{h}}$

$\eta = \frac{W}{W + Q_{c}}$

$0.26 = \frac{W}{W + 4700}$

(0.26)(W + 4700) = W

0.26W + 1222 = W

0.74W = 1222

W = 1651.35 J

Work done by the engine in one second = 1651.35 J