Answer
(a) $\epsilon = 23 \%$
(b) $𝑄_𝐶 = 12400 J $
(c) $m = 0.35 g$
(d) $P = 222 kW$
$P =298 hp$
Work Step by Step
(a) Thermal efficiency,
$\epsilon = \frac{W}{|Q_H|} \times 100 $
$\epsilon = \frac{3700J }{16,100 J} \times 100 $
$\epsilon = 23 \%$
(b) Heat is discarded in each cycle,
$ 𝑄_𝐶 = 𝑄_𝐻 - W $
$𝑄_𝐶 = 16100 J - 3700J $
$𝑄_𝐶 = 12400 J $
(c) Mass of fuel is burned in each cycle,
$m = \frac{Q_H}{L_C} $
$m = \frac{16100 J }{4.60 \times 10^4 J/g} $
$m = 0.35 g$
(d) The power output in kilowatts when it goes through 60 cycles
$P = \frac{W}{t}$
$P = \frac{60 \times 3700J }{1 s}$
$P = 222 kW$
The power output in horsepower?
$P = (2.22×10^5 W)(1 hp/746 W) = 298 hp$
