Answer
(a) $W = 90.2 J$
(b) $Q_C = -319.8 J $
(c) $ T_C = 318. 2 K \approx 45.2 ^o C$
(d) $\Delta S = 0$
(e) $ m = 0.263 kg $
Work Step by Step
(a) The work done is calculated using the efficiency equation
$\epsilon = \frac{W}{Q_H} $
$W = \epsilon Q_H $
$W = (0.220) (410 J) $
$W = 90.2 J$
(b) heat the engine waste for each cycle, $Q_C$ is
$Q_C = W - Q_H$
$Q_C = 90.2 - 410 J $
$Q_C = -319.8 J $
(c) The temperature of cold reservoir, $T_C$ is
$\frac{Q_C}{Q_H} = - \frac{T_C}{T_H}$
$ T_C = - \frac{Q_C}{Q_H} T_H $
$ T_C = - \frac{(-319.8J)}{410 J } (408K) $
$ T_C = 318. 2 K \approx 45.2 ^o C$
(d) Entropy change in each cycle, $\Delta S$
$ \Delta S = \frac{|Q_C|}{T_C} - \frac{|Q_H|}{T_H} $
$ \Delta S = \frac{|319.8 J|}{318. 2 K} - \frac{|410 J |}{408 K} $
$ \Delta S = 0.000126 J/K \approx 0 J/K$
It indicates an irreversible process, $\Delta S = 0$
(e) To find the mass of water, we use weight-gravity equation
$W = mgh$
$ m = \frac{W}{gh}$
$ m = \frac{90.2 J}{(9.80 m/s^2 )(35.0 m ) }$
$ m = 0.263 kg $
