Answer
(a) $Q_C = - 121 J$
(b) $3.8 \times 10^3 \space cycles$
Work Step by Step
(a) Heat delivered to the cold reservoir is
$Q_C = -Q_H (\frac{T_C}{T_H})$
$Q_C = -(250J) (\frac{373 K}{773K})$
$Q_C = - 121 J$
(b) $| W| = 250 J - 121 J $
$| W| = 129 J$ This is the work done in one cycle
$W_{total} = F.s$
$W_{total} = (ma)(s)$
$W_{total} = (500 kg)(9.8 m/s^2)(100m) $
$W_{total} = 4.90 \times 10^5 J$
So the cycles are
$\frac{W_{total}}{|W|} = \frac{4.90 \times 10^5 J}{129 J} = 3.8 \times 10^3 \space cycles$
