Answer
See explanation.
Work Step by Step
(a) In one year, the freezer operates for 5 hours every day. The freezer operates for $ 5 h \times 365 \space days = 1825 h $
To find the operating power,
$P = \frac{P_{kWh} }{t}$
$P = \frac{730 kWh }{1825 h }$
$P = 0.4 kW = 400 W$
(b) The maximum coefficient
$K = \frac{268 K }{293 K- 268 K}$
$K = 10.72$
(c) $|W| = pt$
$|W| = (400 W) (3600s) $
$|W| = 1.44 \times 10^6 J$
$|Q_C| = K|W| $
$|Q_C| = 10.72(1.44 \times 10^6 J) $
$|Q_C| = 1.54 \times 10^7 J$
$Q = mC_W \Delta T + mL_f$
$Q = m(C_W \Delta T + L_f)$
$m = \frac{1.54 \times 10^7 J}{(4190 J/kg K) (20K) + (334 \times 10^3 J/kg)}$
$m = 36.86 kg $
