Answer
(a) $751KJ$
(b) $194KJ$
(c) $945KJ$
Work Step by Step
(a) We can find the required heat removed as follows:
$Q=mC_w(T_w-0C^{\circ})+mL_f+mC_{ice}[0C^{\circ}-(-5C^{\circ})]$
We plug in the known values to obtain:
$Q=175Kg[(4186J/KgC^{\circ})(20^{\circ}C)+33.5\times 10^4J/Kg+(2090J/Kg.C^{\circ})(5C^{\circ})]$
$Q=751KJ$
(b) We know that
$W=\frac{Q_c}{Coefficient \space of \space performance}$
We plug in the known values to obtain:
$W=\frac{751KJ}{3.88}$
$W=194KJ$
(c) We know that
$Q_h=W+Q_c$
We plug in the known values to obtain:
$Q=194KJ+751KJ$
$Q=945KJ$
