Answer
$1.5\times 10^5J$
Work Step by Step
We can find the required electrical energy as follows:
$Q_c=mC_w(\Delta T_1)+mL_f+mC_{ice}(\Delta T_2)$
We plug in the known values to obtain:
$Q_c=1.5Kg[(4186J/Kg.C^{\circ})(15^{\circ}C-0^{\circ})+33.5\times 10^4J/Kg.C^{\circ}+(2090J/Kg.C^{\circ})(-5C^{\circ})]$
$Q_c=62.65\times 10^4J$
Now $W=\frac{Q_c}{Coefficient \space of \space performance }$
We plug in the known values to obtain:
$W=\frac{62.65\times 10^4J}{4}$
$W=1.5\times 10^5J$
