Answer
a) $40.7
b) 1400kg
Work Step by Step
a) $P=4.4kWh+2.4kWh+3.0kWh+2.0kWh=11.8kWh$
$11.8kWh\frac{$0.115}{kWh}=\$1.4$
$\$1.4\times30=\$40.7$
b) $\frac{11.8kWh}{1day}\times\frac{365days}{1year}=4307kWh/year$
$\frac{7500kcal}{1kg}=\frac{8.7kWh}{1kg}$
At 100% efficiency, $495kg$ of coal.
At 35% efficiency, $1400kg$