Answer
(a) $1700MW$
(b)$0.43$
(c) $500K$
Work Step by Step
(a) We know that
$P_c=mc\frac{\Delta T}{dt}$
We plug in the known values to obtain:
$P_c=(2.8\times 10^4)(4184)(8.5)=996MW$
and $P_h=P_c+W$
$\implies P_h=996+750=1700MW$
(b) $e=1-\frac{Q_c}{Q_h}$
We plug in the known values to obtain:
$e=1-\frac{996}{1746}=0.43$
(c) $e=1-\frac{T_c}{T_h}$
We plug in the known values to obtain:
$0.43=1-\frac{273+15}{T_h}$
This simplifies to:
$T_h=500K$
