## Essential University Physics: Volume 1 (3rd Edition)

(a) $1700MW$ (b)$0.43$ (c) $500K$
(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: $T_h=500K$