Chapter 18 - The Laws of Thermodynamics - Problems and Conceptual Exercises - Page 646: 49

(a) $1.16GW$ (b) $1.71GW$

(a) We know that $\eta=\frac{Work\space done}{Heat \space input}$ We plug in the known values to obtain: $0.32=\frac{838}{Heat\space input}$ $\implies Heat\space input=2618.75MW$ Now $Heat \space provided=Work\space done+Heat\space discarded$ $\implies Heat\space discarded=Heat \space provided-Work\space done$ We plug in the known values to obtain: $Heat \space discarded=2618.75-838=1780.75MW=1.16GW$ (b) As $\eta=\frac{Work\space done}{Heat\space input}$ We plug in the known values to obtain: $0.32=\frac{838}{Heat\space input}$ $\implies Heat\space input=2618.75MW=1.71GW$