Step 1: Calculate the actual annual energy output of the power plant. The total capacity is $915 \text{ MW}$, and the load factor is $72.5\%$. There are $8760$ hours in a year ($365 \text{ days} \times 24 \text{ hours/day}$). First, calculate the actual average power output: $$P_{actual} = P_{capacity} \times \text{Load Factor}$$ $$P_{actual} = 915 \text{ MW} \times 0.725 = 663.375 \text{ MW}$$ Next, calculate the annual energy output: $$E_{out} = P_{actual} \times \text{Hours in a year}$$ $$E_{out} = 663.375 \text{ MW} \times 8760 \text{ hours/year}$$ $$E_{out} = 5811390 \text{ MWh/year}$$ Convert MWh to MJ (since $1 \text{ MWh} = 3600 \text{ MJ}$): $$E_{out} = 5811390 \text{ MWh/year} \times 3600 \text{ MJ/MWh}$$ $$E_{out} = 2.0921004 \times 10^{10} \text{ MJ/year}$$ Step 2: Calculate the annual heat input required from the coal. The plant efficiency ($\eta$) is $40\%$. $$\eta = \frac{E_{out}}{Q_{in}}$$ $$Q_{in} = \frac{E_{out}}{\eta}$$ $$Q_{in} = \frac{2.0921004 \times 10^{10} \text{ MJ/year}}{0.40}$$ $$Q_{in} = 5.230251 \times 10^{10} \text{ MJ/year}$$ Step 3: Calculate