Answer
See answers.
Work Step by Step
a. The minimum energy required to pump the water up to the lake is the gravitational potential energy of the water.
$$mgh=(1.00\times10^5kg/s)(10.0h)(9.80m/s^2)(115m)=1.127\times10^9 W\cdot h\approx 1.13\times10^6 kWh$$
b. When this energy is released at 75 percent efficiency, and spread out over 14 hours, find the average power output.
$$\frac{(0.75) 1.13\times10^6 kWh}{14h}=6.0\times10^4kW=60MW$$
