Question

A water reservoir contains 100 tons of water at an average elevation of \(60 \mathrm{m} .\) The maximum amount of electric power that can be generated from this water is (a) \(8 \mathrm{kWh}\) \((b) 16 \mathrm{kWh}\) \((c) 1630 \mathrm{kWh}\) \((d) 16,300 \mathrm{kWh}\) \((e) 58,800 \mathrm{kWh}\)

Short answer

Answer: (b) 16 kWh

Step-by-step solution

Calculate the potential energy of the water

The formula for the potential energy is PE = mgh. In this case, m = 100,000 kg, g = 9.81 m/s^2, and h = 60m. So, the potential energy is: PE = (100,000 kg)(9.81 m/s^2)(60 m)

Evaluate the potential energy

Multiply the values to find the potential energy value in joules: PE = 58,860,000 J

Convert joules to kilowatt-hours

To convert the potential energy into kilowatt-hours (kWh), divide the potential energy by 3,600,000 (since there are 3,600,000 J in 1 kWh): Potential energy in kWh = 58,860,000 J / 3,600,000 J/kWh = 16.35 kWh

Round the result and compare to the options

Round the result to the nearest whole number, which is 16 kWh. This matches option (b). The maximum amount of electric power that can be generated from this water is approximately 16 kWh. The correct answer is (b).