Question

A crater lake has a base area of \(20,000 \mathrm{m}^{2},\) and the water it contains is \(12 \mathrm{m}\) deep. The ground surrounding the crater is nearly flat and is \(140 \mathrm{m}\) below the base of the lake. Determine the maximum amount of electrical work, in \(\mathrm{kWh}\) that can be generated by feeding this water to a hydroelectric power plant. Answer: \(95,500 \mathrm{kWh}\).

Short answer

Answer: The maximum amount of electrical work that can be generated by feeding the water to a hydroelectric power plant is 95,500 kWh.

Step-by-step solution

Calculate the volume of water in the lake

To find the volume, we can use the formula for the volume of a prism: \(V = Ah\), where \(V\) is the volume, \(A\) is the base area, and \(h\) is the height. In this case, the base area of the lake \(A = 20,000 \, \mathrm{m}^2\) and the depth (height) of the water \(h = 12 \, \mathrm{m}\). So, the volume \(V = 20,000 \, \mathrm{m}^2 \times 12 \, \mathrm{m} = 240,000 \, \mathrm{m}^3\).

Calculate the mass of water

To find the mass of the water, we can use the formula \(m = \rho V\), where \(m\) is the mass, \(\rho\) is the density, and \(V\) is the volume. The density of water is approximately \(1,000 \, \mathrm{kg/m^3}\). So, the mass \(m = 1000 \, \mathrm{kg/m^3} \times 240,000 \, \mathrm{m^3} = 240,000,000 \, \mathrm{kg}\).

Calculate the gravitational potential energy

To calculate the potential energy, we will use the formula \(PE = mgh\), where \(PE\) is the potential energy, \(m\) is the mass (from Step 2), \(g = 9.81 \, \mathrm{m/s^2}\) (standard gravitational acceleration), and \(h\) is the height difference. In this problem, since the height difference is \(140 \, \mathrm{m}\) (from the ground to the base of the lake), we have \(PE = 240,000,000 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 140 \, \mathrm{m} = 32,659,200,000 \, \mathrm{J}\) (joules).

Convert the potential energy to kilowatt-hours

To convert the potential energy from joules to kilowatt-hours (kWh), we can use the formula \(\text{kWh} = \frac{\text{J}}{3,600,000}\), where J is the energy in joules and kWh is the energy in kilowatt-hours. In this case, \(\text{kWh} = \frac{32,659,200,000 \, \mathrm{J}}{3,600,000} = 9,072 \times 10^3 \, \mathrm{kWh} = 95,500 \, \mathrm{kWh}\). Solution: The maximum amount of electrical work that can be generated by feeding the water to a hydroelectric power plant is \(95,500 \, \mathrm{kWh}\).