Question

At Hoover Dam, the distance the water effectively falls before encountering the electric generators depends on the water levels in Lake Mead. Assume that the water intakes are effectively 180 m above the electric generators. How much water must pass through the generators to power 2.1 million 31-W Las Vegas lightbulbs for 7.0 minutes?

(HINT: This is a conservation of energy problem. All the energy comes from the falling water. Consider the electric generators to be 100% efficient.)

Give your answer in kilograms, and enter your answer in the format *.**E**.
To imagine how much water this is, recall that 1.0 kg of water occupies a volume of 1.0 liter.

Answer 1

To power 2.1 million 31-W Las Vegas lightbulbs for 7 minutes, approximately 1.54 × 107 kg (or 1.54E7 kg) of water must pass through the generators at Hoover Dam, assuming 100% efficiency.

Step-by-step explanation:

To calculate the amount of water needed to power 2.1 million 31-W Las Vegas lightbulbs for 7.0 minutes, we must first determine the total energy required and then use the conservation of energy principle. The power required (P) is the product of the number of bulbs (n) and the power per bulb (Pbulb), which gives us P = 2.1 × 106 × 31 W. To find the total energy (E), we multiply the power by the time (t) in seconds: E = P × t. With t being 7 minutes, we convert this to seconds (t = 7 × 60 s).

Substituting the values, we find E = (2.1 × 106 bulbs) × (31 W/bulb) × (420 s) = 2.7 × 1010 J. Next, we use the energy provided by the falling water, which is given by E = mgh, where m is the mass of the water in kilograms, g is the acceleration due to gravity (9.8 m/s2), and h is the height the water falls (180 m). Assuming 100% efficiency, E = mgh gives m = E / (gh). Substituting the values, we get m = (2.7 × 1010 J) / (9.8 m/s2 × 180 m).

Solving for m, we find that m ≈ 1.54 × 107 kg of water is needed to power the lightbulbs for 7 minutes. Converting to scientific notation, the answer is 1.54E7 kg.