How Much Energy Does Falling Water Generate at Niagara Falls?

[Qnslaught]

Homework Statement

In the Niagara Falls hydroelectric generating plant, the energy of falling water is converted into electricity. The height of the falls is about 50 meters. Assuming that the energy conversion is highly efficient, approximately how much energy is obtained from one kilogram of falling water? Therefore, approximately how many kilograms of water must go through the generators every second to produce a megawatt of power (10e6 watts)?

The Attempt at a Solution

I used the idea that energy of falling water is found with mgh. I found that 1 kg of water falling 50 meters gives off 490.5 joules, so i used (10e6)/(490.5) and got 20387.35984 kg's. The site keeps telling me that answer is wrong and I have no idea what I'm not doing right.

 

[mgb_phys]

mgh = 1 * 9.8 * 50 = 490J/kg
So 10,000,000 W/ 490J/kg/s = 20 400 kg/s

Seems right, a rather optimistic number of significant figure though,

 

[tomcue]

Your attempt at a solution is correct, however, there may be a mistake in the unit conversion. The answer you obtained, 20387.35984 kg's, is actually the mass of water needed per second to produce a megawatt of power, not the energy obtained from one kilogram of falling water.

To find the energy obtained from one kilogram of falling water, we can use the equation E = mgh, where E is the energy in joules, m is the mass in kilograms, g is the acceleration due to gravity (9.8 m/s^2), and h is the height in meters. Plugging in the values, we get:

E = (1 kg)(9.8 m/s^2)(50 m) = 490 joules

Therefore, from one kilogram of falling water, we can obtain 490 joules of energy. To produce a megawatt of power (10e6 watts), we would need:

(10e6 watts)/(490 joules) = 20408.16327 kg's of water per second.

So, your answer was very close, it just needed some unit conversions to get the correct answer. Keep up the good work!