Calculating Water Volume for Energy Storage in Electric-Power Plants

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Electric-power companies utilize water for energy storage by pumping it from a lower reservoir to a higher one using reversible turbine pumps. To calculate the volume of water needed to store energy produced by a 115 MW power plant in one hour, potential gravitational energy must be considered. The formula used is E = m * g * h, where E is energy, m is mass, g is gravitational acceleration, and h is height. After calculations, the estimated volume of water required is around 7000 cubic meters, but there are discrepancies in the calculations presented. Clarification on the correct approach and values is needed to resolve the confusion.
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Some electric-power companies use water to store energy. Water is pumped by reversible turbine pumps from a low to a high reservoir. To store the energy produced in 1.0 hour by a 115 MW (115 106 W) electric-power plant, how many cubic meters of water will have to be pumped from the lower to the upper reservoir? Assume the upper reservoir is 600 m above the lower and we can neglect the small change in depths within each. Water has a mass of 1000 kg for every 1.0 m3.

This problem has completely confused me. How do I do this?
 
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How much energy is produced in one hour by a 115 MW electric power plant?

Then, consider the energy is being stored as potential gravitational energy. So what is the potential energy of the water? From here, you should be able to solve for the mass of the water
 
We know the potential energy and calculate the mass of the water that has to be transported."1.0 hour by a 115 MW (115 106 W) "

E=115106 W
H=600m
m=?

E=m*g*h <=> m=E/g*h

115106/9.82*600*1000 ~ 7032953/1000 ~7000 m³
 
I tried that and it doesn't work.
 
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