The mass of water that must be raised is 
Explanation:
Since the process is 70% efficiency, the power in output to the turbine can be written as

where
is the power in input.
The power in input can be written as

where
W is the work done in lifting the water
t = 3 h = 10,800 s is the time elapsed
The work done in lifting the water is given by

where
m is the mass of water
is the acceleration of gravity
h = 45 m is the height at which the water is lifted
Combining the three equations together, we get:

Where

And solving for m, we find:

Learn more about power:
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