Question

A rain cloud contains 5.32 × 107 kg of water vapor. The acceleration of gravity is 9.81 m/s 2 . How long would it take for a 2.07 kW pump to raise the same amount of water to the cloud’s altitude of 2.61 km? Answer in units of s.

Answer 1

Answer:

20.85 years

Explanation:

2.61 km = 2610 m

2.07 kW = 2070 W

First we need to calculate the potential energy required to take m = $5.32 * 10^7$ kg of rain cloud to an altitude of 2610 m is

$E = mgh = 5.32 * 10^7 * 9.81 * 2610 = 1.4*10^{12}J$

With a P = 2070 W power pump, this can be done within a time frame of

$t = E/P = \frac{1.4*10^{12}}{2070} = 658037739 s$

or 658037739/(60*60) = 182788 hours or 182788 / 24 = 7616 days or 7616 / 365.25 = 20.85 years