The level of water in an Olympic size swimming pool (50.0 meters long, 25.0 meters wide, and about 2 meters deep) needs to be lowered 5.0 cm. If the water is pumped out at a rate of 4.7 liters per second, how long will it take to lower the water level 5.0 cm?

Answer 1

This is a math problem, not a chemistry problem.
I would convert everything to cm.
50 m = 5000 cm
25 m = 2500 cm
2 m = 200 cm
Volume of pool is width x length x depth = ? cc.
Convert to L. 1000 cc = 1L = initial volume of pool in L.

You want to lower by 5 cm so now the dimensions are
5000 x 2500 x (200-5) = ? cc
Convert that to L = final volume
Take the difference between the two and that gives you the volume that must be pumped out.
Then volume to be pumped in L x 4.7 L/s = # seconds to pump that much water.

Answer 2

Everything is correct above except for the last. It is divide and not multiply.

To get the seconds =(L/4.7)

Also look the units Liter get cancelled and only seconds remain.

Answer 3

You are absolutely correct and I goofed. I didn't do what I preach so much about making sure the units come out right.
I should have written it as
4.7 L/s x ? sec = L to be pumped.
Solve or ? sec.

Answer 4

If you want to lower or deepen the depth, would one not add 5 centimeters rather than subtract, as in to lower the pool? So:(200+5)cm.

Answer 5

It is a chemistry problem, though. When you take chemistry you have to master conversions which is the point of this question.