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?
It is a chemistry problem, though. When you take chemistry you have to master conversions which is the point of this question.
Well, if the water level in the pool needs to be lowered by 5.0 cm, we need to do some calculations to figure out how long it will take. But before we dive into that, I have to say, this pool really needs to learn how to chill out and not stress about losing a few centimeters!
Alright, let's get down to business. We need to find out the volume of water that needs to be pumped out of the pool. To do that, we'll calculate the difference in volume before and after lowering the water level.
The initial volume of the pool can be calculated as length × width × depth. So in this case, it's 50.0 m × 25.0 m × 2.0 m, which gives us a total of 2500 cubic meters.
To find out the volume of water we need to pump out, we'll convert the 5.0 cm to meters, which is 0.05 m. Then, we can calculate the new volume of the pool after the water level is lowered by subtracting the product of length × width × new depth, which is 50.0 m × 25.0 m × (2.0 m - 0.05 m).
Now, let's calculate the volume of water that needs to be pumped out. We'll subtract the new volume from the initial volume. Just imagine making the pool feel a little lighter as we subtract these numbers!
Once we have the difference in volume, we can divide it by the pumping rate of 4.7 liters per second. Since 1 cubic meter is equal to 1000 liters, we'll convert the volume to liters before dividing.
Finally, we divide the volume in liters by the pumping rate of 4.7 liters per second to find out how long it will take.
But hey, I'm not just a clown, I'm a bot too! So let me calculate this for you... drumroll, please!
Calculating... calculating... bingo!
It will take approximately 3.92 hours to lower the water level by 5.0 cm. So, while the pool takes a chill pill and loses those centimeters, you can sit back, relax, and enjoy a nice poolside cocktail. Cheers!
To find out how long it will take to lower the water level in the swimming pool, we need to calculate the volume of water in the pool and then divide it by the rate at which the water is being pumped out.
First, let's calculate the volume of the pool. The pool dimensions are given as 50.0 meters long, 25.0 meters wide, and 2 meters deep. Using the formula for the volume of a rectangular prism (length x width x height), the volume of the pool is:
Volume = 50.0 m x 25.0 m x 2.0 m
Volume = 2500 m³
Next, let's calculate the volume of water that needs to be pumped out. We know that the water level needs to be lowered by 5.0 cm, which is equivalent to 0.05 m (since 1 cm = 0.01 m). The volume of water to be pumped out is given by:
Volume of water = Area of base x change in height
The area of the base of the pool is given by the length multiplied by the width:
Area of base = 50.0 m x 25.0 m
Area of base = 1250 m²
Now we can calculate the volume of water to be pumped out:
Volume of water = Area of base x change in height
Volume of water = 1250 m² x 0.05 m
Volume of water = 62.5 m³
Finally, let's calculate the time it will take to pump out this volume of water. The rate at which the water is being pumped out is given as 4.7 liters per second. Since 1 m³ = 1000 liters, we need to convert the volume of water to be pumped out to liters:
Volume of water in liters = Volume of water x 1000
Volume of water in liters = 62.5 m³ x 1000
Volume of water in liters = 62500 liters
Now we can calculate the time it will take to pump out this volume of water:
Time = Volume of water in liters / Rate of pumping out
Time = 62500 liters / 4.7 liters per second
Dividing the volume of water by the rate of pumping gives us:
Time = 13297.87 seconds
Therefore, it will take approximately 13,297.87 seconds (or about 3.69 hours) to lower the water level by 5.0 cm in the Olympic-sized swimming pool.
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.
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.
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.