A outdoor swimming pool is 3.0 m wide and 10.0 m long. If the weekly evaporation is 4.0 cm, how many laters of water must be added to the pool weekly if it does not rain.????
volume of evap=30dm*100dm*.4 dm
= 300*.4=120 dm^3 and of course 1 dm^3 = i liter.
yes, except 30*100 = 3000
To solve this problem, we need to find the volume of water that evaporates from the swimming pool each week and then determine how many liters of water need to be added to compensate for the evaporation.
First, let's calculate the volume of the pool. The volume is equal to the length multiplied by the width multiplied by the depth:
Volume = Length x Width x Depth
As the depth is not given, we can assume a typical depth of around 2 meters for an outdoor swimming pool. Therefore:
Volume = 10.0 m x 3.0 m x 2.0 m = 60.0 cubic meters
Next, we need to convert this volume to liters. Since 1 cubic meter is equal to 1000 liters:
Volume = 60.0 cubic meters x 1000 liters/cubic meter = 60,000 liters
Now we need to determine the volume of water that evaporates each week. The evaporation depth is given as 4.0 cm, which is equivalent to 0.04 meters. Therefore:
Evaporation Volume = Length x Width x Evaporation Depth
Evaporation Volume = 10.0 m x 3.0 m x 0.04 m = 1.2 cubic meters
Converting this volume to liters:
Evaporation Volume = 1.2 cubic meters x 1000 liters/cubic meter = 1200 liters
Finally, to find out how many liters of water must be added weekly, we subtract the evaporation volume from the pool's volume:
Water to be Added Weekly = Volume - Evaporation Volume
Water to be Added Weekly = 60,000 liters - 1200 liters = 58,800 liters
Therefore, you would need to add 58,800 liters of water to the pool every week to compensate for the evaporation, assuming no rainfall occurs.