If a municipal reservoir containing 489.0 acre-feet of water is drained at a rate of 2000.0 gallons every hour, how many days will it take to completely empty the reservoir
I asked Google to convert 489 acre-feet to gallons and Google came up with 1.593e+8 gallons. So 1.593e8 gallons x 1 hr/2000 gallon x (1 day/24 hr) = ? days.
BTW this look much more like a math problem than a chemistry problem. oobleck should check this out.
1 acre-foot is 43560 ft^3
so,
489*43560 ft^3 * 7.48gal/ft^3 * 1hr/20000gal * 1day/24hr = 331.938 days
1.593e8 gallons x 1 hr/2000 gallon x (1 day/24 hr) = 3319 days.
489*43560 ft^3 * 7.48gal/ft^3 * 1hr/20000gal * 1day/24hr = 331.938 days
oobleck made a typo and entered 20,000 gallons/hr and not 2000 gallons/hr so his corrected answer is 3319 days also.
dang - I missed that pesky decimal point.
Muchas gracias, DrBobxxx
There is nothing like a good ol' pesky decimal point.
To find the number of days it will take to completely empty the reservoir, we need to convert the given measurements to a common unit.
1 acre-foot is equal to 43,560 cubic feet.
So, the total volume of water in the reservoir is 489.0 acre-feet * 43,560 cubic feet/acre-foot = 21,331,440 cubic feet.
To convert cubic feet to gallons, we need to know that 1 cubic foot is equal to 7.48 gallons.
So, the total volume of water in the reservoir is 21,331,440 cubic feet * 7.48 gallons/cubic foot = 159,328,051.2 gallons.
Now, to find how many days it will take to completely empty the reservoir, we divide the total volume of water by the drainage rate:
159,328,051.2 gallons / 2000.0 gallons/hour = 79,664.0256 hours.
Since there are 24 hours in a day, we divide the number of hours by 24:
79,664.0256 hours / 24 hours/day = 3,319.3344 days.
Therefore, it will take approximately 3,319.3344 days to completely empty the reservoir.