Solve Water & Salt Problem in 5 Hrs

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Homework Statement


A tank contains 800 gal of water in which 200 lb of salt is dissolved. Two gallons of fresh water runs in per minute, and 2 gal of the mixture in the tank, kept uniform by stirring, runs out per minute. How much salt is left in the tank after 5 hours ?

Homework Equations





The Attempt at a Solution


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Let ##y## be the amount of salt present in the tank at any given time.

##y' \propto -y ##

##y' = -Cy ##

## \frac {dy}{dt} = -Cy ##

##\int \frac {1}{y} dy = -\int Cdt ##

## y = ke^{-Ct} ##

At ## t=0 , y_{0} = 200 = k##

## y = 200e^{-Ct} ##

I don't know how to find the value of ##C##.
 
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Orodruin said:
What does this tell you?
In the two gallons per minute, how much of it is salt and how much is water ?
 
Orodruin said:
The salt is not notably going to change the volume of the water. (Note that salt water is denser than fresh water!) what does the statement "kept uniform by stirring" tell you?
It has the same concentration of salt throughout the mixture?
 
Orodruin said:
Yes. So what is the rate at which salt is removed?
That is directly proportional to the amount of salt present in the tank at that time... according to my first equation.
 
At t = 0, taking y' = 0.5 lb/min as, in 2 gal/min of the mixture flowing out, there will be 0.5 lb of salt.

## y' = -200Ce^{-c(0)} ##
##0.5 = -200Ce^{-c(0)} ##
## C = -0.0025 ##

at t = 5 hours = 300 min

## y = 200e^{-0.0025(300)} ##

## y = 94.4733 lb ##.

Thanks for your help.
 
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