Finding the salt amount in a draining tank after 5 hours

[Monsterboy]

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 $C$.

 

[Orodruin]

What does this tell you?

2 gal of the mixture in the tank, kept uniform by stirring, runs out per minute.

 

[Monsterboy]

What does this tell you?

In the two gallons per minute, how much of it is salt and how much is water ?

 

[Orodruin]

In the two gallons per minute, how much of it is salt and how much is water ?

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?

 

[Monsterboy]

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]

It has the same concentration of salt throughout the mixture?

Yes. So what is the rate at which salt is removed?

 

[Monsterboy]

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.

 

[Orodruin]

That is directly proportional to the amount of salt present in the tank at that time... according to my first equation.

Yes, but you need the numbers and you can get them from this line of argumentation.

 

[Monsterboy]

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.