Question

(1 point) A tank contains 50 kg of salt and 1000 L of

water. A solution of a concentration 0.025 kg of salt
per liter enters a tank at the rate 5 L/min. The solution
is mixed and drains from the tank at the same rate.
(a) What is the concentration of our solution in the
tank initially?
concentration = .05 (kg/L)
(b) Find the amount of salt in the tank after 1 hours.
O (kg)
amount =
(c) Find the concentration of salt in the solution in the
tank as time approaches infinity.
concentration = .025 (kg/L)

Answer 1

Your answers for (a) and (c) are correct.

(b) Salt flows into the tank at a rate of

`\left(0.025 (\rm kg)/(\rm L)\right) \left(5 (\rm L)/(\rm min)\right) = 0.125 (\rm kg)/(\rm min) = \frac18 (\rm kg)/(\rm min)`

If
`A(t)` is the amount of salt (in kg) in the tank at time
`t` (in min), then the salt flows out of the tank at a rate of

`\left((A(t))/(1000+(5-5)t) (\rm kg)/(\rm L)\right) \left(5 (\rm L)/(\rm min)\right) = (A(t))/(200) (\rm kg)/(\rm min)`

The net rate of change in the amount of salt in the tank at any time is then governed by the linear differential equation

`(dA)/(dt) = \frac18 - (A(t))/(200)`

`(dA)/(dt) + (A(t))/(200) = \frac18`

I'll solve this with the integrating factor method. The I.F. is

`\mu = \exp\left(\displaystyle \int (dt)/(200)\right) = e^(t/200)`

Distributing
`\mu` on both sides of the ODE gives

`e^(t/200) (dA)/(dt) + \frac1{200} e^(t/200) A(t) = \frac18 e^(t/200)`

`\frac d{dt} \left(e^(t/200) A(t)\right) = \frac18 e^(t/200)`

Integrate both sides.

`\displaystyle \int \frac d{dt} \left(e^(t/200) A(t)\right) \, dt = \frac18 \int e^(t/200) \, dt`

`e^(t/200) A(t) = \frac{200}8 e^(t/200) + C`

`A(t) = 25 + Ce^(-t/200)`

Given that
`A(0)=50\,\rm kg`, we find

`50 = 25 + Ce^0 \implies C = 25`

so that

`A(t) = 25 + 25e^(-t/200)`

Then the amount of salt in the tank after 1 hr = 60 min is

`A(60) = 25 + 25e^(-60/200) = \boxed{25 \left(1 + e^(-3/10)\right)}`