Question

(Really Hard Question)
Mixing. Suppose a brine containing 0.3 kilogram
(kg) of salt per liter (L) runs into a tank initially
filled with 400 L of water containing 2 kg of salt. If
the brine enters at 10 L/min, the mixture is kept uniform by stirring, and the mixture flows out at the same rate. Find the mass of salt in the tank after 10 min [Hint: Let A denote the number
of kilograms of salt in the tank at t min after the process begins and use the fact that rate of increase in A  rate of input  rate of exit.

Answer 1

my book said its 28.1 kg please explain!

Answer 2

https://docs.google.com/document/d/1MCidnRDmwE2RxXkZio0ZbCfYEuAqOZxajf6KW7Yrpyo/edit

Answer 3

Sorry give me a minute i seem to have misunderstood the question

Answer 4

this is a separable equation if that helps.

Answer 5

In each minute, 3 kg of salt is being added, and 1/40 of the salt is being removed

\[\frac{dA}{dt} = 3 - \frac{A(t)}{40}\]

Answer 6

Oh wait, i was trying to do a calculus question with basic math =.=
Let me look at it differently

Answer 7

dA/dt = 3 - At/40
??

Answer 8

A(t) not At

A as a function of t rather than A times t

Answer 9

okay gotcha

Answer 10

https://docs.google.com/document/d/1MCidnRDmwE2RxXkZio0ZbCfYEuAqOZxajf6KW7Yrpyo/edit

please

Answer 11

i still don't get it

Answer 12

1- solve the equation dAdt=3 -A(t)40
step by step
dA(t)/dt =1/40(120 - A(t)) tho=40
So A(t) = c*exp(-t/tho) + 120 = c*exp(-t/40) + 120

2- Use the initial condition you have to determine ‘c’
A(0) = c + 120 = 2
so c=-118

So A(t)= -118* exp(-t/40) +120
3- Simply replace t with the final time ie 10min

A(10 min) = -118*exp(-10/40) + 120 = 28.1 kg