Question

1 point) A tank contains 2200 L of pure water. Solution that contains 0.06 kg of sugar per liter enters the tank at the rate 5 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate. (a) How much sugar is in the tank at the beginning

Answer 1

Answer:

Question not complete,

So i will analyse the possible problem

Step-by-step explanation:

A tank contain 2200L

Volume V = 2200L

Solution of 0.06kg/L of sugar

Rate of entry i.e input

dL/dt=5L/min

Let y(t) be the amount of sugar in tank at any time.

But at the beginning there was no sugar in the tank

i.e, y(0)=0, this will be out initial value problem,

The rate of amount of sugar at anytime t is

dy/dt=input amount of sugar - output amount of sugar.

Now,

Then rate of input is

5L/min × 0.06kg/L

Then, input rate= 0.3kg/min

Output rate is

5L/mins × y(t)/2200 kg/L

then, output rate = y(t)/440 kg/min

Now then,

dy/dt=input rate -output rate

dy/dt=0.3-y/440

Cross multiply through by 400

400dy/dt=120-y

Using variable separation

400/(120-y) dy = dt

∫400/(120-y) dy = ∫dt

-400In(120-y)=t +C

In(120-y)=-t/400+C/400

C/400 is another constant, let say B

In(120-y)=-t/400+B

Take exponential of both side

120-y(t)=exp(-t/400+B)

120-y(t)=exp(-t/400)exp(B)

exp(B) is a constant let say C

-y(t)=Cexp(-t/400)-120

y(t)=120-Cexp(t/400)

Now, the initial condition

a. At the start the mass of sugar in the water is 0 because it is just pure water at start.

Therefore y(0)=0,

b. Applying this to y(t)

y(t)=120-Cexp(-t/400)

y=0, t=0

0=120-Cexp(0)

0=120-C

C=120

Therefore,

y(t)=120 - 120exp(-t/400)

Let know the mass rate as t tends to infinity

At infinity

exp(-∞)=1/exp(∞)=1/∞=0

Then,

The exponential aspect tend to 0

Then, y(t)=120 as t tend to ∞