A tank contains 70 kg of salt and 1000 L of water. A solution of a concentration 0.035 kg of salt per liter enters a tank at the rate 8 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 = (kg/L) (b) Set up an initial value problem for the quantity y, in kg, of salt in the tank at time t minutes. dy/dt = (kg/min) y(0) = (kg) (c) Solve the initial value problem in part (b). y(t) = (d) Find the amount of salt in the tank after 5 hours. amount = (kg) (e) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration = (kg/L)

(a) Initial concentration of salt = 0.07 kg/L

(b)

(c)

(d)Therefore the amount of salt after 5 hours is  =38.18 kg

(e) The concentration of salt in the solution in the tank as time approaches infinity is = 0.035 kg/L

Step-by-step explanation:

Given that,

A tank contains 70 kg of salt and 1000 L of water.

(a)

                                =0.07 kg/L

(b)

Let Y(t) be the amount of salt at any instant time t.

Therefore

     =(8×0.035) kg/min

     =0.28 kg/min

Since the rate of water in and out are same , the amount of solution remain constant.

(c)

The above equation can be rewrite as

The coefficient of y is p(t)

The integrating factor of the D.E is  

Multiplying the integrating factor of both sides of D.E

Integrating both sides

At initial when t= 0, y =70

Therefore

⇒C= 70-35

⇒C=35

Therefore

(d)

When t= 5 hour = 300 min

To find the amount of salt after 5 hour , we need to put the value of t in the general solution of D.E

Therefore the amount of salt after 5 hours is =  

                                                                                          = 38.18 Kg

(e)

When t=∞

         = 35 Kg  

Since the amount of water is remain same i.e 1000 L

Therefore the concentration of salt is

                                                                 =0.035 kg/L

y = x + 95

y = 42 + 95

y=137 mm hg