Question

A tank initially contains 50 liters of water with 15kg of salt mixed in. Salt water with concentration 1/3 kg/L runs in at the rate of 2 L/min and the well mixed solution runs out of the tank at the same rate. How much salt is in the tank after 5 min?

Answer 1

Answer:

13.57 kg

Step-by-step explanation:

Let the amount of salt in a tank be =$\frac{salt in the tank}{volume of salt in the tank}$

                                                         =    $\frac{15}{50}$

                                                         =  0.3 kg/L

The concentration of salt coming into the tank = 0.333kg/L

We know that the concentration of salt leaving the tank should be:

amount of salt in the tank/volume of tank = y(t)/50 L

rate of change of salt = rate in - rate out

$\frac{dy}{dt}$ = 0.66 - $\frac{y}{50}$

solving the differential equation gives :

ln y = 0.66/y (t) - 0.02 t + C

Initial conditions, t= 0; y (0) = 15

then ln 15 = C

ln y = $\frac{0.66}{y}$t - 0.02t + ln 15

ln$\frac{y}{15}$ = 0.02t

y = 15e( exp -0.02t)

the amount of salt in 5 mins will be:

y (t) = 15e(exp -0.02 (5)

      = 13.57 kg