Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 L of a dyesolution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water owing in at the rate of 2 L/min, the well-stirred solution owing out at the same rate.

a. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value?

Time, t that it will elapse before the concerntration of the dye in the tank reaches1% of its original value is 460.52 seconds

Step-by-step explanation:

Let Q (t) = amount of dye for all time, t.

Let Q' = rate in - rate out.

But Q/200 = concerntration of dye

Therefore rate out = Q/200 * 2.

Q'/Q = - 1/100

Dividing by Q gives

Ln/Q / + c = - 1/100t + c1

Integrating both sides with respect to t

Ln/Q / = - 1/100t + c2.

c1 and c1 are just another constants

Q = c3e^-t/100

Exponentiating both sides will cause the absolute values to get absorbed into the constants giving rise to a new constant c3

200 = c3-0/100

c3 = 200

2 = 200e^-t/100

Ln1/100 = - t/100

Ln0.01 = - t / 100

Cross multiply

t = Ln0.01 * 100

t = 4.6052 * 100

t = 460.52seconds