Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 700 L of a dye solution with a concentration of 1 g/L. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 7 L/min, the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value. (Round your answer to one decimal place.)

460.52 mins

Step-by-step explanation:

First of, we take the overall balance for the system,

Let V = volume of solution in the tank at any time = 700 L (constant, because flowrate into and out of the tank is the same)

Rate of flow into the tank = Fᵢ = 7 L/min

Rate of flow out of the tank = F = 7 L/min

Component balance for the concentration of dye in the tank at any time

Let the initial concentration of dye in the tank be C₀ = 1 g/L

The rate of flow of dye coming into the tank = 0 g/L. min

Concentration of dye in the tank, at any time = C

Rate of flow of dye out of the tank = (C g/L * 7 L/min) / (V L) = (7C/V) g/L. min

But V = 700 L

Rate of flow of dye out of the tank = 0.01 g/L. min

The balance,

Rate of Change of the concentration of dye in the tank = (rate of flow of dye into the tank) - (rate of flow of dye out of the tank)

(dC/dt) = 0 - 0.01C

dC / (-0.01C) = dt

∫ dC / (0.01 C) = ∫ - dt

Integrating the left hand side from C₀ to C and the right hand side from 0 to t

100 (In C - In C₀) = - t

In (C/C₀) = - 0.01t

Now, we calculate t when C = 1% of C₀

C = 1% C₀ = 0.01 C₀

In (0.01) = - 0.01t

- 0.01t = - 4.6052

t = 460.52 mins