Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 800 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 8 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.)

Answer:t = 100 ln 100Step-by-step explanation:D(t) : The amount of dye (in g) at time t (in min)D(0) = 800 L * 1 g/L = 800 gthe change in D is:[tex]\frac{dD(t)}{dt} =D_{in}- D_{out} \\D_{in}: 0*8\ g/min \\D_{out}: \frac{D(t)}{800} *8\ g/min \\\frac{dD(t)}{dt} = -\frac{1}{100}D(t)[/tex][tex]\frac{dD(t)}{D(t)} =-\frac{1}{100}dt \\\int\limits^{D(t)}_{800} {\frac{1}{D(t)} } \, dD(t) =\int\limits^t_0 {t} \, dt \\ln(\frac{D(t)}{800})=-\frac{1}{100}t \\D(t) = 800e^{-\frac{1}{100}t} \\Solving\ D(t) = 0.01* D(0)=0.01*800 =8 \\8 = 800e^{-\frac{1}{100}t} \\ln (\frac{1}{100})=-\frac{1}{100}t \\100 ln 100 = t[/tex]