A brine solution of salt flows at a constant rate of 8L/min into a large tank that initially held 100L of brine solution in which was dissolved 0.5kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the concentration of salt in the brine entering the tank is 0.05kg/L, determine the mass of salt in the tank after t minutes. When will the concentration of salt in the tank reach 0.02kg/L?

Answer:The mass of salt in the tank after t minutes is [tex]y(t) = 5-4.5e^{-\frac{2t}{25}}[/tex]The concentration of salt in the tank reach 0.02 kg/L when [tex]t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669[/tex]Step-by-step explanation:Let y(t) be the mass of salt (in kg) that is in the tank at any time, t (in minutes).The main equation that we will be using to model this mixing process is:Rate of change of [tex]\frac{dy}{dt}[/tex] = Rate of salt in - Rate of salt outWe need to determine the rate at which salts enters the tank. From the information given we know:The brine flows into the tank at a rate of [tex]8\:\frac{L}{min}[/tex]The concentration of salt in the brine entering the tank is [tex]0.05\:\frac{kg}{L}[/tex]The Rate of salt in = (flow rate of liquid entering) x (concentration of salt in liquid entering)[tex](8\:\frac{L}{min}) \cdot (0.05\:\frac{kg}{L})=0.4 \:\frac{kg}{min}[/tex]Next, we need to determine the output rate of salt from the tank. The Rate of salt out = (flow rate of liquid exiting) x (concentration of salt in liquid exiting)The concentration of salt in any part of the tank at time t is just y(t) divided by the volume. From the information given we know:The tank initially contains 100 L and the rate of flow into the tank is the same as the rate of flow out.[tex](8\:\frac{L}{min}) \cdot (\frac{y(t)}{100} \:\frac{kg}{L})= \frac{2y(t)}{25} \:\frac{kg}{min}[/tex]At time t = 0, there is 0.5 kg of salt, so the initial condition is y(0) = 0.5. And the mathematical model for the mixing process is[tex]\frac{dy}{dt}=0.4-\frac{2y(t)}{25}, \quad{y(0)=0.5}[/tex][tex]\frac{dy}{dt}=0.4-\frac{2y(t)}{25}\\\\\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}\\\\ \frac{1}{2}\frac{dy}{5-y}=\frac{1}{25}dt \\\\\frac{1}{2}\int \frac{dy}{5-y}=\int \frac{1}{25}dt\\\\\frac{1}{2}\left(-\ln \left|5-y\right|+C\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|+C_1\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|=\frac{1}{25}t+C_2\\\\5-y=C_3e^{-\frac{2t}{25} }\\\\y(t) =5-C_3e^{-\frac{2t}{25} }[/tex]Using the initial condition y(0)=0.5[tex]y(t) =5-C_3e^{-\frac{2t}{25} }\\y(0)=0.5=5-C_3e^{-\frac{2(0)}{25}} \\C_3=4.5[/tex]The mass of salt in the tank after t minutes is [tex]y(t) = 5-4.5e^{-\frac{2t}{25}}[/tex]To determine when the concentration of salt is 0.02 kg/L, we solve for t[tex]y(t) = 5-4.5e^{-\frac{2t}{25}}\\\\0.02=5-4.5e^{-\frac{2t}{25}}\\\\5\cdot \:100-4.5e^{-\frac{2t}{25}}\cdot \:100=0.02\cdot \:100\\\\500-450e^{-\frac{2t}{25}}=2\\\\500-450e^{-\frac{2t}{25}}-500=2-500\\\\-450e^{-\frac{2t}{25}}=-498\\\\e^{-\frac{2t}{25}}=\frac{83}{75}\\\\\ln \left(e^{-\frac{2t}{25}}\right)=\ln \left(\frac{83}{75}\right)\\\\\frac{2t}{25}=\ln \left(\frac{83}{75}\right)\\\\t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669[/tex]