Answer:
0.88 s
Step-by-step explanation:
Let x(t) be the volume of nitric acid measured in liters after t minute. Therefore the rate of change in the volume of nitric acid (dx / dt) is:
dx / dt = rate of input of nitric acid - rate of output of nitric acid
rate of input of nitric acid = 6 L/min * 20% nitric acid / 100% solution = 1.2 L/min
rate of output of nitric acid = 8 L/min * x L / 200 L = 0.04x L/min
[tex]\frac{dx}{dt}=1.2-0.04x\\\\\frac{dx}{dt}+0.04x=1.2 \\\\The\ integrating\ factor(IF)=e^{\int\limits {0.04} \, dt }=e^{0.04t}\\\\multiply\ through\ by \ IF:\\\\e^{0.04t} \frac{dx}{t}+e^{0.04dt}(0.04x)=e^{0.04t}(1.2)\\\\integrating:xe^{0.04t}=1.2\int\limits e^{0.04t}dt\\\\xe^{0.04t}=30e^{0.04t}+C\\\\x(t)=30+Ce^{-0.04t}\\\\At\ t=0,x=1\\\\x(0)=30+Ce^{-0.04*0}\\\\1=30+C\\\\C=-29\\\\x(t)=30-29e^{-0.04t}\\\\At \ 10\%,x=10%\ of\ 200L=2\\\\2=30-29e^{-0.04t}\\\\29e^{-0.04t}=28\\\\[/tex]
[tex]29e^{-0.04t}=28\\\\e^{-0.04t}=0.9655\\\\t=0.88\ s[/tex]
The percentage of nitric acid in the tank reach 10% at 0.88 s
