A tank initially contains 200 liters of fresh water. Brine containing 2 N/L dissolved salt enters the tank. At the same rate. Find the salt concentration in the tank after 5 minutes.&nbsp;

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A tank initially contains 200 liters of fresh water. Brine containing 2 N/L dissolved salt enters the tank. At the same rate. Find the salt concentration in the tank after 5 minutes.&amp;nbsp;

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Let x(t) be the amount of salt (in moles) in the tank at time t (in minutes), and let v(t) be the volume of the liquid in the tank at time t.At time t=0, we have x(0)=0 and v(0)=200. The brine enters the tank at a rate of 2 liters per minute, so the volume of liquid in the tank at time t is v(t)=200+2t.The amount of salt in the brine that enters the tank at time t is 2t·2=4t moles. Since the rate of flow of the brine is constant, the concentration of salt in the liquid in the tank is the amount of salt divided by the volume of the liquid, or:C(t)=x(t)v(t)To find C(5), we need to find x(5) and v(5).The amount of salt in the tank at time t satisfies the differential equation:dxdt=4t200+2tTo solve this equation, we first separate the variables and integrate both sides:∫0x(5)dxt=∫054t200+2tdtThe left-hand side evaluates to ln(x(5))−ln(0)=∞, while the right-hand side can be integrated using the substitution u=100+t:∫054t200+2tdt=∫1001104(u−100)udu=4ln(1110)Therefore, we have:ln(x(5))=∞+4ln(1110)x(5)=exp(4ln(1110))=4425The volume of the liquid in the tank at time t=5 is:v(5)=200+2·5=210Therefore, the salt concentration in the tank at time t=5 is:C(5)=x(5)v(5)=44/25210=44525≈0.0838&amp;nbsp;moles/LThus, the salt concentration in the tank after 5 minutes is approximately 0.0838 moles/L.

A tank initially contains 200 liters of fresh

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