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A tank contains V0= 250L of an aqueous solution containing m0=30kg of salt. Water is entering a tank at a rate of 5L/min, and mixture is exiting at a rate of 1L/min. The concentration remains uniform by stirring.

a) Construct a differential equation for the amount of salt in the tank??

I have calculated that volume of solution in the tank to be V(t)=250+4t , but am struggling to involve salt amount into my calculations. Potentially;

Salt amount = Initial salt amount - (Initial density of salt * rate of flow * time since beginning)

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Let amount of salt be $S(t)$ with $S(0)=S_0=30$Kg. Let volume of solution be $V(t)$ with $V(0)=V_0=250$L Let entry rate be $R=5$L/min and exit rate $r=1$L/min.

Then $V(t+\Delta t)=V(t)+(R-r)\Delta t$ to give $\frac{dV}{dt}=(R-r)$ which solves to give $V(t)=V_0+(R-r)t$.

Now $S(t+\Delta t)=S(t)-r\frac{S(t)}{V(t)}\Delta t$. I.e. the amount of salt leaving the system is given by the rate of liquid lost times the concentration. So $\frac{dS}{dt}=-r\frac{S(t)}{V(t)}$. You can then substitute the expression for $V(t)$.

I leave you to solve it but it results in an expression that shows the amount of salt eventually decreases to zero

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