Question
A tank holds 200 L of brine containing 20 kg of dissolved salt. Pure water flows in at 5 L/min, and the well-mixed solution flows out at the same rate. Find the amount of salt in the tank after 40 minutes.
Solution — Step by Step
Let S(t) be the amount of salt (kg) at time t (min).
Rate in = 5 L/min×0 kg/L=0 (pure water).
Rate out = 5 L/min×200S kg/L=40S kg/min.
So dtdS=−40S.
SdS=−40dt.
Integrating: lnS=−t/40+C.
S(t)=S0e−t/40.
At t=0, S=20, so S0=20.
S(t)=20e−t/40.
S(40)=20e−1=20/e≈20/2.718≈7.36 kg.
Final answer: Salt remaining = 20/e≈7.36 kg.
Why This Works
This is the classic mixing problem — a first-order linear ODE in disguise. Volume stays constant (200 L) because inflow rate equals outflow rate, so the concentration in the tank is just S/200. The salt leaves at “outflow rate × concentration”, giving an exponential decay.
The factor 40 in e−t/40 is the time constant — it’s V/r=200/5, the time it would take to replace the entire tank volume.
Alternative Method
Use the integrating-factor approach: rewrite as dS/dt+S/40=0. Integrating factor: et/40. d(Set/40)/dt=0, so Set/40=C, giving S=Ce−t/40. Same answer; useful when the inflow has nonzero salt.
Common Mistake
Students sometimes write rate out = 5×S — forgetting to divide by total volume to get concentration. The unit check catches this: rate out should be in kg/min, so we need L/min×kg/L=kg/min. Concentration is kg per litre, so S/V, not S alone.