A tank initially contains 200 liters of water with 10 kg of dissolved salt. Saltwater with a concentration of 0.1 kg/L is pumped in at 5 L/min, while the well-mixed solution drains at 5 L/min. How much salt will remain after 30 minutes?

["How Much Salt Remains After 30 Minutes in a Mixing Tank?", "In a common fluid dynamics and chemistry problem, understanding the behavior of salt in a well-mixed tank helps model real-world processes like water treatment, brine solutions, and chemical reactors. This article explores a scenario where salt dissolves in water in a tank being continuously fed and drained, and calculates how much salt remains after 30 minutes.", "### The Setup", "- Initial volume: 200 liters\n- Initial dissolved salt: 10 kg\n- Initial concentration: 0.1 kg/L\n- Inflow saltwater concentration: 0.1 kg/L\n- Flow rates: both inflow and outflow at 5 L/min\n- Total volume remains constant (200 L) due to equal inflow and outflow", "We assume perfect mixing — the solution is homogeneous so concentration is uniform throughout at all times.", "---", "### Key Principles", "- Since inflow and outflow rates are equal (5 L/min), the total volume does not change.\n- The amount of salt entering per minute = flow rate × concentration out = ( 5 , \ ext{L/min} \ imes 0.1 , \ ext{kg/L} = 0.5 , \ ext{kg/min} )\n- Let ( Q(t) ) be the salt (in kg) in the tank at time ( t ) (in minutes).\n- Salt leaves at the same rate (5 L/min × concentration = 5 Q(t) kg/min outflow), assuming constant concentration in outflow (valid for perfect mixing).\n- Thus, the rate of change of salt is:\n [\n \frac{dQ}{dt} = \ ext{salt inflow} - \ ext{salt outflow} = 0.5 - 5Q(t)\n ]", "---", "### Solving the Differential Equation", "We solve:", "[\n\frac{dQ}{dt} = 0.5 - 5Q\n]", "This is a first-order linear ODE. Rewrite:", "[\n\frac{dQ}{dt} + 5Q = 0.5\n]", "Using integrating factor ( \mu(t) = e^{\int 5,dt} = e^{5t} ):", "Multiply both sides:", "[\ne^{5t} \frac{dQ}{dt} + 5e^{5t} Q = 0.5 e^{5t}\n]", "[\n\frac{d}{dt} \left( Q e^{5t} \right) = 0.5 e^{5t}\n]", "Integrate both sides:", "[\nQ e^{5t} = \int 0.5 e^{5t} dt = 0.5 \cdot \frac{1}{5} e^{5t} + C = 0.1 e^{5t} + C\n]", "[\nQ(t) = 0.1 + C e^{-5t}\n]", "Apply initial condition ( Q(0) = 10 ) kg:", "[\n10 = 0.1 + C \Rightarrow C = 9.9\n]", "Thus, the solution is:", "[\nQ(t) = 0.1 + 9.9 e^{-5t}\n]", "Now compute ( Q(30) ):", "[\nQ(30) = 0.1 + 9.9 e^{-5 \ imes 30} = 0.1 + 9.9 e^{-150}\n]", "Since ( e^{-150} ) is effectively zero (on the order of ( 10^{-65} )), we approximate:", "[\nQ(30) \approx 0.1 , \ ext{kg}\n]", "---", "### Interpretation", "Despite continuous inflow of saltwater, the tank’s perfectly mixed nature causes salt to rapidly dilute due to efficient outflow. After just 30 minutes, nearly all salt has left the system because the outflow rate matches inflow exactly, and autonomous dilution dominates in well-mixed tanks.", "---", "### Final Answer", "After 30 minutes, approximately 0.1 kg of salt remains in the tank.", "---", "### Why This Matters", "This model illustrates a classic tank with continuous mixing and flow, essential in chemical engineering, environmental science, and industrial process design. Understanding salt balance in such systems helps optimize water desalination, brine management, and liquid storage systems.", "For exact theoretical persistence (if inflow had a slightly different concentration), steady-state analysis shows equilibrium where inflow salt equals outflow salt — reinforcing that time-dependent predictions rely on exponential decay driven by outflow dynamics.", "---", "Keywords: salt tank dynamics, well-mixed tank, differential equation salt balance, continuous inflow and outflow, salt concentration decay, fluid mix model, process control."]