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Module 9 · L17 of 20 ~40 min ⚡ +95 XP available

Mixing Problems

A tank holds 200 L of water with 5 g of salt dissolved. Fresh water flows in at 3 L/min and the well-mixed solution drains out at the same rate. How much salt remains after 30 minutes? The key is building the right differential equation from rate in minus rate out and that is exactly what this lesson teaches.

Today's hook, A 100 L tank contains 10 g of salt. Pure water flows in at 2 L/min and drains at 2 L/min. Without solving the DE, do you think the salt amount approaches zero, a positive constant, or infinity as time goes on? Write your gut answer before continuing.

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Recall, your gut answer first

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A 100 L tank contains 10 g of salt. Pure water flows in at 2 L/min and the well-mixed solution drains at 2 L/min. Without any calculationwhat do you think happens to the salt concentration over time? Write your prediction below.

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The key idea, rate in minus rate out

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Mixing problems model how the amount (or concentration) of a substance in a tank changes over time when solution flows in and out. The governing idea is:

Let $Q(t)$ be the amount of substance (e.g. grams of salt) at time $t$ minutes. Then:

$$\frac{dQ}{dt} = \text{rate in} - \text{rate out}$$

where rate in = (inflow rate) × (concentration in) and rate out = (outflow rate) × (concentration in tank) = (outflow rate) × $\dfrac{Q}{V(t)}$.

$\dfrac{dQ}{dt} = r_{\text{in}} \cdot c_{\text{in}} - r_{\text{out}} \cdot \dfrac{Q}{V}$

Constant volume

When inflow rate equals outflow rate, $V$ stays constant and the DE is separable with an exponential solution.

Changing volume

If rates differ, $V(t)$ changes and the DE is more complex, always state the volume formula before writing the DE.

Concentration vs amount

Keep track of whether the question asks for $Q$ (amount) or $Q/V$ (concentration). Answers differ by a factor of $V$.

What you'll master

Know

Key facts

$\dfrac{dQ}{dt} = \text{rate in} - \text{rate out}$ governs all mixing problems
Rate out $= r_{\text{out}} \times \dfrac{Q}{V}$ when volume $V$ is known
When inflow rate $=$ outflow rate, volume is constant and the DE is linear/separable

Understand

Concepts

Why the substance amount decays exponentially when pure liquid flows in
How initial conditions determine the particular solution
The physical meaning of the limiting concentration as $t \to \infty$

Can do

Skills

Write the DE for a given tank scenario
Solve the DE using separation of variables and apply initial conditions
Find the amount or concentration at a specified time

Key terms

$Q(t)$Amount of substance (e.g. grams of salt) in the tank at time $t$.

Rate inInflow volume rate (L/min) multiplied by the concentration of the inflowing solution (g/L).

Rate outOutflow volume rate (L/min) multiplied by the current tank concentration $Q/V$ (g/L).

Constant-volume assumptionInflow rate equals outflow rate so $V$ does not change. Makes the DE separable.

Initial conditionThe known amount $Q(0) = Q_0$ used to evaluate the constant of integration $C$.

Long-run behaviourAs $t \to \infty$, $Q \to V \cdot c_{\text{in}}$ (the tank reaches the same concentration as the inflow).

Setting up the differential equation

core concept

Consider a tank with volume $V$ litres containing $Q_0$ grams of a substance at $t=0$. A solution of concentration $c_{\text{in}}$ g/L flows in at $r$ L/min, and the well-mixed tank solution drains at the same rate $r$ L/min.

Since inflow = outflow, the volume $V$ stays constant throughout. The rate of change of substance is:

$$\frac{dQ}{dt} = r \cdot c_{\text{in}} - r \cdot \frac{Q}{V} = r\!\left(c_{\text{in}} - \frac{Q}{V}\right)$$

This is a linear first-order DE. It is also separable. Separating variables:

$$\frac{dQ}{c_{\text{in}} - Q/V} = r\,dt \implies \int \frac{dQ}{Vc_{\text{in}} - Q}\cdot V = r\!\int dt$$

After integration and applying $Q(0)=Q_0$:

$$Q(t) = V c_{\text{in}} + (Q_0 - V c_{\text{in}})\,e^{-rt/V}$$

As $t \to \infty$, $Q \to Vc_{\text{in}}$, the equilibrium amount. If pure liquid flows in ($c_{\text{in}}=0$), then $Q(t) = Q_0\,e^{-rt/V}$, simple exponential decay.

Answer to the hook. With pure water ($c_{\text{in}}=0$), $Q(t) = 10\,e^{-2t/100} = 10\,e^{-t/50}$. The salt decays to zero, it never reaches a positive constant.

Pause, copy the general dilution DE: $\frac{dQ}{dt}=c_{\text{in}}r_{\text{in}}-\frac{Q}{V}r_{\text{out}}$ and identify what each term represents physically into your book.

Quick check: A 200 L tank contains 8 g of salt. Pure water (0 g/L) flows in at 4 L/min and exits at 4 L/min. Which differential equation correctly models this?

Solving the DE step by step

core concept

We just saw the dilution DE setup: rate of change = inflow concentration $\times$ inflow rate $-$ outflow concentration $\times$ outflow rate, giving $\frac{dQ}{dt}=c_{\text{in}}\cdot r_{\text{in}}-\frac{Q}{V}\cdot r_{\text{out}}$. That raises a question: for the constant-volume pure-inflow case ($c_{\text{in}}=0$), this simplifies to $\frac{dQ}{dt}=-\frac{r}{V}Q$, how do you solve this and what is its physical interpretation? This card answers it → pure exponential decay $Q(t)=Q_0 e^{-rt/V}$; the concentration halves every $V\ln 2/r$ minutes.

For the constant-volume pure-inflow case, $\dfrac{dQ}{dt} = -\dfrac{r}{V}Q$, which is exponential decay:

Separate: $\dfrac{dQ}{Q} = -\dfrac{r}{V}\,dt$
Integrate: $\ln|Q| = -\dfrac{r}{V}t + C_1$
Exponentiate: $Q = Ae^{-rt/V}$ where $A = e^{C_1}$
Apply IC: $Q(0) = Q_0 \implies A = Q_0$, so $Q(t) = Q_0\,e^{-rt/V}$

For a non-zero inflow concentration $c_{\text{in}}$, let $u = Q - Vc_{\text{in}}$; then $\dfrac{du}{dt} = -\dfrac{r}{V}u$, giving $u = (Q_0 - Vc_{\text{in}})\,e^{-rt/V}$, hence:

$$Q(t) = Vc_{\text{in}} + (Q_0 - Vc_{\text{in}})\,e^{-rt/V}$$

Exam tip. Always re-derive this from scratch in an exam, don't memorise the final formula. Set up $dQ/dt$, separate, integrate, and apply the IC. You earn method marks at every step.

For the constant-volume pure-inflow case, $\dfrac{dQ}{dt} = -\dfrac{r}{V}Q$, which is exponential decay:

Pause, copy the pure-decay case: $\frac{dQ}{dt}=-\frac{r}{V}Q\Rightarrow Q=Q_0 e^{-rt/V}$ with its physical meaning (exponential washout) into your book.

Did you get this? True or false: if pure water flows into a tank initially containing salt, the amount of salt in the tank decays exponentially to zero.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · PURE INFLOW

A 100 L tank contains 10 g of salt. Pure water flows in at 2 L/min and drains at 2 L/min. Find $Q(t)$ and the amount of salt after 50 minutes.

$V = 100$, $r = 2$, $c_{\text{in}} = 0$, $Q_0 = 10$. Rate in $= 0$, rate out $= 2 \cdot \dfrac{Q}{100} = \dfrac{Q}{50}$. DE: $\dfrac{dQ}{dt} = -\dfrac{Q}{50}$

Identify all parameters first. Pure inflow means rate in = 0.

PROBLEM 2 · SALINE INFLOW

A 200 L tank initially contains 50 g of salt. Brine of concentration 0.5 g/L flows in at 4 L/min; the well-mixed solution drains at 4 L/min. Find $Q(t)$.

$V=200$, $r=4$, $c_{\text{in}}=0.5$, $Q_0=50$. Rate in $= 4 \times 0.5 = 2$ g/min. Rate out $= 4 \times \dfrac{Q}{200} = \dfrac{Q}{50}$. DE: $\dfrac{dQ}{dt} = 2 - \dfrac{Q}{50}$.

Equilibrium amount: $Vc_{\text{in}} = 200 \times 0.5 = 100$ g.

PROBLEM 3 · FIND TIME

Using the setup from Problem 1 ($Q(t) = 10e^{-t/50}$), find when the salt amount first falls below 1 g. Give your answer to the nearest minute.

Solve $10e^{-t/50} = 1$. Divide: $e^{-t/50} = 0.1$.

Set $Q(t) = 1$ and isolate the exponential.

Fill the gap: A 50 L tank contains 20 g of salt. Pure water flows in at 5 L/min; outflow rate is 5 L/min. The solution is $Q(t) = 20e^{-t/}$$.$

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting to divide by $V$ in the rate out

Rate out $= r_{\text{out}} \times \dfrac{Q}{V}$, not just $r_{\text{out}} \times Q$. The $Q/V$ term converts the amount to a concentration. Omitting the $V$ is the single most common error in HSC mixing problems.

Trap 02

Assuming constant volume when rates differ

If inflow rate ≠ outflow rate, $V(t) = V_0 + (r_{\text{in}} - r_{\text{out}})t$. You must substitute this time-dependent volume into the rate-out term before separating variables.

Trap 03

Confusing $Q$ (amount) with $Q/V$ (concentration)

If the question asks for concentration, divide your final $Q(t)$ by $V$ (or $V(t)$ if volume is not constant). Check the units: g vs g/L.

Did you get this? True or false: if inflow rate is 3 L/min and outflow rate is 5 L/min, the volume of the tank stays constant.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Write the differential equation (but do not solve it) for: a 500 L tank with 30 g of salt, pure water flowing in at 10 L/min, and solution draining at 10 L/min.

Solve the DE from Activity 1 and find $Q(t)$.

Using your answer to Activity 2, find the time at which 10 g of salt remains. Give your answer to the nearest minute.

A 100 L tank contains 0 g of salt. Brine of 2 g/L flows in at 3 L/min; outflow rate is 3 L/min. Write and solve the DE. State the long-run concentration.

Explain in words why the equilibrium amount of substance in a constant-volume tank equals $Vc_{\text{in}}$.

Odd one out: Three of these statements about mixing problems are correct. Which one is NOT?

Revisit your thinking

Earlier you predicted what would happen to the salt in the 100 L tank when pure water flows in at 2 L/min.

The answer: $Q(t) = 10e^{-t/50}$, which decays exponentially to zero. The key insight is that the outflow continuously removes salt while the inflow adds none, so the salt is always being diluted at a rate proportional to what remains. This is identical in form to radioactive decay.

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Multiple choice

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Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. A 150 L tank contains 6 g of salt. Pure water flows in at 3 L/min; the well-mixed solution drains at 3 L/min. Write the differential equation for $Q(t)$ and state its solution. (2 marks)

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ApplyBand 43 marks

Q2. A 400 L tank initially contains 20 g of salt. Brine of concentration 0.1 g/L flows in at 5 L/min; outflow rate is 5 L/min. Find $Q(t)$ and the concentration (in g/L) after 80 minutes. (3 marks)

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AnalyseBand 53 marks

Q3. For the scenario in Q1, find the time (to the nearest minute) at which the salt amount first falls below 2 g. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\dfrac{dQ}{dt} = -\dfrac{Q}{50}$ 2. $Q(t) = 30e^{-t/50}$ 3. $t = 50\ln 3 \approx 55$ min 4. $\dfrac{dQ}{dt} = 6 - \dfrac{3Q}{100} = 6 - 0.03Q$; $Q(t) = 200(1 - e^{-3t/100})$; long-run: 2 g/L 5. At equilibrium $dQ/dt = 0 \implies r c_{\text{in}} = r \cdot Q/V \implies Q = Vc_{\text{in}}$.

Q1 (2 marks): $\dfrac{dQ}{dt} = -\dfrac{3Q}{150} = -\dfrac{Q}{50}$ [1]. $Q(t) = 6e^{-t/50}$ [1].

Q2 (3 marks): $Vc_{\text{in}} = 400 \times 0.1 = 40$ g. $\dfrac{dQ}{dt} = 0.5 - \dfrac{Q}{80}$. $Q(t) = 40 + (20-40)e^{-t/80} = 40 - 20e^{-t/80}$ [2]. At $t=80$: $Q = 40 - 20e^{-1} \approx 40 - 7.36 = 32.64$ g; concentration $= 32.64/400 \approx 0.082$ g/L [1].

Q3 (3 marks): $6e^{-t/50} = 2 \implies e^{-t/50} = 1/3 \implies -t/50 = -\ln 3 \implies t = 50\ln 3 \approx 55$ min [3].

Boss battle · The Mixing Master

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Five timed questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering mixing problems. Lighter alternative to the boss.

Mark lesson as complete

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← Lesson 16 · Modelling with DEs Lesson 18 · Newton's Law of Cooling →

Module overview · Extension 1 · Checkpoint 4