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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 calculation — what 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

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that 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

earn bronze · silver · gold

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

⚔ Enter the arena

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