Applications of Separable Differential Equations

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Quick Reference

The two model families on this page (a cooling body and a well-stirred tank) are the standard physical payoffs of the separation technique. Use this block to point a lecture or a review session at the exact pages where each rate law is derived.

Before You Start: Prerequisite Check

Try This First

A cup of coffee sits at $190°\text{F}$ in a $70°\text{F}$ room. One minute later it reads $180°\text{F}$.

Before any formula, sketch what the temperature does over the next hour. Predict three things on your sketch:

• Does the curve fall fast at the start or slow at the start?
• Does the curve ever reach the room temperature exactly, or only approach it?
• If the room were colder, say $40°\text{F}$, would the coffee cool faster or slower in that first minute?

Hold your three predictions. By the end of this page you will have a single separable equation that produces all three answers, and you can check each one.

The Quantity That Is Changing

In every application here, the changing output is an amount: a temperature, or the quantity of a dissolved substance in a tank. The input is time. The whole modeling move is to describe, in plain words, how fast that amount changes, then write that sentence as a derivative.

For the coffee: the hotter the coffee is relative to the room, the faster it loses heat. As the gap between coffee and room shrinks, the cooling slows. The rate of change of temperature depends on the difference between the body and its surroundings, not on the temperature alone. That single sentence is the entire model, and it is separable.

Prerequisite Hub

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Cross-course prerequisites

Unlocks

No downstream nodes are registered for this skill yet. It is an applied capstone for the separable-equations strand.

Official Definitions

Heat transfer as a separable model

Heat transfer (Newton’s law of cooling) as a separable model. The temperature of a body in an environment at temperature $T_e$ obeys a rate law in which the rate of change of temperature is proportional to the difference between the body temperature and the environment temperature; this gives a separable first-order equation solved by separation of variables.

Logan, 3rd ed., §1.3.2 Heat Transfer, p.31

In symbols, with body temperature $T(t)$ and constant environment temperature $T_e$,

$$\frac{dT}{dt} = -k\,(T - T_e), \qquad k > 0.$$

The constant $k$ carries units of inverse time and measures how quickly the body exchanges heat with its surroundings. The minus sign records that a body hotter than its environment ($T > T_e$) cools.

Chemical reactor / mixing as a separable model

Chemical reactor / mixing model as a separable model. A well-stirred reactor or mixing tank, where rate-in minus rate-out governs the amount of a substance, leads to a first-order equation that is separable (or linear); the separation technique of 1.3.1 is the engine that solves it.

Logan, 3rd ed., §1.3.3 Chemical Reactors, p.32

For a tank holding a fixed volume $V$ of well-stirred liquid, with $Q(t)$ the amount of dissolved substance, the balance “rate in minus rate out” reads

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

When the inflow carries no solute and the tank is well stirred, the outflow rate is proportional to the current concentration $Q/V$, which makes the equation separable.

The Cooling Model

Setting up the rate law

Start from the plain-language sentence: the rate of change of temperature is proportional to how far the body is above (or below) the room. Writing that as a derivative gives the official equation,

$$\frac{dT}{dt} = -k\,(T - T_e).$$

Predict before solving. When $T$ is much larger than $T_e$, the factor $(T - T_e)$ is large, so $\dfrac{dT}{dt}$ is large and negative: fast cooling. As $T$ falls toward $T_e$, the factor shrinks toward zero, so the cooling slows. This already matches the “fast at the start, then slow” prediction from the opener.

Solving by separation

Separate the variables, putting all of $T$ on one side and $t$ on the other:

$$\frac{dT}{T - T_e} = -k\,dt.$$

Integrate both sides. The left side is a logarithm because the numerator is the derivative of the denominator:

$$\ln|T - T_e| = -kt + C.$$

Exponentiate and absorb the sign into a single constant $A$:

$$T - T_e = A e^{-kt} \quad\Longrightarrow\quad \boxed{\,T(t) = T_e + A e^{-kt}\,}.$$

At $t = 0$ this gives $A = T(0) - T_e$, the initial gap between the body and its environment.

Worked Example 1: the cooling coffee

A cup of coffee starts at $190°\text{F}$ in a $70°\text{F}$ room. After $1$ minute it is $180°\text{F}$. Find the temperature as a function of time and the temperature after $10$ minutes.

Predict first. The first minute dropped the temperature by $10°\text{F}$ out of an initial $120°\text{F}$ gap. That is a little under one tenth of the gap, so $k$ should be small, near $0.09$. The coffee should still be well above room temperature at $10$ minutes.

Step 1: write the solution form. Here $T_e = 70$ and $A = T(0) - T_e = 190 - 70 = 120$, so

$$T(t) = 70 + 120\,e^{-kt}.$$

Step 2: use the one-minute reading to find $k$. Set $T(1) = 180$:

$$180 = 70 + 120\,e^{-k} \quad\Longrightarrow\quad 120\,e^{-k} = 110 \quad\Longrightarrow\quad e^{-k} = \frac{110}{120} = \frac{11}{12}.$$

Taking logarithms,

$$k = -\ln\!\left(\frac{11}{12}\right) = \ln\!\left(\frac{12}{11}\right) \approx 0.0870 \ \text{min}^{-1}.$$

This matches the prediction of a small $k$ near $0.09$.

Step 3: evaluate at $t = 10$. Using $e^{-10k} = (e^{-k})^{10} = \left(\tfrac{11}{12}\right)^{10}$,

$$\left(\frac{11}{12}\right)^{10} \approx 0.4189,$$ $$T(10) = 70 + 120(0.4189) \approx 70 + 50.3 = 120.3°\text{F}.$$

Check the prediction. The coffee is still well above the $70°\text{F}$ room after $10$ minutes, as expected. The curve never reaches $70$ exactly: $120\,e^{-kt}$ is positive for every finite $t$, so $T(t) > 70$ always, and $T \to 70$ only in the limit. That answers the second opener prediction.

Worked Example 2: time of death (a reversed question)

A body is found at $85°\text{F}$ in a room held at $68°\text{F}$. Two hours later the body has cooled to $78°\text{F}$. Assuming the body was at the normal living temperature $98.6°\text{F}$ at the moment of death, estimate how long before discovery death occurred.

Predict first. The body has already lost most of the way from $98.6$ down toward $68$. Cooling is slow once the gap is small, so reaching $85°\text{F}$ from $98.6°\text{F}$ should take a couple of hours, not minutes.

Step 1: solution form. With $T_e = 68$, write $T(t) = 68 + A e^{-kt}$, measuring $t$ from the moment of discovery. At discovery, $T(0) = 85$, so $A = 85 - 68 = 17$ and

$$T(t) = 68 + 17\,e^{-kt}.$$

Step 2: find $k$ from the two-hour reading. Set $T(2) = 78$:

$$78 = 68 + 17\,e^{-2k} \quad\Longrightarrow\quad e^{-2k} = \frac{10}{17}, \qquad k = -\tfrac{1}{2}\ln\!\left(\tfrac{10}{17}\right) \approx 0.2653 \ \text{hr}^{-1}.$$

Step 3: solve backward for the moment of death. Death is the time $t^\ast$ (which will be negative, before discovery) when $T = 98.6$:

$$98.6 = 68 + 17\,e^{-k t^\ast} \quad\Longrightarrow\quad e^{-k t^\ast} = \frac{30.6}{17} \approx 1.800.$$

Then

$$-k t^\ast = \ln(1.800) \approx 0.5878 \quad\Longrightarrow\quad t^\ast = -\frac{0.5878}{0.2653} \approx -2.22 \ \text{hr}.$$

Check the prediction. Death occurred about $2.2$ hours before discovery. The value $t^\ast$ is negative, which correctly places death before the observation window, and the magnitude is the “couple of hours” the prediction anticipated.

The Mixing-Tank Model

Setting up the balance

A tank holds $V = 200$ gallons of well-stirred brine. Pure water (no salt) flows in at $4$ gallons per minute, and the mixture flows out at the same $4$ gallons per minute, so the volume stays constant. Let $Q(t)$ be the pounds of salt in the tank.

Rate in. The inflow carries no salt, so the rate in is $0$.

Rate out. The tank is well stirred, so the concentration leaving equals the tank concentration $\dfrac{Q}{V} = \dfrac{Q}{200}$ pounds per gallon. The outflow is $4$ gallons per minute, so

$$\text{rate out} = 4 \cdot \frac{Q}{200} = \frac{Q}{50} \ \text{lb/min}.$$

The balance “rate in minus rate out” gives the separable equation

$$\frac{dQ}{dt} = 0 - \frac{Q}{50} = -\frac{Q}{50}.$$

Worked Example 3: washing out a tank

The tank above starts with $Q(0) = 60$ pounds of salt. Find $Q(t)$ and the amount of salt after $30$ minutes.

Predict first. With pure water entering, the tank can only lose salt. The amount should decay, never rise, and approach zero. After $30$ minutes (slightly more than half of the $50$-minute time constant) somewhat more than half of the salt should be gone.

Step 1: separate and integrate.

$$\frac{dQ}{Q} = -\frac{1}{50}\,dt \quad\Longrightarrow\quad \ln|Q| = -\frac{t}{50} + C \quad\Longrightarrow\quad Q(t) = Q_0\,e^{-t/50}.$$

Step 2: apply the initial amount. With $Q_0 = 60$,

$$Q(t) = 60\,e^{-t/50}.$$

Step 3: evaluate at $t = 30$.

$$Q(30) = 60\,e^{-30/50} = 60\,e^{-0.6} \approx 60(0.5488) \approx 32.9 \ \text{lb}.$$

Check the prediction. A little over half the salt remains after $30$ minutes, the amount only decreases, and $Q \to 0$ as $t$ grows, since the inflow brings in no salt. All three match the prediction.

Worked Example 4: a tank that fills toward a steady concentration

Now let the inflow carry salt. A tank holds $V = 100$ gallons of pure water. Brine with concentration $0.5$ pounds of salt per gallon flows in at $5$ gallons per minute, and the well-stirred mixture flows out at $5$ gallons per minute. Find $Q(t)$.

Predict first. Salt now enters faster than the dilute tank can release it, so $Q$ should rise. As the tank concentration climbs toward the inflow concentration $0.5$ lb/gal, the net gain slows, and $Q$ should level off near $0.5 \times 100 = 50$ pounds.

Step 1: build the balance. Rate in is $0.5 \cdot 5 = 2.5$ lb/min. Rate out is $5 \cdot \dfrac{Q}{100} = \dfrac{Q}{20}$ lb/min, so

$$\frac{dQ}{dt} = 2.5 - \frac{Q}{20} = -\frac{1}{20}\bigl(Q - 50\bigr).$$

Step 2: separate and integrate. The factor $(Q - 50)$ plays the same role the temperature gap played in cooling:

$$\frac{dQ}{Q - 50} = -\frac{1}{20}\,dt \quad\Longrightarrow\quad \ln|Q - 50| = -\frac{t}{20} + C \quad\Longrightarrow\quad Q(t) = 50 + A\,e^{-t/20}.$$

Step 3: apply the initial condition. With $Q(0) = 0$ (pure water), $A = 0 - 50 = -50$, so

$$Q(t) = 50 - 50\,e^{-t/20} = 50\bigl(1 - e^{-t/20}\bigr).$$

Check the prediction. The amount starts at $0$, rises, and approaches $50$ pounds as $t$ grows, exactly the steady amount predicted. The same separable structure that cooled the coffee toward room temperature now fills the tank toward its steady salt level.

One Structure, Two Faces

Notice that Worked Examples 1, 2, and 4 all produced the same shape,

$$y(t) = (\text{steady value}) + (\text{initial gap})\,e^{-kt}.$$

One way to see why: in each case the rate law had the form $\dfrac{dy}{dt} = -k(y - y_\infty)$, where $y_\infty$ is the value that makes the right side zero. Another way to see it: the quantity $u = y - y_\infty$ measures the gap from steady state, and $u$ satisfies pure exponential decay $\dfrac{du}{dt} = -ku$. Which view do you prefer, and why? A reader who likes algebra may favor the substitution $u = y - y_\infty$; a reader who thinks in pictures may prefer watching the gap shrink.

Common Misconceptions

Practice Problems

Mastery Checklist

Novice (Level 1-2):

• I can write the cooling rate law $\dfrac{dT}{dt} = -k(T - T_e)$ from a worded scenario
• I can identify the steady value a model approaches as the value making the rate zero
• I can solve a given cooling model with a known $k$ and an initial temperature

Competent (Level 3-4):

• I can set up a mixing-tank balance as rate in minus rate out, dividing by the volume for the outflow
• I can solve a pure washout tank and a tank that fills toward a steady concentration
• I can find the cooling constant $k$ from a single later reading

Proficient (Level 5):

• I can recover an unknown constant from two readings and judge whether the data are consistent
• I can explain why cooling and inflow tanks share the form (steady value) plus (gap) times $e^{-kt}$
• I can reverse a cooling model to estimate an earlier time, as in the time-of-death problem

Mental Model

The shrinking gap. In every model on this page, one quantity measures a gap from steady state: the temperature above the room, or the salt amount away from its steady level. That gap decays exponentially. The full quantity is the steady value plus the decaying gap. Once you see the gap, every problem here is the same problem wearing a different coat.

Connections

Looking back:

• Separable Differential Equations supplies the separation technique that solves each model
• Differential Equation Modeling supplies the rate in minus rate out and heat-loss reasoning
• Exponential Growth and Decay supplies the $e^{-kt}$ solution family

Cross-course connections:

• MATH162 introduces newtons-law-cooling and exponential-growth-decay-model as first applications of integration
• MATH161 rates-of-change-interpretation supplies the reading of a derivative as a rate

Real-world connections:

• Forensic estimates of time of death use the cooling law in reverse
• Industrial reactors and pollution-flushing models use the well-stirred tank balance
• Drug concentration in the bloodstream follows the same washout structure as the tank

Resources

Last updated: 2026-06-16