Mixing and cooling look like different physical processes, but their equations have the same shape: the current state moves toward a stable equilibrium, and the distance from that equilibrium decays exponentially.

A constant-volume mixing model

Suppose a tank initially contains $100$ gallons of well-stirred solution and $Q_0$ pounds of dissolved salt. To track the salt over time, define:

• $Q(t)$ as the amount of salt, in pounds, after $t$ minutes;
• $r>0$ as both the inflow and outflow rate, in gallons per minute;
• $\frac14$ lb/gal as the salt concentration of the incoming solution.

Because liquid enters and leaves at the same rate, the volume stays fixed at $100$ gallons. The amount of salt changes according to

$$\text{rate of change}=\text{rate in}-\text{rate out}.$$

The incoming salt rate is concentration times flow rate:

$$\left(\frac14\frac{\mathrm{lb}}{\mathrm{gal}}\right) \left(r\frac{\mathrm{gal}}{\mathrm{min}}\right) =\frac r4\frac{\mathrm{lb}}{\mathrm{min}}.$$

Because the mixture is continuously stirred, the concentration is uniform throughout the tank at any instant. The outgoing fluid therefore carries salt at that same concentration:

$$\frac{Q(t)}{100}\frac{\mathrm{lb}}{\mathrm{gal}}.$$

Thus the outgoing salt rate is

$$\left(\frac{Q(t)}{100}\frac{\mathrm{lb}}{\mathrm{gal}}\right) \left(r\frac{\mathrm{gal}}{\mathrm{min}}\right) =\frac r{100}Q(t)\frac{\mathrm{lb}}{\mathrm{min}}.$$

Subtracting the two rates gives the initial-value problem

$$\boxed{ Q'+\frac r{100}Q=\frac r4, \qquad Q(0)=Q_0 }.$$

Keeping the units visible is a useful error check: every term in the differential equation must have units of pounds per minute.

Solving the mixing equation

Because this is a first-order linear differential equation, begin with the integrating factor:

$$\mu(t)=e^{\int r/100\,dt}=e^{rt/100}.$$

After multiplying the equation by $\mu(t)$, the left side becomes a product derivative:

$$\frac d{dt}\left(e^{rt/100}Q(t)\right) =\frac r4e^{rt/100}.$$

Integrating both sides gives

$$\begin{aligned} e^{rt/100}Q(t) &=\frac r4\int e^{rt/100}\,dt+C\\ &=25e^{rt/100}+C. \end{aligned}$$

Dividing by the integrating factor,

$$Q(t)=25+Ce^{-rt/100}.$$

The initial condition determines the constant:

$$Q_0=Q(0)=25+C, \qquad C=Q_0-25.$$

Therefore,

$$\boxed{ Q(t)=25+(Q_0-25)e^{-rt/100} }.$$

The number $25$ is not an arbitrary constant. At equilibrium, the tank has the same concentration as the incoming solution, so

$$100\,\mathrm{gal}\left(\frac14\frac{\mathrm{lb}}{\mathrm{gal}}\right) =25\,\mathrm{lb}.$$

For $r>0$, the exponential factor tends to zero. If $Q_0<25$, the amount rises toward $25$; if $Q_0>25$, it falls toward $25$; and if $Q_0=25$, it stays there.

Stable salt equilibrium at 25 pounds

• Q(0) = 50 lb
• Q(0) = 25 lb
• Q(0) = 10 lb

How fast does the tank approach equilibrium?

Take $Q_0=50$ lb and $r=3$ gal/min. Then

$$Q(t)=25+25e^{-3t/100}.$$

Being within $2\%$ of the $25$-lb equilibrium means

$$|Q(t)-25|\le0.02(25)=0.5.$$

Since this solution approaches equilibrium from above, it first enters the tolerance band when $Q(T)=25.5$. Solving,

$$\begin{aligned} 25+25e^{-3T/100}&=25.5\\ e^{-3T/100}&=\frac1{50}\\ T&=\frac{100}{3}\ln50\\ &\approx130.4\ \mathrm{min}. \end{aligned}$$

Now consider the inverse design problem: what flow rate is needed to reach the same tolerance within $45$ minutes? The threshold case satisfies

$$25+25e^{-45r/100}=25.5.$$

Therefore,

$$\boxed{ r_{\min}=\frac{100}{45}\ln50 \approx8.69\ \mathrm{gal/min} }.$$

Any $r\ge r_{\min}$ reaches the tolerance band no later than $45$ minutes.

Newton's law of cooling

Thermal transfer follows the same stable-equilibrium pattern. Let $T(t)$ be an object's temperature and let $T_{\mathrm{room}}$ be the constant temperature of its surroundings. Newton's law of cooling states that the rate of temperature change is proportional to the temperature difference:

$$\boxed{ \frac{dT}{dt}=-k\left(T-T_{\mathrm{room}}\right), \qquad k>0 }.$$

The minus sign makes the model move toward room temperature:

• If $T>T_{\mathrm{room}}$, then $T'<0$, so the object cools.
• If $T<T_{\mathrm{room}}$, then $T'>0$, so the object warms.
• If $T=T_{\mathrm{room}}$, then $T'=0$, so the temperature is at equilibrium.

The equation is both separable and linear. Separating variables gives

$$\int\frac{dT}{T-T_{\mathrm{room}}} =\int-k\,dt,$$

so

$$\ln|T-T_{\mathrm{room}}|=-kt+C.$$

After exponentiating and absorbing the sign into the arbitrary constant,

$$T(t)=T_{\mathrm{room}}+C_1e^{-kt}.$$

If $T(0)=T_0$, then $C_1=T_0-T_{\mathrm{room}}$. Hence

$$\boxed{ T(t)=T_{\mathrm{room}} +\left(T_0-T_{\mathrm{room}}\right)e^{-kt} }.$$

The temperature difference $T(t)-T_{\mathrm{room}}$ decays exponentially, so $T(t)$ approaches $T_{\mathrm{room}}$ without crossing it under this model.

Example: cooling coffee

A cup of coffee is poured at $90^\circ\mathrm C$ into a room at $20^\circ\mathrm C$. After one minute, its temperature is $85^\circ\mathrm C$. The model is

$$T(t)=20+70e^{-kt}.$$

Using $T(1)=85$,

$$85=20+70e^{-k}, \qquad e^{-k}=\frac{65}{70},$$

so

$$k=-\ln\left(\frac{65}{70}\right) \approx0.0741\ \mathrm{min}^{-1}.$$

To find when the coffee reaches $60^\circ\mathrm C$, solve

$$60=20+70e^{-kt}.$$

This gives

$$\boxed{ t=-\frac1k\ln\left(\frac{40}{70}\right) \approx7.55\ \mathrm{min} }.$$