Cooling and mixing

What you will be able to do

• write and solve Newton's law of cooling, and find $k$ from two readings;
• set up a mixing problem for a tank of constant volume and for a tank whose volume changes;
• find the steady state of a model and say what it means.

Newton's law of cooling

A cup of tea cools fast when it is very hot and slowly when it is nearly at room temperature. Newton's idea was that the rate of cooling is proportional to the gap between the temperature $T$ of the object and the temperature $T_s$ of its surroundings:

The sign matters. If $T > T_s$ the right side is negative and the object cools. If $T < T_s$ the right side is positive and the object warms. In both cases $T$ moves towards $T_s$.

To solve, let $u = T - T_s$. Since $T_s$ is constant, $\dfrac{du}{dt} = \dfrac{dT}{dt} = -ku$. This is the decay equation, so $u = u_0 e^{-kt}$. Putting back $T$:

The gap shrinks by the same factor in equal times. The law is a good model when the temperature difference is small, or when air or water moves past the object. It is a poor model for a very hot object that also loses heat by radiation.

Example 1. A cup of coffee at $85\ ^\circ\text{C}$ stands in a room at $21\ ^\circ\text{C}$. The constant is $k = 0.05\ \text{min}^{-1}$. Find its temperature after $12$ minutes.

1. $T = 21 + (85 - 21)e^{-0.05 \times 12} = 21 + 64e^{-0.6}$.
2. $e^{-0.6} = 0.5488$, so $T = 21 + 35.12 = 56.1\ ^\circ\text{C}$.

Example 2. The constant is not always given. A thermometer reads $95\ ^\circ\text{C}$ at $t = 0$ and $75\ ^\circ\text{C}$ at $t = 5\ \text{min}$ in a room at $20\ ^\circ\text{C}$. When does it read $40\ ^\circ\text{C}$?

1. The gap falls from $75$ to $55$ in $5$ minutes. So $e^{-5k} = \dfrac{55}{75}$ and $k = \dfrac{1}{5}\ln\dfrac{75}{55} = 0.0620\ \text{min}^{-1}$.
2. For $40\ ^\circ\text{C}$ the gap is $20$. Then $e^{-kt} = \dfrac{20}{75}$ and $t = \dfrac{1}{k}\ln\dfrac{75}{20} = 21.3\ \text{min}$.

Two readings are enough. The program below does the same work.

Mixing problems

A tank holds a well-stirred liquid. Brine with a known concentration flows in. The mixture flows out. We follow the amount $Q$ of salt in the tank, in kilograms. The idea is a balance:

Salt in each minute is the inflow rate times the concentration of the brine that enters. Salt out each minute is the outflow rate times the concentration in the tank. The concentration in the tank is $\dfrac{Q}{V}$, because the liquid is stirred. So

Constant volume. If the inflow and outflow rates are both $r$, the volume $V$ does not change. The equation is $\dfrac{dQ}{dt} = rc - \dfrac{r}{V}Q$. With $u = Q - cV$ this becomes $\dfrac{du}{dt} = -\dfrac{r}{V}u$. So

The salt moves towards the steady state $cV$, the amount that makes the tank as salty as the brine. The ratio $\dfrac{V}{r}$ is the time scale: a bigger tank settles more slowly.

Example 3. A tank holds $400\ \text{L}$ of pure water. Brine of $0.25\ \text{kg/L}$ enters at $8\ \text{L/min}$, and the mixture leaves at $8\ \text{L/min}$. Find the salt after $30$ minutes, and the time to reach $80\ \text{kg}$.

1. Here $cV = 0.25 \times 400 = 100\ \text{kg}$ and $\dfrac{r}{V} = 0.02\ \text{min}^{-1}$. So $Q = 100\left(1 - e^{-0.02t}\right)\ \text{kg}$.
2. At $t = 30$: $Q = 100(1 - e^{-0.6}) = 45.12\ \text{kg}$.
3. For $80\ \text{kg}$: $e^{-0.02t} = 0.2$, so $t = 50\ln 5 = 80.5\ \text{min}$.

When the volume changes

If the inflow is $r_{\text{in}}$ and the outflow is $r_{\text{out}}$, the volume is $V = V_0 + (r_{\text{in}} - r_{\text{out}})t$. The concentration $\dfrac{Q}{V}$ now has a changing denominator, and the equation is linear:

Multiply by an integrating factor (the method of the first-order course) and integrate. The tank in the program has $V_0 = 200\ \text{L}$, $6\ \text{L/min}$ of $0.5\ \text{kg/L}$ brine entering, $4\ \text{L/min}$ leaving and $20\ \text{kg}$ of salt at the start. Its solution is

Check it. At $t = 0$, $Q = 100 - 80 = 20\ \text{kg}$, as required. Its derivative is $1 + \dfrac{12.8 \times 10^{6}}{(200 + 2t)^3}$. The right side of the equation is $3 - \dfrac{4Q}{200 + 2t}$, which equals $3 - 2 + \dfrac{12.8 \times 10^{6}}{(200 + 2t)^3}$ after the substitution. The two sides agree.

The same equation in other places

A capacitor charged through a resistor follows $\dfrac{dq}{dt} = \dfrac{E - q/C}{R}$. A drug given at a steady rate $I$ and removed at a rate proportional to the amount $D$ in the body follows $\dfrac{dD}{dt} = I - kD$. Both have the form $\dfrac{dy}{dt} = a - by$, the same shape as cooling, and they settle to the steady state $\dfrac{a}{b}$.

Practise with new numbers

Try it in Python

The program finds $k$ from two readings of a cooling object, predicts other readings and finds the time to reach a target.

Now follow a tank step by step. The next program uses Euler's method for the tank with the changing volume and compares it with the exact formula.

Common mistakes

• Using $T$ instead of the gap. The rate is proportional to $T - T_s$, not to $T$.
• Mixing up the sign of $k$. In $\dfrac{dT}{dt} = -k(T - T_s)$ the constant $k$ is positive, and the minus sign makes the object move towards $T_s$.
• Dividing by the wrong volume. The concentration in the tank is $\dfrac{Q}{V}$. When the volume changes, $V$ is a function of time.
• Using the inflow concentration for the outflow. The liquid that leaves has the concentration of the tank, not of the brine that enters.
• Mixing units. Keep litres with litres and minutes with minutes in $k$ and in the rates.

Remember these

• Newton's law: $\dfrac{dT}{dt} = -k(T - T_s)$ gives $T = T_s + (T_0 - T_s)e^{-kt}$.
• Two readings give $k$: the gap changes by the same factor in equal times.
• Mixing: $\dfrac{dQ}{dt} = r_{\text{in}}c_{\text{in}} - r_{\text{out}}\dfrac{Q}{V}$.
• Constant volume gives $Q = cV + (Q_0 - cV)e^{-rt/V}$. The steady state is $cV$.
• A changing volume is $V = V_0 + (r_{\text{in}} - r_{\text{out}})t$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A metal block at $150\ ^\circ\text{C}$ is placed in air at $30\ ^\circ\text{C}$. After $10$ minutes it is at $90\ ^\circ\text{C}$. Find $k$, and the temperature after $20$ minutes.

2. A tank holds $500\ \text{L}$ of pure water. Brine of $0.2\ \text{kg/L}$ enters at $10\ \text{L/min}$ and the mixture leaves at $10\ \text{L/min}$. How much salt is in the tank after $50$ minutes?

3. A tank of $300\ \text{L}$ has brine of $0.4\ \text{kg/L}$ flowing through it at a steady rate, with equal inflow and outflow. What amount of salt does the tank approach?

4. A bottle at $5\ ^\circ\text{C}$ is put into a room at $25\ ^\circ\text{C}$. After $6$ minutes it is at $11\ ^\circ\text{C}$. Find $k$.

5. True or false: if more liquid enters a tank than leaves, the volume stays the same.

References

• OpenStax, Calculus Volume 2, Chapter 4, section 4.3 Separable Equations (mixing, with the cooling problems) and section 4.5 First-order Linear Equations.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the sections on differential equations and their applications.