Applications: Newton's law of cooling and electrical circuits

40 min

From a physical law to an equation

The methods of lessons 1 to 4 pay off when a physical law is written as a differential equation. In each application below the work has three steps: translate the law into an equation, solve it with the right method, and read the answer back as physics (a temperature, a current, a time).

Definition

Newton's law of cooling: the rate at which a body's temperature $T$ changes is proportional to the difference between $T$ and the constant temperature $T_{s}$ of its surroundings:

\frac{dT}{dt} = -k(T - T_s), \qquad k \gt 0 .

The minus sign makes a hot body ( $T \gt T_s$ ) cool and a cold body ( $T \lt T_s$ ) warm. The law holds well when the temperature difference is not too large and the surroundings are well mixed.

Solution: the equation is separable (and linear). With $T (0) = T_{0}$ ,

\int \frac{dT}{T - T_s} = -\int k\,dt \quad\Longrightarrow\quad T = T_s + (T_0 - T_s)\,e^{-kt} .

Worked example 1: a cooling cup of tea

A cup of tea at $90^\circ\mathrm{C}$ is left in a room at $20^\circ\mathrm{C}$ . After 10 minutes it is at $60^\circ\mathrm{C}$ . (a) Find $T (t)$ . (b) Find its temperature after 30 minutes. (c) When will it reach $40^\circ\mathrm{C}$ ?

Solution

$T = 20 + 70e^{-kt}$ . At $t = 10$ : $60 = 20 + 70e^{-10k}$ , so $e^{-10k} = \frac{4}{7}$ and $k = \frac{1}{10}\ln\frac{7}{4} \approx 0.0560$ per minute.
$T(30) = 20 + 70\left(e^{-10k}\right)^3 = 20 + 70\left(\frac{4}{7}\right)^3 \approx 20 + 13.1 = 33.1^\circ\mathrm{C}$ .
$40 = 20 + 70e^{-kt}$ gives $e^{-kt} = \frac{2}{7}$ , so $t = \frac{\ln 3.5}{k} = \frac{10\ln 3.5}{\ln 1.75} \approx 22.4$ minutes.

Newton's law of cooling: the further the tea is above room temperature, the faster it cools, so the curve is steep at first and flattens as T approaches 20 °C.

Warning

Use the temperature difference, not the temperature. The exponential decays in $T - T_{s}$ , not in $T$ . Writing $T = 90e^{-kt}$ would make the tea cool towards $0^\circ\mathrm{C}$ , colder than the room, which is impossible.

Electrical circuits

In a series circuit the same current $i$ (in amperes) flows through every component, and the charge $q$ (in coulombs) on the capacitor satisfies $i = \frac{dq}{dt}$ . Each component takes a voltage drop:

component	constant (unit)	voltage drop
resistor	resistance $R$ (ohm, $\Omega$ )	$R i$
inductor (coil)	inductance $L$ (henry, H)	$L\frac{di}{dt}$
capacitor	capacitance $C$ (farad, F)	$\frac{q}{C}$

Kirchhoff's voltage law says the drops around the loop add up to the source voltage $E (t)$ . So

\text{RL: } L\frac{di}{dt} + Ri = E, \qquad \text{RC: } R\frac{dq}{dt} + \frac{q}{C} = E, \qquad \text{RLC: } L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = E(t) .

The first two are linear first order equations (lesson 2); the third is a second order equation with constant coefficients (lessons 3 and 4), the electrical twin of the damped spring.

A series RLC circuit. Leave out the capacitor for an RL circuit, or the inductor for an RC circuit. Kirchhoff's voltage law: the voltage drops across R, L and C add up to the source voltage E.

Worked example 2: switching on an RL circuit

A 12 V battery, a $4\,\Omega$ resistor and a 2 H coil are connected in series, and the switch is closed at $t = 0$ , when $i = 0$ . Find $i (t)$ , the current after 0.5 s, and the time for the current to reach 90% of its final value.

Solution $2\frac{di}{dt} + 4i = 12$ , that is $\frac{di}{dt} + 2i = 6$ . The integrating factor is $e^{2t}$ : $\frac{d}{dt}\left(e^{2t}i\right) = 6e^{2t}$ , so $e^{2t}i = 3e^{2t} + C$ and $i = 3 + Ce^{-2t}$ . From $i (0) = 0$ , $C = - 3$ :

i = 3\left(1 - e^{-2t}\right)\ \mathrm{A}.

At $t = 0.5\,\mathrm{s}$ : $i = 3(1 - e^{-1}) \approx 1.90\,\mathrm{A}$ . The current rises towards $\frac{E}{R} = 3\,\mathrm{A}$ ; it reaches $2.7\,\mathrm{A}$ when $e^{-2t} = 0.1$ , at $t = \frac{\ln 10}{2} \approx 1.15\,\mathrm{s}$ . The coil delays the current: the larger $L$ , the slower the rise. The quantity $\frac{L}{R} = 0.5\,\mathrm{s}$ is the circuit's time constant.

Worked example 3: charging through an RLC circuit

A series circuit has $L = 1\,\mathrm{H}$ , $R = 6\,\Omega$ , $C = 0.1\,\mathrm{F}$ and a 20 V battery. The capacitor is uncharged and no current flows when the switch is closed at $t = 0$ . Find the charge $q (t)$ and the current $i (t)$ .

Solution $q^{''} + 6 q^{'} + 10 q = 20$ , with $q (0) = 0$ and $q^{'} (0) = 0$ . The characteristic equation $r^{2} + 6 r + 10 = 0$ has roots $-3 \pm i$ , so $q_c = e^{-3t}(C_1\cos t + C_2\sin t)$ . A constant trial gives $q_{p} = 2$ . From $q (0) = 0$ , $C_{1} = - 2$ ; from $q^{'} (0) = - 3 C_{1} + C_{2} = 0$ , $C_{2} = - 6$ . Hence

q = 2 - e^{-3t}(2\cos t + 6\sin t)\ \mathrm{C}, \qquad i = \frac{dq}{dt} = 20e^{-3t}\sin t\ \mathrm{A}.

The charge settles at $CE = 0.1 \times 20 = 2\,\mathrm{C}$ , and the current rises to a peak (about $2.41\,\mathrm{A}$ , at $t = \tan^{-1}\frac{1}{3} \approx 0.32\,\mathrm{s}$ ) and then dies away. The roots are complex, so the circuit is underdamped, but the damping $e^{-3t}$ is so strong that the oscillation is hardly visible.

Tip

Check every circuit answer against common sense: at $t = 0$ the initial conditions must hold, and as $t \to \infty$ the transient terms (with $e^{-at}$ ) vanish, leaving the steady state. In an RL circuit with a battery the steady current is $\frac{E}{R}$ ; in an RC or RLC circuit with a battery the capacitor's charge tends to $C E$ and the current to 0.

Houston Math Prep: Newton's Law of Cooling (Separable Differential Equations). Watch how the law is written as an equation, separated and solved, and how one measurement fixes the constant k, as in Worked example 1.

✦

Practice

Give answers to 3 significant figures. Questions 7 and 8 are harder.

A metal rod at $200^\circ\mathrm{C}$ cools in air at $25^\circ\mathrm{C}$ . After 5 minutes it is at $150^\circ\mathrm{C}$ . Find its temperature after 15 minutes, and when it reaches $50^\circ\mathrm{C}$ .
A bottle of juice at $4^\circ\mathrm{C}$ is taken from a fridge into a room at $24^\circ\mathrm{C}$ . After 15 minutes it is at $10^\circ\mathrm{C}$ . When will it reach $18^\circ\mathrm{C}$ ?
Solve $\frac{dT}{dt} + kT = kT_s$ with an integrating factor, and show that you get the same solution as by separating the variables.
A circuit has $E = 24\,\mathrm{V}$ , $R = 8\,\Omega$ and $L = 4\,\mathrm{H}$ , with $i (0) = 0$ . Find $i (t)$ , the time constant, and the current after 1 s.
A $0.002\,\mathrm{F}$ capacitor, initially uncharged, is charged from a 10 V source through a $1000\,\Omega$ resistor. Find $q (t)$ , and the time for the capacitor's voltage $\frac{q}{C}$ to reach 9 V.
An LC circuit (no resistance, no source) has $L = 0.5\,\mathrm{H}$ and $C = 0.02\,\mathrm{F}$ , with $q(0) = 0.1\,\mathrm{C}$ and $i (0) = 0$ . Find $q (t)$ and the angular frequency of the oscillation.
A series RLC circuit with no source has $L = 1\,\mathrm{H}$ and $C = \frac{1}{3}\,\mathrm{F}$ . Is it overdamped, critically damped or underdamped when (a) $R = 4\,\Omega$ (b) $R = 2\,\Omega$ ? For which $R$ is it critically damped?
A series circuit has $L = 1\,\mathrm{H}$ , $R = 2\,\Omega$ , $C = 0.5\,\mathrm{F}$ and a 10 V battery, with $q (0) = 0$ and $i (0) = 0$ . Find $q (t)$ and $i (t)$ .

Show answersHide answers

$T = 25 + 175e^{-kt}$ with $e^{-5k} = \frac{5}{7}$ : $T(15) = 25 + 175\left(\frac{5}{7}\right)^3 \approx 88.8^\circ\mathrm{C}$ ; $50^\circ\mathrm{C}$ when $e^{-kt} = \frac{1}{7}$ , at $t = \frac{5\ln 7}{\ln 1.4} \approx 28.9$ min.
$T = 24 - 20e^{-kt}$ , $e^{-15k} = 0.7$ ; $18^\circ\mathrm{C}$ when $e^{-kt} = 0.3$ : $t = \frac{15\ln 0.3}{\ln 0.7} \approx 50.6$ min.
$\mu = e^{kt}$ : $\frac{d}{dt}\left(e^{kt}T\right) = kT_se^{kt}$ , so $T = T_s + Ce^{-kt}$ , and $T (0) = T_{0}$ gives $C = T_{0} - T_{s}$ .
$4 i^{'} + 8 i = 24$ : $i = 3\left(1 - e^{-2t}\right)$ ; time constant $\frac{L}{R} = 0.5\,\mathrm{s}$ ; $i(1) \approx 2.59\,\mathrm{A}$ .
$RC = 2\,\mathrm{s}$ : $q = 0.02\left(1 - e^{-\frac{t}{2}}\right)$ ; $\frac{q}{C} = 10\left(1 - e^{-\frac{t}{2}}\right) = 9$ at $t = 2\ln 10 \approx 4.61\,\mathrm{s}$ .
$0.5 q^{''} + 50 q = 0$ , so $q^{''} + 100 q = 0$ : $q = 0.1\cos 10t$ ; $\omega = \frac{1}{\sqrt{LC}} = 10\,\mathrm{rad\,s^{-1}}$ .
$q^{''} + R q^{'} + 3 q = 0$ , $\Delta = R^2 - 12$ : (a) $\Delta = 4 \gt 0$ , overdamped (b) $\Delta = -8 \lt 0$ , underdamped; critical when $R = 2\sqrt{3} \approx 3.46\,\Omega$ .
$q^{''} + 2 q^{'} + 2 q = 10$ : $q = 5 - 5e^{-t}(\cos t + \sin t)$ and $i = 10e^{-t}\sin t$ .

1From a physical law to an equation

2Electrical circuits

From a physical law to an equation

Electrical circuits