Irodov 3.195: Heating Toward a Joule Balance – Amit & Deepali Sinha Foundation

Problem Irodov 3.195

A conductor has a resistance $R$ that does not depend on temperature and a total heat capacity $C$ . At time $t = 0$ a constant voltage $V$ is applied across it. The conductor loses heat to its surroundings at the rate $q = k(T - T_0)$ , where $k$ is a constant and $T_0$ is the temperature of the surroundings, which is also the conductor’s temperature at $t = 0$ . Find how the conductor’s temperature $T$ depends on time.

Given

resistance $R$ (independent of $T$ )
heat capacity $C$
voltage $V$ from $t = 0$
heat loss $q = k(T - T_0)$
$T(0) = T_0$

Find

the temperature

T(t)

Solution

Step 1The heat balance

In a time $dt$ the current releases the Joule heat $(V^2/R)\,dt$ ; part of it leaves as $q\,dt$ and the rest raises the conductor’s temperature by $dT$ . Writing the temperature rise as $\theta = T - T_0$ :

C\,\frac{d\theta}{dt} = \frac{V^2}{R} - k\theta, \qquad \theta(0) = 0 \tag{1}

Step 2Integrating the balance

The rise stops at the value $\theta_\infty$ where the right side of (1) vanishes:

\frac{V^2}{R} - k\theta_\infty = 0, \qquad \theta_\infty = \frac{V^2}{kR}

With it, (1) reads $C\,d\theta/dt = k(\theta_\infty - \theta)$ . Separating the variables and integrating from the start:

\begin{aligned} \frac{d\theta}{\theta_\infty - \theta} &= \frac{k}{C}\,dt\\[2pt] \int_0^{\theta}\frac{d\theta'}{\theta_\infty - \theta'} &= \int_0^{t}\frac{k}{C}\,dt'\\[2pt] -\ln\frac{\theta_\infty - \theta}{\theta_\infty} &= \frac{kt}{C} \end{aligned}

Step 3The temperature against time

Solving for $\theta$ and putting back $\theta = T - T_0$ and $\theta_\infty = V^2/kR$ :

\theta = \theta_\infty\left(1 - e^{-kt/C}\right)

T = T_0 + \frac{V^2}{kR}\left(1 - e^{-kt/C}\right)

(2)

The temperature rise against time: it climbs from zero towards the dotted asymptote, its steady value, reaching 1 - 1/e of it at t = C/k. — The temperature rise against time: it climbs from zero towards the dotted asymptote, its steady value, reaching $1 - 1/e$ of it at $t = C/k$ .

The temperature approaches $T_0 + V^2/kR$ , where the Joule power is all carried away, with the time constant $C/k$ .

Check

Dimensions: $k$ is a power per kelvin, so $V^2/kR$ is in kelvin and $C/k$ (joule per kelvin over watt per kelvin) is a time. At $t = 0$ , (2) gives $T = T_0$ ; for $t \ll C/k$ it gives $T - T_0 \approx V^2t/RC$ , the heating of a conductor that has not yet begun to lose heat; for $t \to \infty$ it gives the steady state where $k(T - T_0)$ equals $V^2/R$ .

Remark

The conductor is a thermal RC circuit: the heat capacity $C$ plays the capacitor, $1/k$ the resistance to the surroundings, and the constant Joule power a constant current source, so the temperature charges up exactly as a capacitor’s voltage does, with time constant $C\cdot(1/k)$ . If $R$ grew with temperature, as a real metal’s does, the heating power $V^2/R$ would fall as the conductor warmed and the approach would be faster and lower.