I wanted to offer a simple model for students to work with about resistive heating.

Since $P=\frac{V^2}{R}$ we can calculate how much energy will be put into a wire over time, if we know the voltage and resistance of the wire.

So for a short wire of resistivity $\rho$ we can use $R = \frac{\rho L}{A}$ where A is the cross-sectional area and L is the length.

Well and good, and then all I have to do is apply the equation for power and I can figure out how much energy in joules gets applied to the wire for any given voltage.

The thing is, when I then try some simple model like using specific heat (I used the old thing from first year chemistry $Q = mc\Delta T$)I get rather large numbers for temperature increase, on the order of thousands of degrees for a 120 V circuit.

Wait a minute, I thought. The current drawn isn't going to be arbitrary; there's only so much that any circuit will provide (this is why when I short something I can blow a fuse, right?).

So let's say I limit the current to say, 15 A. Then I can use $P= I^2R$ and I get a relatively small number, which makes more sense. In fact here as resistance goes up the energy deposited into the wire goes up, and it heats more, which also makes more intuitive sense.

I also know that if I hook up a wire heating element to a circuit the energy from the current doesn't all get deposited in the heating element; the element radiates heat away (like in a toaster, that is what it is supposed to do). A small wire radiates a lot of heat, as it has a high surface-to-volume ratio. That is probably why we use them for heating elements in toasters, as they radiate a good chunk of the energy away into your toast.

Anyhow I am curious how I might set this model up. The point is to offer a kind of back-of-the-envelope model to demonstrate conservation of energy and resistive heating without using calculus and giving an idea of what the upper limits are.

  • $\begingroup$ There is also convective cooling. The wire heats air nearby. Hot air rises, bringing more cool air near the wire. $\endgroup$
    – mmesser314
    Commented May 21, 2021 at 13:58
  • $\begingroup$ Also, remember that the resistance changes with temperature. $\endgroup$
    – R.W. Bird
    Commented May 21, 2021 at 15:01
  • 1
    $\begingroup$ You can't neglect heat transfer out of the wire into ambient. That's what determines the actual final temperature of the wire. And that always makes these kinds of calculations tricky since its partial differential equations. $\endgroup$
    – DKNguyen
    Commented May 21, 2021 at 18:22

4 Answers 4


This is a great opportunity to teach your students about performing an energy balance, about various heat transfer modes, about solving differential equations (or approximating them with finite-difference equations), and about model simplification.

For any small section of the wire (length $dx$, cross-sectional area $A$), we can perform an energy balance with the following components:

  • Volumetric heat generation: $J^2 rA\,dx$, where $J=I/A$ is the current density and $r$ is the wire material resistivity, a material property.

  • Conductive heat transfer: $kA\frac{\partial^2 T(x)}{\partial x^2}dx$, as derived here, where $k$ is the thermal conductivity of the wire, a material property, and $T$ is the temperature in that wire section. This term implies that conduction tends to eliminate curvature (the second derivative) in temperature distributions.

  • Convective heat loss: $-hP[T(x)-T_\infty]dx$, where $h$ is the convective coefficient, characterizing the natural or forced airflow past the wire, $P$ is the perimeter length (e.g., pi times the diameter), and $T_\infty$ is the ambient temperature.

  • Radiative heat loss: $-\sigma\epsilon P[T(x)^4-T_\infty^4]dx$, where $\sigma=5.67\times 10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ is the Stefan-Boltzmann constant and $\epsilon$ is the emissivity of the wire surface.

Add these together and divide by the section volume to obtain the general heat equation:

$$k\frac{\partial^2 T(x)}{\partial x^2}+J^2r-\frac{hP}{A}[T(x)-T_\infty]-\frac{\sigma\epsilon P}{A} [T(x)^4-T_\infty^4]=\rho c\frac{\partial T(x)}{\partial t},$$

where $\rho$ is the wire material density and $c$ is its specific heat capacity. The term on the right indicates that the result of the energy balance on the left (whether positive or negative) goes to change the wire temperature over time $t$.

This equation can be hard to solve. Let's try some simplifying assumptions. If the wire is very long and thin, then we might consider conduction along its length to be unimportant (this is equivalent to assuming that it remains at the same temperature, something you can evaluate by looking at the glow when it gets very hot). Furthermore, at very high temperatures, the $T(x)^4$-containing radiation term dominates over the $T(x)$-containing convection term. Plus, let's change differentials to finite differences to obtain some general scaling laws:

$$J^2r-\frac{\sigma\epsilon P}{A} [T(x)^4-T_\infty^4]=\rho c\frac{\Delta T}{\Delta t}.$$ From this, we can, for example, estimate the time required for melting, which you noted was of interest to you:

$$t\approx\frac{\rho c T_m}{J^2r-\sigma\epsilon P T_m^4/A}.$$

Here, I've assumed that the ambient or room temperature $T_\infty$ is small compared to the melting temperature $T_m$ of the wire material. Does this make sense? Try plugging in your specific conditions!

  • $\begingroup$ +1, but does radiation dominate? The sun is cooled by radiation, and the surface is very hot. Modern computer chips generate a lot of heat. I have heard they give off more heat than an equal size patch of Sun. Chips are much cooler because of very aggressive cooling. A light bulb filament gives off a lot of radiation, but a typical wire gives off a lot less. $\endgroup$
    – mmesser314
    Commented May 21, 2021 at 19:26
  • 1
    $\begingroup$ The dominant mode can be determined by comparing the $k\frac{\partial^2 T(x)}{\partial x^2}$, $\frac{hP}{A}[T(x)-T_\infty]$, and $\frac{\sigma\epsilon P}{A} [T(x)^4-T_\infty^4]$ terms. Radiation predominates for sufficiently long, hot wires. Processors are cooled by making sure they're not long, i.e., that they're compact and close to a thermal sink of some type. $\endgroup$ Commented May 21, 2021 at 19:43
  • $\begingroup$ I like this one as well, though I suspect for my students the math would be a little challenging. But it's a good way to look at time-to-melt, and I think I'd use it as a way to demonstrate why you can't leave a toaster on all night. $\endgroup$
    – Jesse
    Commented May 23, 2021 at 16:08

The thing is, when I then try some simple model like using specific heat (I used the old thing from first year chemistry $Q = mc\Delta > T$)I get rather large numbers for temperature increase, on the order of thousands of degrees for a 120 V circuit.

That is the correct equation for calculating the energy delivered to the wire, but it doesn't take into account the fact that while energy is delivered to the wire raising its temperature, heat is transferring from the wire to the cooler environment by conduction, convection and/or radiation, depending on the nature of the environment.

In short, your model has to reflect a balance between the steady state transfer of energy from the electrical source (power) to the wire and the steady state transfer of energy in the form of heat from the wire to its environment.

Hope this helps.


When you pass a current through a wire, for a fraction of a second (a time $\Delta t$, say) almost all the electrical work ($I^2 R\Delta t$) goes into internal thermal energy in the wire (that is, as you say, $mc\Delta T$). But as the wire gets hotter more and more of the energy being supplied is lost as heat to the surroundings. Let's suppose that the wire never reaches its melting point. Instead it will reach a maximum temperature, when it gives off heat at the same rate that energy is being supplied electrically.

How do we calculate this maximum temperature? If the wire gets red hot, or not far short of it, much of the heat – my limited researches suggest more than 80% for temperatures over 600°C – will be lost as infrared radiation. In this case we have $$I^2 R=\alpha\sigma 2\pi r l(T^4-T_0^4)$$ The right hand side is the net rate of heat loss by radiation. $\alpha$ is a number between 1 and 0, dependent on the nature of the surface of the wire. Putting $\alpha=1$ isn't usually too bad an approximation. $\sigma\ (\text{the Stefan-Boltzmann constant})=5.67\times10^{-8}\ \text{Wm}^{-2} \text{K}{^{-4}}$, $2\pi rl$ is the surface area of the wire, $T$ is its temperature and $T_0$ is room temperature. $T_0^4$ may well be negligible compared with $T^4$. Remember to use kelvin temperatures.

Hence you can find a theoretical value for the wire's final temperature, $T$. If it works out to be more than the wire's melting point – then we predict that the wire will melt!

Note that this is all rather approximate. $\alpha$ is difficult to put a precise figure on, and we have neglected to consider the heat loss rate by convection (which is not easy to estimate, as it depends on whether the wire is vertical or horizontal and on the presence of surrounding objects).

If you consider a wire made of the special alloy constantan, you remove one complication. Constantan (aka Eureka) is is designed so that its resistivity changes very little with temperature. Its resistivity is 0.49 $\mu \Omega \ \text m$ and its melting point is 1210 °C = 1483 K. [Using these figures I found that a constantan wire of diameter 0.50 mm would need a current of 13 A to reach its melting point. Not unreasonable?]

  • $\begingroup$ Thanks, this one is especially nice because I don't need to use calc and can plug in a lot of this as constants which we can take as read - the point is to just show what happens and how say, a toaster or electric stove works. $\endgroup$
    – Jesse
    Commented May 23, 2021 at 13:07

The time it takes for a copper wire to melt under short circuit conditions can determined from ``Onderdonk's equation for fusing current''. The following are two good references with derivations, assumptions, and some discussion of the shortcomings of the equation:



Typically it is used to size conductors for a power supply that short circuits, but the protection circuit has a finite delay time before it reacts.

The following is a simplified derivation for Onderdonk's equation for fusing current:

Consider a solid wire with a current passing through it.

$i$ = current (A)
$t$ = time (s)
$t_0$ = reference time (s)
$\Delta t = t - t_0$ = time difference (s)
$T$ = average wire temperature (°C)
$T_0$ = reference temperature (°C)
$L$ = wire length (m)
$A$ = wire cross sectional area (m$^2$)
$\rho$ = mass density of the wire (kg / m$^3$)
$c_P$ = specific heat (J / Kg K)
$R$ = electrical resistance (Ω)
$\alpha$ = temperature coefficient of resistivity (1 / K)
$\rho_0$ = electrical resistivity at the reference temperature (Ω m)


  1. Current is applied to the conductor at time $t_0$.
  2. Before application of current, the temperature of the conductor is $T_0$.
  3. The outside of the wire is perfectly insulated.
  4. The wire is very long.
  5. The wire is very thin.
  6. The resistivity of the wire is linear with respect to temperature.

As a current passes through the wire, the increase in thermal energy must equal the heat dissipated by Joule heating:

$c_p \, \rho \, A \, L \, \left(T - T_0\right) = R \, i^2 \, \Delta t$

We are assuming there is no thermal conduction down the length of the wire due to assumption 5. Because of assumptions 3 and 4 we can neglect thermal conduction perpendicular to the wire. This means that $T$ represents an average wire temperature.

The electrical resistance is given by (assumption 6):
$R=\frac{ \rho_0 \, \left(1 + \alpha \, \left(T - T_0\right) \right) \, L} {A}$

Combining the equations, eliminating $L$, and rearranging:
$ \frac{\left(T - T_0\right)} {\Delta t} = \frac{i^2 \, \rho_0} {c_p\, \rho \, A^2} \left(1 + \alpha \, \left(T - T_0\right) \right)$

Turning the left hand side into a differential:
$ \frac{d \, \left(T - T_0\right)} {d \, \Delta t} = \frac{i^2 \, \rho_0} {c_p\, \rho \, A^2} \left(1 + \alpha \, \left(T - T_0\right) \right)$


$T - T_0 = C \, e^{\frac{i^2 \, \rho_0 \alpha \Delta t} {c_p\, \rho \, A^2}} -\frac{1} {\alpha}$

where $C$ is an unknown which can be solved for by assumptions 1 and 2 ($T - T_0 = 0$ at $\Delta t = 0$). This yields $C = \frac{1} {\alpha}$

Solving for the change in time:

$\Delta t = \frac{c_p\, \rho \, A^2} {i^2 \, \rho_0 \, \alpha} \text{ln}\left( 1 + \alpha \left(T - T_0\right)\right) $

How long it takes for an insulated wire to melt for a given current can be found by replacing $T$ with the melt temperature.

  • $\begingroup$ Please try to make this answer as self-contained as possible (which doesn't preclude providing links). As it is, it's almost a link-only answer, and those tend to be deleted. Thank you. $\endgroup$
    – Miyase
    Commented Jul 15, 2022 at 22:42

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