Speed of heat through an object

Question

According to the Heat equation (the PDE), heat can travel infinitely fast, which doesn't seem right to me. So I was wondering, at what speed does heat actually propogate through an object?

For example, if I have a really long iron rod at a constant temperature (say 0 Celsius), and one end of it instantenously becomes hot (e.g. 1000 Celsius), how far down the rod will the temperature have changed in 1 second? I don't care how much the temperature changes, only how far a temperature change (however minuscule) happened.

Would changing the material (e.g. steel instead of iron) or the initial temperatures change the answer?

My gut tells me the answer should be the speed of sound for the material, because that's the speed at which movement in the atoms can affect each other.

Answer 1

TL;DR: Heat can't travel instantaneously because relativity disallows it. Some microscopic models of heat transfer predict that heat travels in a specific formulation of the speed of the sound. So its kinda the "heat speed of sound" in the specific material

Heat can't travel instantaneously because it is limited through relativity by the speed of light. Yes, the Solution to the transient fourier equation allows for any small time that the temperature at the finite end of an object to increase, even if its just a tiny number, but is still nonzero, so in a sense, the fourier equation allows heat with unlimited speed, but that's not possible.

Its hardly a problem practically, though, because in practice the materials we utilize to measure heat transfer are small enough so that you can ignore relativistic effects. Only in, say , a bar that connects the earth and the sun, you would have to consider the speed of light limit.

But, there are some models that modify the fourier equation to be compatible with relativity. You can see them here: https://en.wikipedia.org/wiki/Relativistic_heat_conduction (of course, only to give you a very rough idea)

Answer 2

Here it is with no math:

Heat in a solid is not transferred at infinite speed, it diffuses through the solid the way that the molecules in a spoon of sugar syrup diffuse slowly into an unmixed drink. This is a relatively slow process and there are equations you can use to solve for the amount of time it takes for heat to diffuse from one end of a rod, for example, to the other. Gert's answer furnishes that equation and solves a sample problem for you.

Answer 3

The Heat Conduction Paradox and the Cattaneo Equation

Firstly I want to apologise to the OP for having been too dismissive of his claims. Mea culpa.

The so-called Heat Conduction Paradox is hardly a new problem and it has gotten loads of attention, just search the term to get an idea.

One rather pragmatic resource here provides useful information.

It focuses on 'fixing' Fourier's Heat Equation by means of a 'patch' called the Cattaneo Equation. As regards a discussion regards 'instantaneous heat conduction' (or not), some of it can be found in the link.

$$\boxed{\tau T_{tt}+T_{t}=\alpha T_{xx}}\tag{1}$$

here for $1D$.

where $\tau$ is a constant. Obviously for very small or zero values of $\tau$, Cattaneo reduces to Fourier.

This paper calls $\tau$ a 'relaxation time' and claims that the 'associated temperature propagation speed', $c_T$, is given by:

$$c_T=\sqrt\frac{\alpha}{\tau}$$

Also:

It is clear that in the above, that as $c_T \longrightarrow \infty$ (i.e. the limiting cases of zero relaxation time, $\tau \longrightarrow 0$), which is the case of infinite speed of heat propagation, the hyperbolic heat conduction (HHC) equation reverts back to the classical parabolic heat conduction (PHC) equation. Furthermore, at steady state, the Cattaneo model reverts to the Fourier model, although the relaxation parameter, $\tau \neq 0$. As a consequence, the temperature results for the two models differ only during the transient state.

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My purpose in this answer is not to prove (or disprove) Cattaneo avoids the paradox but rather to show a simple solution of the equation and compare it to the solution obtained by Fourier.

I summarise again the BCs and IC in the diagram below, for the sample problem:

I'll kind of gloss over the mechanics of solving $(1)$, cos' boring.

$$T=X\Theta$$ $$\tau X\Theta''+X\Theta'=\alpha \Theta X''$$ $$\frac{1}{\alpha}\left(\tau\frac{\Theta''}{\Theta}+\frac{\Theta'}{\Theta}\right)=\frac{X''}{X}-m^2$$ $$\frac{X''}{X}=-m^2$$ $$m=\frac{n\pi}{L}\text{ for }n=1,2,3....$$ $$X_n(x)=A_n\sin\left(\frac{n\pi x}{L}\right)$$ $$\frac{1}{\alpha}\left(\tau\frac{\Theta''}{\Theta}+\frac{\Theta'}{\Theta}\right)=-m^2$$ $$\tau\Theta''+\Theta'+m^2\alpha \Theta=0$$ The determinant of this ODE's Characteristic Equation is (for $\alpha =1$): $$\Delta=\sqrt{1-4 \tau m^2}$$ So we have: $$\Theta_n(t)= C_n\exp{\left(\frac{-t (\Delta +1)}{2\tau}\right)}+D_n\exp{\left(\frac{t (\Delta -1)}{2\tau}\right)}$$ Because $\Theta_n(t)$ must asymptotically reach some value for $t \to +\infty$, this means that $D_n$ must be $0$, so we obtain: $$T_n=E_n\exp{\left(\frac{-t (\Delta +1)}{2\tau}\right)}\sin\left(\frac{n \pi x}{L}\right)$$ For $t=0$ with the Fourier expansion, we get: $$E_n=\frac{4T_0}{n\pi}$$ $$T(x,t)=\sum\limits_{n = 1}^{\infty}E_n\exp{\left(\frac{-t (\Delta +1)}{2\tau}\right)}\sin\left(\frac{n \pi x}{L}\right)$$ Note that for $\alpha=1$ and $L=1$, $\Rightarrow m=n\pi$.

Superposition:

$$T(x,t)=\sum\limits_{n = 1}^{\infty}E_n\exp{\left(\frac{-t \left[(\sqrt{(1-4\tau n^2\pi^2 )}+1\right]}{2\tau}\right)}\sin\left(n \pi x\right)\tag{2}$$ In order for the determinant to be positive: $$1-4\tau n^2\pi^2 \geq 0 \Rightarrow \tau \leq \frac{1}{4n^2\pi^2}\tag{3}$$

For the same conditions, Fourier yields:

$$T_F(x,t)=\sum\limits_{n = 1}^{\infty}E_n\exp{(-n^2\pi^2 t)}\sin\left(n \pi x\right)\tag{4}$$

A bit of function analysis reveals that $(2)$ and $(4)$ differ in the sense that:

$$\lim_{\tau\to 0}\frac{(\Delta +1)}{2\tau}=+\infty$$

from both sides. So $(1)$ (Cattaneo) does reduce to FHE for $\tau \to 0$ but not for this simple application.

We can compare $(2)$ and $(4)$ somewhat 'sloppily' but simply (and avoiding bulky spreadsheets to produce the plots) by using the first term approximation ($n=1$ only) because then:

$$\frac{T_1(x,t)}{T_{1,F}(x,t)}=\frac{\exp{\left(\frac{-t \left[(\sqrt{(1-4\tau \pi^2 )}+1\right]}{2\tau}\right)}}{\exp{(-\pi^2 t)}}$$

$$\frac{T_1(x,t)}{T_{1,F}(x,t)}=\exp{\left[\pi^2 t-\left(\frac{t \left[(\sqrt{(1-4\tau \pi^2 )}+1\right]}{2\tau}\right)\right]}$$

I chose $\tau=0.02\,\mathrm{s^2}$, only just compliant with $(3)$.

So, for this specific example and its first term approximation, $\frac{T_1(x,t)}{T_{1,F}(x,t)}$ starts off at unity ($t=0$) but then Fourier gradually 'takes over'.

Answer 4

Heat equation for a $1D$ bar: $$T_t=\alpha T_{xx}\tag{1}$$ Note that heat losses like radiation or convection are excluded from this analysis.

Boundary conditions (BC, chosen) $$T(0,t)=0$$ $$T(L,t)=0$$ Initial condition (chosen) $$T(x,0)=f(x)$$ We're looking for a function:

$$T(x,t)$$ Which we believe can be represented by the Ansatz (assumption): $$T(x,t)=X(x)\Theta(t)\tag{2}$$ Insert $(2)$ into $(1)$, so we get:

$$\frac{1}{\alpha}X\Theta'=\Theta X''$$

Divide by $(2)$, to obtain:

$$\frac{\Theta'}{\alpha \Theta}=\frac{X''}{X}$$

So separation of variable has been achieved and we can write:

$$\frac{\Theta'}{\alpha \Theta}=\frac{X''}{X}=-m^2$$

where $m$ is a Real separation constant.

This gives us two ODEs:

$$\frac{\Theta'}{\alpha \Theta}=-m^2\tag{3}$$ $$\frac{X''}{X}=-m^2\tag{4}$$

$(4)$ solves easily to:

$$X(x)=A\sin mx+B\cos mx$$

The first BC gives us:

$$0=A\sin 0+B\cos 0 \Rightarrow B=0$$

So this leaves us with:

$$X(x)=A\sin mx$$

So with the second BC:

$$0=A_n\sin mL$$

Which means that:

$$mL=n\pi\text{ for }n=1,2,3...$$ So that:

$$X_n(x)=A\sin\left(\frac{n \pi x}{L}\right)$$

From $(3)$ we glean:

$$\Theta_n(t)=C_n\exp{(-m^2 \alpha t)}$$

So that (combining $A_n$ and $C_n$)

$$T_n(x,t)=D_n\exp{(-m^2 \alpha t)}\sin\left(\frac{n \pi x}{L}\right)$$

And with the superposition principle we get the temperature evolution of the $1D$ bar:

$$T(x,t)=\sum\limits_{n = 1}^{\infty}D_n\exp{(-m^2 \alpha t)}\sin\left(\frac{n \pi x}{L}\right)\tag{5}$$

The coefficients $D_n$ are obtained from the Fourier expansion:

$${D_n} = \frac{2}{L}\int_{{\,0}}^{{\,L}}{{f\left( x \right)\sin \left( {\frac{{n\,\pi x}}{L}} \right)\,\mathrm{d}x}}\,\,\,\,\,\,\,n = 1,2,3, \ldots$$

So for example if $T(x,0)=100$ and at $t=0$ we plunge both ends in a large reservoir at $T=0$, then the bar will cool down. The speed of cooling will depend mainly on $L$ and $\alpha$.

Answer 5

If instead of one metal rod, we have $2$ of them, from different metals joined at one end, a voltage is produced between the other ends, when the join is heated. In this case, the velocity of the propagation is the velocity of the electric current, close to the speed of light.

If we have only one metal rod, the heating of one of the ends can not of course generates an steady electric current, but the modification in the occupied states of the conduction band due to the heating spreads along the rod. And the velocity is again close to the speed of light.

I don't say that it is the main mechanism of heat transfer by conduction. But for a really very small $\Delta T$, a transient current carries some heat by Joule effect.

Answer 6

It seems to me that what you are looking for is the thermal diffusivity, as this is the coefficient that balances the rate and area of the temperature change: $$\partial_tT=\alpha\nabla^2T\implies\alpha=\frac{\partial_tT}{\nabla^2T} $$ and has units of area per time ($\partial_t$ has units of inverse time, $\nabla^2$ has units of inverse area, hence area per time).

It is also empirically measured by a method called laser flashing in which a material of thickness $d$ is heated by a laser on one side and the temperature measured on the other side. The thickness and the time to half-maximum temperature, are then used in the formula, $$ \alpha=\eta\frac{d^2}{t_{1/2}}\tag{1} $$ where $\eta$ is some small constant. An example I found online of such a test shows a very nice chart of a finite temperature change,

(source)

So it seems to me that one could invert the relationship in Eq (1) for the distance, $$d\sim\left(\alpha\,t\right)^{1/2}$$ and find at least a first-order approximation. For the example given of the iron rod, using the Wikipedia entry for the diffusivity, after one second the heat should have traveled about 5 mm.

Answer 7

The "speed" of heat should be close to the speed of sound. What is a bit surprising to me, is that this holds even for metals, which are known to have a thermal conductivity determined mostly by free electrons rather than phonons. To give you an idea about the speed of those electrons, in good conductors like Cu, it is about c/100 (the Fermi velocity), however they scatter (for example with phonons), randomizing their momentum and direction, and I guess this explains why the speed of heat is likely smaller than the Fermi velocity.

Here is a passage from PW Bridgman I find particularly enlightning:

Imagine a centimeter cube of copper between the opposite faces of which there is a temperature difference of 100°. The thermal flux is approximately 100 cal ./sec. The volume density of energy corresponding to this flux is such that its product into the velocity of flux is equal to 100. For the velocity we may take, in accordance with the Debye picture of thermal conduction, the velocity of sound, which for copper is about $3.5 \times 10^6$ cm/sec. The space density of energy is therefore $100/3.5 \times 10^6 = 3 \times 10^{-5}$ gm cal./cm3. The heat capacity of 1 cm3 of copper is about 0.8 gm cal. This means, therefore, that if a copper cube in which a thermal current of 100 cal ./sec is flowing is suddenly isolated from the source and sink of heat flow, its final equilibrium temperature will be about $4\times > 10^{-5}$ °C higher than its average temperature during the flow.

This, of course, would be very difficult to detect. It is interesting, however, that if a different velocity were assumed, as for example a velocity of the order of a few cm per sec, which is the order of the apparent velocity with which the maxima or minima of ordinary periodic thermal disturbances sink into the metal, a temperature effect of the order of many degrees would have been found. This affords rather direct confirmation of the correctness of the Debye point of view. The experiment might be worth making to find how far the velocity limit could be pushed.

Note that this invalidates some of the answers you had gotten here thus far. (Many of which completely misunderstood your question, and still don't realize it even when showed evidence).