# Fourier's law and time taken for heat transfer?

Fourier's law states that

"The time rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area."

A mathematical description of this law is given as

$$\frac{dQ}{dt}=-KA \frac{dT}{dL} \, . \tag{1}$$

where $K$ is the thermal conductivity of the substance in question, $A$ is the area of the substance normal to the direction of flow of heat current and $L$ is the lateral length of the substance.

However, we may write $dQ/dt=mC(dT/dt)$ -where $m$ is the mass of the substance and $C$ is the specific heat capacity of the substance-to obtain

$$m C \frac{dT}{dt} = K A \frac{dT}{dL} \, . \tag{2}$$

This would allow me to cancel the $dT$ on both sides of the equation to obtain

$$\frac{mC}{dt} = \frac{K A}{dL} \, , \tag{3}$$

which suggests that the time taken for the heat transfer to complete, between any two temperatures is a constant given by

$$dt = \frac{dL m C}{KA} \, . \tag{4}$$

However, this is not at all the case. What mistakes have I made in writing the above steps?

• I believe your top equation should be $dQ/dt=-KAdT/dL$ – Cort Ammon Sep 16 '16 at 3:20
• Please define all symbols. What are $A$, $K$, $m$, and $C$? – DanielSank Sep 16 '16 at 3:27
• The problem here is cancelling the $dT$'s, which are not independent algebraic quantities. $dT/dt$ is a single object, the time rate of change of the temperature. In some cases you can cancel differentials and get a meaningful result, but I don't think that works here because you have partial differential equations. Can't say anything more useful than that yet... – DanielSank Sep 16 '16 at 3:39
• $\dot{Q}=-k\frac{\partial T}{\partial x}$ (one dimension) is the equation you're looking for, where $T=T(x,t)$. – Gert Sep 16 '16 at 7:38

There are a few subtle things here.

You've concluded that the time for 'heat transfer to complete' (for example, the time for a bar with ends initially at different temperatures to relax to a uniform temperature) doesn't depend on the overall temperature differences. This is true! If you double every temperature difference, you double the amount of heat you need to transfer, but you also double the heat transfer rate. So the total time stays the same; we say heat diffusion has a characteristic timescale.

However, the method you've used is incorrect. For example, your result says that $$t \propto L$$ where $L$ is the length of the bar. However, a full analysis shows that $t \propto L^2$ instead; this is a general property of diffusion.

The reason your method is incorrect is that the differentials in this problem mean something slightly different from usual. Here, the temperature $T$ is not a single quantity $T(t)$, but a function $T(x, t)$. And the two $dT$'s in your equation can't be cancelled, because they mean different things.

• The $dT$ on the left, inside the time derivative, is comparing $T(x, t)$ to $T(x, t+dt)$.
• The $dT$ on the right, inside the space derivative, is comparing $T(x, t)$ to $T(x+dx, t)$.

If $T$ were just a function $T(t)$, then canceling the $dT$'s would have been okay, because $dT$ can only possibly be comparing $T(t)$ and $T(t+dt)$. But the heat equation is more complicated than that.

• I think I've understood this now. Thanks; your efforts are MUCH appreciated :-) – user106570 Sep 16 '16 at 7:43
• Am I right that $dT$ is incorrect notation here, and a more rigorous notation involving $\partial T$ would be better? – John Dvorak Sep 16 '16 at 10:16
• @Jan Dvorak: I agree. – user106570 Sep 16 '16 at 10:22

When you use the equation $\dot Q = - kA \dfrac {dT(x)}{dx}$ one of the assumptions which has been made is that you have steady state conditions.
That is the temperature $T$ only depends on position $x$ and not on the time $t$.

What you are trying to deal with is a situation where the temperature at a position $x$ also depends on time so the equation to use looks similar

$\dot Q = - kA \dfrac {\partial T(x,t)}{\partial x}$

but notice the subtle difference in that because the temperature depends on two variables, partial derivative need to be used.

This derivation shows you the sort of equation which needs to be used in such circumstances.

Imagine a slab of cross sectional area $A$ and thickness $\Delta x$ with the temperature on one side being $T(x)$ and on the other side $T(x+\Delta x)$ as shown in the diagram below

Unlike steady state conductivity problems here the amount of heat entering the slab $\dot Q(x)$ differs from the amount of heat leaving the slab $\dot Q(x+\Delta x)$, that difference being responsible for the temperature of the slab changing with time.

So the equation which balances the flow energy is

$\dot Q(x) - \dot Q(x+\Delta x) = \dfrac{ A \;\Delta x \; \rho\; c \;[T(t+ \Delta t - T(t)]}{\Delta t}$

where $\rho$ is the density of the slab and $c$ is the specific heat capacity of the slab.

Rearranging this gives

$- \dfrac 1 A \dfrac {\dot Q(x+\Delta x)-\dot Q(x)}{\Delta x} = \rho \;c\; \dfrac{T(t+ \Delta t - T(t)}{\Delta t}$

Now allowing $\Delta x$ and $\Delta t$ to tend to zero gives

$- \dfrac 1 A \dfrac {\partial \dot Q}{\partial x} = \rho \;c\; \dfrac{\partial T}{\partial t} \Rightarrow - \dfrac 1 A \dfrac {\partial }{\partial x} {\left( -k\;A\; \dfrac{\partial T}{\partial x}\right) }= \rho \;c\; \dfrac{\partial T}{\partial t} \Rightarrow k \;\dfrac {\partial^2 T(x,t)}{\partial x^2}= \rho \;c\; \dfrac{\partial T(x,t)}{\partial t}$

assuming that the thermal conductivity $k$ is constant.

You are equating the temperature change of an object due to heat loss (or gain) to the heat current across an unrelated interface. You should have written the equation (2) with partial derivatives, on the left side with respect to time, on the right side with respect to the coordinate. Therefore you cannot simply cancel the ∂T in this equation. One is a time change of temperature of an extended body with spatially constant T and the other is the spatial change of T along a heat conduction path unrelated to this body.