Suppose I have surface A in contact with surface B, if I apply Fourier's law of heat transfer, which $K$ should I use, $K_a$ or $K_b$?

Essentially asking whether the same block of material heats faster in 300 degree water or 300 degree air or the same.


2 Answers 2


I'm going to add some mathematical detail to what akhmeteli has said.

Let's restrict the discussion to one dimension with coordinate $x$, then Fourier's law in differential form says $$ q(x) = -k(x) T'(x) $$ where $q(x)$ is the local heat flux, $k(x)$ is the conductivity, and $T(x)$ is the temperature gradient. Notice that Fourier's law show that that a given point, the derivative of the temperature is important, but derivatives of a function depend on the value of that function in a neighborhood of that point, not just the value of the function at that particular point. Therefore, the (not entirely explicit) answer to your question is that you need both $k$'s at a point where two materials with different $k$ are in contact. Now let's see the math.

If you are looking at a point $x_0$ at which two materials with different conductivities are joined, (say $k_a$ corresponds to $x<x_0$ and $k_b$ corresponds to $x>x_0$, then $k(x)$ has a jump discontinuity that can be written with the use of the Heaviside step function $\theta(x)$; $$ k(x) = (k_b-k_a)\theta(x-x_0) + k_a $$ Which results in the following differential equation: $$ q(x) = -[(k_b-k_a)\theta(x-x_0) + k_a] T'(x) $$ Which you can attempt to solve in a given case. For example, let's consider a steady-state system in which $q(x) = q_0$ is a constant and for which we want to determine the temperature gradient. Let's suppose that this system consists of metal bars joined at the point $x_0$ and whose endpoints are located at $x_0-L$ and $x_0+L$ respectively. Additionally, we assume that these other two endpoints are kept at a temperature $T_0$ In this case, the differential equation we would want to solve for $T(x)$ is $$ q_0 = -[(k_b-k_a)\theta(x-x_0) + k_a] T'(x) $$ with the boundary data $$ T(x_0-L) = T_0,\qquad T(x_0+L) = T_0 $$ The differential equation we want to solve can be rewritten as a set of two equations, one for $x<x_0$ and another for $x>x_0$; $$ q_0 = -k_a T_a'(x), \qquad q_0 = -k_b T_b'(x) $$ The general solutions are $$ T_a(x) = T_0-\frac{q_0}{k_a} [x-(x_0-L)], \qquad T_b(x) = T_0-\frac{q_0}{k_b} [x-(x_0+L)] $$ and the temperature everywhere except at $x=x_0$ can be written as $$ T(x) = (T_b(x) - T_a(x))\theta(x-x_0) + T_a(x) $$ In particular, notice that there is a jump discontinuity in the temperature at $x=x_0$; $$ T_b(x_0) -T_a(x_0) = q_0 L\left(\frac{1}{k_b}-\frac{1}{k_a}\right) $$ and this discontinuity depends on both $k$ values, not just the value on a particular side. Note further that if $k_a = k_b$, the the discontinuity disappears as you might intuitively expect!

Hope that helps! Let me know of any typos.


  • $\begingroup$ No problem @mezhang. I think this is a great question that should get more attention to be honest. It was also educational for me! $\endgroup$ Feb 4, 2013 at 2:06
  • $\begingroup$ This answer seems wrong. First of, even if kb==ba your example results in a discontinuity, as the last equation has a sign error in it. More importantly, you're effectively solving both sides as independent problems, not guaranteeing that the law holds at x0. That is how you can solve the example problem in the frist place, even though it should be impossible: Both ends have the same temperature, and yet you assume that heat is transferred. This would require thermal super conductivity. $\endgroup$
    – Stefan
    Jul 30, 2015 at 22:32
  • $\begingroup$ @Stefan I agree. There must be an error. I haven't looked at this in a long time, so I'll need to re-read this answer when I have a bit of time to see what I did wrong. $\endgroup$ Jul 30, 2015 at 22:54
  • $\begingroup$ No hurry, it took me 2+ years to point it out to you :) I'm pretty sure in an ideal setting T is continuous, but in real live the contact of two bodies is not perfect. There will be a poorly conductive third material (air, glue,...) partically inbetween, causing a steep temperature change microscopically, resulting in a macroscopic discontinuity. I think this is called "Thermal contact conductance". Wikipedia has a nice article on it. $\endgroup$
    – Stefan
    Jul 31, 2015 at 10:55

You should use the Fourier law in its differential form and use the actual temperature in each point. In your case, it is not important that temperature, strictly speaking, is not unambiguously defined at the contact surface (initially). You can still solve the equation of heat transfer (using distributions).


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