thermodynamics - Heat transfer between two surfaces

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.

Answer

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 $xx_0$, 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$xx_0$;$$q_0 = -k_a T_a'(x), \qquad q_0 = -k_b T_b'(x)$$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.

Cheers!