How to describe heat transfer between two solid materials? A general equation for dealing with heat transfer between one material and a region of insulating material. I've seen basic heat transfer equations for one material, but I'd love to see an explanation of how to do two. 
 A: The one-dimensional heat equation for a solid can be written as:
$$ 
  \rho C_p\frac{\partial T}{\partial t}= -\frac{\partial}{\partial x} \left( k\frac{\partial T}{\partial x} \right) +\sigma 
$$
where $\sigma$ is the source term and $ \dot q =-k\frac{\partial T}{\partial x}$ is the diffusive heat flux. At the boundary the temperature and the flux must be continuous (if we consider contact resistance negligible, otherwise a gap in temperature could be possible) that is:
$$T_1=T_2$$ $$\dot q_1=\dot q_2 $$
A: Basically you can use Fourier's Law
$$ 
  q = -k\frac{dT}{dx}
$$
with the appropriate boundary conditions between the two materials. The basic issue is that at the interface between the two materials, there is a jump discontinuity in the value of the thermal conductivity, and you have to take this into account in solving the equation.
I did a very detailed calculation in a related post Heat transfer between two surfaces that you might find useful in this regard.
Also, if you want to actually calculate something about something in the real world, you may find this list of thermal conductivities useful.
Addendum. In response to comments below Lorenzo's response.
Consider two bars of length $L$ and of uniform (but unequal) thermal conductivities $k_a$ and $k_b$.  Let the heat flow $q_0>0$ be constant along the bars, then Fourier's Law shows that the temperatures $T_a$ and $T_b$ of bars $a$ and $b$ satisfy
$$
  T_a(x) = -\frac{q_0}{k_a} x + C_a, \qquad T_b(x) = - \frac{q_0}{k_b}x+C_b
$$
for some constants $C_a$ and $C_b$.  Now suppose that the left end of bar $a$ is located at $x=-L$ and the right end of bar $b$ is at $x=L$ so that they are joined at $x=0$.  Suppose additionally that the left end of bar $a$ is at temperature $T_L$ and the right end of bar $b$ is at $T_R$, then we have the boundary conditions
$$
  T_a(-L) = T_L, \qquad T_b(L) = T_R
$$
which tells us that
$$
  C_a = T_L -\frac{q_0}{k_a} L, \qquad C_b = T_R+\frac{q_0}{k_b}L
$$
So that
$$
  T_a(x) = T_L - \frac{q_0}{k_a}(x+L), \qquad T_b(x) = T_R -\frac{q_0}{k_b}(x-L)
$$
In particular, at $x=0$ we find
$$
  T_b(0)-T_a(0) = (T_R-T_L)+q_0\left(\frac{L}{k_a}-\frac{L}{k_b}\right)
$$
In particular, there is, in general a jump discontinuity in the temperature at the interface between the two materials unless the temperatures at which the ends of the bars are being kept are related by
$$
  T_R - T_L = q_0\left(\frac{L}{k_b}-\frac{L}{k_a}\right)
$$
