I have one block of material 1, and one block of material 2. They are in contact with each other, sharing an interface of area A.
Material 1 behaves as a lumped capacitance, with temperature T1(t). Material 2 has a temperature distribution of T2(x,t). At time t = 0, T1 > T2, with T2 initially being constant for all x.
How do I determine Q'(t), the rate of heat transfer from material 1 to material 2 due to conduction, in terms of T1(t) and/or T2(x,t)?
Assume that k2, the thermal conductivity of material 2, is known. (Do I need to know k1 or any other constants to solve the problem as well? If so, please let me know and include them in your answer!)
I know that the usual equation for conductive heat transfer through a single body is
$Q'(x,t) = k*A*\frac{dT(x,t)}{dx}$
So I initially assumed that the rate of heat transfer from material 1 to material 2 was simply
$Q'(t) = k2*A*\frac{dT2(x = 0,t)}{dx}$
However, this doesn't seem quite right. At t = 0, $\frac{dT2}{dx} = 0$ (since T2 is constant for all x at the beginning), so by the formula I suggested, Q'(0) = 0. And yet T1 > T2 at t = 0, so by the laws of thermodynamics, we must have Q'(0) > 0. There's a contradiction here.
What am I not taking into account? I'm assuming there's some factor I need to add to my formula for Q'(t), but I don't know quite enough about conductive heat transfer to know what I'm missing.
Thank you!