Hard to say since your image is way too small to read
#3
Wledig
69
1
Perfect thermal contact occurs at x=1, so the flow coming from the right is equal to the flow coming from the left. Now recall that the flow is given by ##\phi=-K_0\frac{\partial u}{\partial x}##, that is Fourier's law. The derivatives on both sides are constants as it's shown on your work so we're just left with: $$K_{1}\dfrac{\partial u_1}{\partial x} = K_{2}\dfrac{\partial u_2}{\partial x} \\ -c = -3d \\ -c+3d=0$$ I took the liberty to introduce some notation since it's a little confusing in your solution, but essentially all you have to do is substitute the conductivities given in the exercise and the solutions you found for u.
