Plane wall with temperature dependent conductivity

Derivation

For steady one-dimensional flow, $Q$ is constant along $x$:

$$Q = -kA\frac{dT}{dx} = -k_0(1+bT)A\frac{dT}{dx}$$

Separate the variables and integrate from $x = 0$ ($T_1$) to $x = L$ ($T_2$):

$$Q\int_0^L dx = -k_0A\int_{T_1}^{T_2}(1+bT)\,dT$$ $$QL = k_0A\left[(T_1 - T_2)+\frac{b}{2}\left(T_1^2-T_2^2\right)\right]$$ $$Q = \frac{k_0A}{L}\left[1+\frac{b}{2}(T_1+T_2)\right](T_1-T_2) = \frac{k_mA\,(T_1-T_2)}{L}$$

So the heat flow is the same as for constant conductivity equal to the mean value $k_m = k_0\left[1+b\,\dfrac{T_1+T_2}{2}\right]$, the value of $k$ at the average temperature.

Shape of temperature profile

Integrating only up to a general $x$ gives a quadratic in $T$: $T + \dfrac{b}{2}T^2 = T_1+\dfrac{b}{2}T_1^2 - \dfrac{Q x}{k_0A}$. Since it is quadratic in $T$, $T$ is not linear in $x$. Where $k$ is large the gradient is small, and where $k$ is small the gradient is steep. For $b > 0$ the hot side has the larger $k$, so the profile is flatter near the hot face and steeper near the cold face (concave downward).

Numerical value

$$\begin{aligned} k_m &= 0.8\left[1 + 0.0006\times\frac{800+100}{2}\right] = 0.8(1.27) = 1.016\ \text{W/m K}\\ q &= \frac{k_m\,(T_1-T_2)}{L} = \frac{1.016\times700}{0.25} = 2845\ \text{W/m}^2 \end{aligned}$$

Answer: $q \approx 2.85\ \text{kW/m}^2$ with $k_m = 1.016$ W/m K.