Step 1: Steady-state heat conduction in a plane wall.
For steady-state heat conduction through a plane wall with constant thermal conductivity, the temperature distribution is linear. This follows from Fourier's law of heat conduction:
\[
q = -k \frac{dT}{dx}
\]
Where:
- \( q \) is the heat flux (constant in steady-state),
- \( k \) is the thermal conductivity (constant),
- \( \frac{dT}{dx} \) is the temperature gradient.
The solution to this equation, given constant properties and boundary conditions, results in a linear temperature distribution across the plane wall.
Final Answer: \[ \boxed{\text{linear}} \]
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