Steady-state temperature profiles for the same heat flux out of two different plane walls with the same thickness and temperature on the left face are shown. For the same heat flux out of each wall:
A. Wall 1 and Wall 2 have the same thermal conductivity.
B. Wall 1 has a higher thermal conductivity than Wall 2.
C. It is not possible to tell which wall has the higher thermal conductivity.
D. Wall 2 has a higher thermal conductivity than Wall 1.
E. None of the above.
The Correct Answer and Explanation is:
Correct Answer: B. Wall 1 has a higher thermal conductivity than Wall 2.
Explanation:
To understand why Wall 1 has a higher thermal conductivity, we must examine how thermal conductivity affects the steady-state temperature profile in a material.
At steady-state, the rate of heat transfer qq through a plane wall is governed by Fourier’s Law: q=−kdTdxq = -k \frac{dT}{dx}
Where:
- qq is the heat flux (W/m²),
- kk is the thermal conductivity (W/m·K),
- dTdx\frac{dT}{dx} is the temperature gradient (K/m).
If two walls (Wall 1 and Wall 2) have the same thickness and the same steady-state heat flux, and the left face of both walls is held at the same temperature, we can compare their temperature profiles to infer differences in thermal conductivity.
Suppose the temperature drop across Wall 1 is less steep (i.e., more gradual) than across Wall 2. That means the temperature gradient dTdx\frac{dT}{dx} in Wall 1 is smaller than in Wall 2.
Since the heat flux qq is the same for both walls, but Wall 1 has a smaller temperature gradient, it must compensate with a higher thermal conductivity kk to maintain the same qq.
This is directly implied from rearranging Fourier’s Law: k=q∣dTdx∣k = \frac{q}{|\frac{dT}{dx}|}
So, for the same qq, if dTdx\frac{dT}{dx} is smaller (a flatter temperature curve), then kk must be larger. Therefore, Wall 1—with the flatter temperature gradient—has the higher thermal conductivity.
Thus, the correct answer is B: Wall 1 has a higher thermal conductivity than Wall 2.
