Step 1: Understanding the Question:
The question asks for the mathematical shape of the temperature distribution profile across a flat (plane) wall experiencing steady-state, one-dimensional conductive heat transfer with constant thermal conductivity and no internal heat source.
Step 2: Key Formula or Approach:
Fourier's Law of heat conduction in one dimension is:
\[ q = -k A \frac{dT}{dx} \]
where:
$q$ = heat transfer rate ($\text{W}$),
$k$ = thermal conductivity ($\text{W/m}\cdot\text{K}$),
$A$ = cross-sectional area perpendicular to heat flow ($\text{m}^2$),
$\frac{dT}{dx}$ = temperature gradient in the direction of distance $x$.
Step 3: Detailed Explanation:
• Steady-State Conditions: Under steady-state conditions, the temperature at any position $x$ is constant over time, meaning the heat transfer rate ($q$) passing through the wall remains constant.
• No Heat Generation: With no internal heat generation and assuming constant thermal conductivity ($k$) and cross-sectional area ($A$), we can rearrange Fourier's Law as:
\[ \frac{dT}{dx} = -\frac{q}{k A} = \text{Constant} \]
• Integration: Integrating this differential equation with respect to distance $x$:
\[ \int dT = \int \left( -\frac{q}{k A} \right) dx \]
\[ T(x) = C_1 x + C_2 \]
where $C_1 = -\frac{q}{k A}$ and $C_2$ is the integration constant defined by boundary conditions.
• Profile Analysis: This is the equation of a straight line ($y = mx + c$). Consequently, the temperature decreases uniformly and linearly with distance through the plane wall.
Step 4: Final Answer:
Under steady-state, one-dimensional conduction with no heat generation, the temperature profile across a plane wall is linear, making option (A) the correct choice.