Consider a laterally insulated rod of length L and constant thermal conductivity. Assuming one-dimensional heat conduction in the rod, which of the following steady-state temperature profile(s) can occur without internal heat generation?

The correct temperature profile is represented by the straight line with a negative slope, where the temperature decreases linearly from $x=0$ to $x=L$ (corresponding to the option showing a linear profile with a downward slope).

Step-by-Step Mathematical Derivation:

For one-dimensional heat conduction in a rod, the general heat conduction equation is given by Fourier's law and the conservation of energy:
$$\frac{\partial }{\partial x}\left(k\frac{\partial T}{\partial x}\right)+{\dot{q}}_g=\rho C_p\frac{\partial T}{\partial t}$$
where:
• $T$ is the temperature as a function of position $x$ and time $t$,
• $k$ is the thermal conductivity of the rod,
• ${\dot{q}}_g$ is the rate of internal heat generation per unit volume,
• $\rho$ is the density, and
• $C_p$ is the specific heat capacity.

Simplifying the Governing Equation:
Based on the conditions specified in the problem statement:
1. Constant Thermal Conductivity: Since $k$ is constant, it can be taken outside the spatial derivative.
2. Steady-State Conditions: In steady-state, temperature does not change with time, meaning $\frac{\partial T}{\partial t}=0$.
3. No Internal Heat Generation: There is no heat generated within the rod, so ${\dot{q}}_g=0$.

Substituting these conditions simplifies the heat conduction equation to:
$$\frac{d^{2}T}{d{x}^2}=0$$

Integrating the Simplified Equation:
Integrating once with respect to $x$ yields:
$$\frac{dT}{dx}=C_1$$
where $C_1$ is the first constant of integration representing a constant temperature gradient.

Integrating a second time with respect to $x$ yields the temperature profile:
$$T\left(x\right)=C_{1}x+C_2$$
where $C_2$ is the second constant of integration.

Physical Interpretation of the Profile:
The equation $T\left(x\right)=C_{1}x+C_2$ describes a straight line with a constant slope of $C_1$.
• If heat is conducted from left to right along the rod (from $x=0$ to $x=L$), the temperature must decrease in the direction of heat flow according to the second law of thermodynamics (heat flows from high to low temperature). This results in a negative slope ( $C_1<0$).
• The correct option shows exactly this case: a straight line starting at a higher temperature at $x=0$ and decreasing linearly to a lower temperature at $x=L$.
• Non-linear temperature profiles (such as parabolic curves or profiles with internal local extrema) require either internal heat generation (which would result in non-zero second-order derivatives) or varying thermal conductivity, neither of which is present here.