Question

A rod of length L with sides fully insulated is of a material whose thermal conductivity varies with temperature as

$K = \frac{\alpha}{T}$, where $\alpha$ is a constant. The ends of the rod are kept at temperature $T_1$ and $T_2$. The temperature T at x, where x is the distance from the end whose temperature is $T_1$ is

• A. $T_1 (\frac{T_2}{T_1})^{\frac{x}{L}}$
• B. $\frac{x}{L} \ln \frac{T_2}{T_1}$
• C. $T_1 e^{\frac{T_2x}{T_1L}}$
• D. $T_1 + \frac{T_2-T_1}{L}x$

Answer 1

Answer: (A)

Option (A) $T_1 (\frac{T_2}{T_1})^{\frac{x}{L}}$

Answer 2

Given the steady-state conduction equation, the heat flux $q$ is constant:

$$q = -K \frac{dT}{dx} = -\frac{\alpha}{T} \frac{dT}{dx}.$$

Rearrange:

$$\frac{dT}{T} = -\frac{q}{\alpha} \, dx.$$

Integrate from $x=0$ ($T=T_1$) to $x$ ($T=T$):

$$\int_{T_1}^{T} \frac{dT'}{T'} = -\frac{q}{\alpha} \int_{0}^{x} dx' \quad \Longrightarrow \quad \ln\frac{T}{T_1} = -\frac{q}{\alpha}x.$$

At $x=L$ ($T=T_2$):

$$\ln\frac{T_2}{T_1} = -\frac{q}{\alpha}L \quad \Longrightarrow \quad \frac{q}{\alpha} = -\frac{1}{L} \ln\frac{T_2}{T_1}.$$

Substitute back:

$$\ln\frac{T}{T_1} = \frac{x}{L} \ln\frac{T_2}{T_1}.$$

Exponentiating:

$$T = T_1 \left(\frac{T_2}{T_1}\right)^{\frac{x}{L}}.$$

Answer: Option (A)

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Explanation:

• Write Fourier's law with $K=\frac{\alpha}{T}$ and set up the ODE.
• Integrate after separating variables.
• Determine the constant using boundary conditions.
• Simplify to obtain $T = T_1 \left(\frac{T_2}{T_1}\right)^{\frac{x}{L}}$.