Consider steady state, one-dimensional heat conduction in an infinite slab of thickness 2L (L = 1 m) as shown in the figure. The conductivity (k) of the material varies with temperature as k = CT, where T is the temperature in K, and C is a constant equal to 2 W.m^-1K^-2. There is a uniform heat generation of 1280 kW/m³ in the slab. If both faces of the slab are maintained at 600 K, then the temperature at x = 0 is _______ K (in integer).

The correct answer is 1000.

Problem Breakdown and Analysis:
We are given a steady-state, one-dimensional heat conduction problem inside an infinite slab of thickness 2L (where L = 1 m).
Based on the provided figures:
- The boundaries of the slab are located at x = -L and x = L, both maintained at a surface temperature of T_s = 600 K.
- The thermal conductivity of the material varies linearly with temperature: k = C T, where C = 2 W.m^-1K^-2.
- There is a uniform volumetric heat generation rate within the slab: q = 1280 kW/m³ = 1.28 * 10⁶ W/m³.

1. Governing Equation:
The steady-state, one-dimensional heat conduction equation with internal heat generation is expressed as:

$$\frac{d}{dx}\left(k\frac{dT}{dx}\right)+q=0$$

Substituting k = C T into the equation gives:

$$\frac{d}{dx}\left(CT\frac{dT}{dx}\right)=-q$$

2. First Integration:
Recall that T (dT/dx) = (1/2) * d(T²)/dx. Using this relation, we can rewrite the equation as:

$$\frac{d}{dx}\left[\frac{C}{2}\frac{d\left(T^2\right)}{dx}\right]=-q$$

Integrating once with respect to x:

$$\frac{C}{2}\frac{d\left(T^2\right)}{dx}=-qx+C_1$$

Due to the symmetry of the slab boundary conditions at x = -L and x = L, the temperature profile is symmetric about the midplane x = 0. Therefore, the temperature gradient is zero at the center: dT/dx = 0 at x = 0.
Substituting x = 0 and dT/dx = 0 into the integrated equation yields:

$$C_1=0$$

This simplifies the integrated relation to:

$$\frac{d\left(T^2\right)}{dx}=-\frac{2q}{C}x$$

3. Second Integration:
Integrating once more with respect to x gives:

$$T^2=-\frac{q}{C}{x}^2+C_2$$

4. Applying Boundary Conditions:
We apply the boundary condition at the outer surface, x = L = 1 m, where T = 600 K. We also plug in C = 2 W.m^-1K^-2 and q = 1280 * 10³ W/m³:

$${600}^2=-\frac{1280*{10}^3}{2}{\left(1\right)}^2+C_2$$

$$360000=-640000+C_2$$

Solving for C₂:

$$C_2=360000+640000=1000000={10}^6$$

5. Calculating the Temperature at the Center (x = 0):
Now we substitute x = 0 into the temperature profile equation:

$${\left[T\left(0\right)\right]}^2=-\frac{q}{C}{\left(0\right)}^2+C_2$$

$${\left[T\left(0\right)\right]}^2=C_2={10}^6$$

Taking the square root of both sides:

$$T\left(0\right)=\sqrt{{10}^6}=1000\text{K}$$