Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are $$\theta_2$$ and $$\theta_1$$ respectively, $$(\theta_2 > \theta_1)$$. The temperature at the interface is:

Step-by-Step Solution

1. Principle of Heat Conduction in Series

When two conducting layers are joined in series to form a composite slab under steady-state conditions, the rate of heat flow ($$\frac{Q}{t}$$) through both materials must be identical.

The rate of heat conduction is governed by Fourier's Law:

$$\frac{Q}{t} = \frac{KA(\Delta\theta)}{x}$$

Where:

• $$K$$ = Coefficient of thermal conductivity
• $$A$$ = Cross-sectional area
• $$\Delta\theta$$ = Temperature difference across the layer
• $$x$$ = Thickness of the layer
• For Slab 1: Thermal conductivity is $$3K$$, thickness is $$d$$, and the temperature difference is $$(\theta_2 - T)$$.
• For Slab 2: Thermal conductivity is $$K$$, thickness is $$3d$$, and the temperature difference is $$(T - \theta_1)$$.

2. Set Up the Heat Flow Balance Equation

Let the unknown interface temperature be $$T$$.

Equating the two heat flow rates:

$$\frac{3KA(\theta_2 - T)}{d} = \frac{KA(T - \theta_1)}{3d}$$

3. Simplify and Solve for $$T$$

Cancel out the common parameters ($$K, A,$$ and $$d$$) from both sides:

$$3(\theta_2 - T) = \frac{1}{3}(T - \theta_1)$$

Multiply the entire equation by $$3$$ to eliminate the fraction:

$$9(\theta_2 - T) = T - \theta_1$$

Expand the left side:

$$9\theta_2 - 9T = T - \theta_1$$

Group the terms containing $$T$$ onto one side:

$$9\theta_2 + \theta_1 = T + 9T$$

$$10T = 9\theta_2 + \theta_1$$

$$T = \frac{9\theta_2 + \theta_1}{10}$$

Separating the terms yields:

$$T = \frac{\theta_1}{10} + \frac{9\theta_2}{10}$$

Final Answer

The temperature at the interface is $$\frac{\theta_1 + 9\theta_2}{10}$$ (or $$\frac{\theta_1}{10} + \frac{9\theta_2}{10}$$).