A slab consists of two parallel layers of copper and brass of the same thichness and having thermal conductivities in the ratio `1:4` If the free face of brass is at `100^(@)C` anf that of copper at `0^(@)C` the temperature of interface is .

Options

A. (a)`80^@C`
B. (b)`20^@C`
C. (c)`60^@C`
D. (d)`40^@C`

Answer

To solve the problem step by step, we will use the concept of thermal conduction through two layers of materials with different thermal conductivities.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Thermal conductivity of copper (K_c) = 1K
- Thermal conductivity of brass (K_b) = 4K
- Temperature of brass face (T_b) = 100°C
- Temperature of copper face (T_c) = 0°C
- Let the thickness of each layer be 'd'.

2. **Set Up the Temperature Gradient:**
- Let the temperature at the interface between the two materials be θ (in °C).
- The temperature difference across the brass layer is (T_b - θ) = (100 - θ)°C.
- The temperature difference across the copper layer is (θ - T_c) = (θ - 0)°C = θ°C.

3. **Apply the Heat Conduction Equation:**
- According to Fourier's law of heat conduction, the heat flow (Q) through each layer must be equal since they are in steady state.
- For brass:
\[ Q = \frac{K_b \cdot A \cdot (T_b - \theta)}{d} = \frac{4K \cdot A \cdot (100 - \theta)}{d} \]
- For copper:
\[ Q = \frac{K_c \cdot A \cdot (\theta - T_c)}{d} = \frac{1K \cdot A \cdot \theta}{d} \]

4. **Equate the Heat Flow:**
- Since both expressions for Q are equal:
\[ \frac{4K \cdot A \cdot (100 - \theta)}{d} = \frac{1K \cdot A \cdot \theta}{d} \]
- The area (A) and thickness (d) cancel out, leading to:
\[ 4(100 - \theta) = \theta \]

5. **Solve for θ:**
- Expanding the equation:
\[ 400 - 4\theta = \theta \]
- Rearranging gives:
\[ 400 = 5\theta \]
- Therefore, solving for θ:
\[ \theta = \frac{400}{5} = 80°C \]

6. **Conclusion:**
- The temperature at the interface is θ = 80°C.