The outer faces of a rectangular slab made of equal thickness of iron and brass are maintained at `100^(@)C` and `0^(@)C` respectively. The temperature at the interface is (Thermal conductivity of iron and brass are 0.2 and 0.3 respectively)

Options

A. `100^(@)C`
B. `40^(@)C`
C. `50^(@)C`
D. `70^(@)C`

Answer

To find the temperature at the interface of the rectangular slab made of iron and brass, we can use the principle of steady-state heat conduction. The heat flow through both materials must be equal at the interface.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a slab made of two materials: iron and brass.
- The outer face of iron is maintained at \(100^\circ C\) and the outer face of brass at \(0^\circ C\).
- Let \(T\) be the temperature at the interface.

2. **Using Fourier's Law of Heat Conduction**:
- The rate of heat transfer through a material is given by:
\[ Q = k \cdot A \cdot \frac{\Delta T}{\Delta x} \]
- Where:
- \(Q\) = rate of heat transfer
- \(k\) = thermal conductivity of the material
- \(A\) = area of the slab
- \(\Delta T\) = temperature difference across the material
- \(\Delta x\) = thickness of the material

3. **Setting Up the Equations**:
- For iron:
\[ Q = k_I \cdot A \cdot \frac{100 - T}{\Delta x} \]
- For brass:
\[ Q = k_B \cdot A \cdot \frac{T - 0}{\Delta x} \]
- Here, \(k_I = 0.2\) (thermal conductivity of iron) and \(k_B = 0.3\) (thermal conductivity of brass).

4. **Equating Heat Flow**:
- Since the heat flow through both materials is the same at steady state:
\[ k_I \cdot (100 - T) = k_B \cdot T \]

5. **Substituting the Values**:
- Substitute the values of \(k_I\) and \(k_B\):
\[ 0.2 \cdot (100 - T) = 0.3 \cdot T \]

6. **Expanding and Rearranging**:
- Expanding the equation:
\[ 20 - 0.2T = 0.3T \]
- Rearranging gives:
\[ 20 = 0.2T + 0.3T \]
\[ 20 = 0.5T \]

7. **Solving for T**:
- Dividing both sides by \(0.5\):
\[ T = \frac{20}{0.5} = 40^\circ C \]

### Final Answer:
The temperature at the interface is \(T = 40^\circ C\).

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