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 solve the problem, we need to find the temperature at the interface of two materials (iron and brass) that are in thermal contact, given their thermal conductivities and the temperatures at their outer faces.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Temperature at the outer face of iron, \( T_1 = 100^\circ C \)
- Temperature at the outer face of brass, \( T_2 = 0^\circ C \)
- Thermal conductivity of iron, \( K_i = 0.2 \, \text{W/m°C} \)
- Thermal conductivity of brass, \( K_b = 0.3 \, \text{W/m°C} \)
- Thickness of both materials is equal, denoted as \( \Delta x \).

2. **Set Up the Heat Flow Equation:**
Since the heat flow through both materials is the same at steady state, we can write:
\[ \frac{K_i A (T_1 - T)}{\Delta x} = \frac{K_b A (T - T_2)}{\Delta x} \]
Here, \( A \) is the cross-sectional area, and \( T \) is the temperature at the interface.

3. **Simplify the Equation:**
Since the area \( A \) and thickness \( \Delta x \) are the same for both materials, they can be canceled out from the equation:
\[ K_i (T_1 - T) = K_b (T - T_2) \]

4. **Substitute the Known Values:**
Plugging in the values of \( K_i \), \( K_b \), \( T_1 \), and \( T_2 \):
\[ 0.2 (100 - T) = 0.3 (T - 0) \]

5. **Expand and Rearrange the Equation:**
Expanding both sides gives:
\[ 20 - 0.2T = 0.3T \]
Rearranging the equation:
\[ 20 = 0.2T + 0.3T \]
\[ 20 = 0.5T \]

6. **Solve 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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