A compound slab is made of two parallel plates of copper and brass of the same thickness and having thermal conductivities in the ratio `4:1`. The free face of copper is at `0^(@)C`. The temperature of the internal is `20^(@)C`. What is the temperature of the free face of brass?

Options

A. `0^(@)C`
B. `20^(@)C`
C. `40^(@)C`
D. `100^(@)C`

Explanation

To solve the problem, we will use the concept of thermal conductivity and the steady-state heat transfer through the compound slab made of copper and brass.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Thermal conductivity ratio: \( K_{Cu} : K_{Br} = 4 : 1 \)
- Temperature of the free face of copper: \( T_{Cu} = 0^\circ C \)
- Internal temperature (interface between copper and brass): \( T_{int} = 20^\circ C \)

2. **Let the Thermal Conductivities be:**
- \( K_{Cu} = 4k \)
- \( K_{Br} = k \)
Here, \( k \) is a constant representing the thermal conductivity of brass.

3. **Assume Thickness of Each Plate:**
- Let the thickness of both copper and brass be \( d \).

4. **Using the Heat Transfer Equation:**
The heat transfer \( Q \) through each material can be expressed as:
\[ Q = \frac{K \cdot A \cdot (T_1 - T_2)}{d} \]
Since the area \( A \) and thickness \( d \) are the same for both materials, we can simplify our calculations by focusing on the temperature differences and thermal conductivities.

5. **Set Up the Heat Flow Equations:**
For copper:
\[ Q = \frac{K_{Cu} \cdot (T_{Cu} - T_{int})}{d} = \frac{4k \cdot (0 - 20)}{d} = \frac{-80k}{d} \]
For brass:
\[ Q = \frac{K_{Br} \cdot (T_{int} - T_{Br})}{d} = \frac{k \cdot (20 - T_{Br})}{d} \]

6. **Since the heat flow is the same through both materials (steady state), we can equate the two expressions:**
\[ \frac{-80k}{d} = \frac{k \cdot (20 - T_{Br})}{d} \]
We can cancel \( k \) and \( d \) from both sides (assuming they are not zero):
\[ -80 = 20 - T_{Br} \]

7. **Solve for \( T_{Br} \):**
\[ T_{Br} = 20 + 80 = 100^\circ C \]

### Final Answer:
The temperature of the free face of brass is \( 100^\circ C \).

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