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. `80^(@)C`
B. `20^(@)C`
C. `60^(@)C`
D. `40^(@)C`

Explanation

To solve the problem step by step, we will use the concept of thermal conductivity and the principle of heat flow through the two layers of the slab.

### Step 1: Understand the setup
We have a slab made of two materials: copper and brass. The thermal conductivities of copper (Kc) and brass (Kb) are in the ratio of 1:4. The temperatures at the two ends of the slab are given: the temperature of the brass side (T1) is 100°C and the temperature of the copper side (T3) is 0°C.

### Step 2: Define the variables
Let:
- T2 = Temperature at the interface between copper and brass.
- Kc = Thermal conductivity of copper.
- Kb = Thermal conductivity of brass.
- The thickness of both layers is the same, which we will denote as "d".

Given the ratio of thermal conductivities:
\[ \frac{Kc}{Kb} = \frac{1}{4} \]
This implies:
\[ Kb = 4Kc \]

### Step 3: Write the heat flow equations
In a steady state, the heat flow (Q) through both materials must be equal:
\[ \frac{T1 - T2}{R1} = \frac{T2 - T3}{R2} \]

Where R1 and R2 are the thermal resistances of copper and brass respectively, given by:
\[ R1 = \frac{d}{Kc \cdot A} \]
\[ R2 = \frac{d}{Kb \cdot A} \]

### Step 4: Substitute the resistances
Substituting R1 and R2 into the heat flow equation:
\[ \frac{100 - T2}{\frac{d}{Kc \cdot A}} = \frac{T2 - 0}{\frac{d}{Kb \cdot A}} \]

### Step 5: Simplify the equation
Since the area (A) and thickness (d) are the same for both materials, they can be cancelled out:
\[ \frac{100 - T2}{\frac{1}{Kc}} = \frac{T2}{\frac{1}{Kb}} \]

This simplifies to:
\[ (100 - T2) \cdot Kc = T2 \cdot Kb \]

### Step 6: Substitute Kb in terms of Kc
Using the relation \( Kb = 4Kc \):
\[ (100 - T2) \cdot Kc = T2 \cdot (4Kc) \]

### Step 7: Cancel Kc from both sides
Assuming Kc is not zero, we can divide both sides by Kc:
\[ 100 - T2 = 4T2 \]

### Step 8: Solve for T2
Rearranging gives:
\[ 100 = 5T2 \]
\[ T2 = \frac{100}{5} = 20°C \]

### Step 9: Conclusion
The temperature at the interface (T2) is 20°C.