A wall consists of two layer, one of thickness 3 cm and other of 6 cm. the thermal conductivities of the materials of these layers are K and 3K respectively. The temperature of the outer surfaces of the two layers are `-5^(@)C` and `20^(@)C` respectively. determine the temperature of the surface of contact of the two layers in the steady state.

Answer

To determine the temperature at the surface of contact of the two layers in the steady state, we can follow these steps:

### Step 1: Understand the setup
We have two layers of a wall:
- Layer A: Thickness = 3 cm, Thermal conductivity = K
- Layer B: Thickness = 6 cm, Thermal conductivity = 3K
- Outer surface temperatures: T1 = -5°C (for Layer A) and T2 = 20°C (for Layer B)

### Step 2: Set up the heat flow equation
In steady state, the rate of heat flow through both layers must be equal. We denote the temperature at the interface of the two layers as T.

The heat flow through Layer A (dQ/dt)_A can be expressed as:
\[ \frac{dQ}{dt} = \frac{K \cdot A \cdot (T - T_1)}{L_A} \]
Where:
- \(T_1 = -5°C\)
- \(L_A = 3 \text{ cm} = 0.03 \text{ m}\)

The heat flow through Layer B (dQ/dt)_B can be expressed as:
\[ \frac{dQ}{dt} = \frac{3K \cdot A \cdot (T_2 - T)}{L_B} \]
Where:
- \(T_2 = 20°C\)
- \(L_B = 6 \text{ cm} = 0.06 \text{ m}\)

### Step 3: Equate the heat flows
Since the heat flow is the same through both layers, we can set the two equations equal to each other:
\[ \frac{K \cdot A \cdot (T - (-5))}{0.03} = \frac{3K \cdot A \cdot (20 - T)}{0.06} \]

### Step 4: Simplify the equation
We can cancel \(K\) and \(A\) from both sides, as they are common factors:
\[ \frac{T + 5}{0.03} = \frac{3(20 - T)}{0.06} \]

Next, simplify the right side:
\[ \frac{T + 5}{0.03} = \frac{3 \cdot 20 - 3T}{0.06} \]
\[ \frac{T + 5}{0.03} = \frac{60 - 3T}{0.06} \]

Multiply both sides by 0.06 to eliminate the denominators:
\[ 2(T + 5) = 60 - 3T \]

### Step 5: Solve for T
Expanding the left side:
\[ 2T + 10 = 60 - 3T \]

Rearranging gives:
\[ 2T + 3T = 60 - 10 \]
\[ 5T = 50 \]
\[ T = 10°C \]

### Conclusion
The temperature at the surface of contact of the two layers in the steady state is **10°C**.

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