Heat is conducted through a slab composed of paralel layers of two different materials of conductivities `134.4` SI units and 58.8 SI units and of thickness `3.6cm` and `4.2cm` respectively. The temperature of the outer faces of the compound slab are `96^(@)C` and `8^(@)C`. Find (i) the temperature of the interface, (ii) temperature gradient in each section of the slab.

Answer

To solve the problem, we will use the concept of thermal conduction through materials and the formula for temperature gradient. The temperature gradient (dT/dx) is defined as the change in temperature (ΔT) over the change in distance (Δx).

### Given Data:
- Conductivity of material 1 (k1) = 134.4 SI units
- Conductivity of material 2 (k2) = 58.8 SI units
- Thickness of material 1 (L1) = 3.6 cm = 0.036 m
- Thickness of material 2 (L2) = 4.2 cm = 0.042 m
- Temperature at the outer face of the slab (T1) = 96°C
- Temperature at the outer face of the slab (T2) = 8°C

### Step 1: Calculate the temperature at the interface (T)
For steady-state conduction through parallel layers, the heat flow (Q) through each layer is equal. We can express this as:

\[ \frac{T1 - T}{L1} \cdot k1 = \frac{T - T2}{L2} \cdot k2 \]

Substituting the known values:

\[ \frac{96 - T}{0.036} \cdot 134.4 = \frac{T - 8}{0.042} \cdot 58.8 \]

### Step 2: Rearranging the equation
Multiply both sides by \(0.036 \cdot 0.042\) to eliminate the denominators:

\[ (96 - T) \cdot 134.4 \cdot 0.042 = (T - 8) \cdot 58.8 \cdot 0.036 \]

### Step 3: Simplifying the equation
Now, we will simplify both sides:

Left side:
\[ 134.4 \cdot 0.042 = 5.6448 \]
So,
\[ (96 - T) \cdot 5.6448 = (T - 8) \cdot 2.1168 \]

Right side:
\[ 58.8 \cdot 0.036 = 2.1168 \]

### Step 4: Expanding the equation
Expanding both sides gives us:

\[ 96 \cdot 5.6448 - T \cdot 5.6448 = T \cdot 2.1168 - 8 \cdot 2.1168 \]

### Step 5: Collecting like terms
Now, we will collect all terms involving T on one side:

\[ 96 \cdot 5.6448 + 8 \cdot 2.1168 = T \cdot (5.6448 + 2.1168) \]

### Step 6: Solving for T
Now we can calculate T:

1. Calculate \(96 \cdot 5.6448\) and \(8 \cdot 2.1168\).
2. Add these results together.
3. Divide by \(5.6448 + 2.1168\) to find T.

### Step 7: Calculate the temperature gradients
Now that we have T, we can calculate the temperature gradients in each section.

#### For Material 1:
\[ \text{Temperature Gradient (G1)} = \frac{T1 - T}{L1} = \frac{96 - T}{0.036} \]

#### For Material 2:
\[ \text{Temperature Gradient (G2)} = \frac{T - T2}{L2} = \frac{T - 8}{0.042} \]

### Final Calculation
1. Substitute the value of T into the equations for G1 and G2.
2. Calculate the gradients.

### Summary of Results:
- (i) The temperature at the interface (T).
- (ii) The temperature gradients in each section (G1 and G2).