A cubical box of volume `216cm^(3)` is made up of `0.1cm` thick wood, The inside is heated electrically by 100W heater. It is found that the temperature difference between the inside and the outside surface `5^(@)C` in steady state . Assuming that the entire electrically energy spent appears as heat, find the thermal conductivity of the material of the box.

Explanation

To solve the problem step by step, we need to find the thermal conductivity of the material of the cubical box using the given data.

### Step 1: Determine the dimensions of the cube
Given the volume of the cubical box is \( V = 216 \, \text{cm}^3 \).

Since the volume of a cube is given by \( V = a^3 \), where \( a \) is the length of an edge, we can find \( a \) as follows:

\[ a = \sqrt[3]{V} = \sqrt[3]{216 \, \text{cm}^3} = 6 \, \text{cm} \]

### Step 2: Convert dimensions to meters
The thickness of the wood is given as \( 0.1 \, \text{cm} \). We need to convert this to meters for consistency in SI units:

\[ \text{Thickness} (B) = 0.1 \, \text{cm} = 0.001 \, \text{m} \]

### Step 3: Calculate the surface area of the cube
The surface area \( A \) of a cube is given by:

\[ A = 6a^2 \]

Substituting \( a = 0.06 \, \text{m} \):

\[ A = 6 \times (0.06 \, \text{m})^2 = 6 \times 0.0036 \, \text{m}^2 = 0.0216 \, \text{m}^2 \]

### Step 4: Set up the heat conduction equation
The rate of heat transfer \( \frac{\Delta Q}{\Delta t} \) through the material is given by Fourier’s law of heat conduction:

\[ \frac{\Delta Q}{\Delta t} = k \cdot A \cdot \frac{\Delta T}{B} \]

Where:
- \( k \) = thermal conductivity (W/m·°C)
- \( A \) = surface area (m²)
- \( \Delta T \) = temperature difference (°C)
- \( B \) = thickness of the material (m)

### Step 5: Substitute known values
We know:
- \( \frac{\Delta Q}{\Delta t} = 100 \, \text{W} \)
- \( \Delta T = 5 \, \text{°C} \)
- \( B = 0.001 \, \text{m} \)
- \( A = 0.0216 \, \text{m}^2 \)

Substituting these values into the equation:

\[ 100 = k \cdot 0.0216 \cdot \frac{5}{0.001} \]

### Step 6: Solve for \( k \)
Rearranging the equation to solve for \( k \):

\[ k = \frac{100 \cdot 0.001}{0.0216 \cdot 5} \]

Calculating the right-hand side:

\[ k = \frac{0.1}{0.108} \approx 0.9259 \, \text{W/m·°C} \]

### Final Answer
The thermal conductivity of the material of the box is approximately:

\[ k \approx 0.9259 \, \text{W/m·°C} \]

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