A 100 W heater is placed in a cubical container of edge length `6xx10^(-2)m.` The wall thick ness of the container is 1 mm. If inside and outside temperature in steady state are `30^(@)C.` and `25^(@)C,` find the thermal conductivity of the material of the box.

Explanation

To find the thermal conductivity of the material of the box, we can follow these steps:

### Step 1: Understand the Problem
We have a heater of power \( P = 100 \, \text{W} \) placed in a cubical container with edge length \( L = 6 \times 10^{-2} \, \text{m} \) and wall thickness \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \). The inside temperature is \( T_1 = 30^\circ C \) and the outside temperature is \( T_2 = 25^\circ C \). We need to find the thermal conductivity \( k \) of the container material.

### Step 2: Calculate the Temperature Difference
The temperature difference \( \Delta T \) between the inside and outside of the container is:
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