Question

A cubical box of volume 216 cm3 is made up of 0.1 cm thick wood. The inside is heated electrically by a 100 W heater. It is found that the temperature difference between the inside and the outside surface is 5°C in steady state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box.

Answer 1

The thermal conductivity of the material of the box is $\mathrm{K}=0.9259 \mathrm{W} / \mathrm{m}^{\circ} \mathrm{C}$

Explanation:

Step 1:

Given,  

volume of cube =$216 \mathrm{cm}^{3}$

Volume = $(side )^{3}$

Side of cube = $6 \mathrm{cm}$

Surface area of cube $=6(\text { side })^{2}=216 \times 10^{-4} \mathrm{m}^{2}$

Thickness of wood of box         $\mathrm{I}=0.1 \times 10^{-2}$

Temperature difference between inside and outside surface     $\Delta T=5^{\circ} \mathrm{C}$

Electrical energy or the rate of heat flow   $\mathrm{P}=100 \mathrm{W}$

Step 2:

The entire electrical energy spent appear as heat flow given condition in the question so

$\frac{\Delta Q}{\Delta t}=100$

$\frac{\Delta Q}{\Delta t}=\frac{K A \Delta T}{l}$

$\frac{K A \Delta T}{l}=100$

Step 3:

By substituting the values of given quantities in above equation we get thermal conductivity

$\mathrm{K}=0.9259 \mathrm{W} / \mathrm{m}^{\circ} \mathrm{C}$