Q. 12

A cubical box of volume 216 cm3 is made up to 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 :


Given:
Volume of the box : V = 216 cm3 = 216× 10-6 m3.
Thickness of the box: x = 0.1 cm = 0.001 m
Power of the heater : P = 100 W
Temperature difference : ΔT = 5 °C

Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
Volume of the cube is a3. Where a is the side of the cube.
a = (216× 10-6)1/3 = 0.06 m
As heat will be transferred from all the sides of the cube,
Surface area of the cube is : A = 6a2
A = 6× (0.06)2 = 0.0216 m2.
We know that,
Power = Energy per unit time
Thus,

Substituting we get,





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