A cubical box of volume 216 cm^{3} 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.

#### Solution

Volume of the cube*, V* = *a*^{3} = 216 cm^{3}

Edge of the cube*, a *= 6 cm

Surface area of the cube = 6a^{2 }

= 6 (6 × 10^{−2})^{2}

= 216 × 10^{−4 }m2^{}Thickness, l = 0.1 cm = 0.1 × 10^{–2} m

Temperature difference, Δ T = 5°C

The inner surface of the cube is heated by a 100 W heater.

∴ Power, P = 100 W

Power = Energy per unit time

∴ Rate of flow of heat inside the cube, R = 100 J/s

Rate of flow of heat is given by^{}`(DeltaQ)/(Deltat) = (DeltaT)/(l/(KA)`

`100 = Kxx216xx10^-4xx5/(0.1xx10^-2)`

`K = 0.9259 ` W/ m° C