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.
Solution
Volume of the cube, V = a3 = 216 cm3
Edge of the cube, a = 6 cm
Surface area of the cube = 6a2
= 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
