A cubical box with a volume of 216 cm ^3 is constructed from wood of thickness 0.1 cm . Its interior is electrically heated using a 100 W heater. During steady state, the temperature difference between the inside and outside surfaces is observed to be 5^ C . Assuming all the consumed electrical energy is converted into heat, determine the thermal conductivity of the box's material.

Step-by-step solution

Volume of the cubical box is V = 216 cm ^3 Side length of the cubical box is a = V^ 1/3 = 6 cm = 0.06 m Total surface area of the cubical box is A = 6a^2 = 6 (0.06)^2 = 0.0216 m ^2 Thickness of the wood is d = 0.1 cm = 10⁻³ m In steady state, the rate of heat flow equals the power of the heater, so dQ dt = 100 W Using the formula for thermal conduction: dQ dt = KA T d Substituting the given values: 100 = K 0.0216 5 10⁻³ K = 100 10⁻³ 0.0216 5 K = 0.1 0.108 0.926 W m ⁻¹ ^ C ⁻¹ Rounding to two decimal places, we get 0