(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is 120 m2 and their inside surface is at 18.0 oC, while their outside surface is at 5.0 oC. (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction

a. 960 W b. One 1 kW room heater

Explanation:

a. The rate of heat conduction P = kA(T₂ - T₁)/d where k = 2 × 0.040 W/m-K =  0.080 W/m-K since the thermal conductivity of glass wool is 0.040 W/m-K and that of the material is twice the thermal conductivity of glass wool, A = area of walls = 120 m², T₁ = outside surface temperature = 5.0 °C, T₂ = inside surface temperature = 18.0 °C and d = thickness of wall = 13.0 cm = 0.13 m

P = kA(T₂ - T₁)/d

= 0.080 W/m-K × 120 m²(18.0 °C - 5.0 °C)/0.13 m

= 9.6 Wm/K × 13 K/0.13 m

= 124.8 Wm/0.13 m

= 960 W

b. The number of 1 kW room heater required will be

n = rate of heat conduction/power of one room heater = 960 W/ 1 kW = 960 W/1000 W = 0.96 ≅ 1

So we need only one 1 kW room heater.