(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivi
ty 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. 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
A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two.
there is not enough information about the liquid to know the force required for each. ie. stirring a cup of water is different than stirring a cup of pudding.
