The concept required to solve this problem is linked to the expression of the rate of heat conduction through a material.
This rate can be calculated according to the Area, the coefficient of thermal expansion, the temperature and the thickness where the heat flows. Mathematically it is described as
[tex]\frac{Q}{t} = \frac{kA\Delta T}{d}[/tex]
[tex]\frac{Q}{t} = \frac{kA(T_2-T_1)}{d}[/tex]
Where
k = Thermal Coefficient Constant
A = Surface Area
d = Thickness
[tex]\Delta T =[/tex] Change of temperature across the material.
Our values are given as
[tex]k_g = 0.042J/s\cdot m \cdot \°C \rightarrow[/tex] Thermal conductivity of the Glass wool
[tex]k = 2*0.042J/s\cdot m \cdot \°C \rightarrow[/tex] Thermal conductivity of material is twice of the glass wool
[tex]A = 120m^2[/tex]
[tex]d = 0.13m[/tex]
[tex]T_1 = 5\°C[/tex]
[tex]T_2 = 18\°C[/tex]
Replacing we have,
[tex]\frac{Q}{t} = \frac{2(0.042)(120)(18-5)}{0.13}[/tex]
[tex]\frac{Q}{t} = 1.008*10^3W[/tex]
Therefore the rate of heat conduction through house walls is [tex]1.008*10^3W[/tex]
