Question

Calculate the rate of heat conduction (W) 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 18.00C, while their outside surface is at 5.000C

Answer 1

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

`(Q)/(t) = (kA\Delta T)/(d)`

`(Q)/(t) = (kA(T_2-T_1))/(d)`

Where

k = Thermal Coefficient Constant

A = Surface Area

d = Thickness

`\Delta T =` Change of temperature across the material.

Our values are given as

`k_g = 0.042J/s\cdot m \cdot \°C \rightarrow` Thermal conductivity of the Glass wool

`k = 2*0.042J/s\cdot m \cdot \°C \rightarrow` Thermal conductivity of material is twice of the glass wool

`A = 120m^2`

`d = 0.13m`

`T_1 = 5\°C`

`T_2 = 18\°C`

Replacing we have,

`(Q)/(t) = (2(0.042)(120)(18-5))/(0.13)`

`(Q)/(t) = 1.008*10^3W`

Therefore the rate of heat conduction through house walls is
`1.008*10^3W`