To solve this problem we will apply the concept of heat transfer and conduction depending on the Area, the temperature change and the thermal resistance, this can be expressed as
[tex]Q = \frac{A(HDD)}{R}[/tex]
Where,
A = Surface Area
HDD=Heating Degree Day
R = Thermal Resistance
According to the information the Area of the wall can be calculated with the given dimensions
[tex]A = 2 ((13*8) + (12*8))[/tex]
[tex]A= 400 ft^2[/tex]
Heating Degree Day would be
[tex]HDD = 65-25 = 40 \°F[/tex]
Finally the heat transferred is
[tex]Q = \frac{(400ft^2)(40 \°F)(\frac{24 hours}{1 day})}{13\frac{\° F \cdot ft^2 hr}{BTU}}[/tex]
[tex]Q = 29538.46BTU/day[/tex]
Therefore the heat that is lost through the walls by conduction on that day is 29538.46BTUs
