Explanation:
The given data is as follows.
Area = [tex]5 ft \times 2.5 ft[/tex] = 12.5 [tex]ft^{2}[/tex]
L = 0.25 in = 0.0208 ft (as 1 inch = 0.0833 ft)
[tex]T_{s, in} = 60^{o}F[/tex], [tex]T_{s, out} = 30^{o}F[/tex]
k (thermal conductivity) = 0.8 [tex]Btu/hr-ft-^{o}F[/tex]
Formula to calculate the heat loss will be as follows.
Q = [tex]k \times A \times \frac{T_{s, in} - T_{s, out}}{L}[/tex]
Putting the given values into the above formula as follows.
Q = [tex]k \times A \times \frac{T_{s, in} - T_{s, out}}{L}[/tex]
= [tex]0.8 Btu/hr-ft-^{o}F \times 12.5 ft^{2} \times \frac{60^{o}F - 30^{o}F}{0.0208 ft}[/tex]
= 14285.7 Btu/hr
So, [tex]Q_{net} = 4Q[/tex]
= [tex]4 \times 14285.7[/tex]
= 57142.8 Btu/hr
Thus, we can conclude that heat loss through the windows is 57142.8 Btu/hr.
