Answer:
the percentage of heat lost by the window is 93.18%
Explanation:
Given the data in the question;
Area of glass [tex]A_{glass[/tex] = 0.10 m²
Thickness of glass [tex]t_{glass[/tex] = 8 mm = 0.008 m
Area of Styrofoam [tex]A_{styrofoam[/tex] = 11 m²
Thickness of Styrofoam [tex]t_{styrofoam[/tex] = 0.15 m
we know that;
Thermal conductivity of glass [tex]k_{glass[/tex] = 0.80 J/smC°
Thermal conductivity of Styrofoam [tex]k_{styrofoam[/tex] = 0.010 J/smC°
Now, temperature difference between outside and inside the walls and window is ΔT
So, In time t, heat lost due to conduction in the window will be;
[tex]Q_{glass[/tex] = [[tex]k_{glass[/tex] × [tex]A_{glass[/tex] × ΔTt] / [tex]t_{glass[/tex]
we substitute
[tex]Q_{glass[/tex] = [ 0.80 × 0.10 × (ΔT)t] / 0.008
[tex]Q_{glass[/tex] = [ 0.80 × 0.10 × (ΔT)t] / 0.008
[tex]Q_{glass[/tex] = 10(ΔT)t J
Also, the heat lost due to conduction in the wall be;
[tex]Q_{styrofoam[/tex] = [[tex]k_{styrofoam[/tex] × [tex]A_{styrofoam[/tex] × ΔTt] / [tex]t_{styrofoam[/tex]
we substitute
[tex]Q_{styrofoam[/tex] = [ 0.010 × 11 × ΔTt] / 0.15
[tex]Q_{styrofoam[/tex] = 0.7333(ΔT)t J
Now, Net heat lost in the wall and window is;
Q = [tex]Q_{glass[/tex] + [tex]Q_{styrofoam[/tex]
Q = 10(ΔT)t J + 0.7333(ΔT)t J
Q = 10.7333(ΔT)t J
So, the percentage of heat lost by the windows will be;
% of heat lost = [tex]Q_{glass[/tex] / Q
= 10(ΔT)t J / 10.7333(ΔT)t J
= 0.93167
= ( 0.93167 × 100 )%
= 93.18%
Therefore, the percentage of heat lost by the window is 93.18%
