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the percentage of heat lost by the window is 93.18%

Explanation:

Given the data in the question;

Area of glass = 0.10 m²

Thickness of glass = 8 mm = 0.008 m

Area of Styrofoam = 11 m²

Thickness of Styrofoam = 0.15 m

we know that;

Thermal conductivity of glass =  0.80 J/smC°

Thermal conductivity of Styrofoam   =  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;

= [ ×  × ΔTt] /

we substitute

= [ 0.80 × 0.10 × (ΔT)t] / 0.008

= [ 0.80 × 0.10 × (ΔT)t] / 0.008  

= 10(ΔT)t J

Also, the heat lost due to conduction in the wall be;

=  [ × × ΔTt] /

we substitute

=  [ 0.010 × 11 × ΔTt] / 0.15

=  0.7333(ΔT)t J

Now, Net heat lost in the wall and window is;

Q = +

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 = / 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%