To solve this problem it is necessary to apply the concepts related to rate of thermal conduction
[tex]\frac{Q}{t} = \frac{kA\Delta T}{d}[/tex]
The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, [tex]\Delta T[/tex], T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.
The change made between glass and air would be determined by:
[tex](\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}[/tex]
[tex] k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}[/tex]
[tex]\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}[/tex]
[tex]\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}[/tex]
[tex]\Delta T_{air} = 44.9 \Delta T_{glass}[/tex]
There are two layers of Glass and one layer of Air so the total temperature would be given as,
[tex]\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}[/tex]
[tex]\Delta T = 2\Delta T_{glass} +\Delta T_{air}[/tex]
[tex]20\°C = 46.9\Delta T_{glass}[/tex]
[tex]\Delta T_{glass} = 0.426\°C[/tex]
Finally the rate of heat flow through this windows is given as,
[tex]\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}[/tex]
[tex]\Delta {Q}{t} = 0.84*24*10 -3*0.426[/tex]
[tex]\Delta {Q}{t} = 179W[/tex]
Therefore the correct answer is D. 180W.
