The thermal conductivity of the glass window is taken to be #k = 0.8 Wm^-1"^@C^-1#. Let #DeltaT# be the temperature difference between the inner and outer surfaces
Heat conducted through the window #Q_g=kA(DeltaT" ")/"Thickness of glass"#
#=>Q_g=(0.8xx2)/(0.75xx10^-2)DeltaT#
The thermal conductivity of the walls is given to be twice of glass wool #k_w = 2xx0.04 Wm^-1"^@C^-1#.
Therefore, heat conducted through the walls #Q_w=k_wA_w(DeltaT" ")/"Thickness of wall"#
#=>Q_w=(0.08xx10)/(13xx10^-2)DeltaT#
Window to Wall Ratio#=Q_g/Q_w=((0.8xx2)/(0.75xx10^-2)DeltaT)/((0.08xx10)/(13xx10^-2)DeltaT)#
Given is the same temperature difference between them. Above expression reduces to
#Q_g/Q_w=((0.8xx2)/0.75)/((0.08xx10)/13)#
#=>Q_g/Q_w=((0.8xx2)/0.75)xx(13/(0.08xx10))#
#=>Q_g/Q_w=34.dot6approx35#
