A room has a window fixed with a pane of area `1.2m^(s)` The glass has thickness `2.2mm` If the temperature outside the room is `36^(@)C` and the temperature inside is `26^(@)C`
(a) calculate the heat flowing into the room every hour
(b) If the same single pane window is replaced by double paned window with an air gap of `0.50cm` between the two panes calculate the heat flowing into the room every hour
`K_(g) =0.80 Wm^(-1)K^(-1),K_(air)=0.0234Wm^(-1)K^(-1)` .

Explanation

We assume the one side of the pane is at `36^(@)C` and the other side (inside the room) is at `26^(@)C` Given Thickness of the window pane
`d = 2.2mm = 2.2 xx 10^(-3)m`
Area of window pane ` A =1.2m^(2)`
`(Q)/(t)=K(A(theta_(2)-theta_(1)))/(d)= (0.8xx1.2xx10)/(2.2xx10^(-3)) =4364J//s`
Therefore heat flown into the room per hour is
`Q = 4364 xx 3600 =1.57 xx 10^(2)J`
(b) When single pane widow is replaced by a duble paned widow we have two layers of glss and one layer of air between them Thermal resistacne for glass
`R_(a) =(d_(a))/(K_(a)A_(a)) = (0.5xx10^(-2))/(0.0234xx1.2) =178xx10^(3)K//W`
Net thermal resistance `R_(T) =R_(g) +R_(a)+R_(g)`
` =(2.29 xx 10^(-3)+178 xx 10^(-3)+2.29 xx 10^(-3))`
`=182.6 xx 10^(-3)K//W`
`P=(Deltatheta)/(R_(T)) = (10)/(182.6xx10^(-3)) =55J//5`
Therefore heat flown into the room per hour is
`Q =P xx t =1.98 xx 10^(5) J` .