A double pan window used for insulating a room thermally from outside consists of two glass sheets each of area 1 `m^(2)` and thickness `0.01m` separated by `0.05` m thick stagnant air space. In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of `27^(@)C` and `0^(@)C` respectively. The thermal conductivity of glass is `0.8 Wm^(-1)K^(-1)` and of air `0.08 Wm^(-1)K^(-1)`. Answer the following questions.
(a) Calculate the temperature of the inner glass-air interface.
(b) Calculate the temperature of the outer glass-air interface.
(c) Calculate the rate of flow of heat through the window pane.

Answer

(i) 41.5
(ii) 26.48, 0.52

Explanation

Let `theta_(1)andtheta_(2)` be the temperatures of the two interfaces as shown in figure
Thermal resistance, `R=(l)/(KA)`
`thereforeR_(1)=R_(3)=((0.01))/((0.8)(1))=0.0125K//Wor.^(@)C//W`
and `R_(2)=((0.05))/((0.08)(1))=0.625^(@)C//W`
Now, the rate of heat flow `((dQ)/(dt))` will be equal from all

the three sections and since rate of heat flow is given by
`(dQ)/(dt)=("Temperature difference")/("Them alresistance")and((dQ)/(dt))_(1)=((dQ)/(dt))_(2)=((dQ)/(dt))_(3)`
Therefore , `(27-theta_(1))/(0.0125)=(theta_(1)-theta_(2))/(0.625)=(theta_(2)-0)/(0.0125)`
Solving this equation, we get `theta_(1)=26.48^(@)Candtheta_(2)=0.52^(@)C`
and `(dQ)/(dt)=(27-theta_(1))/(0.0125)rArr(dQ)/(dt)=((27-26.48))/(0.0125)=41.5W`