Q. 28

A room has a window fitted with a single 1.0 m × 2.0 m glass of thickness 2 mm.

(a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and that outside is 10°C.

(b) The glass is now replaced by two glass panes, each having a thickness of 1 mm and separated by a distance of 1 mm.

Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0 J s–1m–1 °C–1and that of air = 0.025 J s–1m–1 °C–1.

Answer :

The diagram is shown –


(a) Given

Thickness, l = 2 mm = 0.0002 m

Temperature inside the room = 32°C

outside = 10°C

Dimensions of wall = 1.0 m × 2.0 m

Rate of flow of heat



∆T = is change in temperature between the two sides of the window.

A= Area of cross section of the window

K = thermal conductivity of the window

L = length of the window


= 8000J/s

(b). Resistance of glass-

The equivalent circuit for the two glass panes and air becomes


Here resistance of glass Rg =

And of air Rair =

Since, these are connected in series, equivalent resistance becomes

Reqv =

Thermal conductivity of window glass = 1.0 J s–1m–1 °C–1

And of air, = 0.025 J s–1m–1 °C–1.

Substituting values

Reqv = ( + )

= ( + )

= 0.021

Now, rate of heat flow


= 380.95

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