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.

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

=

Where,

∆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 R_{g =}

And of air R_{air =}

Since, these are connected in series, equivalent resistance becomes

R_{eqv =}

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

R_eqv = ( + )

= ( + )

= 0.021

Now, rate of heat flow

=

= 380.95