Question

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.16m^2 and whose thickness is 2.0 mm. Treat the wall as a slab of the insulating material styrofoam whose are and thickness are 18m ^ 2 and 0.10 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window?

Answer 1

Given that

Area of single pane glass window  Ag = 0.16 m^2

thickness of pane glass window  tg = 2.0 mm= 0.002 m

Area of styrofoam wall  As = 18m^2

Thickness of styrofoam wall ts = 0.10 m

heat loss due to window conduction

Qg =( Kg *Ag *T ) / tg

where

Kg = thermal conductivity of glass = 0.80 J

T = temperature difference between the inside and outside of the wall and window

Qg = (0.80 *0.16*T ) / 0.002

Qg = 64 T J ---------------------------------------(1)

heat loss due to wall conduction

Qs = ( Kst *Ast *T ) / ts

where Ks = thermal conductivity of styrofoam = 0.010 J

Qs = (0.010*18*T ) / 0.10

Qs= 1.8 J ---------------------------------------(2)

Now the Total  heat lost by the wall and window will be

Q= Qg + Qs = 64+1.8= 65.8 J -----------------------------(3)

So now %heat loss by the window will be as

Q % = Qg / Q *100

Q%= (64T / 65.8 T) *100

Q% = 97.26 %

Good luck !!