Question

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.11 m2 and whose thickness is 4 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.20 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

Step-by-step explanation:

Given that,

The area of glass
`A_g` =
`0.11m^2`

The thickness of the glass
`t_g=4mm=4*10^-^3m`

The area of the styrofoam
`A_s=11m^2`

The thickness of the styrofoam
`t_s=0.20m`

The thermal conductivity of the glass
`k_g=0.80J(s.m.C^o)`

The thermal conductivity of the styrofoam
`k_s=0.010J(s.m.C^o)`

Inside and outside temperature difference is ΔT

The heat loss due to conduction in the window is

`Q_g=(k_gA_g\Delta T t)/(t_g) \\\\=((0.8)(0.11)(\Delta T)t)/(4.0* 10^-^3)\\\\=(22\Delta Tt)j`

The heat loss due to conduction in the wall is

`Q_s=(k_sA_s\Delta T t)/(t_g) \\\\=((0.010)(11)(\Delta T)t)/(0.20)\\\\=(0.55\Delta Tt)j`

The net heat loss of the wall and the window is

`Q=Q_g+Q_s\\\\=(k_gA_g\Delta T t)/(t_g)+(k_sA_s\Delta T t)/(t_g)\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j`

The percentage of heat lost by the window is

`=(Q_g)/(Q)* 100\\\\=(22\Delta T t)/(22.55\Delta T t)* 100\\\\=97.6 \%`