Question

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 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

The percentage loss of the window is
`E = 97.3`%

Step-by-step explanation:

From the question we are told that

The area of pane of glass is
`A = 0.15 m^2`

The thickness is
`d = 5mm = (5)/(1000) = 0.005m`

The thickness of the wall is
`D = 0.15m`

The area of the wall is
`a = 10m^2`

Generally the heat lost as a result of conduction of the window is

`Q_(window) = (j_(glass) * A * (\Delta T) )/(d)`

Where
`j_(glass)` is the thermal conductivity of glass which has a constant value of

`j_(glass) = 0.80 J/(s \cdot m \cdot C^o)`

Substituting values

`Q_(window) = ( 0.80 * 0.15 * (\Delta T) )/(0.005)`

`Q_(window) = 24 \Delta T`

Generally the heat lost as a result of conduction of the wall is

`Q_(wall) = (j_(styrofoam) * A * (\Delta T) )/(d)`

`j_(styrofoam)` s the thermal conductivity of Styrofoam which has a constant value of
`j_(styrofoam) = 0.010J / (s \cdot m \cdot C^o)`

Substituting values

`Q_(wall) = ( 0.010 * 10 * (\Delta T) )/(0.15)`

`Q_(wall) = 0.667 \ \Delta T`

Now the net loss of heat is

`Q_(net) = Q_(window) + Q_(wall)`

Substituting values

`Q_(net) = 24 + 0.667`

`Q_(net) = 24.667`

Now the percentage loss by the window is

`E = (Q_(window) )/(Q_(net)) * 100`

Substituting value

`E = (24)/(24 .667) * 100`

`E = 97.3`%