In example `3, K_(1) = 0.125 W//m-.^(@)C, K_(2) = 5K_(1) = 0.625 W//m-.^(@)C` and thermal conductivity of the unknown material is `K = 0.25 W//m^(@)C L_(1) = 4cm, L_(2) = 5L_(1) = 20cm`. If the house consists of a single room of total wall area of `100 m^(2)`. then find the power of the electric heater being used in the room.

Explanation

`R_1=R_2=((4xx10^2 m))/((0.125 w//m -""^@ C)(100 m^2 ))`
`=32xx10^(-4)""^@ C//w`
`R=((10xx10^(-2) m ))/((0.25 W//m-""^@ C ) (100 m^2))=40xx10^(-4^@) C//w`
the equivalent thermal resistance of the entire wall
`=R_1+R_2+2R=144xx10^(-4^@ )C//W`
`therefore` Net heat current, i.e. amount of heat flowing out of the house per second `=(T_H-T_C)/(R)`
`=(25^@ C-(-20^@ C))/(144xx10^(-4^@)C//w)=(45xx10^4)/(144)` watt
Hence the heater must supply 3.12 kW to compensate for the outflow of heat.