Question

The walls of a closed cubical box of edge `50cm` are made of a material of thickness `1mm` and thermal conductivity `4xx10^(-4) cm^(-1)(.^@C)^(-1)`. The interior of the box is maintained `100^(@)C` above the outside temperature by a heater placed inside the box and connected across `420V` d.c. calculate the resistance of copper.

Answer 1

`A` = total surface area of box
`= 6xx` area of each face
`= 6xx50xx50xx10^(-4)m^(2) = 1.5m^(2)`
`K = 4xx4.2xx10^(-2)J//sm.^(@)C`
`theta_(2)-theta_(1) = 100^(@)C, L = 1 mm = 10^(-3)m`
The rate of heat transfer through walls
`P = Q/t = (KA(theta_(2)-theta_(1)))/(L) = (4xx4.2xx10^(-2)xx1.5xx100)/(10^(-3))`
`=25200W`
Heat produced by `"heater"//"sec" = (V^2)/(R)`
`P = (V^2)/(R)`
`R = (V^2)/(P) = (420xx420)/((4xx4.2xx10^(-2)xx1.5xx100)//10^(-3))`
`=7Omega`.