Question

The walls of a closed cubical box of edge` 40 cm` are made of a material of thickness `1 mm` and thermal conductivity `4 xx 10^(-4) cals^(-1@) C^(-1)`. The interior of the box is maintained at `100 ^(@) C` above the outside temperature by a heater placed inside the box and connected across `400 V dc` . Calculate the resistance of the heater.

Answer 1

Here, `l = 50 cm , d = 1 mm = 0.1 cm`,
`K = 4 xx 10^(-4) cal s^(-1) cm^(-1) .^(@)C^(-1)`,
`theta_(1) - theta_(2) = 100^@C, V= 400 V`
Area, `A = 6 xx` area of each face
` = 6 xx (50 xx 50) cm^(2)`
When temperature inside the box is maintianed, then
`V^(2)/(R xx 4.2) = (KA (theta_(1) - theta_(2)))/(d)`
or `R = (V^(2)d)/(4.2 KA (theta_(1) - theta_(2)))`
`= ((400)^(2) xx 0.1)/(4.2 xx (4 xx 10^(-4)) xx (6 xx 50 xx 50) xx 100)`
`= 6.35 Omega`