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.

Explanation

Heat produced = heat loss or `(V^(2))/(R ) = (K A Delta T)/(l)`
Here `V = 400 V , K = 4 xx 10^(-4) cal//cms^(@) C = 16.8 xx 10^(4) J//cms^(@) C`
`Delta T = 100 , l = 1 mm = 0.1 cm , A = 6 (40)^(2) cm^(2)`
`(400^(2))/( R ) = ( 16.8 xx 10^(-4) xx 6 xx (40)^(2) xx 100 )/( 0.1) or R = 9.92 Omega`