An insulated container has a wooden lid at the top whose conductivity is `0.15 J//m""^@Cs` , thickness is 5 mm and emissivity is 0.5. The temperature of the top of the wooden lid is maintained at `125^@C` and the ambient temperature is `27^@C` . Hot liquid is now circulated through the container as shown in the figure. Determine the rate of loss of heat per unit area due to radiation from the wooden lid and temperature of the oil.

Take `sigma = 5.66 xx 10^(-8) Wm^(-2) K^(-4)`

Explanation

The rate of loss of heat per unit area from the lid is
` (dQ)/(dt) = sigma e [T_(u)^(4) - T_(a)^(4) ]`
` = 5.66 xx 10^(-8) xx 0.5 [ 398^4 - 300^4]`
` = 480.9 Wm^(-2)`
` = 481 Wm^(-2)`
If `T_1` is the temperature of hot liquid , then
`(KA (T_i - T_u) )/(t ) = 481 A`
` rArr (0.15A(T_i - 398) )/(5 xx 10^(-3) ) = 481A`
` rArr T_i - 398 = (481 xx 5 xx 10^(-3) )/(0.15) `
` rArr T_i = 414.03 K = 414 K`