Question

Hot oil is circulated through a insulated container with a wooden lid at the top whose conductivity $K=0.149$ J$/(m{-}^{o}C$, thickness $t=5$mm, emissivity $=0.6$. Temperature of the top of the lid of maintained at $T_l={127}^{o}C$. If the ambient temperature $Ta={27}^o$. Calculate rate of heat loss per unit area due to radiation from the lid in $J{s}^{-1}{m}^{-2}$..

Answer 1

solution:

The rate of heat loss per unit area per second due to
radiation is given by Stefan's-Boltzmann law

$E$ = $e\sigma (T^4-T{o}^4)=0.6$ x $\frac{17}{3}$ x${10}^{-8}\left[\right(400{)}^4-\left(300{)}^4\right]$

$=595wa/m^2$

hence the correct answer is 595