$(a)\;12W\qquad(b)\;22W\qquad(c)\;32 W\qquad(d)\;24 W$

Answer : 22 W

Explanation :

Area , $\;A = 10^{-2} m^{2}$

$T_{1}=227^{0}C=500 K$

$T_{2}=127^{0}C=400K$

Radiation emitted at $\;A \sigma T_{1}^{4}\;$

Radiation absorbed at $\;A \sigma T_{2}^{4}$

Therefore , The power needed = $\; A \sigma(T_{1}^{4}-T_{2}^{4})$

$=10^{-2}\times 6 \times 10^{-8} ((500)^{4}-(400)^{4})W$

$=2.214 \times 10^{11-10}W$

$=22.14 W$

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