Question #268192

The energy emitted by radiation by a blackbody of surface area 410m2 and temperature of 155c


1
Expert's answer
2021-11-18T15:09:16-0500

Explanations & Calculations


  • The equations involved here is R=ϵσAT4\small R=\small \epsilon \sigma A T^4, in which R\small R is the power of energy emission, ϵ=1\small \epsilon =1 for a perfect black body and σ=5.67×108Wm2K4\small \sigma= 5.67\times10^{-8}Wm^{-2}K^{-4} .
  • Then,

W=5.67×108×410m2×(155+273)4=7.8×105W\qquad\qquad \begin{aligned} \small W&=\small 5.67\times10^{-8}\times410m^2\times(155+273)^4\\ &=\small 7.8\times10^5\,W \end{aligned}

  • To find any energy emitted, one needs to know the time for the emission did last.

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