Answer to Question #268192 in Mechanics | Relativity for Ali Carton

Question

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

Answer 1

Explanations & Calculations

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

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

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