Utilize a lei de Stephan, R(T)=\sigma T^{4}, \sigma=5,67 \times 10^{-8} \mathrm{~W} /\left(\mathrm{m}^{2} \cdot \mathrm{~K}^{4}\right), para determinar quais as taxas de emissão de energia, por metro quadrado, de corpos negros às temperaturas listadas no exercício anterior.
2,7 \mathrm{~K} 273 K 310 K 373 K 973 K
\left(-270,3^{\circ} \mathrm{C}\right) \left(0^{\circ} \mathrm{C}\right) \left(37^{\circ} \mathrm{C}\right) \left(100^{\circ} \mathrm{C}\right) \left(700^{\circ} \mathrm{C}\right)

Passo 1

O enunciado pede que a gente use a lei de Stefan–Boltzmann
R(T)=\sigma\,T^{4},\quad \sigma=5,67\times10^{-8}\,\mathrm{W/(m^{2}\,K^{4})}
para calcular a potência radiada por metro quadrado de um corpo negro às temperaturas
2,7\:\mathrm{K}, 273\:\mathrm{K}, 310\:\mathrm{K}, 373\:\mathrm{K} e 973\:\mathrm{K}.

Passo 2

Para T=2,7\:\mathrm{K} e T=273\:\mathrm{K} calculamos
(2,7)^{4}=53{,}14,\quad R(2,7)=5,67\times10^{-8}\times53{,}14\approx3,01\times10^{-6}\:\mathrm{W/m^{2}}
(273)^{4}\approx5{,}55\times10^{9},\quad R(273)=5,67\times10^{-8}\times5,55\times10^{9}\approx3,15\times10^{2}\:\mathrm{W/m^{2}}

Passo 3

Para T=310\:\mathrm{K} e T=373\:\mathrm{K} temos
(310)^{4}\approx9,24\times10^{9},\quad R(310)=5,67\times10^{-8}\times9,24\times10^{9}\approx5,24\times10^{2}\:\mathrm{W/m^{2}}
(373)^{4}\approx1,94\times10^{10},\quad R(373)=5,67\times10^{-8}\times1,94\times10^{10}\approx1,10\times10^{3}\:\mathrm{W/m^{2}}

Passo 4

Para T=973\:\mathrm{K} calculamos
(973)^{4}\approx8,96\times10^{11},\quad R(973)=5,67\times10^{-8}\times8,96\times10^{11}\approx5,09\times10^{4}\:\mathrm{W/m^{2}}

Resposta

T=2,7\:\mathrm{K}:\;R\approx3,0\times10^{-6}\:\mathrm{W/m^{2}}
T=273\:\mathrm{K}:\;R\approx3,15\times10^{2}\:\mathrm{W/m^{2}}
T=310\:\mathrm{K}:\;R\approx5,24\times10^{2}\:\mathrm{W/m^{2}}
T=373\:\mathrm{K}:\;R\approx1,10\times10^{3}\:\mathrm{W/m^{2}}
T=973\:\mathrm{K}:\;R\approx5,09\times10^{4}\:\mathrm{W/m^{2}}

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