Given: d = 0.2 m ; P = 192 W . d = 0.2~\text{m};~P = 192~\text{W}. \\ d = 0.2 m ; P = 192 W .
Stefan-Boltzmann Law: P = σ A T 4 ; P = {\sigma}AT^4; \\ P = σ A T 4 ;
where σ ≈ 5.670367 ∗ 1 0 − 8 W m 2 K 4 \sigma \approx 5.670367*10^{−8}~\frac{\text{W}}{\text{m}^2\text{K}^4}~ σ ≈ 5.670367 ∗ 1 0 − 8 m 2 K 4 W is the Stefan–Boltzmann constant.
Area of a sphere surface is expressed via diameter as A = π d 2 . A = {\pi}d^2.\\ A = π d 2 .
Using this knowledge and numerical values, the temperature in absolute (Kelvin) scale is
T = P σ A 4 = P σ π d 2 4 = 192 W 5.670367 ∗ 1 0 − 8 W m 2 K 4 ∗ 3.14 ∗ ( 0.2 m ) 2 4 ≈ 405.21 K . T = \sqrt[4]{\frac{P}{{\sigma}A}} = \sqrt[4]{\frac{P}{{\sigma}{\pi}d^2}} = \sqrt[4]{\frac{192~\text{W}}{5.670367*10^{−8}~\frac{\text{W}}{\text{m}^2\text{K}^4}~*~3.14~*~(0.2~\text{m})^2}} \approx 405.21~\text{K}. T = 4 σ A P = 4 σ π d 2 P = 4 5.670367 ∗ 1 0 − 8 m 2 K 4 W ∗ 3.14 ∗ ( 0.2 m ) 2 192 W ≈ 405.21 K .
Comments