Answer to Question #250045 in Mechanics | Relativity for John

Question

Calculate the total radiant exitance (W∙m-2) for the following blackbody temperatures:

a. 250 K

b. 13°C

c. 981 K

d. -28°C

e. 273.15 K 

Answer 1

According to the Stefan-Boltzmann Law, radiant exitance of a black body is given by

$M_e = \sigma T^4$

T = temperature of black body

$\sigma =$ Stefan-Boltzmann constant

$\sigma = 5.67 \times 10^{-8} \;Wm^{-2}K^{-4}$

a. T=250 K

$M_e = 5.67 \times 10^{-8} \times 250^4 = 221.48 \;W \cdot m^{-2}$

b. $T= 13 + 273.15 = 286.15 \;K$

$M_e = 5.67 \times 10^{-8} \times 286.15 = 380.15 \;W \cdot m^{-2}$

c. T = 981 K

$M_e = 5.67 \times 10^{-8} \times 981 = 52.51 \times 10^3 \;W \cdot m^{-2}$

d. $T= -28 + 273.15 = 245.15 \;K$

$M_e = 5.67 \times 10^{-8} \times 245.15 = 204.79 \;W \cdot m^{-2}$

e. T = 273.15 K

$M_e = 5.67 \times 10^{-8} \times 273.15 = 315.63 \;W \cdot m^{-2}$