Answer:
a) The rate at which the cube emits radiation energy is 704.48 W
b) The spectral blackbody emissive power is 194.27 W/m²μm
Explanation:
Given data:
a = side of the cube = 0.2 m
T = temperature = 477°C
Wavelength = 4 µm
a) The surface area is:
[tex]A_{s} =6a^{2} =6(0.2)^{2} =0.24m^{2}[/tex]
According Stefan-Boltzman law, the rate of emission is:
[tex]E=\sigma T^{4} A_{s} =5.67x10^{-8} *(477)^{4} *0.24=704.48W[/tex]
b) Using Plank´s distribution law to get the spectral blackbody emissive power.
[tex]E=\frac{C_{1} }{\lambda ^{5}(exp(\frac{C_{2} }{\lambda T}) -1 )} =\frac{3.743x10^{8} }{4^{5}(exp(\frac{1.4387x10x^{4} }{4*477})-1) } =194.27W/m^{2} \mu m[/tex]
