Question

Consider a 20-cm X 20-cm X 20-cm cubical body at 477°C suspended in the air. Assuming the body closely approximates a blackbody, determine
(a) the rate at which the cube emits radiation energy, in W and
(b) the spectral blackbody emissive power at a wavelength of 4 µm.

Answer 1

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:

$A_{s} =6a^{2} =6(0.2)^{2} =0.24m^{2}$

According Stefan-Boltzman law, the rate of emission is:

$E=\sigma T^{4} A_{s} =5.67x10^{-8} *(477)^{4} *0.24=704.48W$

b) Using Plank´s distribution law to get the spectral blackbody emissive power.

$E=\frac{C_{1} }{\lambda ^{5}(exp(\frac{C_{2} }{\lambda T}) -1 )} =\frac{3.743x10^{8} }{4^{5}(exp(\frac{1.4387x10x^{4} }{4*477})-1) } =194.27W/m^{2} \mu m$