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

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

Step-by-step 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=(C_(1) )/(\lambda ^(5)(exp((C_(2) )/(\lambda T)) -1 )) =(3.743x10^(8) )/(4^(5)(exp((1.4387x10x^(4) )/(4*477))-1) ) =194.27W/m^(2) \mu m`