Question

A sphere of radius 0.500 m, temperature 27.0 oC and emissivity 0.850 is located in an environment of temperature 77.0 oC. At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere’s net rate of energy exchange (J/s)?

Answer 1

Use Stefan-Boltzmann law.

P = A*ε*σ*T^4

where

A - surface area of the object

ε - emissivity

σ - Stefan-Boltzmann constant σ = 5.6704 x 10-8W/m^2K^4

T - absolute temperature in K

(a)

Surface area of a sphere is: A= 4*π*R^2

for a sphere of radius R

Hence rate of emission of energy is:

P_emitted = 4π*R^2*ε*σ*T_sphere^4

P_emitted = 4π*(0.500m)^2*0.850*(5.6704 x 10-8 W/m^2K^4)*(300K)^4

P_emitted = 1226.41 W

(b)

The absorptivity for any greay body is equal to its emissivity.

Assuming surrounding acts as black body you find for the rate of energy absorbed by the sphere:

P_absorbed = 4π*R^2*ε*σ*T_surrounding^4

P_absorbed = 4π*(0.500m)^2*0.850*(5.6704 x 10-8 W/m^2K^4)*(350K)^4

P_absorbed = 2272.1 W

(c)

Net rate of energy exchanged is the difference of energy absorbed and energy emitted:

P = P_absorbed - P_emitted

p = 1045.672 J/s or W