Question

A sphere of radius 0.500 m, temperature 26.1°C, and emissivity 0.765 is isolated in an environment of temperature 77.0°C.

(a) At what rate does the sphere emit thermal radiation?
W

(b) At what rate does the sphere absorb thermal radiation?
W

(c) What is the sphere's net rate of energy exchange?
W

Answer 1

Given , r = 0.5 m
T = 26.1 C = 299.1 K
e = 0.765

Total energy flow fro a grey body per unit time is given by:
P = A·ε·σ·T⁴

where
A - surface area of the object = 4*π*r^2
ε - emissivity = 0.765
σ - Stefan-Boltzmann constant σ= 5.6704×10-8W/m²K⁴
T - absolute temperature in K

a)
Surface area of a sphere is:
A= 4·π·R²
Rate of emission of energy is:
Pemitted = 4·π·R² *ε*σ*Tsphere^4
Pemitted = 4 * π * 0.5^2 * 0.765 * 5.67 * 10^-8 * (299.1)^4
Pemitted =1090.5 W

(b)

Pabsorbed = 4·π·R²·ε·σ·Tsurrounding^4
Pabsorbed = 4 * π * 0.5^2 * 0.765 * 5.67 * 10^-8 * (77.0 + 273)^4
Pabsorbed = 2044.9 W

(c)
Net rate of energy exchanged is the difference of energy absorbed and energy emitted:
P = Pabsorbed - Pemitted
P = 2044.9 - 1090.5 W
P = 954.4 W