The rate urad at which an object emits energy via Electromagnetic radiation depends on objects surface area A
and temperature T in kelvin of that area and is given by urad=σϵAT4 ------(1)
Where σ=5.6703×10−8W/m2K4
is Stefan Boltzmann constant and ϵ is emissivity of object's surface with value between 0 and 1.
Black - Body radiator has emissivity of 1.0 which is an ideal limit and does not occur in nature.
The rate uabs at which an object absorbs energy via thermal radiation from its environment with temperature Tenv (in kelvin) is uabs=σϵAT4env ---(2)
Where ε is same as in equation 1
Since an object radiate energy to the environment and absorbs energy from environment its net energy exchange due to thermal radiation is u=urad−uabs u=σϵA(T4−T4env) ----(3)
u is positive if net energy is being lost via radiation. and negative if it is being absorbed via radiation
Solved Example
Question 1
A body of emissivity (.5) and of surface area 10cm2 is heated to 400 K and is suspended in a room at temperature 300 K. Calculate the initial rate of loss of heat from the body to the room? Solution
Now u=σϵA(T4−T4env)
Here ϵ=.5 T=400K Tenv=300K A=10×10−4m2 σ=5.6703×10−8W/m2K4
Substituting all these values u=.495W
Question 2
The surface of a body has a emissivity of .55 and area of 1.5 m2
Find out the following
a. What rate of heat is radiated from the body if the temperature is 50°C
b. At what rate is radiation absorbed by the radiator when the surrounding temperature is 22°C
c What is the net rate of radiation from the body
Given σ=5.67 *10-8 Solutions
a) Rate of radiation radiated=eσATb4=(.55)(5.67 *10-8)(1.5)(323)4=509W
b) Rate of radiation absorbed=eσATs4=(.55)(5.67 *10-8)(1.5)(295)4=354W
c)Net =Rate of radiation radiated-Rate of radiation absorbed=155 W