physicscatalyst.com logo




Stefan Boltzmann law






Stefan Boltzmann law


  • 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
    $u_{rad} = \sigma \epsilon AT^4$ ------(1)
    Where
    $\sigma = 5.6703 \times 10^{-8} W/m^2K^4 $
    is stefan boltzmann constant and $\epsilon$ 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 $u_{abs}$ at which an object absorbs energy via thermal radiation from its environment with temperature $T_{env}$ (in kelvin) is
    $u_{abs} = \sigma \epsilon AT_{env}^4$ ---(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
    $u=u_{rad}-u_{abs}$
    $u = \sigma \epsilon A(T^4 - T_{env}^4)$ ----(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 $10 \; cm^2$ 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
$u = \sigma \epsilon A(T^4 - T_{env}^4)$
Here $\epsilon=.5$
$T=400K$
$T_{env}=300K$
$A=10 \times 10^{-4} m^2$
$\sigma = 5.6703 \times 10^{-8} W/m^2K^4 $
Substituting all these values
$u=.495 W$

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



link to this page by copying the following text


Class 11 Maths Class 11 Physics Class 11 Chemistry





Note to our visitors :-

Thanks for visiting our website.
DISCLOSURE: THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY DISCLOSURE FOR MORE INFO.