Stefan Boltzmann Law: Overview, Questions, Preparation

All matter like solid, liquid and gas with certain temperatures emits electromagnetic radiation. This radiation falls on other bodies, is partly absorbed and partly reflected. The amount of heat absorbed by the body depends on the colour of the body. Black bodies absorb and emit radiant energy better than other colours. 

The Stefan Boltzmann Law, also known as Stefan’s Law states that the energy radiated per unit surface area of a black body in unit time, H (black body irradiance or emissive power) is directly proportional to the fourth power of the black body’s thermodynamic temperature T (absolute temperature).

                                                          H = eσT⁴

• The SI unit of the irradiance is joules per second per square meter.
•  The SI unit of the absolute temperature is the Kelvin.
•  Є is the emissivity of the black body.
• The constant of proportionality σ called the Stefan-Boltzmann constant. Its value in SI units is 5.67 × 10⁻⁸ j s⁻¹m⁻² K⁻⁴.
• The radiated energy increases as temperature increases. We can observe that the object glows brighter as temperature increases.

The total absolute power of energy radiated for an object when the surface area included, A(in m²) 

                                                          H = AeσT⁴

A body at temperature T, with surroundings at temperatures Ts, emits and receives energy. For a perfect radiator, the net rate of loss of radiant energy is

                                                          H = σA (T⁴ – Ts⁴ ). 

For a body with emissivity, it changes to H = eσA (T⁴ – Ts⁴)

It is the law of radiation having importance to calculate the energy output of an object and the temperature of radiation sources. At higher levels, you will come across the derivation of the law using differential equations. The weightage of this topic is around 5%. 

1. A body of surface area 20 cm² has emissivity 0.6. When the body is heated to 500K and suspended at 300K, what will be the initial rate loss of heat?

 H = eσA (T⁴ – Ts⁴)

Given, e =0.6

             T= 500K

             Ts=300K

             A = 20cm² =20 × 10⁻⁴ m²

H = 0.6 (5.67 × 10⁻⁸)(20 × 10⁻⁴)(500-300)⁴ =0.108 W

2. The surface of a black body has an area of 2.0m² and an emissivity of 0.5. Find out a) rate of heat radiated from the body when the temperature is 75⁰c. b) rate of heat absorbed when the surrounding temperature is 25⁰c. c)net rate of radiation.
3. a) H = eσAT⁴ = 0.5(5.67 × 10⁻⁸)(2.0)(348)⁴ = 831W
4. b) H = eσATs⁴ = 0.5(5.67 × 10⁻⁸)(2.0)(298)⁴ =447W

c)Net rate = 831- 447 = 434W

3. The filament of a light bulb is cylindrical with length l = 20 mm and radius r = 0.05 mm. The filament is maintained at a temperature T = 5000 K. What is the total power radiated by filament?

P = AσT⁴ ; Area = 2πrl

P = 2πrlσT⁴ =2(1.44)(0.05 × 10⁻³)(0.02)(5.67 × 10⁻⁸)(5000)⁴ =220W

Q: What are the applications of Stefan Boltzmann Law?

Q: What are the Laws of Radiation?

Q: What are black body radiation sources?

Q: How can we derive the Stefan Boltzmann law?

Q: What are the characteristics of a perfect black body?