## Stefan Boltzmann Law - Definition and Equation

Josef Stefan was an Austrian physicist. In 1879, he formulated a law about the radiant energy of a blackbody. According to the law, a theoretical object that absorbs all the radiation that falls on it is proportional to the fourth power of its temperature. His law was one of the first important steps toward the understanding of the blackbody radiation, from which raised the quantum idea of radiation.

Stefan rose from a lecturer in the subject of mathematical physics in 1858 to a professor Ordinarius of physics in 1863 and to the director of the Physical Institute in 1866, at the University of Vienna. Five years later he derived his law empirically. The same was derived theoretically by Ludwig Bolzmann of Austria and hence became known as the Stefan-Boltzmann Law.

### Blackbody Radiation

A theoretical object that can be perfectly efficient at absorbing and emitting radiation is known as a blackbody, and it therefore emits what's known as blackbody radiation.

A blackbody is black at room temperature, hence the name. However, at higher temperatures, it can actually glow at visible wavelengths. Therefore, you should be aware that astronomers and physicists use the term 'blackbody' to also refer to objects that glow.

Naturally, we may tend to think that blackbody radiation is given off only by hot objects like that hot metal rod or an extremely hot star. Not so. The radiation that is emitted by a heated body is termed as Blackbody radiation. But 'heated' is a relative term. What I mean is, it's heated when compared to absolute zero - 0 Kelvin, -273.2 Celsius or -459.7 Fahrenheit.

Knowing this, it won't surprise you to learn that blackbody radiation is emitted by everything from stars and light bulbs to ice cubes. That's because at temperatures above absolute zero, particles in an object still have thermal energy and thus radiate photons. An ice cube may be as cool as a cucumber but compared to absolute zero, it's as agitated and hot as you are when your computer freezes or the car doesn't start.

### Why Does Hotter = Brighter?

Now we need to learn how the temperature and the colour of the glow of an object interact. Surely you've seen how as metal is heated more and more, it actually changes the colour of light it emits from red, to orange, to yellow, as it gets hotter. Now you'll learn why it changes colour and gets brighter as it gets hotter, not just why it glows.

Firstly, the hotter an object is, the more blackbody radiation is emitted by it. Why? Well, the madder you are, the more agitated you are, and the more steam you have to blow off. The hotter an object, the more collisions there are, the more violent those collisions, and the more radiation is emitted. This is why as things get hotter, they get brighter.

### What is Absolute Temperature?

The absolute temperature of an object is the temperature on a scale, where 0 is taken as absolute zero. This is also known as thermodynamic temperature; absolute temperature scales are Kelvin, degree units Celsius and Rankine degree unit Fahrenheit.

Absolute zero is the temperature at which a system is in the state of lowest possible energy i.e., minimum energy. When molecules approach this temperature, their movements drop towards zero. It is the lowest temperature that a gas thermometer can measure. You know electronic devices will not work at this temperature. Finally, the Kinetic Energy of the molecules becomes negligible or zero.

Common temperatures as seen in the absolute scale are:

0°C (freezing point of water) = 273.15 K

25°C (room temperature) = 298.15 K

100°C (boiling point of water) = 373.15 K

0 K (absolute zero) = –273.15 Celsius

232.15 K(equal measures in Celsius and Fahrenheit) = –41 Celsius

To convert from the Celsius scale into the absolute temperature, you have to add 273.15 and change °C to K. To get a temperature on the absolute scale to the Celsius scale, you have to subtract or minus 273.15 and change K to °C. This is generally used in the world of science. Kelvin is used globally. It is one of the 7 base units of the system. The value of Absolute temperature is 0 K.

### Conversion of units of Temperature

Celsius to Kelvin, K = C + 273.15

Kelvin to Celsius, C = K – 273.15

Fahrenheit to Rankine, R = F + 459.67

Rankine to Fahrenheit, F = R – 459.67

### Application of Stefan-Boltzmann Law

The Stephan-Boltzmann Law describes the power radiated by a body that absorbs all radiation that falls on its surface in terms of its temperature. The radiation energy per unit time from a black body is proportional to the fourth power of the absolute temperature and can be expressed as the following formula.

Radiate energy = (Stefan-Boltzmann constant) * (Temperature)4

The equation is: J = σ T^{4} J m^{-2} s^{-1}

J - Energy radiated per unit area by a blackbody per unit time [Units: J m^{-2} s^{-1}]

σ - Stefan-Boltzmann constant [Value: 5.67 × 10^{-8} J s^{-1} m^{-2} K^{-4}]

T - Absolute temperature

Units: K

A body that does not absorb all the radiations that are incident on it are known as a grey body which emits less total energy than a blackbody and is characterised by an emissivity, є < 1.

J = єσ T^{4}

The radiant emittance J has the dimensions of energy flux (energy per time per area), and the SI units of measurement are joules per second per square metre, or equivalently, watts per square metre. Kelvin is the SI unit for absolute temperature T. є is the emissivity of the grey body; if it is a perfect blackbody, є =1.

To find the total radiated power from an object, multiply the formula with its surface area A.

P = J * A = Aєσ T^{4}

Radiation energy/Power can be represented as following:

Radiate energy = (Emissivity) × (Stefan-Boltzmann constant) × (Temperature)^{4} × (Area)

The equation is: P = є σ T^{4} A Watts

P: Radiate energy

σ: The Stefan-Boltzmann Constant

T: Absolute temperature in Kelvin

є: Emissivity of the material

A: Area of the emitting body

### Pictorial representation of Stefan-Boltzmann law

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Check out the graphs below representing blackbody curves to see this concept represented visually. Note that these graphs have a y-axis indicating intensity and an x-axis representing wavelength.

You see larger areas under curves representing hotter objects, where the area under the curve is proportional to the total energy emitted. So, a hotter object emits more radiation (including visible light). This concept refers to the Stefan-Boltzmann Law, which says, in simple terms, that the total energy radiated from a blackbody is proportional to the fourth power of its temperature.

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### What is Stephan-Boltzmann Constant?

The Stefan-Boltzmann constant, symbolised by the lowercase Greek letter sigma (σ), is a physical constant related to black body radiation. The constant defines how much power a black body can emit per unit area, as a function of its thermodynamic temperature. Stefan's law establishes the relationship between the emitted power per unit area in watts per square metre and the thermodynamic temperature in kelvin’s (T) .This means, the power per unit area is directly proportional to the fourth power of the thermodynamic temperature.

Thus, the value of the Stefan-Boltzmann constant is approximately 5.67 x 10^{-8} watt per metre squared per kelvin to the fourth.

**Stephan-Boltzmann Formula Questions:**

**Example 1: **

A body having emissivity of 0.1 and its area is 200 m^{2}, at 500 K. Find at what rate does it radiate the energy?

**Solution**

The energy radiated is given by the formula:

P = є σ T^{4} A

P = 0.1 × 5.67 × 10(-8) W/(m^{2} K^{4}) × (500 K)^{4} × 200 m^{2}

P = 7.08×10^{(4)} W

**Example 2:**

A metal ball of radius 3 cm is heated to 5000°C. If its emissivity is 0.5, at what rate does it radiate the energy?

**Solution**

The temperature in Kelvin is (5000°C + 273°C) K/°C = 5273 K.

The surface of the sphere is 4 π r^{2} = 4 π (0.03m)^{2} = 0.011 m^{2}.

The energy radiated is given by the formula:

P = є σ T^{4} AP = 0.5 × 5.67 × 10^{(-8)} W/(m^{2} K^{4}) × (5273 K)^{4} × 0.011 m^{2}

P = 2.4 × 10^{(5)} W

**Example 3:**

Assuming the Sun is a blackbody and its temperature is 6000 K. How much energy is it emitting per one square unit?

**Solution**

The energy radiated is given by the formula:

P = є σ T^{4} A

P = 1 × 5.67 × 10^{(-8)} W/(m^{2} K^{4}) × (6000K) × 1 m^{2}

P = 73,483,200 W

## FAQs on Stefan Boltzmann Law

**1.How was the Stefan-Boltzmann constant discovered?**

The Stefan-Boltzmann constant is named after Joseph Stefan and Ludwig Boltzmann. While Joseph found this constant, Ludwig derived it from the first principles.

In 1864, John Tyndall derived the measurements of the infrared emission by a platinum filament and the corresponding colour of the filament. In 1879, on the basis of Tyndall's measurements, Josef Stefan, deduced the proportionality to the fourth power of the absolute temperature.

In 1884, Ludwig Boltzmann presented the law’s derivation from theoretical considerations. He drew upon the work of Adolfo Bartoli, who had some years earlier proved from the principles of thermodynamics that radiation pressure exists. Following the findings of Bartoli, Boltzmann used electromagnetic radiation in place of an ideal gas as a working matter to consider a heat engine. The law was, almost immediately, verified experimentally.

**2. What is Blackbody Radiation?**

The amount of radiation a surface emits at a given wavelength depends on the material of the body, the surface’s condition and temperature. Therefore, different materials emit different amounts of radiation energy. A body which absorbs all the radiation falling on it and emitting huge amounts of heat at its absolute temperature is called a blackbody.

It is a physical body with certain properties and has the emissivity of ε = 1.0.

The surface of the blackbody emits radiation at 448 watts per square metre (approximately) at room temperature.

Real objects cannot radiate as much heat as a perfect black body does. Therefore they are called grey bodies.

Real objects with emissivities less than 1.0 emit radiate at lower rates.

Since an object’s emissivity and absorptivity are interconnected as per the Kirchhoff’s Law of thermal radiation, a blackbody therefore is also an ideal absorber of electromagnetic radiation.