Stefan Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in thermodynamics that describes the relationship between the temperature of an object and the total energy it radiates. The law applies to idealized blackbodies, objects that absorb and emit radiation perfectly. It plays a crucial role in understanding various physical phenomena, such as the heat emission from stars and the analysis of thermal radiation in various fields, including astrophysics, climate science, and engineering. 

1.0Statement Of Stefan Boltzmann Law

The amount of radiation emitted per second per unit area by a black body is directly proportional to the fourth power of its absolute temperature.

$\text{Amount of Radiation Emitted (E)}\propto T^4$

$E=\sigma T^4$ (T=Temperature of ideal black body in Kelvin)

• This law is true for only ideal black body
• SI Unit-$E=watt/m^2$
• $\sigma$ =Stefan Constant =$5.67✕10^{-8}watt/m^2{K}^4$
• Dimensions of $\sigma =[M^1{L}^0{T}^{-3}{\theta }^{-4}]$
• Total radiation energy emitted out by surface of area A in time t : If T is constant.

       Ideal black body $Q_{IBB}=\sigma A{T}^{4}t$ and for any other body  $Q_{GB}=e_r\sigma A{T}^{4}t$

       e→ emissivity of surface. It is property of surface

       0 ≤ e ≤ 1

        e=1 for Black Body

2.0Derivation of Stefan Boltzmann Law (Using Planck Radiation)

• Radiation emitted by black body is a function of wavelength and temperature and Energy spectral density is given as,

$\rho (\lambda ,T)$=For overall Wavelength

$E(T)=\sigma T^4$

$E(T)={\int }_0^{\infty }\rho (\lambda ,T)d\lambda$$={\int }_0^{\infty }\frac{8\pi hc}{{\lambda }^5(e^{[\frac{hc}{\lambda KT}]}-1)}d\lambda$$=8\pi hc{\int }_0^{\infty }\frac{d\lambda }{{\lambda }^3{\lambda }^2(e^{[\frac{hc}{\lambda KT}]}-1)}$

$u=\frac{hc}{\lambda kT},\lambda =\frac{hc}{ukT};\{\lambda \to 0,u\to \infty \},\{\lambda \to \infty ,u\to 0$

$\frac{du}{d\lambda }=\frac{d}{d\lambda }\left(\frac{hc}{\lambda kT}\right)=\frac{hc}{kT}\cdot \frac{d}{d\lambda }\left(\frac{1}{\lambda }\right)=\frac{hc}{kT}\left(-\frac{1}{{\lambda }^2}\right)\Rightarrow -\frac{kT}{hc}du=\frac{d\lambda }{{\lambda }^2}$

$E=8\pi hc{\int }_{\infty }^0{\left(\frac{ukT}{hc}\right)}^3\cdot \frac{1}{(e^u-1)}\cdot \frac{-kT}{hc}du$

$E=-\frac{8\pi k^4}{(hc{)}^3}{T}^4{\int }_{\infty }^0\frac{u^3}{(e^u-1)}du$ $(\therefore {\int }_{\infty }^0\frac{u^3}{(e^u-1)}.du=[-\frac{{\pi }^4}{15}])$

$E(T)=(\frac{8{\pi }^5{k}^4}{15(hc{)}^3})T^4$ $(\therefore \frac{8{\pi }^5{k}^4}{15(hc{)}^3}=\sigma )$

$E=\sigma T^4\Rightarrow$Stefan Boltzmann Law

3.0Key Points On Stefan Radiation Law

• A blackbody absorbs all radiation, from visible to infrared, and emits more power per unit area than any real object at the same temperature.
• The rate at which a blackbody emits radiation per unit surface area is proportional to the fourth power of the absolute temperature, $P=\sigma A{T}^4$. A  is the surface area and T is the surface temperature of the blackbody in kelvins.
• Kelvins must be used in Stefan's law because it is based on absolute temperature, not temperature difference. Therefore, Celsius (°C) cannot be used as a substitute.
• The universal constant  $\sigma$ (Greek letter sigma) is called Stefan’s constant $\sigma =5.670✕10^{-8}watt/(m^2{K}^4)$
• The fourth-power temperature dependence implies that the power emitted is extremely sensitive to temperature changes. 
• Since real objects are not perfect absorbers and emit less radiation than a blackbody, emissivity (e) is defined as the ratio of the emitted power of the object to the emitted power of a blackbody at the same temperature. Stefan’s law becomes,$P=e\sigma A{T}^4$
• Emissivity values range from0 to 1, with e = 1 representing a perfect radiator and absorber (a blackbody), and e = 0 indicating a perfect reflector. For example, polished aluminum, a highly efficient reflector, has an emissivity of around 0.05, while soot (carbon black) has an emissivity of approximately 0.95.

4.0Calculating the Sun's Temperature Using Stefan's Law   

Solar Constant 's'

• The Sun emits radiant energy continuously in space of which an insignificant part reaches the Earth. The solar radiant energy received per unit area per unit time by a black surface held at right angles to the Sun's rays and placed at the mean distance of the Earth (in the absence of atmosphere) is called solar constant.
• The solar constant S is taken to be $1340watts/m^2$ or $1.937Cal/c{m}^2$-minute

Temperature of the Sun

• Let R be the radius of the Sun and 'd' be the radius of Earth's orbit around the Sun. Let E be the energy emitted by the Sun per second per unit area. The total energy emitted by the Sun in one second $=EA=E\times 4\pi R^2$

(This energy is falling on a sphere of radius equal to the radius of the Earth's orbit around the Sun i.e., on a sphere of surface area $4\pi d^2$).

So, the energy falling per unit area of Earth $=\frac{4\pi R^2\times E}{4\pi d^2}=\frac{E{R}^2}{d^2}$

$R=7\times 10^{8}m,d=1.5\times 10^{11}m,s=5.7✕10^{-8}W{m}^{-2}K-4$

Solar constant, $S=\frac{E{R}^2}{d^2}$

By Stefan's Law,$E=\sigma T^4$

$S=\frac{\sigma T^4{R}^2}{d^2}$

$\Rightarrow T=[\frac{s\times d^2}{\sigma \times R^2}{]}^{\frac{1}{4}}$$=[\frac{1340\times (1.5\times 10^{11}{)}^2}{5.7\times 10^{-8}\times (7\times 10^8{)}^2}{]}^{\frac{1}{4}}=5732K$

5.0Sample Questions On Stefan Boltzmann Law

Q-1. If temperature of the ideal black body is increased by 50%, what will be a percentage increase in the quantity of radiation emitted from its surface.

Solution:   

$E\propto T^4$

$E\propto (1.5{)}^4{T}^4\propto {\left(\frac{15}{10}\right)}^4{T}^4\propto {\left(\frac{3}{2}\right)}^4{T}^4\propto \frac{81}{16}{T}^4\approx 5T^4$

$\frac{E^{\prime }-E}{E}\times 100\%=\left[\frac{5T^4-T^4}{T^4}\right]\times 100\%=400\%$

Q-2. Determine the temperature at which a perfect black body emits radiation at a specific rate of $5.67Wc{m}^{-2}$. Stefan's constant is $5.67✕10^{-8}J{s}^{-1}{m}^{-2}{K}^{-4}$

Solution:

$E=5.67Wc{m}^{-2}=5.67\times 10^{+4}W{m}^{-2},\sigma =5.67\times 10^{-8}J{s}^{-1}{m}^{-2}{K}^{-4}$

$E=\sigma T^4\Rightarrow T^4=\frac{E}{\sigma }$

$T={\left[\frac{E}{\sigma }\right]}^{\frac{1}{4}}={\left[\frac{5.67\times 10^{14}}{5.67\times 10^{-8}}\right]}^{\frac{1}{4}}={\left(10^{12}\right)}^{\frac{1}{4}}=1000K$

Q-3. A black body is at 27°C. Its surface area is equal to $\frac{1}{5.67}$ Find power emitted by it ?

Solution:

$P=\epsilon \sigma A{T}^4\Rightarrow (1)\frac{1}{5.67}\times 5.67\times 10^{-8}\times (300{)}^4$$\Rightarrow 10^{-8}\times 81\times 10^8=81W$