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Stefan’s Law | Stefan’s Boltzmann Law | Stefan’s Fourth Power Law

Stefan’s Law | Stefan’s Boltzmann Law | Stefan’s Fourth Power Law :- Stefan’s Law was experimentally formulated by Josef Stefan in 1879 and theoretically derived by Ludwig Boltzmann in 1884, using thermodynamics and Maxwell’s electromagnetic theory.

Stefan’s Law states that The total energy radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of its absolute temperature.

Mathematically :

$\displaystyle E=\sigma {{T}^{4}}$

Where :

= Energy emitted per second per unit area (W/m²)
= Absolute temperature in kelvins (K)
= Stefan–Boltzmann constant

$\displaystyle \sigma =5.67\times {{10}^{-8}}W{{m}^{-2}}{{K}^{-4}}$

Dimensions of

For an Ideal Black Body (IBB) (emissivity ε $= 1$ )

$\displaystyle E=\sigma {{T}^{4}}$

For a Real Body/Grey Body (GB) (0 < emissivity ε ≤ 1)

$\displaystyle E=\varepsilon \sigma {{T}^{4}}$

Total Radiant Power Emitted by a Body

If a body has surface area , then the radiant power emitted is :

$\displaystyle P=\varepsilon \sigma A{{T}^{4}}$

Total Radiation Energy Emitted out by surface of area A in time t :

For ideal black body, $\displaystyle {{Q}_{IBB}}=\sigma A{{T}^{4}}t$
For any other body, $\displaystyle {{Q}_{GB}}=\varepsilon \sigma A{{T}^{4}}t$

Let us apply Prevost’s theory of exchanges to Stefan’s law to determine the net loss of heat radiation from a body. Every body at a temperature above 0 K is constantly emitting thermal radiation, regardless of the surroundings. At the same time, it’s also absorbing radiation from its surroundings. The body’s net heat exchange is the difference between what it emits and what it absorbs.

Let a blackbody be placed in surroundings at temperature T₀, where (T₀ < T)

Rate of emission of radiation from ideal black body surface, $\displaystyle {{E}_{1}}=\sigma {{T}^{4}}$

Rate of emission of radiation from surroundings, $\displaystyle {{E}_{2}}=\sigma T_{0}^{4}$

Net rate of loss of radiation from ideal black body surface is :

$\displaystyle E={{E}_{1}}-{{E}_{2}}=\sigma ({{T}^{4}}-T_{0}^{4})$

Net loss of radiation energy from entire surface area A in time t is

$\displaystyle {{Q}_{IBB}}=\sigma A({{T}^{4}}-T_{0}^{4})t$

For any other body,

$\displaystyle {{Q}_{GB}}=\varepsilon \sigma A({{T}^{4}}-T_{0}^{4})t$

If in time dt the net heat energy loss for ideal black body is dQ and because of this its temperature falls by dθ, then

Rate of loss of heat/emitted power (R_H) :

$\displaystyle {{R}_{H}}=\frac{dQ}{dt}=\varepsilon \sigma A({{T}^{4}}-T_{0}^{4})$

Also,

$\displaystyle \frac{dQ}{dt}=m\,s\,\frac{d\theta }{dt}$

$\displaystyle \Rightarrow \frac{d\theta }{dt}=\frac{1}{m\,s}\frac{dQ}{dt}$

Hence

Rate of fall in temperature (Rate of cooling) (R_F) :

$\displaystyle {{R}_{F}}=\frac{d\theta }{dt}=\frac{1}{ms}\frac{dQ}{dt}=\frac{\varepsilon \sigma A}{ms}({{T}^{4}}-T_{0}^{4})$

Note :

(i) If two bodies of different shapes have identical parameters — T, T₀, m, s, V and ρ, then

and

will be maximum for the body with the larger surface area.

(ii) Let us compare

and

for a sphere (radius = R), a cube (side = R) and a cylinder (radius = R, length = R) :

Stefan's Law | Stefan's Boltzmann Law | Stefan's Fourth Power Law - Curio Physics

(iii) If a solid and hollow sphere are taken with all the parameters same then hollow sphere will cool down at faster rate (If mass is the same but one object is hollow and the other is solid, then to keep the masses equal you must make the hollow sphere bigger in size having more surface area).

(iv) Rate of fall of temperature,

$\displaystyle {{R}_{F}}\propto \frac{d\theta }{dt}\propto \frac{1}{s}$

$\displaystyle \Rightarrow dt\propto s$

If three bodies having different specific heats (), starts cooling at the same temperature and end at the same lower temperature, then body with larger specific heat needs more heat removal to achieve the same temperature drop and therefore ().

Example 1.

A body with a surface area of 5 cm² and a temperature of 727 °C radiates 300 J of energy every minute. What is its emissivity ?

(Stefan-Boltzmann constant σ = 5.67 × 10^-8 W m^-2 K^-4)

Solution :

$\displaystyle P=\varepsilon \sigma A{{T}^{4}}$

Here $\displaystyle P=\left( \frac{300}{60} \right)J/s=5J{{s}^{-1}}$ and $\displaystyle T=273+727=1000K$

$\displaystyle \Rightarrow \varepsilon =\frac{P}{\sigma A{{T}^{4}}}=\frac{5}{\left( 5.67\times {{10}^{-8}} \right)\times \left( 5\times {{10}^{-4}} \right)\times {{\left( 1000 \right)}^{4}}}$

$\displaystyle \therefore \varepsilon =0.18$

Example 2.

The operating temperature of a tungsten filament in an incandescent lamp is 2000 K and its emissivity is 0.3. Find the surface area of the filament of a 25 watt lamp. Stefan’s constant σ = 5.67 × 10^–8 Wm^–2 K^–4.

Solution :

Rate of emission = wattage/power of the lamp, hence

$\displaystyle A=\frac{P}{\varepsilon \sigma {{T}^{4}}}=\frac{25}{\left( 0.3 \right)\times \left( 5.67\times {{10}^{-8}} \right)\times {{\left( 2000 \right)}^{4}}}=0.918c{{m}^{2}}$

Example 3.

The filament of an incandescent lamp is at a temperature of and emits of energy. The surface area of the filament is . If the room temperature is , calculate the emissivity of the filament. Stefan’s constant σ = 5.67 × 10^–8 Wm^–2 K^–4

Solution :

$\displaystyle P=\varepsilon \sigma A({{T}^{4}}-T_{0}^{4})$

$\displaystyle \Rightarrow \varepsilon =\frac{P}{\sigma A({{T}^{4}}-T_{0}^{4})}$

Given,

T = 1727 + 273 = 2000 K

T_o = 27+273 = 300 K

Power emitted, P = (2000J/60sec) = (100/3) W

A =

Here,

$\displaystyle {{T}^{4}}={{(2000)}^{4}}=1.6\times {{10}^{13}}$

$\displaystyle T_{0}^{4}={{(300)}^{4}}=8.1\times {{10}^{9}}$

$\displaystyle \Rightarrow T_{0}^{4}<<{{T}^{4}}$

$\displaystyle \Rightarrow ({{T}^{4}}-T_{0}^{4})\cong {{T}^{4}}=1.6\times {{10}^{13}}$

$\displaystyle \Rightarrow \varepsilon =\frac{P}{\sigma A({{T}^{4}}-T_{0}^{4})}\cong \frac{P}{\sigma A{{T}^{4}}}=\frac{100/3}{\left( 5.67\times {{10}^{-8}} \right)\times \left( 1\times {{10}^{-4}} \right)\times 1.6\times {{10}^{13}}}$

$\displaystyle \therefore \varepsilon \approx 0.37$

Example 4.

If two bodies A and B, with emissivities and respectively, radiate energy at the same rate with equal temperatures , and , what is the ratio of their radii assuming both are spherical ?

Solution :

According to question,

$\displaystyle {{P}_{A}}={{P}_{B}}$

$\displaystyle \Rightarrow {{\varepsilon }_{A}}\sigma {{A}_{A}}T_{A}^{4}={{\varepsilon }_{B}}\sigma {{A}_{B}}T_{B}^{4}$

$\displaystyle \Rightarrow \frac{{{\varepsilon }_{A}}}{{{\varepsilon }_{B}}}=\frac{{{A}_{B}}}{{{A}_{A}}}=\frac{4\pi r_{B}^{2}}{4\pi r_{A}^{2}}=\frac{r_{B}^{2}}{r_{A}^{2}}$

$\displaystyle \therefore \frac{{{r}_{A}}}{{{r}_{B}}}=\sqrt{\frac{{{\varepsilon }_{B}}}{{{\varepsilon }_{A}}}}=\sqrt{\frac{3}{2}}$

Example 5.

If the temperature of an ideal black body is increased by 50 %, what will be percentage increase in quantity of radiation emitted from its surface ?

Solution :

For an ideal black body,

$\displaystyle E=\sigma {{T}^{4}}$

$\displaystyle \Rightarrow {{E}_{1}}=\sigma T_{1}^{4}$

New temperature,

$\displaystyle {{T}_{2}}={{T}_{1}}+\frac{{{T}_{1}}}{2}=\frac{3}{2}{{T}_{1}}$

$\displaystyle \Rightarrow {{E}_{2}}=\sigma T_{2}^{4}=\sigma {{\left( \frac{3}{2}{{T}_{1}} \right)}^{4}}=\sigma \frac{81}{16}T_{1}^{4}$

Percentage increase in quantity of radiation emitted,

$\displaystyle \frac{\Delta E}{{{E}_{1}}}\times 100=\frac{{{E}_{2}}-{{E}_{1}}}{{{E}_{1}}}\times 100=\left( \frac{\sigma \frac{81}{16}T_{1}^{4}-\sigma T_{1}^{4}}{\sigma T_{1}^{4}} \right)\times 100\cong 400%$

Example 6.

If the temperature of an ideal black body is decreased from to , calculate the percentage loss in its rate of emission of radiation.

Solution :

$\displaystyle {{E}_{1}}=\sigma T_{1}^{4}$

New temperature,

$\displaystyle {{T}_{2}}=\frac{{{T}_{1}}}{2}$

$\displaystyle \Rightarrow {{E}_{2}}=\sigma T_{2}^{4}=\sigma {{\left( \frac{{{T}_{1}}}{2} \right)}^{4}}=\sigma \frac{1}{16}T_{1}^{4}$

Percentage loss in emission rate,

$\displaystyle \frac{\Delta E}{{{E}_{1}}}\times 100=\frac{{{E}_{2}}-{{E}_{1}}}{{{E}_{1}}}\times 100=\left( \frac{\sigma \frac{1}{16}T_{1}^{4}-\sigma T_{1}^{4}}{\sigma T_{1}^{4}} \right)\times 100$

$\displaystyle \Rightarrow \frac{\Delta E}{{{E}_{1}}}\times 100=-93.75%$

Example 7.

Calculate the temperature (in K) at which a perfect black body radiates energy at the rate of 5.67 W cm⁻² . Given σ = 5.67 × 10^–8 Wm^–2 K^–4 .

Solution :

Given,

$\displaystyle E=5.67Wc{{m}^{-2}}=5.67\times {{10}^{4}}W{{m}^{-2}}$

Using,

$\displaystyle E=\sigma {{T}^{4}}$

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

Example 8.

A man, the surface area of whose skin is 2 m², is sitting in a room where the air temperature is 20°C. If his skin temperature is 28 °C, find the rate at which his body loses heat. The emissivity of his skin = 0.97.

Solution :

Absolute room temperature (T₀) = 20 + 273 = 293 K

Absolute skin temperature (T) = 28 + 273 = 301 K

Rate of heat loss :

$\displaystyle {{R}_{H}}=\frac{dQ}{dt}=\varepsilon \sigma A({{T}^{4}}-T_{0}^{4})$

$\displaystyle \Rightarrow {{R}_{H}}=0.97\times 5.67\times {{10}^{-8}}\times 2\times \left[ {{\left( 301 \right)}^{4}}-{{\left( 293 \right)}^{4}} \right]$

$\displaystyle \therefore {{R}_{H}}=92.2W$

Example 9.

Compare the rate of loss of heat from a metal sphere of the temperature 827 °C, with the rate of loss of heat from the same sphere at 427 °C, if the temperature of surroundings is 27 °C.

Solution :

According to Stefan’s Law of radiation, rate of loss of heat (R_H) :

$\displaystyle {{R}_{H}}=\frac{dQ}{dt}=\varepsilon \sigma A({{T}^{4}}-T_{0}^{4})$

Given : T₁ = 827 °C = 1100 K, T₂ = 427 °C = 700 K and T₀ = 27 °C = 300 K

$\displaystyle \therefore \frac{{{\left( \frac{dQ}{dt} \right)}_{1}}}{{{\left( \frac{dQ}{dt} \right)}_{2}}}=\frac{\left( T_{1}^{4}-T_{0}^{4} \right)}{\left( T_{2}^{4}-T_{0}^{4} \right)}=\frac{\left[ {{\left( 1100 \right)}^{4}}-{{\left( 300 \right)}^{4}} \right]}{\left[ {{\left( 700 \right)}^{4}}-{{\left( 300 \right)}^{4}} \right]}=6.276$

$\displaystyle {{\left( \frac{dQ}{dt} \right)}_{1}}:{{\left( \frac{dQ}{dt} \right)}_{2}}=6.276:1$

Example 10.

Two spheres of the same material have radii 6 cm and 9 cm respectively. They are heated to the same temperature and allowed to cool in the same enclosure. Compare their initial rates of emission of heat and initial rates of fall of temperature.

Solution :

Given : r₁ = 6 cm, r₂ = 9 cm,

$\displaystyle \frac{{{r}_{1}}}{{{r}_{2}}}=\frac{6}{9}=\frac{2}{3}$

According to Stefan’s Law of radiation, rate of loss/emission of heat (R_H) :

$\displaystyle {{R}_{H}}=\frac{dQ}{dt}=\varepsilon \sigma A({{T}^{4}}-T_{0}^{4})$

$\displaystyle \Rightarrow {{R}_{H}}\propto A$

As, A = 4πr²,

$\displaystyle \therefore {{R}_{H}}\propto {{r}^{2}}$

$\displaystyle \Rightarrow \frac{{{\left( {{R}_{H}} \right)}_{1}}}{{{\left( {{R}_{H}} \right)}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}={{\left( \frac{2}{3} \right)}^{2}}=\frac{4}{9}$

Rate of fall in temperature (Rate of cooling) (R_F)

$\displaystyle \Rightarrow {{R}_{F}}\propto \frac{A}{V}$

$\displaystyle \Rightarrow {{R}_{F}}\propto \frac{4\pi {{r}^{2}}}{\frac{4}{3}\pi {{r}^{3}}}$

$\displaystyle \Rightarrow {{R}_{F}}\propto \frac{1}{r}$

$\displaystyle \therefore \frac{{{\left( {{R}_{F}} \right)}_{1}}}{{{\left( {{R}_{F}} \right)}_{2}}}=\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{3}{2}$

Example 11.

A sphere of ice at 0°C having initial radius $R$ is placed in an environment having ambient temperature $> 0°C$ . The ice melts uniformly, such that shape remains spherical. After a time ‘t’ the radius of the sphere has reduced to r. Which graph best depicts r(t) :

Solution :

According to Stefan’s Law, rate of heat loss by the environment or rate of heat gain by the sphere :

$\displaystyle {{R}_{H}}=\frac{dQ}{dt}=\sigma A({{T}^{4}}-T_{0}^{4})$ …..(a)

Rate of heat absorption by ice can also be written as :

$\displaystyle \frac{dQ}{dt}=\frac{d(mL)}{dt}=L\frac{dm}{dt}$ …..(b)

Now, the rate of melting of ice (mass loss per unit time) :

$\displaystyle \frac{dm}{dt}=\frac{d}{dt}\left[ \frac{4}{3}\pi {{r}^{3}}\rho \right]=\frac{4}{3}\pi \rho \times 3{{r}^{2}}\left( -\frac{dr}{dt} \right)$ …..(c)

-ve is because the radius of the sphere decreases with time.

Using equation (c) in equation (b), we get

$\displaystyle \frac{dQ}{dt}=L\left[ \frac{4}{3}\pi \rho \times 3{{r}^{2}}\left( -\frac{dr}{dt} \right) \right]=-4\pi \rho {{r}^{2}}L\frac{dr}{dt}$ …..(d)

Now from equation (a) and equation (d),

$\displaystyle -4\pi \rho {{r}^{2}}L\frac{dr}{dt}=\sigma A({{T}^{4}}-T_{0}^{4})$

$\displaystyle \Rightarrow \frac{dr}{dt}=-\frac{\sigma A({{T}^{4}}-T_{0}^{4})}{4\pi \rho {{r}^{2}}L}$

$\displaystyle \Rightarrow \frac{dr}{dt}=-\frac{\sigma \left( 4\pi {{r}^{2}} \right)({{T}^{4}}-T_{0}^{4})}{4\pi \rho {{r}^{2}}L}$

$\displaystyle \Rightarrow \frac{dr}{dt}=-\frac{\sigma ({{T}^{4}}-T_{0}^{4})}{\rho L}$

$\displaystyle \Rightarrow \frac{dr}{dt}=-constant$

The relationship derived indicates that $r (t)$ is a linear function with a negative slope. This means that as time increases, the radius decreases linearly. The graph of $r (t)$ will be a straight line that starts at $R$ and decreases over time. Hence correct option is (2).

Stefan’s Law | Stefan’s Boltzmann Law | Stefan’s Fourth Power Law

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