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Stefan-Boltzmann & Wien's Law Reference cheat sheet - grade 11-12

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Physics Grade 11-12

Stefan-Boltzmann & Wien's Law Reference Cheat Sheet

A printable reference covering Stefan-Boltzmann law, Wien's displacement law, emissivity, blackbody radiation, radiant power, and peak wavelength for grades 11-12.

TopicsThermodynamicsHeatLightQuantum PhysicsPhysics
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This cheat sheet covers the two key laws used to describe thermal radiation from hot objects: the Stefan-Boltzmann law and Wien's displacement law. Students need these laws to connect temperature with total emitted power and the color or wavelength of peak emission. These ideas are essential in topics such as stars, infrared imaging, heat transfer, and blackbody radiation.

The reference is designed to help students quickly identify the correct formula, units, and physical meaning.

The Stefan-Boltzmann law says that total radiated power increases with the fourth power of absolute temperature, so small temperature changes can cause large power changes. Wien's law says that hotter objects emit their strongest radiation at shorter wavelengths. Emissivity, written as ee, adjusts blackbody formulas for real materials.

All temperatures in these formulas must be measured in kelvins, not degrees Celsius.

Key Facts

  • The Stefan-Boltzmann law for total emitted power is P=eσAT4P = e\sigma AT^4, where PP is power, ee is emissivity, AA is surface area, and TT is absolute temperature.
  • The Stefan-Boltzmann constant is σ=5.67×10−8 W m−2 K−4\sigma = 5.67 \times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}.
  • The emitted power per unit area is j=PA=eσT4j = \frac{P}{A} = e\sigma T^4.
  • The net radiated power between an object and its surroundings is Pnet=eσA(T4−Tenv4)P_{\text{net}} = e\sigma A\left(T^4 - T_{\text{env}}^4\right).
  • Wien's displacement law is λmax⁡T=b\lambda_{\max}T = b, where b=2.90×10−3 m Kb = 2.90 \times 10^{-3}\ \mathrm{m\,K}.
  • The peak wavelength from Wien's law is λmax⁡=bT\lambda_{\max} = \frac{b}{T}, so increasing TT shifts the peak toward shorter wavelengths.
  • A perfect blackbody has emissivity e=1e = 1, while real surfaces have 0<e<10 < e < 1.
  • Temperature must be converted to kelvins using TK=T∘C+273.15T_{K} = T_{^{\circ}\mathrm{C}} + 273.15 before using radiation laws.

Vocabulary

Blackbody
A blackbody is an ideal object that absorbs all incoming radiation and emits the maximum possible thermal radiation at each temperature.
Emissivity
Emissivity, written as ee, is a number from 00 to 11 that compares a real surface's radiation to that of a perfect blackbody.
Stefan-Boltzmann Law
The Stefan-Boltzmann law states that radiated power is proportional to surface area and the fourth power of absolute temperature, P=eσAT4P = e\sigma AT^4.
Wien's Displacement Law
Wien's displacement law states that the peak wavelength of thermal radiation satisfies λmax⁡T=b\lambda_{\max}T = b.
Peak Wavelength
The peak wavelength λmax⁡\lambda_{\max} is the wavelength at which a hot object emits the greatest intensity of radiation.
Absolute Temperature
Absolute temperature is temperature measured in kelvins, where 0 K0\ \mathrm{K} represents absolute zero.

Common Mistakes to Avoid

  • Using Celsius in radiation formulas is wrong because P=eσAT4P = e\sigma AT^4 and λmax⁡T=b\lambda_{\max}T = b require absolute temperature in kelvins.
  • Forgetting the fourth power in the Stefan-Boltzmann law is wrong because doubling TT makes emitted power increase by a factor of 24=162^4 = 16, not by a factor of 22.
  • Treating emissivity as greater than 11 is wrong because ordinary real surfaces cannot emit more thermal radiation than a perfect blackbody at the same temperature.
  • Confusing total power with power per unit area is wrong because P=eσAT4P = e\sigma AT^4 includes area, while j=eσT4j = e\sigma T^4 does not.
  • Using T−TenvT - T_{\text{env}} for net radiation is wrong because the correct net expression is Pnet=eσA(T4−Tenv4)P_{\text{net}} = e\sigma A\left(T^4 - T_{\text{env}}^4\right).

Practice Questions

  1. 1 A metal plate has A=0.50 m2A = 0.50\ \mathrm{m^2}, e=0.80e = 0.80, and T=600 KT = 600\ \mathrm{K}. Find the total radiated power using P=eσAT4P = e\sigma AT^4.
  2. 2 A star has a peak wavelength of λmax⁡=5.0×10−7 m\lambda_{\max} = 5.0 \times 10^{-7}\ \mathrm{m}. Estimate its surface temperature using λmax⁡T=2.90×10−3 m K\lambda_{\max}T = 2.90 \times 10^{-3}\ \mathrm{m\,K}.
  3. 3 An object at 300 K300\ \mathrm{K} is heated to 600 K600\ \mathrm{K} with the same area and emissivity. By what factor does its emitted power change?
  4. 4 Explain why a hotter star appears bluer than a cooler red star using Wien's displacement law.

Understanding Stefan-Boltzmann & Wien's Law Reference

Thermal radiation is electromagnetic energy produced by the random motion of particles inside matter. At higher temperatures, atoms and electrons move more vigorously. Their changing electric fields release radiation across a wide range of wavelengths.

A hot object does not emit one single wavelength. It produces a continuous spectrum with a broad hump. The peak of that hump marks the wavelength carrying the greatest intensity, not the only wavelength present.

This distinction matters when interpreting a graph of intensity against wavelength. The area under the whole curve represents the total energy emitted, while the location of the peak describes where the spectrum is strongest.

A blackbody is an ideal model, not usually a real object. It absorbs all incoming radiation and emits the maximum possible thermal radiation at a given temperature. A small hole in the wall of a heated cavity behaves nearly like a blackbody.

Radiation entering the hole is repeatedly absorbed and reflected inside, so very little escapes. The radiation that comes out depends mainly on the cavity temperature. Real surfaces differ because their emissivity depends on material, surface finish, and sometimes wavelength.

Matte black paint often has high emissivity. Shiny polished metal often has low emissivity because it reflects much of the incoming infrared radiation.

Radiation becomes especially important when objects are separated by empty space, where conduction and convection cannot transfer heat. The Sun warms Earth mainly through radiation. Satellites need reflective coatings and radiators to control their temperature in space.

Infrared cameras detect radiation emitted by people, animals, engines, and buildings. A warm person can be detected in darkness because human skin emits strongly in infrared wavelengths.

The body does not glow visibly at normal temperature. Visible red or orange glow appears only at much higher temperatures, such as in a heating element or molten metal.

For heat transfer problems, focus on the difference between emitted power and net power. Every object above absolute zero emits radiation. Surroundings emit radiation toward it too.

A cooler object can still emit energy, but it gains more energy than it loses when placed near a hotter environment. Net radiation tells the direction of energy transfer. This prevents the common mistake of treating a cool object as if it emits nothing.

Students should convert every temperature before doing any fourth power or wavelength calculation. They should keep area in square metres, power in watts, and wavelength units consistent with the displacement constant. It is useful to estimate first.

A higher temperature should give much greater total emission and a shorter peak wavelength. If a calculation gives the opposite trend, the temperature conversion or unit conversion is probably wrong.