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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=5.67×108 Wm2K4\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(T4Tenv4)P_{\text{net}} = e\sigma A\left(T^4 - T_{\text{env}}^4\right).
  • Wien's displacement law is λmaxT=b\lambda_{\max}T = b, where b=2.90×103 mKb = 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=TC+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 λmaxT=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 λmaxT=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 TTenvT - T_{\text{env}} for net radiation is wrong because the correct net expression is Pnet=eσA(T4Tenv4)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=5.0×107 m\lambda_{\max} = 5.0 \times 10^{-7}\ \mathrm{m}. Estimate its surface temperature using λmaxT=2.90×103 mK\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.