Physics Lab
JEE/Unit 7: Thermal Physics & Thermodynamics/Radiation: Stefan and Wien Laws

Radiation — Stefan and Wien Laws

Every body at temperature TT emits electromagnetic radiation. The Stefan-Boltzmann law gives the total power; Wien's law gives the wavelength of peak emission.

Concept

Stefan-Boltzmann law (black body): P=σAT4P = \sigma A T^4 where σ=5.67×108\sigma = 5.67\times 10^{-8} W/m²·K⁴.

For a body with emissivity ε\varepsilon (gray body): P=εσAT4P = \varepsilon \sigma A T^4.

Net radiation when a body at TT is in surroundings at T0T_0: Pnet=εσA(T4T04)P_{net} = \varepsilon\sigma A(T^4 - T_0^4)

Wien's displacement law: λmaxT=b\lambda_{max}\,T = b, b=2.898×103b = 2.898\times 10^{-3} m·K.

Newton's law of cooling (small ΔT\Delta T): dT/dt=k(TT0)dT/dt = -k(T-T_0); cooling is exponential to ambient.

Derivation

For small temperature excess θ=TT0\theta = T - T_0 with θT0\theta \ll T_0: T4T044T03θT^4 - T_0^4 \approx 4T_0^3 \theta

So Pnet4εσAT03θP_{net} \approx 4\varepsilon\sigma A T_0^3 \theta. Energy balance: mcdθ/dt=Pnetmc\,d\theta/dt = -P_{net}, giving dθdt=4εσAT03mcθ=kθ\frac{d\theta}{dt} = -\frac{4\varepsilon\sigma A T_0^3}{mc}\theta = -k\theta

— this is Newton's law of cooling with k=4εσAT03/(mc)k = 4\varepsilon\sigma A T_0^3/(mc). Solution: θ(t)=θ0ekt\theta(t) = \theta_0 e^{-kt}.

JEE Worked Example

Problem: The sun (T=5800T = 5800 K, R=7×108R = 7\times 10^8 m) radiates as a black body. Find peak wavelength and total power.

Solution: λmax=2.898×103/5800500\lambda_{max} = 2.898\times 10^{-3}/5800 \approx 500 nm (green).

Power: P=σ4πR2T4=5.67×1084π(7×108)2(5800)43.9×1026P = \sigma\cdot 4\pi R^2 \cdot T^4 = 5.67\times 10^{-8}\cdot 4\pi(7\times 10^8)^2\cdot(5800)^4 \approx 3.9\times 10^{26} W.

Traps

  • TT must be in kelvin (never Celsius) in Stefan's law.
  • Wien's bb refers to peak of spectral radiance vs. wavelength, not vs. frequency.
  • Newton's law of cooling is valid only for small temperature differences.
  • Emissivity ε\varepsilon equals absorptivity (Kirchhoff's law) at thermal equilibrium.
  • The T4T^4 scaling means doubling temperature gives 16×16\times power.

Key Takeaways

  • P=εσAT4P = \varepsilon\sigma A T^4; net T4T04\propto T^4 - T_0^4.
  • λmaxT=b\lambda_{max} T = b (Wien).
  • Newton's cooling: exponential decay of excess temperature.
  • Hotter bodies emit more and peak at shorter wavelengths.

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