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Black and grey body emission numerical

Short answer

λ_max = 2898/1500 = 1.93 μm; E_b = σT^4 = 5.67e-8 × 1500^4 = 287 kW/m^2; for grey body ε=0.8, Q = ε E_b A = 0.8 × 287e3 × 0.5 = 114.8 kW. Sun at 5800 K peaks at 0.50 μm.

A tungsten filament furnace surface is at 1500 K. Calculate (a) the wavelength at which the monochromatic emissive power is maximum, (b) the total emissive power if it behaves as a black body, and (c) the heat radiated per second from a 0.5 m^2 surface if it is a grey body with emissivity 0.8. Compare the peak wavelength with that of the sun, taken as a black body at 5800 K.

(a) Peak wavelength (Wien's law)

λmax=2898T=28981500=1.932 μm\lambda_{max} = \frac{2898}{T} = \frac{2898}{1500} = 1.932\ \mu\text{m}

This is in the near-infrared; only a small tail lies in the visible range.

(b) Black-body emissive power (Stefan-Boltzmann law)

Eb=σT4=5.67×10−8×(1500)4=2.870×105 W/m2E_b = \sigma T^4 = 5.67\times10^{-8}\times(1500)^4 = 2.870\times10^{5}\ \text{W/m}^2

(c) Grey body radiation

E=εσT4=0.8×2.870×105=2.296×105 W/m2Q=E A=2.296×105×0.5=1.148×105 W\begin{aligned} E &= \varepsilon\sigma T^4 = 0.8\times2.870\times10^5 = 2.296\times10^5\ \text{W/m}^2\\ Q &= E\,A = 2.296\times10^5\times0.5 = 1.148\times10^{5}\ \text{W} \end{aligned}

Comparison with the sun

λmax,sun=28985800=0.4997 μm\lambda_{max,sun} = \frac{2898}{5800} = 0.4997\ \mu\text{m}

The sun peaks in the visible (green), at about 0.5 μ\mum, about one quarter of the wavelength of the 1500 K surface.

Answer: λmax=1.93 μ\lambda_{max} = 1.93\ \mum; Eb=287 kW/m2E_b = 287\ \text{kW/m}^2; Q=114.8Q = 114.8 kW; sun peaks at 0.50 μ0.50\ \mum.

✓ Answer checked by Claude Sonnet 5.5

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