Find the emitted power per square meter and wavelength of peak intensity for a 3000 K object that emits thermal radiation.

Question

Find the emitted power per square meter and wavelength of peak intensity for a 3000 K object that emits thermal radiation.

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Neala 5 years 2021-07-17T00:02:41+00:00 1 Answers 1109 views 0

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  1. thienthanh
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    2021-07-17T00:04:17+00:00
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    Answer:

    power per square meter = 4.593 × 10^(6) W/m²

    Wavelength of peak intensity = 9.67 × 10^(-7) m

    Explanation:

    From Stefan-Boltzmann law, total emitted power per square meter is given as;

    P/A = eσT⁴

    where;

    P is power

    A is surface area

    σ = Stefan-Boltzmann constant = 5.67 × 10^(-8) W/m².k⁴

    T = temperature of the body = 3000 K

    e = emissivity of the substance (for ideal radiation, it has a value = 1)

    Thus, Plugging in the relevant values we have;

    P/A = 1 × 5.67 × 10^(-8) × (3000)^(4)

    P/A = 4.593 × 10^(6) W/m²

    Let’s find the wavelength of peak intensity.

    From wiens displacement law, we know that;

    λ_m × T = b

    where;

    λ_m = maximum wavelength

    T = Temperature

    b is Wien’s displacement constant = 2.9 × 10^(−3) m/K

    thus;

    λ_m = b/T = (2.9 × 10^(−3))/3000 = 9.67 × 10^(-7) m

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