how to calculate blackbody temperature of the earth?

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  • 1 decade ago
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    I found a website that describes how to calculate the blackbody temperature of the Earth.

    http://www.math.duke.edu/education/prep02/teams/pr...

    The equation you need to use says that the blackbody temperature is equal to the fourth root of the following: ratio of solar constant and four times the Stefan-Boltzmann constant multiplied by the fraction of solar radiation absorbed.

    I found the values you need to use by looking in one of my textbooks.

    solar constant = 1,368 Watts per squared meter

    Stefan-Boltzmann constant = 5.67*10^-8 Watts per squared meter Kelvin to the fourth

    average global albedo = 0.3

    Let's plug in the numbers.

    T = ((1,368 W/m² / (4 * 5.67*10^-8 W/m²K^4)) * (1 - 0.3))^(1/4)

    ≈ 255°K

  • 4 years ago

    Blackbody Calculator

  • Anonymous
    5 years ago

    You need to calculate the total solar energy intercepted by the Earth and the energy reradiated. Given: Surface temperature of Sun, Ts = 6000 K Radius of Sun Rs = 7x10^8 m Stefan - Boltzmann law js = σTs^4 j = energy flux density σ=5.67x10^-8 J s^1 m^2 K-^4 The total energy emitted by the Sun is Es =(σTs^4)(4πRs^2) Next, calculate the energy density Jd (J/m^2) at a distance d = 1.5x10^11 m Jd=Es/4πd^2 The amount of total energy intercepted by Earth is (Jd)(πRe^2) where Re is the radius of the Earth. In thermal equilibrium, the amount of energy intercepted is equal to the amount of energy reradiated: (Jd)(πRe^2) = (σTe^4)(4πRe^2) Notice that the energy intercepted is proportional to the Earth's cross sectional area while the emitting area is the full Earth's surface (thermal equilibrium condition) Te is the temperature we are after, so Te = [(Jd)(πRe^2)/σ(4πRe^2)]^1/4 Te = [{(σTs^4)(Rs/d)^2}/4σ]^1/4 = Ts √(Rs/d√2) =344 K The actual temperature is a bit less since neither the Earth nor the Sun are ideal black bodies with absorptivities and emissivities of 1.

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