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The intensity of radiation in normal direction of a diffused surface at 600°C with emissivity of 0.15 (in W/m2sr is__________ . (Round off to two decimal places)
    Correct answer is '1573'. Can you explain this answer?
    Most Upvoted Answer
    The intensity of radiation in normal direction of a diffused surface ...
    T = 600oC = 873 K
    E = πin
    ⇒ in = E/π
    = εEb
    = εσT4
    Free Test
    Community Answer
    The intensity of radiation in normal direction of a diffused surface ...
    Given:
    - Temperature (T) = 600°C = 600 + 273.15 = 873.15 K
    - Emissivity (ε) = 0.15

    To Find:
    - Intensity of radiation in the normal direction of the diffused surface

    Solution:

    Stefan-Boltzmann Law:
    The Stefan-Boltzmann law states that the total radiant heat energy emitted per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature (in Kelvin). Mathematically, it can be expressed as:
    E = σ * ε * T^4
    Where,
    - E is the radiant heat energy emitted per unit area (in W/m^2)
    - σ is the Stefan-Boltzmann constant (5.67 × 10^-8 W/m^2K^4)
    - ε is the emissivity of the surface (dimensionless)
    - T is the absolute temperature (in K)

    Calculating the Intensity:
    Substituting the given values into the Stefan-Boltzmann law equation:
    E = σ * ε * T^4
    = (5.67 × 10^-8) * 0.15 * (873.15)^4
    ≈ 1572.83 W/m^2

    Rounding off to two decimal places, the intensity of radiation in the normal direction of the diffused surface at 600°C with an emissivity of 0.15 is approximately 1573 W/m^2.

    Therefore, the correct answer is 1573.

    Note:
    The intensity of radiation is calculated assuming that the surface is a diffused surface, which means it is not a perfect black body. The emissivity value (ε) represents the ability of the surface to emit radiation compared to a perfect black body. A perfect black body has an emissivity of 1.
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    The intensity of radiation in normal direction of a diffused surface at 600°C with emissivity of 0.15 (in W/m2sr is__________ . (Round off to two decimal places)Correct answer is '1573'. Can you explain this answer?
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