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23 March, 17:02

The outer surface of a skier's clothes of emissivity 0.7000.700 is at a temperature of 5.505.50 °C. Find the rate of radiation if the skier has a surface area of 1.601.60 m2 and the surroundings are at - 20.0-20.0 °C.

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  1. 23 March, 17:08
    0
    122.1 W.

    Explanation:

    Thermal radiation is defined as the electromagnetic radiation generated by the thermal motion of particles in matter.

    Mathematically,

    q = σ * ε * A * (Ts^4 - Ta^4)

    where,

    q = heat transfer per unit time (W)

    σ = The Stefan-Boltzmann Constant = 5.6703 x 10^-8 (W/m^2. K^4)

    Ts = absolute temperature of the surroundings in kelvins (K)

    ε = emissivity of skier's clothes = 0.70

    Ta = absolute temperature of the object in kelvins (K)

    A = area of the emitting body (m2)

    = 5.6703 x 10^-8 * 0.7 * 1.6 * (278.65^4 - 253.15^4)

    = 122.1 W.
  2. 23 March, 18:00
    0
    121.0 W

    Explanation:

    We use the equation for rate of heat transfer during radiation.

    Q/t = σεA (T₂⁴ - T₁⁴)

    Since temperature of surroundings = T₁ = - 20.0°C = 273 + (-20) = 253 K, and temperature of skier's clothes = T₂ = 5.50°C = 273 + 5.50 = 278.5 K.

    Surface area of skier, A = 1.60 m², emissivity of skier's clothes, ε = 0.70 and σ = 5.67 * 10⁻⁸ W/m²K⁴.

    Therefore, the rate of heat transfer by radiation Q/t is

    Q/t = σεA (T₂⁴ - T₁⁴) = (5.67 * 10⁻⁸ W/m²K⁴) * 0.70 * 1.60 m² * (278.5⁴ - 253⁴) = 6.3054 * (1918750544.0625) * 10⁻⁸ W = 1.2098 * 10² W = 120.98 W ≅ 121.0 W
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