Physics question

A naked person with a surface area of 1.55 meters squared and a skin temperature of 37.3 degrees is in a sauna room at T = 82.0 degrees. The person's skin has an emissivity of 0.910. She evaporates sweat from her body to balance the rate of heat absorption by radiation.
(i) Neglecting any heat production due to metabolism, what is the net rate of heat absorption by the person due to radiation? in W
(ii) How much sweat must be evaporated per hour to balance a net rate of heat absorption of H = 0.850 × 103 W? in kg/hr

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asked by Sarah
  1. Radiation power absorbed
    = Area*sigma*emissivity*(T2^4 - T1^4)
    T1 and T2 are the skin and sauna wall temperatures, respectively, in Kelvin.
    sigma is the Stefan-Boltzmann constant. (Look it up).

    ii) Divide the heat absorption rate (W or J/s) by the heat of vaporization of water, 2.26*10^6 J/kg. Then convert kg/s to kg/hr

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    posted by drwls

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