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What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22.0ºC . The person has a normal skin temperature of 33.0ºC and a surface area of 1.50 m2. The emissivity of skin is 0.97 in the infrared, where the radiation takes place.

Respuesta :

Answer:

[tex] 5.45\times 10^{-4}[/tex] W

Explanation:

[tex]T_{r}[/tex] = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

[tex]T_{s}[/tex] = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

[tex]A[/tex] = Surface area = 1.50 m²

[tex]\epsilon[/tex] = emissivity = 0.97

[tex]\sigma[/tex] = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

[tex]R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})[/tex]

[tex]R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})[/tex]

[tex]R = 5.45\times 10^{-4}[/tex] W

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