What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of t

Question

A person is standing outdoors in the shade where the temperature is 26 °C

A person is standing outdoors in the shade where the temperature is 26 °C. (a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is 180 cm2 and its emissivity is 0.85. (b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were 0.61?

Answer 1

Power/area = e*s*T^4 - total power radiated by a gray body

But assuming the person doen't heat up, he has to radiate whatever he absorbs at temperture T

Let T = 26C = 299K
s = stefan-boltzmann constant = 5.67e-8 W/m^2/K^4 = 5.67e-12 W/cm^2/K4
e = emissivity of hairy head = 0.8

The Power/area = 3.625e-4 W/cm^2

Area of head = 160 cm^2 so total power is P = 0.006W

If emissivity is 0.6 instead then Power = 0.6/0.8*0.006W = 0.0043W

Answer 2

The greenhouse result's while the radiant power passes with the aid of something sparkling to it ( like glass of the ambience), hits the floor and gets switched over into thermal power with the radiant section being of a degraded frequency so as that the glass ( and greenhouse gases, of which water vapor is overwhelmingly the main severe ) is now no longer sparkling to it. to that end the greenhouse ( and the earth ) heats up. N.B. it isn't any longer an endorsement of human-led to worldwide warming ( notice the emphasis on water vapor and not CO2 ) and the undeniable fact that the result's self-proscribing ( i.e. extra CO2 does no longer mean extra opacity to pondered radiation ).