The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin tempe

Question

The emissivity of the human skin is 97.0 percent. Use 35.0 °C for the skin temperature and approximate the human body by a rectangular block with a height of 1.65 m, a width of 43.5 cm and a length of 22.0 cm.

A.) Calculate the power emitted by the human body.

B.)What is the wavelength of the peak in the spectral distribution for this temperature?

C.)Fortunately our environment radiates too. The human body absorbs this radiation with an absorbance of 97.0 percent, so we don't lose our internal energy so quickly. How much power do we absorb when we are in a room where the temperature is 22.0 °C?

D.) How much energy does our body lose in one second?

Explanation / Answer

emissivity is e = 0.97

T = 35+273 = 308 K

area through which the power is emitting A = (1.65*0.435)+(1.65*0.22) +(0.22*0.435) = 1.17645 m^2

power emiitted is P = e*sigma*A*T^4

P = 0.97*5.67*10^-8*1.17645*(308)^4 = 582.27 W

B) lamda = 2.9*10^-3 /T = 2.9*10^-3 / 308 = 9.41*10^-6 m

C) Pnet = e*sigma*A*(T^4-Ts^4)

Ts = 22+273 = 295 K

Then Pnet = 0.97*5.67*10^-8*1.17645*(308^4-295^4) = 92.25 W

D) energy lost is E = 92.25 J

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