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

Question

The emissivity of the human skin is 97.0 percent. Use 35.0 degree C for the skin temperature and approximate the human body by a rectangular block with a height of 1.70 m, a width 38.5 cm a length of 25.5 cm. Calculate the power emitted by the human body. Fortunately our environment radiates too. The human body absorbs this radiation with an absorption of 97.0 percent, so we don't lose our internal energy to quickly. How much power do we absorb when we are in a room where the temperature is 26.0 degree C? How much energy does our body lose in one second?

Explanation / Answer

Total surface area of the human body =2(lb+bh+hl) =2*1.7*0.305 + 2*1.7*0.255 + 2*.255*0.305 = 2.0595 m^2. Using the Stefan-Boltzmann equation
T = 273+35 = 308 K
E = T^4 = 0.97*5.67 x 10^-8*308^4 = 495 W/m^2
Power of men body = E*A

P = 1019.477 W

B) E = T^4 = 0.97*5.67 x 10^-8*299^4 = 439.58 W/m^2

The power absorb P =E * A = 905.31 W

C) The energy our body loses in one second = (1019.477 905.31) = 114.16 J

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