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 cm, a width of 38.5 cm, and 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 absorbance of 97.0 percent, so we don't lose our energy so quickly. How much power do absorb when we are in where the temperature is 26.0 degree C? How much energy does our body lose in one second?

Explanation / Answer

Question 1

Power emmited=emmisivity*area*sigma*T^4

T is in kelvin

emmisivity=0.97

so Area=0.385*0.255=0.098 m^2

so power=48.5 W

2

power by environment=0.97*0.098*sigma*(26+273)^4=43.07 W

so power absorbed=43.07 W

3

net power=5.42 W

net energy in 1 sec =5.42 J

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