Radiant Energy Absorbed by Person's Head with Hair

[grandprix]

Homework Statement

A person is standing outdoors in the shade where the temperature is 28 degrees C. 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 160 cm^2 and its emissivity is 0.85.

Homework Equations

Qrad = emiss*area*sigma*T^4

The Attempt at a Solution

.85 is the emissivity, so the absorptivity is .15

The temperature of the head originally is 98.6 degrees, or 310 K

so

Qrad/T = (.16 m^2)(.15)(5.67 X10^-8)(310^4 - 301^4)

Not sure if this is right?? I am coming with an answer of 13.9, but the book says the answer is 6.3 J/s

 

[ideasrule]

The absorbed energy per second is given by P=emiss*area*sigma*T^4. T is the temperature of the surroundings, which is 28 degrees in this case; it is not the temperature of the body doing the absorption!

You apparently thought P=absorptivity*area*sigma*(temperature of head)^4, which is not correct.

 

[Redbelly98]

Note that absorptivity = emissivity.

Also, double-check the conversion from cm^2 to m^2.