Much of the of heat lost from a cross-country skier's body is radiated from the head, since it is often uncovered. Treat a head as a 20-cm-diameter, 20-cm-tall cylinder with a flat top. If the surface temperature of a body is 35 C, what is the net rate of heat loss on a frosty 5 C day? All skin, regardless of color in the visible wavelengths of light, is effectively black in the infrared where the radiation occurs. Therefore, assume an emissivity of 0.97.
Q/Δt = eσA(T4-T04)
A=Surface area of cylinder, SA=2∏r2+2∏rh, = .188 m2
The Attempt at a Solution
Seems like straight plug n' chug.
Q/Δt = (.97)(5.67*10-8)(.188)(3084-2784)
However, this isn't right. Can anyone tell me what I did wrong?