(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Q/Δt = eσA(T^{4}-T_{0}^{4})

where:

e=.97

σ=5.76*10^{-8}W/m^{2}k^{4}

A=Surface area of cylinder, SA=2∏r^{2}+2∏rh, = .188 m^{2}

T=308 K

T_{0}=278K

3. The attempt at a solution

Seems like straight plug n' chug.

Q/Δt = (.97)(5.67*10^{-8})(.188)(308^{4}-278^{4})

=31.29 W

However, this isn't right. Can anyone tell me what I did wrong?

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# Homework Help: Heat lost from face due to black body radiation

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