- #1

nlingraham

- 16

- 0

## Homework Statement

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.

## Homework 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

## 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?