Heat lost from face due to black body radiation

AI Thread Summary
The discussion revolves around calculating the net rate of heat loss from a skier's head due to black body radiation, using the formula Q/Δt = eσA(T^4 - T0^4). The initial calculation mistakenly used the wrong surface area for the cylindrical head, leading to an incorrect heat loss value. After clarifying the surface area formula, which should include only the curved surface and top, the correct surface area was determined to be 0.754 m². This adjustment resulted in a recalculated heat loss of 125.5 W, resolving the initial confusion. The importance of accurately determining the surface area in thermal calculations is emphasized.
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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(T4-T04)

where:
e=.97
σ=5.76*10-8 W/m2k4
A=Surface area of cylinder, SA=2∏r2+2∏rh, = .188 m2
T=308 K
T0=278K

The Attempt at a Solution



Seems like straight plug n' chug.

Q/Δt = (.97)(5.67*10-8)(.188)(3084-2784)
=31.29 W

However, this isn't right. Can anyone tell me what I did wrong?
 
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Why 2\pi r^2?
 
So would the SA just be SA=2∏r+2∏rh since the cylinder doesn't have a bottom? Which would make the SA = .754 m2, Which would give an answer of 125.5 W?
 
So I think sometimes I may be retarded. So it should have been ∏r2+2∏rh. I got the correct answer now. Thank you
 
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