Heat lost from face due to black body radiation

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SUMMARY

The discussion centers on calculating the net rate of heat loss from a skier's head due to black body radiation, modeled as a cylinder with a diameter and height of 20 cm. Using the formula Q/Δt = eσA(T^4 - T0^4), where emissivity (e) is 0.97, the correct surface area (A) is determined to be 0.754 m². The final calculation yields a heat loss of 125.5 W, correcting an initial miscalculation that resulted in 31.29 W. The participant clarified the surface area formula, confirming the absence of a bottom in the cylinder.

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  • Understanding of black body radiation principles
  • Familiarity with the Stefan-Boltzmann law
  • Basic geometry for calculating surface area of cylinders
  • Knowledge of thermal emissivity and its significance
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Physics students, engineers, and sports scientists interested in thermal dynamics and heat loss calculations in outdoor activities.

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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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