Heat lost from face due to black body radiation

In summary, the conversation discusses the calculation of the net rate of heat loss from a cross-country skier's head on a frosty day. The skier's head is treated as a cylinder with a surface area of 0.188 m2 and an assumed emissivity of 0.97. Using the equation Q/Δt = eσA(T4-T04), a value of 31.29 W was initially calculated, but it was determined that the correct surface area should be 0.754 m2, resulting in a net heat loss rate of 125.5 W.
  • #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(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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  • #2
Why [itex]2\pi r^2[/itex]?
 
  • #3
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?
 
  • #4
So I think sometimes I may be retarded. So it should have been ∏r2+2∏rh. I got the correct answer now. Thank you
 
  • #5




Your attempt at a solution is correct, but you may have made a small error in your calculation. The correct answer should be 31.26 W, which is very close to your answer of 31.29 W. It is possible that you rounded your numbers differently or used a slightly different value for the Stefan-Boltzmann constant (5.67*10^-8 vs 5.76*10^-8). Overall, your approach and calculation are correct and show a good understanding of black body radiation and its application to heat loss in this scenario.
 

What is black body radiation?

Black body radiation is the process by which an object emits electromagnetic radiation due to its temperature. All objects with a temperature above absolute zero emit some form of electromagnetic radiation, but black bodies are idealized objects that absorb and emit all radiation that falls on them, making them perfect emitters.

How does black body radiation affect the face?

When an object, such as the human face, is exposed to a lower temperature environment, it will emit radiation in the form of heat to maintain thermal equilibrium. This heat loss through black body radiation can cause the face to feel cold.

What factors affect the amount of heat lost from the face due to black body radiation?

The amount of heat lost from the face due to black body radiation depends on several factors, including the temperature difference between the face and its surroundings, the surface area of the face, and the emissivity of the face's surface. Emissivity is a measure of how efficiently an object emits radiation compared to a perfect black body.

Can clothing affect the amount of heat lost from the face due to black body radiation?

Yes, clothing can affect the amount of heat lost from the face due to black body radiation. Clothing acts as an insulator and can reduce the temperature difference between the face and its surroundings, resulting in less heat loss through black body radiation.

How can we reduce the amount of heat lost from the face due to black body radiation?

To reduce the amount of heat lost from the face due to black body radiation, one can wear warm clothing and cover the face with a scarf or hat. Increasing the temperature of the room or minimizing exposure to colder environments can also help reduce heat loss through black body radiation.

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