Problem involving the human body and rate of heat transfer

In summary, the body loses energy by radiation to the night sky at a rate of 543W when lying on the ground at night in a dry climate with thin clothing providing little insulation. The equation used is Q/Δt=eσAT4, where e is .97, σ is 5.67x10-8W/m2K4, and A is 1.8m2. The temperature difference used is between the body temperature of 30°C and the sky temperature of -40°C.
  • #1
chinnie15
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Homework Statement


If you lie on the ground at night with no cover, you get cold rather quickly. Much of this is due to energy loss by radiation. At night in a dry climate, the temperature of the sky can drop to -40°C. If you are lying on the ground with thin clothing that provides little insulation, the surface temperature of your skin and clothes will be about 30°C.

Estimate the net rate at which your body loses energy by radiation to the night sky under these conditions.
Hint: What area should you use?

Homework Equations


Q/Δt=(kA/L)ΔT

The Attempt at a Solution



So far, I have looked up the k of skin is .50W/mK. I also have the change of temperature, converted to Kelvin, as being 70K. What I have no idea of, though, is what to use for Length and Area? I am not getting the 'hint', lol. It's bugging me because I know how to solve it, I just need to figure out what to use for these values?

Any help would be amazing!
Thank you!
Brittney
 
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  • #2
You are trying to use the equation for conductive heat transfer to solve a problem which involves radiative heat transfer. Look up the equation for radiative heat transfer.
 
  • #3
Oh geez, can't believe I did that. Thanks! So, the equation I need is: Q/Δt=eσAT4

Area, I found out is 1.8m2, e is .97, and σ is 5.67x10-8W/m2K4

My question is... what is T? Is it the temperature of the surroundings? Of the skin? It doesn't say delta T, so I am assuming it's one or the other?

Thank you!
Brittney
 
  • #4
chinnie15 said:
Oh geez, can't believe I did that. Thanks! So, the equation I need is: Q/Δt=eσAT4

Area, I found out is 1.8m2, e is .97, and σ is 5.67x10-8W/m2K4

My question is... what is T? Is it the temperature of the surroundings? Of the skin? It doesn't say delta T, so I am assuming it's one or the other?

Thank you!
Brittney

There are two terms. The heat leaving is at the body temperature of 30C = 303K. The heat arriving from space is at the sky temperature of - 40C = 233K. You need to calculate the net of these two heat flow rates.
 
  • #5
Ok, for 'T' I ended up calculating 5.48x109K. After plugging everything in, I am getting the answer of 543W, but it's telling me it's wrong? What could I still be doing wrong?
 
  • #6
If you're lying flat on the ground, how much of your surface area is facing the sky?
 
  • #7
Thank you! I got it now!
 
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