Problem involving the human body and rate of heat transfer

AI Thread Summary
The discussion revolves around estimating the net rate of energy loss by radiation for a person lying on the ground at night. The correct equation for radiative heat transfer, Q/Δt = eσAT^4, is emphasized, with key variables identified as area (1.8 m²), emissivity (0.97), and the Stefan-Boltzmann constant (5.67 x 10^-8 W/m²K⁴). The temperatures used are the body temperature (30°C or 303K) and the sky temperature (-40°C or 233K). The user initially miscalculated the temperature input, leading to an incorrect energy loss estimate of 543W, which was later resolved by correctly identifying the effective surface area exposed to the sky. Accurate calculations are crucial for determining the net energy loss in this scenario.
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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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.
 
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
 
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.
 
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?
 
If you're lying flat on the ground, how much of your surface area is facing the sky?
 
Thank you! I got it now!
 
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