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.
nlingraham
Messages
16
Reaction score
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?
 
Physics news on Phys.org
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
 
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
View full post »

Similar threads

Black Body Radiation at a Distance from Emitter
Replies
6
Views
2K
Thermodynamics Heat Radiation Problem
Replies
1
Views
2K
Modeling Heat Transfer from a Flame to a Copper Tube using ANSYS
Replies
3
Views
3K

Hot Threads

Recent Insights

Back
Top