Calculate Body Heat Loss & Time to Compensate with Help from Physics Experts

  • Thread starter mlee
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In summary, the skier would need to sit outside for 3 hours to compensate for the candy bar. If her clothes are wet, she would need to sit outside for 6 hours.
  • #1
mlee
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Dear people,

Who can help me with this physics problem?

A skier is sitting on top of the mountain on a windless day. The sky is cloudy and the principal loss of her body heat is due to conduction. Her body temperature is 33 degrees of celsius and the air temperature at the surface of her clothes is one degrees of celsius. Her weight is 60kg, her body surface area is 1.8m^2 and the thickness of her ski clothes is 1.2 cm. The thermal conductivity of her ski clothing is 0.040 W/(mK)

Question a:
Calculate the rate (in watts) at which body heat flows out through her clothing?

Question b:
How would this rate change if, after a fall, her clothes became totally wet so that the thermal conductivity of her ski clothing became comparable to that of water, that is, 0.5 W/(mK)

Question c:
To compensate for her loss of heat, the skier eats a candy bar which has 200 calories. How many minutes must she sit outside to compensate for the candy bar if her clothes are dry? How many minutes if her clothes are wet?

thanxxx!
 
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  • #2
About the two first a) and b) you have all the parameters needed for the Fourier's Law. You only have to substitute them inside the formula.

When you have done a) and b) we'll talk again. :smile:
 
  • #3
a) Q= kA(T_H-T_L)/h, where k=0.040W/(mK),A= 1.8m^2, h=1.2x10^-2m, T_H= 33, T_L=1
Q= 1830 Watts

b) Q= 22875 Watts

IS this right>??
 
  • #4
It seems to be right. But check your answer if you have a solution. I'm not going to switch on the calculator. I'm a bit lazy.¡

For question c), the simplest solution I see is the conservation of the flux rate:

[tex]\frac{dQ}{dt}=1830W=\frac{200cal}{t}[/tex]

work out t and tell me what happens. Be careful with the units and converse it to joules.
 
  • #5
thank you very much
 

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