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MattMark'90
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Homework Statement
A naked person, whose skin area is 1.7 m2, sits in a sauna that has a wall
temperature of 61oC. If the person’s skin temperature is 37oC, find the net rate at
which the person absorbs heat by radiative transfer (assume an emissivity e of 1).
How much liquid must the person consume after 30 minutes to replace the sweat
evaporated as a result of this heat absorption. (assume the latent heat of vaporisation
of sweat is 2427 kJ/kg at 37oC.
Homework Equations
Stefan-Boltzmann Law: P = e[tex]\sigma[/tex]AT4
The Attempt at a Solution
Plugging in the numbers and using the the temperature of the sauna walls and human body termperature to give the NET rate of absorption:
P = 1 x 5.67x10-8 x 1.7 x (614 - 374)
= 1.15 W
Multiplying this power by the 30 minute time period will give the heat energy, Q, absorbed by the person:
P x t = 1.15 x 30 x 60 = 2077J
We are given the latent heat of vaporisation of sweat so dividing Q by this value will give the mass of water lost:
M = 2077 / 2427x103 = 0.9 g
This mass of water seems too small to be reasonable. Although this calculation will only give an approximate value I would nevertheless expect it to be above 10g at the very least. Having checked my calculation I cannot find any obvious mistakes, my only real doubt is my calculation of the net rate of absorption but I cannot see how else I would calculate this value.
Thanks for the help.