# Black Body Net Heat Absorption Problem

1. May 10, 2010

### MattMark'90

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

Stefan-Boltzmann Law: P = e$$\sigma$$AT4

3. 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.

2. May 10, 2010

### Redbelly98

Staff Emeritus
Welcome to PF.
You need to use absolute temperatures here, not the °C values. The rest of your method looks pretty good, though note that water's heat of vaporization is around 2300 J/g, not 2300x103J/g.

3. May 11, 2010

### MattMark'90

Thanks for the reply. I made the absolute temperature correction and got 229g/229mL which seems much more reasonable.

Thanks again!

4. May 11, 2010

### Redbelly98

Staff Emeritus
You're welcome! Yup, 8 ounces, or a small glass of water, definitely reasonable.