Black Body Net Heat Absorption Problem

[MattMark'90]

Homework Statement

A naked person, whose skin area is 1.7 m², sits in a sauna that has a wall
temperature of 61^oC. If the person’s skin temperature is 37^oC, 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 37^oC.

Homework Equations

Stefan-Boltzmann Law: P = e_$$\sigma$$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.

 

[Redbelly98]

Welcome to PF.

Homework Equations

Stefan-Boltzmann Law: P = e_$$\sigma$$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

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.

 

[MattMark'90]

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

Thanks again!

 

[Redbelly98]

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

 

[rdl]

It is important to note that the calculations provided are based on several assumptions, such as the person's skin being a perfect black body and the sauna walls being at a constant temperature. In reality, the human body and sauna walls are not perfect black bodies and their temperatures can vary. Additionally, factors such as air temperature and humidity can also affect the rate of heat absorption and sweat evaporation.

To improve the accuracy of the calculation, it may be helpful to take into account the specific heat capacity of human skin and the surrounding air, as well as the rate of heat transfer through convection. It may also be beneficial to conduct experiments to measure the actual rate of heat absorption and sweat evaporation in a sauna environment.

Furthermore, it is important to note that the amount of liquid a person needs to consume to replace sweat lost during a sauna session can vary greatly depending on individual factors such as body size, level of activity, and hydration status. It is always recommended to drink plenty of water before, during, and after a sauna session to prevent dehydration.