Luminosity and Surface Area

1. Feb 10, 2014

Sastronaut

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

1)How much energy does the average human emit every second
2)What is another term to use for expressing a “Joule per second”?
3)From your answers to (1) and (2), you just determined the “luminosity” of the average human. If the average body temperature is 98.6℉. What is the area over which a human body emits its energy?
4)Look up the surface area of human skin. How does your result from (3) compare to this value? Why is there a difference?

2. Relevant equations

3. The attempt at a solution
1. 2500 Calories/day | 4184 Joules/1Calorie | 1day/24hrs | 1hr/60min | 1min/60sec | = 121.06 J/s

2. Watt

3. L=(5.67×〖10〗^(-8) W/(m^2 K^4 ))(A)(T^4 )

121.06W = (5.67x10^-8 W/m^2K^4)(A)(310K)^4 

A= .2312 m^2

4. average adult skin surface area = 1.8 m^2.

****If I did my math correctly....could I please have some opinions on what would create the difference in my answers in part 3 and 4? Thanks PF! I though perhaps it relates to the fact that different parts of the body radiate at different rates?

Last edited: Feb 10, 2014
2. Feb 10, 2014

barryj

For one thing, remember that when a body radiates energy, it also accepts energy from its environment.
Also, the body may experience cooling from evaporation of water on the skin.

3. Feb 10, 2014

barryj

Attached is an example from a physics book

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4. Feb 18, 2014

Sastronaut

Thank you very much for your help!