Luminosity and Surface Area

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

A typical adult burns about 2500 Calories in one day.
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 )

120.06W = (5.67×10^-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?

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