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Luminosity and Surface Area

  1. Feb 10, 2014 #1
    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 )

    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. jcsd
  3. Feb 10, 2014 #2
    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.
     
  4. Feb 10, 2014 #3
    Attached is an example from a physics book
     

    Attached Files:

  5. Feb 18, 2014 #4
    Thank you very much for your help!
     
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