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Rate of heat radiated from a person

  1. Apr 19, 2007 #1
    1. The problem statement, all variables and given/known data
    What is the percentage increase in the rate of heat radiated from a person with a surface skin temperature of 34.0 °C compared with the same person with a skin temperature of 33 °C?


    2. Relevant equations
    Stefan's Law of emission:
    P = σAeT^4

    - P = rate of energy transfer (Watts)
    - σ = 5.6696 x 10^–8 W m^–2 K^–4
    - A = surface area of the object
    - e = emissivity (varies from 0 to 1)
    - T = temperature (Kelvins)

    3. The attempt at a solution

    (34+273)^4/(33+273)^4
    = 1.013 % (3sf)

    Actual answer is 1.31%
     
  2. jcsd
  3. Apr 19, 2007 #2

    Hootenanny

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    That isn't a percentage, thats a ratio. You need to multiply by 100 to obtain a percentage.
     
  4. Apr 19, 2007 #3
    Oh, thanks lol.


    Is this right:-

    (34+273)^4/(33+273)^4 * 100
    = 101.3136113 %

    the increase being:
    = 101.3136113 % - 100
    = 1.3136113%
     
  5. Apr 19, 2007 #4

    Hootenanny

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    Looks good to me :approve:
     
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