Radiation - solving for Q - Stefan-Boltzmann law

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SUMMARY

The discussion focuses on calculating radiant energy absorbed by a person's head using the Stefan-Boltzmann law. The correct emissivity values are 0.85 for a person with hair and 0.66 for a bald person. The surface area of the hair is 160 cm², which converts to 0.016 m², not 1.6 m² as initially stated. The final calculations yield 625.86 J/s for the person with hair and 485.96 J/s for the bald person when using the correct area conversion and temperature in Kelvin.

PREREQUISITES
  • Understanding of the Stefan-Boltzmann law
  • Knowledge of emissivity and its significance in thermal radiation
  • Ability to convert units between square centimeters and square meters
  • Familiarity with temperature conversion from Celsius to Kelvin
NEXT STEPS
  • Review the Stefan-Boltzmann law and its applications in thermal physics
  • Learn about emissivity and how it affects heat transfer
  • Practice unit conversions, particularly between cm² and m²
  • Explore temperature conversion techniques and their relevance in scientific calculations
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone interested in understanding thermal radiation and energy transfer principles.

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Homework Statement



A person is standing outdoors in the shade where the temperature is 27°C.
(a) What is the radiant energy absorbed per second by his head when it is covered with hair? The surface area of the hair (assumed to be flat) is 160 cm2 and its emissivity is 0.85.

(b) What would be the radiant energy absorbed per second by the same person if he were bald and the emissivity of his head were 0.66?

e = .85 for a, .66 for b
sigma = 5.67*10^-8
A = 1.6m squared
T = 300.15K (27 + 273.15)

Homework Equations



q = e*sigma*T^4*A*t


The Attempt at a Solution



By plugging in the known variables, I got

Q/(1 sec) = .85*(5.67x10^-8)*((300.15) ^4)*1.6m squared

Q =625.86 J/s

Q/(1 sec) = .66*(5.67x10^-8)*((300.15) ^4)*1.6m squared

Q = 485.96 J/s


I was told by webassign these answers were wrong. Am I not solving just for Q here? I can't think of any other way besides solving for Q. Also, I'm not sure on the temperature. Do I just take the 27 and convert to kelvin or do I have to do something else? thanks in advance for the help
 
Physics news on Phys.org
A meter squared is 100 cm by 100 cm = 10 000 cm².
So 160 cm² is not 1.6 m².
 
Delphi51 said:
A meter squared is 100 cm by 100 cm = 10 000 cm².
So 160 cm² is not 1.6 m².

thank you. By converting it right I got the right answer
 

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