Rate of heat radiated from a person

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SUMMARY

The discussion centers on calculating the percentage increase in the rate of heat radiated from a person with a skin temperature of 34.0 °C compared to 33.0 °C using Stefan's Law of emission. The correct calculation involves using the formula P = σAeT^4, where the temperatures are converted to Kelvin. The final percentage increase is determined to be approximately 1.31% after correcting the initial miscalculation that presented the result as a ratio instead of a percentage.

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  • Understanding of Stefan's Law of emission
  • Knowledge of temperature conversion to Kelvin
  • Familiarity with basic algebra for manipulating equations
  • Concept of emissivity in thermal radiation
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Homework Statement


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?


Homework 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)

The Attempt at a Solution



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

Actual answer is 1.31%
 
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Hendrick said:
(34+273)^4/(33+273)^4
= 1.013 % (3sf)
That isn't a percentage, that's a ratio. You need to multiply by 100 to obtain a percentage.
 
Hootenanny said:
That isn't a percentage, that's a ratio. You need to multiply by 100 to obtain a percentage.

Oh, thanks lol.


Is this right:-

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

the increase being:
= 101.3136113 % - 100
= 1.3136113%
 
Looks good to me :approve:
 

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