Temperature of a body exposed to sunlight

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SUMMARY

The discussion focuses on calculating the temperature of a black-painted aluminum object exposed to sunlight at an ambient temperature of 27ºC and receiving 1 kW-hr/m² of solar radiation. It is established that the equilibrium temperature can be determined using Stefan's law, E = σT^4, leading to an approximate temperature of 90ºC. Participants emphasize the importance of considering the cross-sectional area of the aluminum and the variables affecting heat absorption, including painting absorption and aluminum's heat capacity.

PREREQUISITES
  • Understanding of Stefan's law (E = σT^4)
  • Knowledge of solar radiation units (kW-hr/m²)
  • Familiarity with heat capacity concepts
  • Basic principles of thermal equilibrium
NEXT STEPS
  • Research the application of Stefan's law in thermal calculations
  • Learn about the heat capacity of aluminum and its implications in thermal dynamics
  • Explore methods for calculating energy absorption based on surface area
  • Investigate free online simulators for thermal modeling of materials
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Engineers, physicists, and anyone involved in thermal analysis or material science, particularly those interested in the effects of solar radiation on different materials.

jc.int
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hello,
If I have a black painted aluminium exposed to sunlight (being it and the atmosfere at 27ºC), and I have the total solar radiation it is exposed to (lets say 1 ° KW-hr/m²), how can I know the evolution of its temperature?
(if there is any free simulator on the internet I would be glad if someone could give it, but I haven't found any).

Thank you,
Jaimie
 
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If you only consider heat loss by radiation then I would say that the blackened Aluminium would reach an equilibrium temperature when it is radiating 1kW/m^2
Then use Stefan's law
E = σT^4 to find T (in Kelvin)
I got about 90C using your figures...seems reasonable
 
Hi Jaimie,

I don't understand this unit here below. What you will need to do is find the cross sectional area of your aluminum bucket and use that to find the energy per unit time being absorbed by the bucket.

jc.int said:
(lets say 1 ° KW-hr/m²)

The other answer given to you will be very approximate because the entire bucket will be radiating heat, while only the side facing the sun will be acquiring heat.

You have a few assumptions to make.
 
I mean that the solar energy received in one m2 in 1hour is 1 ° KW (metric units for solar energy radiatio, see http://www.raws.dri.edu/cgi-bin/rawMAIN.pl?nmXHAC , equivalent to the Langley, ºly). What I'd like to know what is the relation between the painting absorbion, aluminum heat capacity, etc

Thank you,
Jaimie
 
I was confused previously because your first unit was not a rate.

I think it would help for you to write down a list of variables that you have to consider. That way you can decide which ones are "negligible" and which ones you'd like to make assumptions for. I assure you that there are more variables than absorption and heat capacity.
 

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