if we measure the surface temperature of PVC tube exposed to the sun at 60 ° C for example, we find that the tube surface temperature is higher than that at the exposure temperature.
Simply because the absorption of the PVC pipe is higher than that of air.
I think the dielectric properties of the pipe is higher than the air. Thus, heating rate in the pipe is faster.
@Khaoula Hidouri: Stefan-Boltzmann gives you the temperature dependent wavelength of a black body radiation. This has not really anything to do with the question.
@Roland Macana: Dielectric properties have something to do with heat transport within an object or between two objects not with temperature rise of the pipe.
It is really the higher absorption rate of electromagnetic radiation of the pipe, especially for infrared radiation.
In my opinion there are different reasons, being the increase of the temperature due to three factors: heat absorbed, heat dispersed, heat capacity., The most important is, as Joannes said, is
- the rate of absorption of a photon is proportional to Xsection*Density. The density of a solid is much higher than a gas, therefore the radiative heat flux absorbed by the surface of a solid is much higher than that of air.
There is also a dependence on the material related to the cross section and to the reflectivity capabilities.
As a secondary effect, there is much dispersion in air, due to regions not exposed to the sun, which means that the volume of the air to heat is higher than that exposed, therefore air has an higher heat capacity. If the air is closed in a bottle, the temperature inside will be higher than outside.
Hi all,
Mr. Amar did find an exact solution for your question?
What temperature you considered for the PVC pipe exposed to radiation?
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