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How does heat transfer through vacuum?

seo-qna
Last updated date: 01st May 2024
Total views: 365.1k
Views today: 6.65k
Answer
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Hint: In order to understand this condition, it is important to understand the principles of heat transfer. The principle behind the mode of heat transfer among the three modes of heat transfer viz. conduction, convection and radiation, that is responsible for this condition must be understood to explain the reason.

Complete answer:
While the subject of thermodynamics gives us the estimation of the heat transferred, the subject of heat transfer deals with the mode in which the heat transfer.
There are three principal modes of heat transfer, viz. Conduction, Convection and Radiation.
Conduction of heat occurs through transfer of energy through vibration of the particles of the medium without actual displacement of the particles of the medium. It mainly occurs in elastic media like solids.
Convection of heat occurs through the actual movement of particles of the medium from regions of high temperature to lower temperature thereby, conveying the heat energy in the process. It occurs in fluidic media like liquids and gases.
Radiation of heat occurs through emission of heat energy in the form of electromagnetic radiation directly from the surface of the body.
Vacuum is defined as the space that is completely devoid of any matter. Vacuum is created in a region of closed space, by extracting all the air in the space through vacuum pump equipment.
The processes conduction and convection take place with the help of the molecules, they are not possible in vacuum since there is no existence of any molecules to transfer the heat. Since, radiation is in the form of electromagnetic waves, it does not require any medium and hence, the waves can pass through vacuum.

Hence, the heat transfer through vacuum happens through radiation.

Note: The region of the electromagnetic spectrum that is responsible for the radiation of heat is known as infrared region. They are invisible waves with wavelengths higher than that of red, i.e., above $800nm$.
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