Does heat transfer cause molecules to cool down?

AI Thread Summary
Heat transfer occurs through conduction when molecules collide, transferring energy from one end of an object to another. When molecules at one end transfer energy to nearby molecules at the other end, they lose energy and thus cool down, assuming no external heat is supplied. If heat is continuously supplied, both ends will remain hot, but the end in contact with the heat source will be at a higher temperature than the other end. The temperature of the heat source and the first end will be nearly the same. Overall, energy conservation principles confirm that the initial end cools down after energy transfer.
gracy
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One quick question about heat transfer.
during conduction heat transfer occur when molecules collide with each other .heat from one end of the object is transferred to other end .when molecules from end 1 collide with near by molecules of end 2 they transfer their energy to those nearby molecules.My question is after transferring energy do the molecules of end 1 get colder (provided that no constant supply of heat.)
 
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gracy said:
(provided that no constant supply of heat.)
If you're adding no energy to the system, and the energy of the system is conserved, the obvious conclusion is ... ?
 
Energy, even as heat, is conserved.
 
Yes.
 
Bystander said:
If you're adding no energy to the system, and the energy of the system is conserved, the obvious conclusion is ... ?
the molecules of end 1 get colder,right?
 
Yes.
 
Bystander said:
Yes.
And if heat is supplied continuously both ends would be hot but at any given instant end 1 which is in direct contact with heat source would be at higher temperature than end 2,right?
 
So long as the source is at a higher T than end 2.
 
Bystander said:
So long as the source is at a higher T than end 2.
you mean source and end 1 would be at same temperature?
 
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Or very close.
 
  • #11
Bystander said:
Or very close.
Thanks a lot for answering all my question.Can you please have a look at my thread moment of inertia ,this would be a great help.
 
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