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Physics 23 Online
OpenStudy (anonymous):

according to the 2nd law of thermodynamics, temperature will only go down in a vacuum, by theoretically, in a true vacuum with only one thing in it, the temperature would not be able to go down because there is nothing to transfer energy to. Is this a loophole in the 2nd law of thermodynamics or is energy able to transfer without anything?

OpenStudy (anonymous):

what does the second law of thermodynamics actually say, in your understanding ?

OpenStudy (anonymous):

In any closed system, the entropy of the system will either remain constant or increase. physics.about.com

OpenStudy (anonymous):

so why do you think the temperature should go down ?

OpenStudy (anonymous):

in a vacuum, if I understand this correctly, the temperature should only go down as stated by the 2nd law of thermodynamics,

OpenStudy (anonymous):

that's what you said to start with, but why do you think that ?

OpenStudy (anonymous):

that is what i am trying to understand, but I don't see any known reason why it would go down. unless it was a heated body that was emitting light

OpenStudy (anonymous):

if a heated body is emitting light that is lost from the system, then it is not a closed system

OpenStudy (anonymous):

this is just theoretical, because a true vacuum is imposable

OpenStudy (anonymous):

I'm not really sure what your reasoning is, but I'm confident that there are no loopholes in the second law of thermodynamics

OpenStudy (anonymous):

why would the entropy of the system increase then?

OpenStudy (anonymous):

If the isolated system is in equilibrium to start with, then it will just stay in a state of equilibrium and it's entropy will remain constant. If, however, the system is not in equilibrium to start with, it is possible that the system might move towards equilibrium through irreversible changes and this is the origin of the increase in entropy. For example, if a system had finite differences in temperature within its different parts, over time the temperature would equalise and the entropy would increase in the process.

OpenStudy (anonymous):

ok, thanks for the help, i have a better understanding now

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