(Side note: I've long thought you can derive Carnot efficiency based on information-theoretic arguments about the information contained in knowing only the temperature difference between a heat source and a sink.)
>Does it imply the computer would not give off heat? Or that a significant amount of the heat generated by a computer is due to information loss?
Basically, yes. You can't avoid the heat emission from setting a 0 to a 1 on that computer's storage medium, but you would avoid the loss from any of the intermediate computations that typically make CPUs so hot.
But I just meant that while it's easy to make a reversible circuit, no tech we have today can make it as energy efficient as might be theoretically possible.
It does, and there is a very strong (but not universally accepted) case that they are the same thing: https://en.wikipedia.org/wiki/Entropy_in_thermodynamics_and_...
Great lay article about it: http://lesswrong.com/lw/o5/the_second_law_of_thermodynamics_...
(Side note: I've long thought you can derive Carnot efficiency based on information-theoretic arguments about the information contained in knowing only the temperature difference between a heat source and a sink.)
>Does it imply the computer would not give off heat? Or that a significant amount of the heat generated by a computer is due to information loss?
Basically, yes. You can't avoid the heat emission from setting a 0 to a 1 on that computer's storage medium, but you would avoid the loss from any of the intermediate computations that typically make CPUs so hot.