Quote:
Originally Posted by Old Tele man
H2+½O2→H2O(liquid) <> ΔHf° = -285.8kJ/mol
H2+½O2→H2O(gas) <> ΔHf° = -241.8kJ/mol
...multiplied by 50% (at best) times efficiency of alternator (35%) means something like 1382kJ/mol are needed to dissociate water at the engine, that's less the 18% efficiency:
% = 100%*(P.out/P.in)
% = 100*(241.8kJ/mol / 1382kJ/mol) = 100%*(0.175) < 18%
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Your facts look correct (tho I've not checked- I assume you did), and can give a person some insight into one of the many, many processes that occur in the internal combustion engine. I this case, though, they tend to be of limited utility in open systems such as the one as we have been discussing.
But I personally am proud to even be part of such discussions that inspire people to think, learn, teach, and participate with others. Good job!