Posted by Anonymous on Saturday, January 29, 2011 at 11:19pm.
Divide his "useful work" by the energy value of the food eaten, but converted to the same units as the work (Joules)
3000 kcal of food = 1.254*10^7 J
I get about 20% efficiency
Ty I was converting the J to kcal no wonder it wasn't working! Ty for the help.
Tezara
You could convert the J to kCal also, as long as numerator and denominator units are the same. Either both kcal or both Joules, for example.
The efficiency ratio you get should be the same.
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