Asked by Anonymous

What is the efficiency of an athlete who consumes 3000kcal of food and does 2.5 x 10*6 J of useful work?
Answered by drwls
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
Answered by Anonymous
Ty I was converting the J to kcal no wonder it wasn't working! Ty for the help.

Tezara
Answered by drwls
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.

Answered by shine
1,791,686.574?
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