Physics
posted by Anonymous on .
What is the efficiency of an athlete who consumes 3000kcal of food and does 2.5 x 10*6 J of useful work?

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. 
1,791,686.574?