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What is the efficiency of an athlete who consumes 3000kcal of food and does 2.5 x 10*6 J of useful work?

  • Physics -

    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

  • Physics -

    Ty I was converting the J to kcal no wonder it wasn't working! Ty for the help.


  • Physics -

    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.

  • Physics -


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