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(a) What is the efficiency of an out of condition professor who 2.10e5 J of useful work while metabolizing 500 kcal of food energy?

(b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of 25%?

## Homework Equations

OE + Wnc = OEf

η = Work_out / Work_in

Work_in = η * work_out

## The Attempt at a Solution

For (a) I converted the calories to Joules

1 kcal = 4184 J

500 kcal (4184J/1 kcal) = 2.1e6 J

η = W_out/W_in

η = (2.10e5J) / (2.10e6J)

η = 10%

(b) I assume I could input the 25% into the equation and work backwards but it's not giving me a correct anwser. I've tried the following:

0.25 = (2.10e5J) / (kcal)

kcal = (0.25)(2.10e5J)

kcal = 52500J

Convert the Joules to kcal and get something like 12kcal

But that's wrong also.