How Many Food Calories with Efficiency %

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The discussion focuses on calculating the efficiency of energy use in physical work, specifically comparing an out-of-condition professor and a well-conditioned athlete. The professor's efficiency is determined to be 10% after converting 500 kcal to joules and calculating the useful work done. For the athlete, with an assumed efficiency of 25%, the correct calculation shows that they would need to metabolize approximately 201 kcal to perform the same amount of work. The initial attempts at solving the problem included errors in the equations used, which were later corrected. Ultimately, the final answer for the athlete's caloric requirement is established as 201 kcal.
Edemardil
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I figured out (a) but I am having trouble with (b):
(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.
 
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I realize that I wrote my equation wrong at the end I said that it should be η*work_out but it is work_out/η . Here is how I finalized the problem:

η = work_out / work_in

0.25 = (2.10e5) / (kcal)

kcal = (2.10e5) / (0.25)

kcal = 840e3 J

J to kcal 1kcal = 4184 J

840e3 / 4184 = 201 kcal
 
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