- #1
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%?
OE + Wnc = OEf
η = Work_out / Work_in
Work_in = η * work_out
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.
(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.