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?
To calculate the efficiency of an athlete, we can use the formula:
Efficiency = (Useful work output / Energy input) x 100
In this case, the useful work output is given as 2.5 x 10^6 J.
To find the energy input (food consumed), we need to convert 3000 kcal to joules since kcal is a unit of energy.
To convert from kcal to joules, we know that 1 kilocalorie (kcal) is equal to 4184 joules (J).
So, 3000 kcal = 3000 * 4184 J = 12,552,000 J.
Now, we can substitute the values into the efficiency formula:
Efficiency = (2.5 x 10^6 J / 12,552,000 J) x 100
Efficiency = (0.1992) x 100
Efficiency ≈ 19.92%
Therefore, the efficiency of the athlete is approximately 19.92%.