Question
(a) What is the efficiency of an out-of-condition professor who does 2.20 ✕ 10^5 J of useful work while metabolizing 510 kcal of food energy? (Enter your answer as a percentage. Enter a number.) (b) How many food calories (in kcal) would a well-conditioned athlete metabolize in doing the same work with an efficiency of 20%? (Enter a number.)
Answer
4.7
(1 Votes)
Upton
Elite · Tutor for 8 years
Answer
## Answer### Part (a)First, we need to convert the food energy from kilocalories (kcal) to joules (J) because the work done is given in joules. We know that 1 kcal = 4184 J. So, 510 kcal = 510 * 4184 J = 2.1344 x 10^6 JThe efficiency (η) of a system is defined as the ratio of useful output to the total input. In this case, the useful output is the work done and the total input is the food energy metabolized. So, η = (Work done / Energy input) * 100%Substituting the given values, we get:η = (2.20 x 10^5 J / 2.1344 x 10^6 J) * 100% = 10.31%### Part (b)In this case, we are given the efficiency and the work done, and we need to find the energy input (food calories metabolized). We can rearrange the formula for efficiency to find the energy input:Energy input = Work done / (Efficiency / 100%)Substituting the given values, we get:Energy input = 2.20 x 10^5 J / (20 / 100) = 1.10 x 10^6 JFinally, we need to convert this energy from joules back to kilocalories. So, Energy input = 1.10 x 10^6 J / 4184 J/kcal = 263 kcalSo, a well-conditioned athlete would metabolize 263 kcal to do the same work with an efficiency of 20%.