Question

(a) What is the efficiency of an out-of-condition professor who does 1.90 ✕ 105 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%? kcal

Answer 1

a) The energy efficiency of the out-of-condition professor is 9.082 %.

b) The food calories needed by the well-conditioned athlete is 181.644 kilocalories.

Step-by-step explanation:

a) The energy efficiency of the food metabolization (
`\eta`), no unit, is defined by following formula:

`\eta = (W)/(E)* 100\,\%` (1)

Where:

`W` - Useful work, in joules.

`E` - Food energy, in joules.

If we know that
`W = 1.90* 10^(5)\,J` and
`E = 2.092* 10^(6)\,J`, the energy efficiency of the food metabolization is:

`\eta = (1.90* 10^(5)\,J)/(2.092* 10^(6)\,J) * 100\,\%`

`\eta = 9.082\,\%`

The energy efficiency of the out-of-condition professor is 9.082 %.

b) If we know that
`W = 1.90* 10^(5)\,J` and
`\eta = 25\,\%`, then the quantity of food energy is:

`E = (W)/(\eta)* 100\,\%`

`E = 1.90* 10^(5)\,J* (100\,\%)/(25\,\%)`

`E = 7.60* 10^(5)\,J`

`E = 181.644\,kcal`

The food calories needed by the well-conditioned athlete is 181.644 kilocalories.