(a) What is the efficiency of an out-of-condition professor who does 2.10×105 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 20%?

Solution 47PE

a.)

Step 1 of 4</p>

The efficiency of the out of condition professor can be calculated by using the definition of the efficiency which is defined as the ratio of the work output to work input.

Efficiency (=

Workoutput is the amount of useful work done by the professor is 2.10105 J .

Workoutput = 2.10105 J

Workinput is the amount of energy in 500 kcal of food .

1 kcal = 4184 J

So, 500 kcal = 500 kcal

= 2092000 J

Workinput = 2092000 J

Step 2 of 4</p>

The efficiency () can be calculated as

Efficiency(=

=

= 0.10

or= 10%

Therefore, the efficiency of an out-of -condition professor who does 2.10×105 J of useful work while metabolizing 500 kcal of food energy is 10 %.

b.)