Suggested languages for you:

Americas

Europe

Q47 PE

Expert-verifiedFound in: Page 262

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) What is the efficiency of an out-of-condition professor who does \(2.10 \times {10^5}{\rm{ J}}\) of useful work while metabolizing \(500{\rm{ kcal}}\) of food energy? **

**(b) How many food calories would a well-conditioned athlete metabolize in doing the same work with an efficiency of \(20\% \)?**

(a) The efficiency of an out-of-condition professor is 10%.

(b) The amount of food calories a well-conditioned athlete metabolizes in doing same work with an efficiency of 20% is \(250.96{\rm{ kcal}}\).

**The ****number**** of labor or energy saved during a process is measured by efficiency.**

The efficiency is defined as,

\({\rm{Efficiency}} = \frac{{{\rm{Useful work}}}}{{{\rm{Energy consumed}}}} \times 100\% \) (1.1)

(a)

An out-of-condition professor does \(2.10 \times {10^5}{\rm{ J}}\) of useful work while metabolizing 500 kcal of food energy.

The efficiency of an out-of-condition professor can be calculated using equation (1.1).

Putting all known values into equation (1.1),

\(\begin{aligned}{\rm{Efficiency}} &= \frac{{2.10 \times {{10}^5}{\rm{ J}}}}{{500{\rm{ kcal}}}} \times 100\% \\ &= \frac{{2.10 \times {{10}^5}{\rm{ J}}}}{{\left( {500{\rm{ kcal}}} \right) \times \left( {\frac{{4184{\rm{ J}}}}{{1{\rm{ kcal}}}}} \right)}} \times 100\% \\ &= 10\% \end{aligned}\)

Therefore, the required efficiency of an out-of-condition professor is 10%.

(b)

Amount of food calories metabolized if a well-conditioned athlete does same amount of work with 20% efficiency can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the energy consumed.

\({\rm{Energy consumed}} = \frac{{{\rm{Useful work}}}}{{{\rm{Efficiency}}}} \times 100\% \)

Putting all known values,

\(\begin{aligned}{\rm{Energy consumed}} &= \frac{{2.10 \times {{10}^5}{\rm{ J}}}}{{20\% }} \times 100\% \\ &= 1.05 \times {10^6}{\rm{ J}} \times \left( {\frac{{1{\rm{ kcal}}}}{{4184{\rm{ J}}}}} \right)\\ &= 250.96{\rm{ kcal}}\end{aligned}\)

Therefore, the required amount of food calories a well-conditioned athlete metabolizes in doing same work with an efficiency of 20% is \(250.96{\rm{ kcal}}\).

94% of StudySmarter users get better grades.

Sign up for free