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Q47 PE

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Found in: Page 262

### College Physics (Urone)

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 $$250.96{\rm{ kcal}}$$.

See the step by step solution

## Efficiency:

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)

## Efficiency of an out-of-condition professor

(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%.

## Amount of food calories needed for metabolizing an athlete

(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 $$250.96{\rm{ kcal}}$$.