Suggested languages for you:

Americas

Europe

Q55PE

Expert-verified
Found in: Page 263

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# Kanellos Kanellopoulos flew 119 km from Crete to Santorini, Greece, on April 23, 1988, in the Daedalus 88, an aircraft powered by a bicycle-type drive mechanism (see Figure 7.42). His useful power output for the 234-min trip was about 350 W. Using the efficiency for cycling from Table 7.2, calculate the food energy in kilojoules he metabolized during the flight.

The energy used in metabolism during flight is $$5872.37{\rm{ kcal}}$$.

See the step by step solution

## Step 1: Efficiency

The efficiency is the ratio of the total useful output by the total input. It is a measure of how efficiently a device uses energy.

Mathematically,

$$\eta \% = \frac{{{P_o}}}{{{P_i}}} \times 100\%$$

Here, $$\eta$$ is the efficiency, $${P_o}$$ is the useful power output, and $$P$$ is the total power input.

## Step 2: Food energy in kilojoules he metabolized during the flight

The total power input of the person can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the total power input.

$${P_i} = \frac{{{P_o}}}{{\eta \% }} \times 100\%$$

Here, $${P_o}$$ is the useful power output $$\left( {{P_o} = 350{\rm{ W}}} \right)$$, and $$\eta$$ is the efficiency of cycling$$\left( {\eta = 20\% } \right)$$.

Putting all known values,

\begin{aligned} {P_i} &= \frac{{\left( {350{\rm{ W}}} \right)}}{{20\% }} \times 100\% \\ &= 1750{\rm{ W}}\end{aligned}

The energy consumed is,

$$E = {P_i}t$$

Here, $${P_i}$$ is the input power $$\left( {{P_i} = 1750{\rm{ W}}} \right)$$, and t is the time of operation $$\left( {t = 234{\rm{ min}}} \right)$$.

Putting all known values,

\begin{aligned} E &= \left( {1750{\rm{ W}}} \right) \times \left( {234{\rm{ min}}} \right)\\ &= \left( {1750{\rm{ W}}} \right) \times \left( {234{\rm{ min}}} \right) \times \left( {\frac{{60{\rm{ sec}}}}{{1{\rm{ min}}}}} \right)\\ &= 24570000{\rm{ J}} \times \left( {\frac{{1{\rm{ kcal}}}}{{4184{\rm{ J}}}}} \right)\\ &= 5872.37{\rm{ kcal}}\end{aligned}

Therefore, the required energy used in metabolism during flight is $$5872.37{\rm{ kcal}}$$.