Chapter 6, Problem 6.18P

Interpretation Introduction

Interpretation:

The time to work off $1.0\text{ lb}$ of body fat by an athlete is to be calculated if he expends $1950\text{kJ}$ energy per hour.

Concept introduction:

Power is generally calculated as work divided by time. The unit of work is watt or joule per sec.

$\text{Power}=\frac{\text{work}\left(\text{J}\right)}{\text{time}\left(\sec \right)}$

Other units of power include $\text{kJ/sec}$ and $\text{kJ/h}$.

The conversion factor to convert $\text{kJ}$ into $\text{J}$ is,

$\text{1}\text{kJ}=1000\text{J}$

The e conversion factor to convert nutritional calorie $\left(\text{Calorie}\right)$ into $\text{cal}$ is,

$\text{1}\text{Calorie}=1000\text{cal}$

The conversion factor to convert $\text{cal}$ into $\text{J}$ is,

$\text{1}\text{cal}=4.184\text{J}$

The conversion factor to convert $\text{lb}$ into $\text{cal}$ is,

$\text{1}\text{lb}=4.1\times {10}^3\text{cal}$