Chapter 10, Problem 67A

Interpretation Introduction

Interpretation: The miles walk at $4.0\text{miles}/\text{h}$ to lose 1 lb of body fat needs to be explained.

Concept Introduction: In metric system, Calorie is the amount of energy (heat) required to increase 1 g of water by $1\text{°C}$ . Calorie likely used to calculate the food energy content.

  $1\text{kcalorie}=1000\text{calorie}$

Answer to Problem 67A

The miles walk at $4.0\text{miles}/\text{h}$ to lose 1 lb of body fat is $34.96\text{miles}$ .

Explanation of Solution

Since,

  $\begin{array}{c}\text{1}\text{g}\text{of body fat}=7.7\text{kcal}\text{of energy}\\ \text{454}\text{g}\text{of body fat}=\frac{\text{454}\text{g}\times 7.7\text{kcal}}{\text{1}\text{g}}\\ =3495.8\text{kcal}\end{array}$$\begin{array}{c}400\text{kcal}\text{are}\text{conjumed}\text{on}\text{walking}\text{at 4}\text{mile/h}\text{for}=1\text{h}\\ 3495.8\text{kcal}\text{are}\text{conjumed}\text{on}\text{walking}\text{at 4}\text{mile/h}\text{for}=\frac{1\text{h}\times 3495.8\text{kcal}}{400\text{kcal}}\\ =8.74\text{h}\end{array}$

The speed of walking is $4\text{miles}/\text{h}$ .

  $\begin{array}{c}\text{So},\text{distance}\text{covered}\text{in}8.74\text{h}=8.74\text{h}\times 4\text{miles}/\text{h}\\ =34.96\text{miles}\end{array}$

Conclusion

The miles walk at $4.0\text{miles}/\text{h}$ to lose 1 lb of body fat is $34.96\text{miles}$ .