Chapter 15, Problem 16P

To determine

The weight(or mass) a person can expect to lose in $1$ year, assuming no change in food intake.

Answer to Problem 16P

5.3 kg

Explanation of Solution

Given:

A person decides to sleeping one hour less per day and use this time for light activity to lose weight.

Formula used:

Energy E , power P , and time T are related as,

  $\mathrm{E}=\mathrm{P}\times \mathrm{T}$

Calculation:

Metabolic rate for sleeping is 70 W while for light activity like eating is 230 W. So, the rate of loss of energy with the change in activity is

  $\begin{array}{c}{\mathrm{P}}_{\mathrm{c}}=230-70\\ =160\text{W}\end{array}$

If a person sleeps one hour less, then there will be increase in one-hour light activity. So, total energy lost in 1 year, by doing the activity 1 hour each day is,

  $\begin{array}{c}{\mathrm{E}}_{\mathrm{c}}={\mathrm{P}}_{\mathrm{c}}\times \mathrm{T}\\ =160\times \left(1\text{yr}\times \frac{365\text{days}}{1\text{yr}}\times \frac{1\times 60\times 60\sec }{1\text{day}}\right)\\ =210240000\text{J}\\ =\text{2}\text{.1024}\times {\text{10}}^8\text{J}\end{array}$

Given that the $1\text{kg}$ of fat stores,

  $\begin{array}{c}\mathrm{E}=40,000\text{kJ}\\ \mathrm{E}=4\times {10}^4\text{kJ}\\ =\text{4}\times {\text{10}}^7\text{J}\end{array}$

So, the fat lose by a person in a year,

  $\begin{array}{c}\mathrm{w}=2.1024\times {10}^8\times \frac{1\text{kg}}{4\times {10}^7}\\ =5.256\text{kg}\\ \approx \text{5}\text{.3}\text{kg}\end{array}$

Conclusion:

Therefore, of mass or weight a person can lose in $1$ year is 5.3 kg.