Question: (II) A 65-kg person decides to lose weight by sleeping one hour less per day, using the time for light activity. How much weight (or mass) can this person expect to lose in 1 year, assuming no change in food intake? Assume that 1 kg of fat stores about 40,000 kJ of energy.

The energy value can be calculated by multiplying the power value with the elapsed time value. Its value is altered linearly to the value of the power.

Given data:

The expected time for losing weight is\(t = 1.0\;{\rm{y}}\).

The energy equivalent to one kilogram of fat is \({\rm{1}}{\rm{.0}}\;{\rm{kg}} = {\rm{4}}{\rm{.0}} \times {\rm{1}}{{\rm{0}}^7}\;{\rm{J}}\).

From table 15-2:

The metabolic rate for engaging in light activity like eating dressing is \(230\;{\rm{W}}\).

The metabolic rate for sleeping is\(70\;{\rm{W}}\).

The rate of loss of energy due to change in activity can be calculated as:

\(\begin{aligned}{l}{P_{\rm{c}}} &= \left( {230\;{\rm{W}}} \right) - \left( {70\;{\rm{W}}} \right)\\{P_{\rm{c}}} &= 160\;{\rm{W}}\end{aligned}\)

The energy lost during one year can be calculated as:

\(\begin{aligned}{l}{E_{\rm{L}}} &= {P_{\rm{c}}}t\\{E_{\rm{L}}} &= \left( {160\;{\rm{W}}} \right)\left( {1.0\;{\rm{y}}} \right)\left( {\frac{{{\rm{365}}\;{\rm{days}}}}{{{\rm{1}}{\rm{.0}}\;{\rm{y}}}}} \right)\left( {\frac{{1.0\;{\rm{h}}}}{{1\;{\rm{day}}}}} \right)\left( {\frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)\\{E_{\rm{L}}} &= 2.1024 \times {10^8}\;{\rm{J}}\end{aligned}\)

Now convert energy lost from joules to kilograms to calculate fat lost by human in one year.

\(\begin{aligned}{l}{E_{\rm{L}}} &= \left( {2.1024 \times {{10}^8}\;{\rm{J}}} \right)\left( {\frac{{1.0\;{\rm{kg}}}}{{4.0 \times {{10}^7}\;{\rm{J}}}}} \right)\\{E_{\rm{L}}} &= 5.3\;{\rm{kg}}\end{aligned}\)

Thus, the required weight loss is \(5.3\;{\rm{kg}}\).