Question

Energy that is not utilized for work or heat transfer is converted to the chemical energy of body fat containing about \(39{\rm{ kJ}}/{\rm{g}}\). How many grams of fat will you gain if you eat \(10,000{\rm{ kJ}}\) (about \(2500{\rm{ kcal}}\)) one day and do nothing but sit relaxed for \(16.0{\rm{ hr}}\) and sleep for the other \(8.00{\rm{ h}}\)? Use data from Table 7.5 for the energy consumption rates of these activities.

Short answer

The amount of fats gained is \(17.89{\rm{ g}}\).

Step-by-step solution

Step 1: Definition of Concepts 

Energy consumed: Energy consumed by any device is given as the product of its power rating and the time of operation of the device.

Mathematically,

\(E = PT\)

Here, P is the power and T is the time.

Calculate the amount of fat will you gain in grams

The energy used in sitting relaxed and sleeping is,

\({E_u} = {P_{sitting}} \times {t_{sleep}} + {P_{sleep}} \times {t_{sleep}}\)

Here, \({P_{sitting}}\) is the power consumed in sitting relaxed \(\left( {{P_{sitting}} = 120{\rm{ W}}} \right)\), \({t_{sitting}}\) is time spent in sitting relaxed \(\left( {{t_{sitting}} = 16.0{\rm{ h}}} \right)\), \({P_{sleep}}\) is the power consumed in sleeping \(\left( {{P_{sleep}} = 83\;{\rm{W}}} \right)\), and \({T_{sleep}}\) is the time spent in sleeping \(\left( {{T_{sleep}} = 8.0{\rm{ h}}} \right)\).

Putting all known values,

\(\begin{array}{c}{E_u} = \left( {120{\rm{ W}}} \right) \times \left( {16.0{\rm{ h}}} \right) + \left( {83{\rm{ W}}} \right) \times \left( {8.0{\rm{ h}}} \right)\\ = \left( {120{\rm{ W}}} \right) \times \left( {16.0{\rm{ h}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ h}}}}} \right) + \left( {83{\rm{ W}}} \right) \times \left( {8.0{\rm{ h}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ h}}}}} \right)\\ = 9302400{\rm{ J}} \times \left( {\frac{{1{\rm{ kJ}}}}{{1000{\rm{ J}}}}} \right)\\ = 9302.4{\rm{ kJ}}\end{array}\)

The unutilized energy is,

\(E = {E_c} - {E_u}\)

Here, \({E_c}\) is the energy consumed \(\left( {{E_c} = 10000{\rm{ kJ}}} \right)\).

Putting all known values,

\(\begin{array}{c}E = \left( {10000{\rm{ kJ}}} \right) - \left( {9302.4{\rm{ kJ}}} \right)\\ = 697.6{\rm{ kJ}}\end{array}\)

The body converts unutilized \(39{\rm{ kJ}}\) of energy into \(1{\rm{ g}}\) body fat. Therefore, the fat gain is,

\(\begin{array}{c}F = \frac{{697.6{\rm{ kJ}}}}{{\left( {39{\rm{ kJ}}/{\rm{g}}} \right)}}\\ = 17.89{\rm{ g}}\end{array}\)

Therefore, the required amount of fats gained is \(17.89{\rm{ g}}\).