91Ó°ÊÓ

Calorific value or calorific power is defined as the amount of energy preset in food or fuel per unit mass. The unit for calorific value is kJ/kg or kcal/g.

(a)

The energy supplied by metabolizing fat is,

$E=em$

Here, e is the calorific value of fat$\left(e=9.3\text{kcal}/\text{g}\right)$ , and m is the mass of fat metabolized$\left(m=0.500\text{kg}\right)$ .

Putting all known values,

$\begin{array}{rcl}E& =& \left(9.3\text{kcal}/\text{g}\right)×\left(0.500\text{kg}\right)\\ & =& \left(9.3\text{kcal}/\text{g}\right)×\left(0.500\text{kg}\right)×\left(\frac{1000\text{g}}{1\text{kg}}\right)\\ & =& 4650\text{kcal}\\ & & \end{array}$

Therefore, the required energy supplied by metabolizing fat is$4650kcal$ .

(b)

The power required to metabolize the fat is,

$P=\frac{E}{t}$

Here, E is the energy supplied by metabolizing 0.5 kg of fat$\left(E=4650\text{kcal}\right)$ , and t is the time$\left(t=2.00\text{hr}\right)$ .

Putting all known values,

$\begin{array}{rcl}P& =& \frac{4650\text{kcal}}{2\text{hr}}\\ & =& \frac{4650\text{kcal}}{\left(2\text{hr}\right)×\left(\frac{60\text{min}}{1\text{hr}}\right)}\\ & =& 38.75\text{kcal}/\text{min}\\ & & \end{array}$

Therefore, the power required is$38.75kcal/min$ .

(c)

The power required is,

$\begin{array}{rcl}P& =& \left(38.75\text{kcal}/\text{min}\right)×\left(\frac{4184\text{J}}{1\text{kcal}}\right)×\left(\frac{1\text{min}}{60\text{s}}\right)\\ & ≈& 2702.17\text{W}\\ & & \end{array}$

Hence, the power consumption in units of watts is 2702.17 W which is 46% greater than that of a professional cyclist. This is unreasonable.

(d)

Assuming that exercise done at 2702.17 W can be sustained for 2 hr is unreasonable.