Question

Body fat is metabolized, supplying 9.30 kcal/g, when dietary intake is less than needed to fuel metabolism. The manufacturers of an exercise bicycle claim that you can lose 0.500 kg of fat per day by vigorously exercising for 2.00 h per day on their machine.

(a) How many kcal are supplied by the metabolization of 0.500 kg of fat?

(b) Calculate the kcal/ min that you would have to utilize to metabolize fat at the rate of 0.500 kg in 2.00 h.

(c) What is unreasonable about the results?

(d) Which premise is unreasonable, or which premises are inconsistent?

Short answer

(a) The energy supplied by metabolizing fat is$4650kcal$.

(b) Power required is$38.75\text{kcal}/\text{min}$.

(c) The result is unreasonable as there is a huge rate requirement.

(d) The premises that are inconsistent is that exercise done at this rate is not possible to performed continuously for 2 hr.

Step-by-step solution

Step 1: Calorific value

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.

Amount of energy supplied by metabolization of fat

(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,

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

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

Calculation of power

(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,

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

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

Power generated

(c)

The power required is,

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

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.

Explanation

(d)

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