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Doughnuts: Breakfast of Champions! A typical doughnut contains \(2.0 \mathrm{~g}\) of protein. \(17.0 \mathrm{~g}\) of carbohydrates, and \(7.0 \mathrm{~g}\) of fat. The average food energy values of these substances are \(4.0 \mathrm{kcal} / \mathrm{g}\) for protein and carbohydrates and \(9.0 \mathrm{kcal} / \mathrm{g}\) for fat. (a) During hcavy exercise, an average person uscs cnergy at a rate of 510 keal \(/\) h. How long would you have to exercise to "work off"one doughnut? (b) If the energy in the doughnut could somehow be converted into the kinetic energy of your body as a whole, how fast could you move after eating the doughnut? Take your mass to be \(60 \mathrm{~kg}_{4}\) and express your answer in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{km} / \mathrm{h}\)

Short Answer

Expert verified
To "work off" the doughnut, exercise for 16.35 minutes. You could move at 139.7 m/s or 503 km/h after eating it.

Step by step solution

01

Calculate Energy from Protein

Each gram of protein provides \(4.0\,\text{kcal/g}\). With \(2.0\,\text{g}\) of protein in a doughnut, the energy from protein is \(2.0\,\text{g} \times 4.0\,\text{kcal/g} = 8.0\,\text{kcal}\).
02

Calculate Energy from Carbohydrates

Each gram of carbohydrate provides \(4.0\,\text{kcal/g}\). With \(17.0\,\text{g}\) of carbohydrates, the energy is \(17.0\,\text{g} \times 4.0\,\text{kcal/g} = 68.0\,\text{kcal}\).
03

Calculate Energy from Fat

Each gram of fat provides \(9.0\,\text{kcal/g}\). With \(7.0\,\text{g}\) of fat, the energy contribution is \(7.0\,\text{g} \times 9.0\,\text{kcal/g} = 63.0\,\text{kcal}\).
04

Total Energy in a Doughnut

Add up the energy contributions: \(8.0\,\text{kcal}\) from protein, \(68.0\,\text{kcal}\) from carbohydrates, and \(63.0\,\text{kcal}\) from fat to get a total of \(139.0\,\text{kcal}\).
05

Time to Work Off the Doughnut

With an exercise rate of \(510\,\text{kcal/h}\), the time needed is \(\frac{139.0\,\text{kcal}}{510\,\text{kcal/h}} = 0.2725\,\text{h}\). Converting to minutes gives \(0.2725\times60 = 16.35\,\text{minutes}\).
06

Convert Energy to Kinetic Energy

Set the doughnut energy \(139.0\,\text{kcal}\) into kinetic energy. Convert \(\text{kcal}\) to \(\text{J}\): \(139.0\,\text{kcal} \times 4184\,\text{J/kcal} = 581576\,\text{J}\).
07

Calculate Speed from Kinetic Energy

Using the kinetic energy formula \(KE = \frac{1}{2}mv^2\), solve for \(v\): \(v = \sqrt{\frac{2\cdot581576}{60}} = 139.7\,\text{m/s}\).
08

Convert Speed to km/h

Convert \(139.7\,\text{m/s}\) to \(\text{km/h}\) by multiplying by \(3.6\): \(139.7\,\text{m/s} \times 3.6 = 503.0\,\text{km/h}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calorimetry
Calorimetry is the science of measuring the heat of chemical reactions or physical processes. In our exercise, we study how much energy, in terms of calories, each component of a doughnut provides.

We calculate the caloric energy based on the macronutrient composition:
  • Proteins: 2.0 g, contributing 8 kcal (2.0 g times 4 kcal/g)
  • Carbohydrates: 17.0 g, providing 68 kcal (17.0 g times 4 kcal/g)
  • Fats: 7.0 g, adding 63 kcal (7.0 g times 9 kcal/g)
Each of these nutrients stores energy, which our bodies can convert to work or heat. Summing up, the total energy from these components equals 139 kcal for a doughnut. This process of energy calculation is essential not only for food labeling but also to understand the potential energy available from consumption.
Kinetic energy
Kinetic energy is the energy an object possesses due to its motion. In physics problems like this, we often transform stored energy, such as that from a doughnut, into kinetic energy to analyze motion.

After eating, if all the energy from the doughnut could convert to kinetic energy, we apply the formula for kinetic energy: \[ KE = \frac{1}{2} m v^2 \] where \( m \) is mass and \( v \) is velocity. For instance, by setting 139 kcal converted to joules (139 kcal * 4184 J/kcal), we get 581,576 J.

With a body mass of 60 kg, solving for velocity \( v \) yields \( v = \sqrt{\frac{2 \times 581576}{60}} = 139.7 \, \text{m/s} \). It illustrates the potential speed if all this energy propels forward motion, albeit theoretical.
Energy Conversion
Energy conversion is a fundamental concept where energy changes from one form to another. Through our exercise, food energy (chemical potential energy) transforms into kinetic energy units that can be executed by the body.

When a person exercises, the caloric energy consumed undergoes conversion to different energy types, mainly kinetic (for movement) and thermal (due to body processes). The given data shows that the total energy in a doughnut is calculated to be 139 kcal, a representation of chemical energy available.

Exercise taps into this stored food energy, gradually converting it based on activity intensity. With a daily activity, understanding energy conversion helps people manage weight, monitor energy expenditure, and optimize physical performance.
Exercise Physiology
Exercise physiology examines how the body responds to physical activity, such as how it processes metabolic energy from foods like doughnuts.

During physical exertion, exercise physiology demonstrates how the body uses energy stores, breaks down nutrients, and converts them to usable forms. The typical energy expenditure rate is around 510 kcal/h. Knowing this, one can calculate how long they need to work out to "burn" a doughnut's energy. For our case, it takes about 16.35 minutes of exercise to expend a single doughnut's energy reserve, illustrating how exercise aligns with caloric intake.

Understanding these intricate processes helps athletes and fitness enthusiasts tailor their diets and workouts for optimal energy use and physical performance.

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Most popular questions from this chapter

Two moles of an ideal gas are compressed in a cylinder at a constant temperature of \(85.0^{\circ} \mathrm{C}\) until the original pressure has tripled. (a) Sketch a \(p V\) -diagram for this process. (b) Calculate the amount of work done.

A monatomic ideal gas expands slowly to twice its original volume, doing 300 \(\mathrm{J}\) of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is (a) isothermal; (b) adiabatic; (c) isobaric.

A Thermodynamic \(\mathbf{P} \mathbf{r} \mathbf{o}-\) cess In an Insect. The Africanbombardier beetle Stenaptinus insignis can emit a jet of defensive spray from the movable tip of its abdomen (Fig, 19.32\()\) . The beetle's body has reservoirs of two different chemicals; when the beetle is disturbed, these chemicals are combined in a reaction chamber, producing compound that is warmed from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 \(\mathrm{m} / \mathrm{s}(68 \mathrm{km} / \mathrm{h})\) , scaring away predators of all kinds. (The beetle shown in the figure is 2 \(\mathrm{cm}\) long.) Calculate the heat of reaction of the two chemicals (in J/kg). Assume that the specific heat capacity of the two chemicall and the spray is the same as that of water, \(4.19 \times 10^{3} \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , and that the initial temperature of the chemicals is \(20^{\circ} \mathrm{C}\) .

A cylinder with a piston contains 0.150 mol of mitrogen at \(1.80 \times 10^{5} \mathrm{Pa}\) and 300 \(\mathrm{K}\) . The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. (a) Show the series of processes in a \(p V\) -diagram. (b) Compute the temperatures at the beginning and end of the adiabatic expansion. (c) Compute the minimum pressure.

On a warm summer day, a large mass of air (atmospheric pressure \(1.01 \times 10^{5} \mathrm{Pa}\) ) is heated by the ground to a temperature of \(26.0^{\circ} \mathrm{C}\) and then begins to rise through the cooler surrounding air. (This can be treated approximately as an adiabatic process; why? Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only \(0.850 \times 10^{5} \mathrm{Pa}\) . Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cool- ing for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of altitude, is called the dry adiabatic lapse rate.)

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