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Bio 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 kcal/g for fat. (a) During heavy exercise, an average person uses energy at a rate of 510 kcal/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 \(60 \mathrm{kg},\) and express your answer in \(\mathrm{m} / \mathrm{s}\) and in \(\mathrm{km} / \mathrm{h} .\)

Short Answer

Expert verified
(a) 16.35 minutes; (b) 139.55 m/s or 502.38 km/h.

Step by step solution

01

Calculate Total Energy of the Doughnut

Calculate the energy from each nutrient. Energy from protein: \(2.0 \, \mathrm{g} \times 4.0 \, \mathrm{kcal/g} = 8.0 \, \mathrm{kcal}\). Energy from carbohydrates: \(17.0 \, \mathrm{g} \times 4.0 \, \mathrm{kcal/g} = 68.0 \, \mathrm{kcal}\). Energy from fat: \(7.0 \, \mathrm{g} \times 9.0 \, \mathrm{kcal/g} = 63.0 \, \mathrm{kcal}\). The total energy in the doughnut is the sum of these values: \(8.0 + 68.0 + 63.0 = 139.0 \, \mathrm{kcal}\).
02

Determine Exercise Duration to Burn the Doughnut

Calculate how long it would take to burn 139 kcal at a rate of 510 kcal/h. Time \(= \frac{139 \, \mathrm{kcal}}{510 \, \mathrm{kcal/h}}\). This simplifies to \(0.2725 \, \mathrm{h}\) or approximately \(16.35 \, \mathrm{minutes}\).
03

Calculate Kinetic Energy from Doughnut

The total energy of the doughnut (139 kcal) needs to be converted to \(\mathrm{Joules}\) for kinetic energy calculations. Since \(1 \, \mathrm{kcal} = 4184 \, \mathrm{J}\), the doughnut energy is \(139 \, \mathrm{kcal} \times 4184 \, \mathrm{J/kcal}\) which equals \(581576 \, \mathrm{J}\).
04

Determine Speed from Kinetic Energy

Use the kinetic energy formula \(\frac{1}{2}mv^2 = \text{energy from doughnut}\). Solve for \(v\), with mass \(m = 60 \, \mathrm{kg}\), and energy \(581576 \, \mathrm{J}\). Rearranging gives \(v = \sqrt{\frac{2 \times 581576 \, \mathrm{J}}{60 \, \mathrm{kg}}}\) which results in \(v \approx 139.55 \, \mathrm{m/s}\).
05

Convert Speed to \\mathrm{km/h}

Convert speed from \(\mathrm{m/s}\) to \(\mathrm{km/h}\) by multiplying by \(3.6\). So, \(139.55 \, \mathrm{m/s} \times 3.6 = 502.38 \, \mathrm{km/h}\).

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Key Concepts

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

Nutritional Energy
Nutritional energy is the fuel our body uses to function and perform activities. It comes from the food and drinks we consume, specifically from proteins, carbohydrates, and fats. Each of these macronutrients provides a different amount of energy when metabolized.
  • Proteins and carbohydrates: Both offer 4 kcal of energy per gram.
  • Fats: Offer 9 kcal per gram, which is more than twice the energy of proteins and carbohydrates.
To determine the total nutritional energy of any food, you can simply multiply the grams of each macronutrient by their respective energy values and sum them up. In the example of the doughnut, this energy calculation is crucial to understanding how much activity is required to "burn off" the energy contained in the food. Such calculations can also provide insight into dietary balances and nutritional planning.
Exercise Physiology
Exercise physiology is the study of functional responses and adaptations to physical activity. One practical application is understanding how our bodies burn energy at different rates during various physical activities. When engaging in exercise, energy expenditure increases, which can help "burn off" the calories consumed.
For example, the doughnut in our problem provides 139 kcal. If a person exercises at a rate that uses up 510 kcal/h, they would need to exercise for around 16.35 minutes to burn off the entire doughnut.
  • This calculation shows the relationship between calorie intake and calorie burning in weight management.
  • It's important to understand that different exercises will affect the body differently, influencing how efficiently calories are burned.
Regular exercise has a myriad of benefits beyond calorie burning, including improved cardiovascular health, increased strength, and enhanced mood, but understanding energy conversion is a key component for managing energy balance related to exercise.
Kinetic Energy Calculation
The concept of converting nutritional energy into kinetic energy can help illustrate the potential energy within food. Kinetic energy is the energy of motion, calculated using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
In our example, transforming the 139 kcal energy from a doughnut into Joules (since \(1 \text{ kcal} = 4184 \text{ J}\)) gives us 581576 J.
This energy can then be used to calculate how fast a 60 kg person could theoretically move if all that energy were converted into kinetic energy:
  • By rearranging the kinetic energy formula, we find velocity: \(v = \sqrt{\frac{2 \times 581576 \text{ J}}{60 \text{ kg}}}\).
  • This results in a theoretical speed of approximately 139.55 m/s, which translates to 502.38 km/h.
While it's an intriguing calculation, in reality, our bodies lose a significant portion of energy as heat rather than converting it into complete kinetic energy. This highlights the complexity of energetic processes in the human body beyond simple theoretical conversions.

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Most popular questions from this chapter

An air pump has a cylinder 0.250 \(\mathrm{m}\) long with a mov- able piston. The pump is used to compress air from the atmosphere (at absolute pressure \(1.01 \times 10^{5}\) Pa) into a very large tank at \(4.20 \times 10^{5}\) Pa gauge pressure. (For air, \(C_{V}=20.8 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\) ) (a) The piston begins the compression stroke at the open end of the cylinder. How far down the length of the cylinder has the piston moved when air first begins to flow from the cylinder into the tank? Assume that the compression is adiabatic. (b) If the air is taken into the pump at \(27.0^{\circ} \mathrm{C}\) , what is the temperature of the compressed air? (c) How much work does the pump do in putting 20.0 mol of air into the tank?

Heat \(Q\) flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?

CALC A cylinder with a frictionless, movable piston like that shown in Fig. 19.5 contains a quantity of helium gas. Initially the gas is at a pressure of \(1.00 \times 10^{5}\) Pa, has a temperature of 300 \(\mathrm{K}\) , and occupies a volume of 1.50 \(\mathrm{L}\) . The gas then undergoes two processes. In the first, the gas is heated and the piston is allowed to move to keep the temperature equal to 300 \(\mathrm{K}\) . This continues until the pressure reaches \(2.50 \times 10^{4}\) Pa. In the second process, the gas is compressed at constant pressure until it returns to its original volume of 1.50 L. Assume that the gas may be treated as ideal. (a) In a \(p V\) -diagram, show both processes. (b) Find th volume of the gas at the end of the first process, and find the pressure and temperature at the end of the second process. (c) Fin the total work done by the gas during both processes. (d) Final would you have to do to the gas to return it to its original pres sure and temperature?

On a warm summer day, a large mass of air (atmospheric pressure \(1.01 \times 10^{5}\) 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 \(0.850 \times 10^{5}\) Pa. Assume that air is an ideal gas, with \(\gamma=1.40\) . (This rate of cooling for dry, rising air, corresponding to roughly \(1^{\circ} \mathrm{C}\) per 100 \(\mathrm{m}\) of altitude, is called the dry adiabatic lapse rate.)

Nitrogen gas in an expandable container is cooled from \(50.0^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\) with the pressure held constant at \(3.00 \times 10^{3}\) Pa. The total heat liberated by the gas is \(2.50 \times 10^{4}\) . Assume that the gas may be treated as ideal. (a) Find the number of moles of gas. (b) Find the change in internal energy of the gas. (c) Find the work done by the gas. (d) How much heat would be liberated by the gas for the same temperature change if the volume were constant?

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