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When you jog, most of the food energy you burn above your basal metabolic rate (BMR) ends up as internal energy that would raise your body temperature if it were not eliminated. The evaporation of perspiration is the primary mechanism for eliminating this energy, Determine the amount of water you lose to evaporation when running for 30 minutes at a rate that uses 400 keal/h above your BMR. (That amount is often considered to be the "maximum fatburning" energy output.) The metabolism of 1 gram of fat generates approximately \(9.0 \mathrm{kcal}\) of energy and produces approximately 1 gram of water. (The hydrogen atoms in the fat molecule are transferred to oxygen to form water.) What fraction of your need for water will be provided by fat metabolism? (The latent heat of vaporization of water at room temperature is \(2.5 \times 10^{6} \mathrm{I} / \mathrm{kg} .\) )

Short Answer

Expert verified
0.067, meaning approximately 6.7% of the water need is provided by fat metabolism.

Step by step solution

01

Calculate the Energy Used

First, let's calculate the energy used in running. The rate is given as 400 kcal/h. So, for 30 minutes, which is half an hour, the energy can be calculated as: \[ Energy = 400 kcal/h \times 0.5h = 200 kcal \]
02

Conversion of Energy into SI unit

Now, convert this energy into SI units (1 kcal = 4184 J): \[ Energy = 200 kcal \times 4184 J/kcal = 836800 J \]
03

Determine the Loss of Mass

Using the latent heat of vaporization formula \(Q= mL\), where \(Q\) is the thermal energy, \(m\) is the mass and \(L\) is the latent heat, we can find the mass, \(m\), of water evaporated. So, \[ m = \frac{Q}{L} = \frac{836800 J}{2.5 \times 10^{6} J/kg} = 0.33 kg \] So, 0.33 kg or 330 g of water is lost due to evaporation.
04

Determine Water Need Supplied by Fat Metabolism

We know fat metabolism generates 9.0 kcal/g of energy and produces 1g of water. Calculate how much fat is metabolized: \[ Energy = Mass \times Metabolic Rate = 200 kcal = Mass \times 9.0 kcal/g \] So, \[ Mass = \frac{200 kcal}{9.0 kcal/g} = 22.2 g \] Hence, 22.2 g of water will be regenerated by the fat metabolism.
05

Calculate the Fraction of Water Need Supplied by Fat Metabolism

The fraction can be calculated as the mass of water obtained by metabolism divided by the mass of water lost: \[ Fraction = \frac{Mass_{metabolism}}{Mass_{lost}} = \frac{22.2g}{330g} = 0.067 \]

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Key Concepts

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

Energy Conversion
Understanding energy conversion is essential when considering the relationship between physical activity and calorie consumption. Energy conversion refers to the process of changing one form of energy into another. In the context of the human body, during exercise, chemical energy from the food we eat is converted into mechanical energy to fuel muscle contractions and thermal energy as a byproduct, which in turn can be dissipated through processes like sweating.

For example, when you jog, the calories burned are initially stored as chemical energy within the molecules of the foods you have consumed. Once you start exercising, your body converts these calories into mechanical energy to keep you moving and thermal energy, which raises your body temperature. To regulate your temperature and prevent overheating, your body expels this excess thermal energy mainly through the evaporation of sweat, a process that involves the conversion of thermal energy into latent heat. The textbook exercise illustrates this beautifully by calculating the amount of water lost through sweat after a run, emphasizing energy conversion in real-world scenarios.
Basal Metabolic Rate
Basal metabolic rate (BMR) is the amount of energy expressed in calories that a person needs to keep the body functioning at rest. This includes energy spent to maintain vital functions such as heart rate, respiration, and maintaining body temperature. BMR can vary depending on a variety of factors, including age, gender, weight, and body composition.

When you jog, the energy you burn above your BMR is primarily used to fuel the extra activity. That energy contributes to the rise in internal body temperature, which your body then seeks to regulate through the evaporation of sweat. The problem provided in the textbook solution involves calculating the energy used above BMR during a 30-minute jog and highlights the role of BMR in understanding total calorie expenditure. By understanding our BMR, we can better gauge the additional energy requirements for various physical activities and the subsequent need for hydration and energy replenishment.
Fat Metabolism
Fat metabolism is the biochemical process by which the human body breaks down fat molecules to generate energy. When you exercise at a rate that uses 400 kcal/h above your BMR, you enter a zone often referred to as 'maximum fat-burning.' This is because moderate-intensity, sustained exercises are efficient at utilizing fat stores for energy.

The metabolic breakdown of 1 gram of fat yields approximately 9 kcal of energy and results in the production of water as a byproduct. In the context of the provided textbook exercise, the fat metabolized during a 30-minute jog not only contributes to the energy needed for the activity but also produces a fraction of water that helps with hydration. The calculation in the solution shows that a modest amount of the water lost through perspiration is replaced by the water produced during fat metabolism, demonstrating the body's complex balance between energy use and hydration.

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Most popular questions from this chapter

[8 The excess internal energy of metabolism is exhausted through a variery of channels, such as through radiation and evaporation of perspiration. Consider another pathway for energy loss: moisture in exhaled breath. Suppose you breathe out \(22.0\) breaths per minute, each with a volume of \(0.600 \mathrm{~L}\). Suppose also that you inhale dry air and exhale air at \(37^{\circ} \mathrm{C}\) containing water vapor with a vapor pressure of \(3.20 \mathrm{kPa}\). The vapor comes from the evaporation of liquid water in your body. Model the water vapor as an ideal gas. Assume its latent heat of evaporation at \(37^{\circ} \mathrm{C}\) is the same as its heat of vaporization at \(100^{\circ} \mathrm{C}\). Calculate the rate at which you lose energy by exhaling humid air.

18 Overall, \(80 \%\) of the energy used by the body must be eliminated as excess thermal energy and needs to be dissipated. The mechanisms of elimination are radiation. evaporation of sweat (2 \(430 \mathrm{~kg} / \mathrm{kg})\), evaporation from the hangs \((38 \mathrm{~kJ} / \mathrm{h})\), conduction, and convection. A person working out in a gym has a metabolic rate of \(2500 \mathrm{~kJ} / \mathrm{h} .\) His body temperature is \(37^{\circ} \mathrm{C}\), and the outside temperature \(24^{\circ} \mathrm{C}\). Assume the skin has an area of \(2.0 \mathrm{~m}^{2}\) and cmissivity of \(0.97\). (a) At what rate is his excess thermal energy dissipated by radiation? (b) If he eliminates \(0.40 \mathrm{~kg}\) of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (c) AL what rate is energy eliminated by evaporation from the lungs? (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

Measurements on two stars indicate that Star \(\mathrm{X}\) has a surface temperature of \(5727^{\circ} \mathrm{C}\) and Star \(\mathrm{Y}\) has a surface temperature of \(11727^{\circ} \mathrm{C}\). If both stars have the same radius, what is the ratio of the luminosity (total power output) of Star Y to the luminosity of Star X? Both stars can be considered to have an emissivity of \(\underline{1.0 .}\)

B. For bacteriological testing of water supplies and in medical clinics, samples must routinely be incubated for \(24 \mathrm{~h}\) at \(37^{\circ} \mathrm{C}\). A standard constant-temperature bath with electric heating and thermostatic control is not suitable in developing nations without continuously operating electric power lines, Peace Corps volunteer and MIT engineenAmy Smith invented a low-cost, low-maintenance incubator to fill the need. The device consists of a foaminsulated box containing several packets of a waxy material that melts at \(37.0^{\circ} \mathrm{C}\), interspersed among tubes, dishes, or bottles containing the test samples and growth medium (food for bacteria). Outside the box, the waxy material is first melted by a stove or solar energy collector. Then it is put into the box to keep the test samples warm as it solidifies. The heat of fusion of the phasechange material is \(205 \mathrm{~kJ} / \mathrm{kg}\). Model the insulation as a panel with surface area \(0.490 \mathrm{~m}^{2}\), thickness \(9.50 \mathrm{~cm}\), and conductivity \(0.0120 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\). Assume the exterior temperature is \(23.0^{\circ} \mathrm{C}\) for \(12.0 \mathrm{~h}\) and \(16.0^{\circ} \mathrm{C}\) for \(12.0 \mathrm{~h}\). (a) What mass of the waxy material is required to conduct the bacteriological test? (b) Explain why your calculation can be done without knowing the mass of the test samples or of the insulation.

1\. A \(55-\mathrm{kg}\) woman cheats on her diet and eats a 540 Calorie (540 kcal) jelly donut for breakfast. (a) How many joules of energy are the equivalent of one jelly doughnut? (b) How many stairs must the woman climb to perform an amount of mechanical work equivalent to the food energy in one jelly doughnut? Assume the height of a single stair is \(15 \mathrm{~cm}\). (c) If the human body is only \(25 \%\) efficient in converting chemical energy to mechanical energy, how many stairs must the woman climb to work off her breakfast?

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