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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 \(400 \mathrm{kcal} / \mathrm{h}\) above your BMR. (That amount is often considered to be the "maximum fat-burning" energy output.) The metabolism of 1 gram of fat gen- erates 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 \(2.5 \times 10^{6} \mathrm{~J} / \mathrm{kg}\).)

Short Answer

Expert verified
During 30 minutes of jogging at a rate of 400 kcal/h, approximately 334.72 g of water will be lost due to evaporation. Fat metabolism provides about 22.22 g of water, which is about 6.6% of the total water requirement.

Step by step solution

01

Calculate Energy Burned

First, calculate the energy burned during 30 minutes of jogging. The rate of energy used is \(400 \mathrm{kcal/h}\). Therefore, the amount of energy burned is \(400 \mathrm{kcal/h} \times 0.5\mathrm{h} = 200 \mathrm{kcal}.\)
02

Convert Energy to Joules

In order to relate the energy burned to the latent heat of vaporization of water, we need to convert the energy from kilocalories into joules by using the equivalence \(1 \mathrm{kcal} = 4184 \mathrm{J}\). So, the energy burned in joules is \(200 \mathrm{kcal} \times 4184 \mathrm{J/kcal} = 836800 \mathrm{J}\).
03

Find Amount of Water Lost to Evaporation

The energy used to evaporate water off your skin would be equal to the extra energy produced while jogging. The amount of water that can be evaporated with this energy can be found by dividing the energy by the latent heat of vaporization (\(2.5 \times 10^{6} \mathrm{J/kg}\)). Therefore, the amount of water lost to evaporation is \(\frac{836800 \mathrm{J}}{2.5 \times 10^{6} \mathrm{J/kg}} = 0.33472 \mathrm{kg} = 334.72 \mathrm{g}.\)
04

Calculate Water Produced by Fat Metabolism

Since the metabolism of 1 gram of fat generates 1 gram of water, we can determine the fat metabolism from the energy burned during jogging. We know that the energy output burned is 200 kcal and 1 gram of fat generates 9.0 kcal of energy. The amount of water produced by fat metabolism is therefore \(\frac{200 \mathrm{kcal}}{9.0 \mathrm{kcal/g}} = 22.22 \mathrm{g}.\)
05

Find Fraction of Water Need Fulfilled by Fat Metabolism

The fraction of your need for water that will be provided by fat metabolism can be found by dividing the amount of water produced by fat metabolism by the total amount of water lost to perspiration. Therefore, the fraction is \(\frac{22.22 \mathrm{g}}{334.72 \mathrm{g}} = 0.066\) or about \(6.6\%\).

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Key Concepts

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

Basal Metabolic Rate (BMR)
Basal Metabolic Rate, commonly referred to as BMR, is the amount of energy your body requires to perform fundamental physiological functions while at rest. These functions include breathing, circulating blood, controlling body temperature, and cell production. Everyone has a different BMR, influenced by several factors:
  • Age: Metabolism generally slows down with age, reducing BMR.
  • Gender: Males often have a higher BMR due to greater muscle mass, compared to females of the same weight and age.
  • Body Composition: More muscle mass increases BMR as muscle tissue is metabolically active.
Understanding your BMR can help you determine how much food energy you need to maintain your current weight when you're not engaging in physical activities. Anything burned above your BMR, such as through exercise, contributes to raising your body's energy expenditure. For example, while jogging for 30 minutes, an individual can burn energy over their BMR—hence using additional calories.
Evaporation and Latent Heat
When you exercise, like jogging, your body produces more internal energy than at rest. This extra energy could increase your body's temperature if not removed effectively. The primary method for energy increase regulation is through the evaporation of sweat.

The process of evaporation absorbs a significant amount of heat due to the latent heat of vaporization. This concept refers to the energy needed to change a substance from liquid to vapor without changing its temperature. For water, this value is very high, around 2.5 million Joules per kilogram, making it effective at removing large amounts of heat from the body.

Sweating, and thereby evaporating water from your skin, helps you maintain your body temperature during physically demanding activities. You might notice losing several grams of water, like when running for 30 minutes, as your body works to keep you cool.
Fat Metabolism
Fat metabolism is a complex process where your body breaks down fat molecules to generate energy. Compared to other nutrients, fats provide a substantial amount of energy. One gram of fat can produce approximately 9.0 kilocalories of energy.

In addition to providing energy, fat metabolism also plays a role in water production within the body. When fat is broken down, the hydrogen atoms in the fat molecule combine with oxygen to form water. For instance, by metabolizing 1 gram of fat, you not only gain 9.0 kcal of energy but also produce 1 gram of water.

This mechanism is crucial during activities like jogging where your body needs extra energy and efficient water replacement. Through fat metabolism, a small fraction of your water needs, about 6.6% in this example, are met as water produced helps in maintaining hydration during intense exercise.

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Most popular questions from this chapter

The thermal conductivities of human tissues vary greatly. Fat and skin have conductivities of about \(0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(0.020 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively, while other tissues inside the body have conductivities of about \(0.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume that between the core region of the body and the skin surface lies a skin layer of \(1.0 \mathrm{~mm}\), fat layer of \(0.50 \mathrm{~cm}\), and \(3.2 \mathrm{~cm}\) of other tissues. (a) Find the \(R\)-factor for each of these layers, and the equivalent \(R\)-factor for all layers taken together, retaining two digits. (b) Find the rate of energy loss when the core temperature both a protective layer of clothing and an insulating layer of unmoving air are absent, and a body area of \(2.0 \mathrm{~m}^{2}\).

In the summer of 1958 in St. Petersburg, Florida, a new sidewalk was poured near the childhood home of one of the authors. No expansion joints were supplied, and by mid-July the sidewalk had been completely destroyed by thermal expansion and had to be replaced, this time with the important addition of expansion joints! This event is modeled here. A slab of concrete \(4.00 \mathrm{~cm}\) thick, \(1.00 \mathrm{~m}\) long, and \(1.00 \mathrm{~m}\) wide is poured for a sidewalk at an ambient temperature of \(25.0^{\circ} \mathrm{C}\) and allowed to set. The slab is exposed to direct sunlight and placed in a series of such slabs without proper expansion joints, so linear expansion is prevented. (a) Using the linear expansion equation (Eq. 10.4), eliminate \(\Delta L\) from the equation for compressive stress and strain (Eq. 9.3). (b) Use the expression found in part (a) to eliminate \(\Delta T\) from Equation \(11.3\), obtaining a symbolic equation for thermal energy transfer \(Q\). (c) Compute the mass of the concrete slab given that its density is \(2.40 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). (d) Concrete has an ultimate compressive strength of \(2.00 \times 10^{7} \mathrm{~Pa}\), specific heat of \(880 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and Young's modulus of \(2.1 \times 10^{10} \mathrm{~Pa}\). How much thermal energy must be transferred to the slab to reach this compressive stress? (e) What temperature change is required? (f) If the Sun delivers \(1.00 \times 10^{3} \mathrm{~W}\) of power to the top surface of the slab and if half the energy, on the average, is absorbed and retained, how long does it take the slab to reach the point at which it is in danger of cracking due to compressive stress?

A \(3.00-\mathrm{g}\) copper coin at \(25.0^{\circ} \mathrm{C}\) drops \(50.0 \mathrm{~m}\) to the ground. (a) Assuming \(60.0 \%\) of the change in gravitational potential energy of the coin-Earth system goes into increasing the internal energy of the coin, determine the coin's final temperature. (b) Does the result depend on the mass of the coin? Explain.

A \(50-\mathrm{g}\) ice cube at \(0{ }^{\circ} \mathrm{C}\) is heated until \(45 \mathrm{~g}\) has become water at \(100^{\circ} \mathrm{C}\) and \(5.0 \mathrm{~g}\) has become steam at \(100^{\circ} \mathrm{C}\). How much energy was added to accomplish the transformation?

A \(1.5-\mathrm{kg}\) copper block is given an initial speed of \(3.0 \mathrm{~m} / \mathrm{s}\) on a rough horizontal surface. Because of friction, the block finally comes to rest. (a) If the block absorbs \(85 \%\) of its initial kinetic energy as internal energy, calculate its increase in temperature. (b) What happens to the remaining energy?

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