Thermodynamics of Biological Systems

Thermodynamics of Biological Systems
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Published Date:02-08-2017
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Thermodynamics of Biological Systems CONTENTS 17.1 Introdução (Introduction) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693 17.2 Living Systems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693 17.3 Thermodynamics of Biological Cells. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695 17.4 Energy Conversion Efficiency of Biological Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 699 17.5 Metabolism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 702 17.6 Thermodynamics of Nutrition and Exercise. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705 17.7 Limits to Biological Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 711 17.8 Locomotion Transport Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714 17.9 Thermodynamics of Aging and Death . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 716 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 721 17.1 INTRODUÇÃO (INTRODUCTION) Over the years, thermodynamics has remained essentially an engineering discipline with only infrequent applications elsewhere. In chemistry, courses entitled Physical Chemistry specialize in applying thermody- namics to chemical systems of the type treated in Chapter 15 of this book. Only recently has the developing field of bioengineering begun to apply the macroscopic mass, energy, and entropy balance concepts of classi- cal thermodynamics to living systems. In this chapter, the results of applying these basic laws of thermody- namics to biological systems are reviewed. The conclusions reached will help you understand how your body functions and give you some insight into the operation of the complex molecular phenomena necessary to sustain life on this or any other planet. In this chapter, the basic thermodynamics of simple living systems (biological cells) is followed by discussions of animal biological energy conversion efficiency, metabolism, nutrition, and exercise. Then, the fascinating subjects of the limits to biological growth and an engineering view of living system mobility are presented. The chapter ends, appropriately, with a thermodynamic discussion of biological aging and death. In this section, an attempt is made to describe how and why living systems age differently from nonliving systems. 17.2 LIVING SYSTEMS Only in the past few years has science begun to realize how the evolution of life is completely compatible with the laws of physics. A key to this understanding has been the entropic explanation of self-organizing systems and the connection between self-organization and energy flow. Self-organization can exist in both living and nonliving systems, but living systems are self-organizing and self-replicating. The origin of life is apparently not694 CHAPTER 17: Thermodynamics of Biological Systems an unusual phenomenon. Fairly complex living microscopic creatures existed on Earth within a few hundred million years of its formation. There is no clear-cut understanding of the scientific concept of life. Living systems have six recognized character- istics: (1) molecular organization, (2) metabolism, (3) growth, (4) adaptation, (5) response to stimuli, and (6) reproduction. But perhaps the best definition available today is that a system is said to be “living” if it sustains its low-entropy molecular complexity (i.e., DNA) that contains hereditary information transmitted to offspring on reproduction, by a metabolic energy transport from a high-energy source (food) to a low-energy sink (waste) via catalytic macromolecules, called enzymes. The different living systems on Earth have numerous items in common. For example, they all use the same class of molecules for energy storage, the nucleotide phosphates. Also, of the billions of chemically possible organic compounds, only about 1500 are actually used by living systems. And all these 1500 compounds are made with less than 50 simpler molecular building blocks, utilizing no more than 24 of the available elements. Hydrogen atoms make up 63% of all the atoms in the human body. Oxygen accounts for 25.5%, carbon 9.50%, nitrogen 1.40%, and the 20 remaining elements essential for mammalian life account for only about 0.60%. Only 3 of the 24 elements known to be essential to life on this planet have atomic numbers greater than 34, and these 3 are needed in only trace amounts. Thus, since living systems are made up of the simplest atomic elements, they can be expected to develop early on any planet that has the proper environmental conditions. Living systems are organized around a cell structure of some kind. A cell is like a small factory whose main func- tion is to carry out its metabolic process, and the cell boundaries appear to exist to provide the high enzyme con- centration necessary for efficient metabolism. The smallest free-living cell known today is a pleuropneumonialike –19 –7 organism that has a mass of only about 5 × 10 kg and has a diameter of only about 10 m (about one fifth of the wavelength of visible light). This cell contains about 100 enzymes and can be seen only with a high-powered 14 –5 electron microscope. The human body contains about 10 cells with an average diameter of 10 m. Each cell typically consists of a central nucleus, with the remaining material being the cytoplasm (see Figure 17.1). The che- mical activity inside the cell is very high, with each enzyme entering into the synthesis of about 100 molecules per second. The oldest remnants of life on Earth are cellular microfossils that are over 3.5 billion years old. Since the age of the Earth is only about 4.5 billion years, the thermal and chemical requirements for the evolution of living sys- tems must have developed remarkably fast. All known living systems on Earth are water based and therefore cannot exist far outside the temperature range from 0 to 100°C. It is amazing that the surface of the Earth had regions in this temperature range for at least 80% of its existence. WHAT IS METABOLISM? Metabolism is the name given to the processes of breakdown and synthesis of large macromolecules within a cell. Nucleus (contains chromosomes) Cell membrane Vacuoles Lysosome Mitochondria FIGURE 17.1 Schematic of a typical living cell.17.3 Thermodynamics of Biological Cells 695 17.3 THERMODYNAMICS OF BIOLOGICAL CELLS It is unlikely that a single energy source was directly responsible for the synthesis of all the organic molecules on the newly formed Earth. In recent decades, laboratory experiments with the elements of carbon, hydrogen, oxy- gen, and nitrogen have shown that basic organic compounds can be synthesized by a variety of energy sources under early Earth conditions. Table 17.1 lists an estimate of the energy rate per unit area available on the surface of the primitive Earth. Though solar radiation was clearly the largest source of energy, the energy contained in long-wavelength (150 to 200 nm) ultraviolet light is so strong that it decomposes absorbing molecules rather than building them. However, the water of the primitive Earth’s oceans protected complex organic molecules from disruptive ultraviolet radiation until the Earth’s ozone layer developed. It was not until this protective atmospheric layer had developed that life forms could leave the oceans and populate the dry land. The most widely used source of energy for the synthesis of primitive organic compounds in the laboratory is an electrical discharge in a mixture of gases. The most common compounds produced by this technique are amino acids, with yields as high as 5%. As concentrations of organic compounds built up in the primitive oceans, biological life processes began to synthesize and replicate molecules. Enzymes and genetic molecules evolved, but reaction rates were limited by the comparatively low concentrations of these essential building blocks. Specialized molecular barriers then evolved that completely enclosed small volumes of fluid containing complex molecular machinery. These bar- riers are called membranes and the resulting enclosure is called a cell. The purpose of biological membranes is to maintain concentration differences that would be advantageous to the molecular operation of the cell. To do this, the membrane must be able to transport certain ions against the concentration gradient (this is called active transport). This requires that the membrane operate as an energy converter, with some of the internal energy of the cell being used to maintain the various concentration gradients across the membrane. Table 17.2 lists some ion concentrations inside and outside common human cells. Table 17.1 Estimates of Energy Rates Available for the Formation of Simple Organic Compounds, Averaged over the Surface Area of Primitive Earth Energy Rates 2 Source per Unit Area KJ/(m ·a) Electric discharge (lightning, etc.) 170 Solar radiation in the 0−150 nm range 71 Thermal quenching of hot gases from Shock waves from meteors and lightning 46 Volcanoes 5.4 Highly ionizing radiation from Radioactivity 1.0 km deep in the Earth 33 Solar wind 8.4 Cosmic rays 0.1 Source: Material drawn from Oró, J., Miller, S. L., Urey, H. C. Energy conversion in the context of the origin of life. In: Buvet, R., Allen, M. J., Massué, J.-P. (Eds.), Living Systems as Energy Converters. North-Holland Publishing, 1977, pp. 7–19, New York. Reprinted by permission of Elsevier Science Publishers (Biomedical Division), Amsterdam, and the authors. Table 17.2 Approximate Ion Concentration Inside and Outside Human Cells 3 Concentration in Osmoles per cm of Water Ion Outside the Cell Inside the Cell + Na 144 14.0 + K 4.1 140 2+ Mg 1.5 31 – Cl 107 4.00 − HCO 27.7 10.0 3 2− 0.5 1 SO 4 2− − 2.0 11 HPO ,H PO 2 4 4 Note: One osmole is the number of gram moles of the substance that do not diffuse or dissociate in solution. Also, pH = 7:4 and outside + −pH pH = 7:0, where the concentration of hydrogen ions ðH Þ in gmoles/Lis10 : inside696 CHAPTER 17: Thermodynamics of Biological Systems Because cell membranes are molecular machines, their Proteins Cell interior 70-100 Å exact structure is not yet completely understood. The most universally accepted model of a membrane is the bimolecular lipid leaflet structure shown schematically in Figure 17.2. In this model, the membrane structure Polar Nonpolar consists of two parallel rows of phospholipid molecules ends ends oriented with their hydrophobic chains pointing inward and their hydrophilic (polar) ends pointing outward. The inner and outer surfaces of the membrane are covered with various protein layers, and the membrane −9 Cell exterior thickness is typically 7 to 10 nanometersð10 mÞ: It is Proteins also felt that the membrane must contain a uniform FIGURE 17.2 distribution of holes, or pores, about 0.8 nm in Schematic of membrane construction. diameter, through which water and certain hydrated ions can pass. Approximately 0.06% of the membrane area is made up of these pores. The concentration of materials inside the cell is determined exclusively by concentration differences across the membrane. Membranes of living cells maintain an electrical potential difference between the inside and outside of the cell. With very small electrodes, a reasonably constant current can be continuously drawn from a cell. A cell can pro- duce electricity in this way only if it has a molecular mechanism for maintaining an unequal ion charge differ- ence across its membrane. Such membranes are known to contain a molecular level ion pump that transports ion species in only one direction (into or out of the cell). How much work is required to pump a charged ion from the solution outside the cell into the solution inside the cell? The answer to this question can be developed by considering the transport process to be car- ried out in two steps. First, consider moving an ion from infinity through a vacuum, through the membrane, andintothecell.Assumeinthisfirststepthatthecellmembranehasnodipolelayer(i.e.,nonetchargeon its surface) and the inside of the cell is electrically neutral. Now, as the charged ion moves from infinity to the membrane, it encounters no resistance, so its transport work is zero. As it moves through the cell mem- brane, it begins to feel electrostatic ion-solvent and ion-ion interactions. We lump all these interactions together and call them chemical effects; therefore, the work done against these interactions in moving the charged ion into the cell is called chemical work. The chemical work done in moving a mole of ions of chemi- cal species i from infinity into an uncharged cell through a dipole layer–free membrane is equal to the molar chemical potential μ of ion species i. i The second step in this process is to allow the membrane to have a dipole layer and allow the cell to have a net internal charge. We call the work required to move the ion into this system of net charges the electrical work; therefore, the total work required to move the ion from infinity through a dipole-layered membrane into a charged cell is the sum of the chemical work plus the electrical work. This total work is called the electrochemical work of the cell. Define ϕ as the electrical potential (in volts) required to transport a unit charge of species i into the cell. Then, k the electrical work required to transport one ion of species i with valence z kgmole of electrons per kgmole of i species i (and thus a charge ze)is zeϕ ,where e is the charge on one electron. The electrical work required to i i ic transport 1 mole of species i into the cell is N zeϕ = zFϕ ,whereN is Avogadro’snumberand F = N e = o i i o o ic ic Faraday’s constant = 96,487 kilocoulombs/kgmole of electrons. LetðÞ w be the electrochemical work required EC ic to transport 1 mole of species i with valence z into a cell. Then, we can write i ðÞ w =ðÞ W /n =μ +zFϕ (17.1) EC EC i ic i ic ic ic Unfortunately, neither μ or ϕ is directly measurable. They were introduced as conceptual quantities for the ic ic purpose of separating chemical effects from electrical effects; however, only their combined effect can be observed in the laboratory. Whatcanwemeasure?Wecanmeasuretheelectricalpotential difference (i.e., voltage) between the inside and outside of the cell. Now, as soon as we introduce an electrode into the cell, we set up a current path, so that the measured potential Δϕ is not the same as the zero current (no electrode) equilibrium potential Δϕ : Again, Δϕ m e e cannot be measured, but we can get around that as follows. From Eq. (17.1), we find that the electrochemical work required to move a mole of ions of species i with valence z from infinity into the solution outside the cell is i ðÞ w =ðÞ W /n =μ +zFϕ (17.2) EC EC i i io io io io17.3 Thermodynamics of Biological Cells 697 and that the electrochemical work required to move that same mole from infinity into the cell is ðÞ w =μ +zFϕ (17.3) EC i ic ic ic then, from Eqs. (17.2) and (17.3), we find that the zero current equilibrium electrical potential difference between the inside and outside of the cell due to the presence of species i is 1 ðÞ Δϕ =ϕ −ϕ = ðÞ w −ðÞ w −ðÞ μ −μ (17.4) EC EC e i ic io ic io ic io zF i The molar chemical potential of species i can be written for isothermal, dilute solutions as o μ =μ +ℜTlnc (17.5) i i i o where μ is the molar chemical potential when c = 1:0,ℜ is the universal gas constant, T is the absolute tem- i i perature, and c is the molar concentration of i. Using Eq. (17.5), we can write Eq. (17.4) as i  1 ℜT c o o io ðÞ Δϕ = ðÞ w −ðÞ w − μ −μ + ln (17.6) EC EC e ic io i ic io c ic zF zF i i  o To simplify the algebra, we call the first term on the right side of Eq. (17.6) Δϕ , which is the value ofðÞ Δϕ e e i i when c = c : Then, Eq. (17.6) becomes ic io  c ℜT io o ðÞ Δϕ = Δϕ + ln (17.7) e i e i c ic zF i  o However, we still cannot measure eitherðÞ Δϕ or Δϕ : At this point, we arbitrarily assign the electrical poten- e i e i tial outside the cell the value zero, and we define the membrane potential E due to species i as i  o E =ðÞ Δϕ − Δϕ i e i e i which is given by Eq. (17.7) as c ℜT io E = ln (17.8) i c ic zF i At 37°C, Eq. (17.8) becomes (recall that 1 coulomb = 1 joule/volt)  . ½ 8314:3J/ðÞ kgmole KðÞ 37:0+273:15K c io EðÞ at37°C = ln i c zðÞ 96,487kilocoulombs/kgmole ic i (17.9)  . 26:7millivoltsðÞ kgmoleelectrons/kgmolei c io = ln c z ic i where z is the valence of species i in kgmole of electrons per kgmole of species i. Note that z can be either posi- i i tive or negative in this equation. WHAT IS SO SPECIAL ABOUT A BODY TEMPERATURE OF 37°C? As the temperature of an organism increases up to about 40ºC, the speed of its enzyme-catalyzed metabolic reactions increases, because the molecules collide more frequently due to thermal agitation. But above 40ºC, the weak bonds that control the functional shape of the enzymes begin to break, and they become ineffective at sustaining metabolism. For many years, it was thought that life as we know it could not exist at temperatures above about 40ºC. However, recently hyperthermophilic (“superheat-loving”) bacteria have been found in high-temperature environments, such as deep sea volcanic hot vents. They grow at temperatures above 80°C and can survive to temperatures up to 113°C. They are very tough life forms, even surviving temperatures as low as −140°C. It seems possible that they could have been carried through space on meteoroids to populate planets.698 CHAPTER 17: Thermodynamics of Biological Systems EXAMPLE 17.1 Using the concentration data provided in Table 17.2, determine the membrane potential in human cells of sodium, potas- sium, and chlorine ions at 37.0°C. Solution Table 17.2 gives the concentration of sodium ions inside a human cell at 37.0°Cas 3 c + = 14:0osmoles/cm Na c while the concentration of sodium ions outside the cell is 3 c + = 144osmoles/cm Na o + The valence of a sodium ion is 1 kgmole electrons/kgmole Na . Then, Eq. (17.9) gives the membrane potential of sodium as + c + 26:7mVðkgmoleelectrons/kgmoleNa Þ Na o + E = ln Na + + c + z kgmoleelectrons/kgmoleNa Na Na c  + 26:7mVðkgmoleelectrons/kgmoleNa Þ 144 = ln = 62:2mV + 1kgmoleelectrons/kgmoleNa 14:0 3 3 For potassium, Table 17.2 gives c + = 140osmoles/cm and c þ = 4:1osmoles/cm : The valence of a potassium ion is also K K c o + 1 kgmole electrons/kgmole K , and Eq. (17.9) gives the membrane potential of potassium in a human cell as + 26:7mVðkgmoleelectrons=kgmoleK Þ 4:1 + E = ln ¼94:3mV K + 140 1kgmoleelectrons=kgmoleK 3 3 − − Finally, Table 17.2 gives c = 4:00osmoles/cm and c = 107osmoles/cm . The valence of a chlorine ion is –1kgmole Cl Cl c o – electrons/kgmole Cl , and Eq. (17.9) gives the membrane potential of chlorine in a human cell as − 26:7mVðkgmoleelectrons/kgmoleCl Þ 107 − E = ln = −87:8mV Cl − −1kgmoleelectrons/kgmoleCl 4:00 Exercises 2+ 1. Determine the membrane potential in human cells of magnesium ions, Mg , at 37.0°C. Answer: E 2+ = –40:4mV: Mg 2– 2. Determine the membrane potential in human cells of sulphate ions, SO , at 37.0°C. Answer: E 2− = 9:3mV: 4 SO4 – 3. Determine the membrane potential in human cells of dihydrogen phosphate ions, H PO , at 37.0°C. 2 4 − Answer: E = 45:5mV: H2PO4 Actual measured potentials are generally in the range of –70 to −90mV and represent the cumulative effect of + + – all the ion species present. However, Na,K,andCl are the primary high-transport ions in most mammal membranes, and their cell potentials, listed earlier, average out to about the measured value. Applying the open system energy rate balance equation to a living cell gives  dU _ _ Q−W +∑me _ −∑me _ = dt cell in out _ Here, Q is the irreversible metabolic heat transfer resulting from the life processes within the cell,∑ me _ is the food in _ _ energy intake,∑ me is the waste product output, and W is the total work done on or by the cell. The food taken out into the cell can be generalized as glucose and molecular oxygen, and the waste products can be generalized as carbon dioxide and water. The total work done on or by the cell is the electrochemical work done in maintaining the chemical differences across the cell membrane,ðÞ w =ðÞ w −ðÞ w , and occasional p-V work done in EC EC EC i ic io enlarging the cell plusγ-A surface tension workdonein generating new membrane surface area. Then,   w μ EC _ _ _ _ _ W =∑m +∑m +γA +pV (17.10) i i M i M i where we have written all terms on a mass rather than a molar basis (using n_ = m _ /M, where M is the molecular i i i i mass of species i) and the intensive properties have been assumed to be constant in time. Thus, the time rate of changeofthe cell’s total internal energy is    w μ dU EC _ _ _ = Q−∑m _ −∑m _ −γA−pV+∑me _ −∑me _ i i dt M M cell i i in outAnaerobic metabolism 17.4 Energy Conversion Efficiency of Biological Systems 699 Most of the various cellular processes that require energy Carbohydrate use adenosine triphosphate (ATP) as the energy source. This compound has about 33 MJ/kgmole of energy stored in each of two phosphate bonds. When these bonds are split by enzyme action to form adenosine diphosphate (ADP), their energy is then made available Carbon dioxide Oxygen for other uses. The cell contains many enzymes that can Pyruvic acid and water catalyze the splitting of the ATP bonds and utilize the liberated energy. Energystoragereactionswithinthecell,ontheother hand, are limited to two basic types: photosynthetic (in plant cells), wherein incoming light is used as the energy source, and metabolism (in animal cells), FIGURE 17.3 wherein the food brought into the cell (generally glu- cose and molecular oxygen) is utilized to reconstitute Energy transport mechanisms in living systems. ATP from ADP, with the production of carbon dioxide and water waste products, which must be expelled Cell membrane from the cell. Figure 17.3 shows how these two energy transport mechanisms are linked together in the life cycle, and Figure 17.4 illustrates the ATP–ADP cycle. Oxygen and ATP High-energy glucose An open system entropy rate balance applied to a release from Cells living cell gives Respiration phosphate needs bonds  ADP _ Q dS _ +∑ms _ −∑ms _ +S = Carbon dioxide P T dt cell b in out and water where T is the temperature of the cell boundary b FIGURE 17.4 (assumed isothermal here). Because the metabolic The ATP–ADP cycle. heat must leave the cell for it to survive, we know that _ _ Q0: Also, S 0 due to the irreversibilities of the life P process within the cell. Since food products are brought into the cell and waste products expelled, ∑ me _ ∑ me _ (as these two flow streams are at the same temperature and the molecular order of the waste out in material is less than that of the food). For a cell to grow and continue to maintain its elaborate internal molecu- lar order, we must haveðÞ dS/dt 0, or cell       _ _ ∑ms _ −∑ms _ +Q/T≥S b P   in out _ which is perfectly reasonable so long as the cell remains alive (i.e., Q0 and∑ me _ ∑ me _ ). in out 17.4 ENERGY CONVERSION EFFICIENCY OF BIOLOGICAL SYSTEMS Metabolism is the name given to all anabolic (constructive) and catabolic (destructive) molecular processes within a living system, and it is a direct measure of the energy used by the system. Because a living system is an open system, it is more convenient to speak of its metabolic rate, that is, its energy usage per unit time. Part of the metabolic energy can appear as physical work done by the system; part of it can appear as an increase in total system internal energy (as in the case of growth); part of it can appear in the creation of high-energy items, such as eggs, seeds, live offspring, and milk; and virtually all of the irreversibilities associated with these pro- cesses appear as heat production within the system. An open system energy rate balance for the life form shown in Figure 17.5 is   2 mgZ dU d mV d _ _ Q − W +∑me _ −∑me _ ⎵ = + + dt dt 2g dt 2g c c in out Metabolic Workdone Food,oxygen Systemchanges heattransfer onorby (17.11) and ð0Þ thesystem waste material Aerobic metabolism Photosynthesis700 CHAPTER 17: Thermodynamics of Biological Systems Heat loss (Q) Work done (W) Waste energy out Food energy in FIGURE 17.5 Energy flows in living systems. _ Life processes all have some degree of irreversibility. Therefore, Q normally is negative since the internal irrever- sibilities generally produce internal heat generation, which must be removed from the system if the system is not to overheat. Classically, the concept of work in thermodynamic analysis has been somewhat ambiguous. As discussed in Chapter 4, during the development of thermodynamics, it was convenient to separate the changes in kinetic and potential energies from the work term. These energy terms are written separately and usually grouped with the system’s total internal energy change, as shown in Eq. (17.11). Thus, the work term in the thermodynamic energy balance encompasses all the work transport of energy into or out of a system except the work associated with changes in the system’s kinetic and potential energy. This can be quite confusing when analyzing biological systems, since one of their major work modes in a social or cultural context is that of mobility, that is, running and climbing, which are the kinetic and potential energy terms we are discussing. Also, whereas a classical ther- modynamic system can either do work or have work done on it, in general, a biological system only does work (i.e., the work term is always negative). As with nonliving work-producing systems, we can define an energy conversion efficiency as Desiredenergyresult Energy conversion efficiency =η (17.12) = E Requiredenergyinput The term energy used in this equation must include relevant kinetic and potential energy changes. For example, the energy conversion efficiency of a human climbing a hill could be calculated by choosing the change in potential energy of the person as the desired energy output, while ignoring other types of energy output simulta- neously performed (such as aerodynamic drag against the atmosphere). This is acceptable, providing the mean- ing of the efficiency is clearly defined in each case. The required energy input part of Eq. (17.12) is more difficult to evaluate. Since, for warm-blooded animals, _ the net Q is always out of the system, it cannot be considered as a source of energy input. Also, one cannot generally input useful energy into a biological system via changes in the system’s kinetic or potential energies. Thus, what remains is dU Requiredenergyinput = − dt Then, we may write Eq. (17.12) as   2 mgZ d mV d _ W + + _ dt 2g dt g c c Q η = = 1+ (17.13) E −dU/dt −dU/dt _ and since both Q and dU/dt are always negative, it is clear that Eq. (17.13) gives an energy conversion efficiency that is always less than 100%.17.4 Energy Conversion Efficiency of Biological Systems 701 On the other hand, in the case of most plants and some animals, there is either direct energy conversion of incoming solar radiation or a metabolic reduction resulting from direct body warming from incoming solar _ radiation. Since radiation is one of the classical heat transfer mechanisms, solar radiation belongs to the Q term. _ In this case, part of the system’s Q is actually incoming and used within the system and must be considered as part of the total energy input. The energy conversion efficiency of plant photosynthesis can be defined as Energy converted to organic molecules by photosynthesisðper unit areaÞ ðη (17.14) Þ = E photosynthesis Solar energy input to earthðper unit areaÞ This is quite low, typically ranging between 0.01 and 1.0%. Part of the reason for the low efficiency is that not all the solar energy incident on a unit area of the Earth is intercepted by a plant. As plants become smaller and more uniformly cover the Earth, this efficiency rises somewhat. For example, in the case of algae (a microscopic one-celled plant), the photosynthetic energy conversion efficiency at a small densely packed test site can be as high as 10%. The energy conversion efficiency of animals can be defined as Rate of food energy stored in the body as complex organic molecules ðÞ η = (17.15) E animal Rate of energy taken into the body as food EXAMPLE 17.2 Everyone has heard about the food chain, but few realize how inefficient it is in nature. The energy conversion efficiency from sunlight to plant growth is only about 1.00%, theenergy conversion efficiency of theplants eaten by grazing herbivores is about 20.0%, and the energyconversion efficiency of the carnivores who hunt and eat the herbivores is only about 5.00%. So the over- −4 all energy conversion efficiency from sunlight to carnivore is about (0.0100)(0.200)(0.0500) = 1.00 × 10 = 0.0100%. If the 2 average daily solar energy reaching the surfaceof the Earth is 15.3 MJ/d · m , then how much land is required to grow the plants needed tofeedthe herbivores eaten bya largecarnivorethatrequires10.0MJ/dtostayalive? Solution Since our hunting carnivore requires 10.0 MJ of food per day, at a 5.00% food energy conversion rate, it must consume 10:0MJ/d = 200:MJ/d 0:0500 of herbivore meat. The food energy conversion rate of the grazing herbivores is 20.0%, so they must consume 200:MJ/d = 1000MJ/d 0:200 in plant food. At a 1.0% energy conversion rate, the plants consumed by the herbivore require 1000MJ/d 5 = 1:00×10 MJ/d 0:0100 2 of solar energy. Since the average solar energy intensity on the surface of the Earth is 15.3 MJ/d·m , 100,000 MJ/d of solar energy require an area of 100,000MJ/d 2 = 6540m . 2 15:3MJ/d m 2 and since 1 acre = 4047 m , then  2 1acre 6540m ¼ 1:62acres 2 4047m of plant food is required to supply the food chain energy required to meet the 10 MJ/d needs on our carnivore. Exercises 4. Using the results of Example 17.2, determine the number of carnivores that can be supported by herbivores living off 1500 acres of plants. Answer: 926 carnivores. 5. If the number of available herbivores in Example 17.2 increases dramatically and the carnivores’ hunting energy expenditure is reduced to the point where their food energy conversion efficiency increases from 5.00% to 12.0%, determine the amount of land required to support one carnivore. Answer: 0.675 acres/carnivore. 6. If the carnivore in Example 17.2 moves to a tropical climate where the solar intensity and the photosynthetic energy conversion efficiency double, how much land would be required to support its food chain? Answer: 0.81 acres.702 CHAPTER 17: Thermodynamics of Biological Systems CRITICAL THINKING The energy conversion efficiencies for plants and animals defined in Eqs. (17.14) and (17.15) are what we define in Chapter 10 as a first law efficiency. Can you use the general definition of a second law efficiency given in Chapter 10 to formulate a second law efficiency for plants and animals? What difficulties are encountered in evaluating this new efficiency? Our conceptual understanding of physiological work is often quite different from our earlier (Chapter 4) definition of thermodynamic work. For example, when an animal walks along a horizontal surface it does no net thermodynamic work. There is no net change in kinetic or potential energy, and there is no appreciable slid- ing friction between the animal’s feet and the ground. Only when the animal moves against an external force (such as hydrodynamic drag or inertia forces) is any classical thermodynamic work done. Walking does involve what we culturally call work, but in thermodynamic jargon the energy associated with constant velocity motion along a horizontal plane (in the absence of hydrodynamic drag) merely involves a net conversion of internal energy into heat. Thus, the thermodynamic efficiency of this type of motion in animals (or machines) is zero. If, instead, an animal walks on a horizontal treadmill, then it does do thermodynamic work. This work appears as friction or electricity, depending in the treadmill design. Part of the friction in this case is external to the animal and is measurable as work in the classical thermodynamic sense. The remaining part of the energy expenditure is internal losses within the animal and appears as metabolic heat. Note, however, that the thermodynamic work efficiency of walking on a treadmill returns to zero if the entire treadmill apparatus is included in the sys- tem with the animal. A key element in understanding the thermodynamics of biological systems is comprehending the role of the heat transfer term in the energy rate balance equation of these systems. Since this equation by itself is useful only if you have just one unknown term, since it is not usually satisfactory to simply ignore or set equal to zero those terms for which we do not have values, and since ðdU/dtÞ is perhaps the most difficult term of all to measure system _ accurately, then it becomes absolutely necessary that a means be found to give accurate measurements of Q: 17.5 METABOLISM The metabolic energy in the resting state is called the basal metabolic rate (BMR). The BMR is essentially the energy required to keep the molecular machinery of life operating at a zero activity level. Similar measure- ments at a higher activity level produce intermediary metabolic rate results. The basal metabolic rate for humans depends on age, sex, height, general health conditions, and the like. Figure 17.6 shows the variation in the average BMR per unit body surface area for human malesandfemalesasafunctionofage.Itisnotuncom- mon to have BMR variations around these normal (or average) values of ±15% for any one individual. Table 17.3 shows the breakdown in energy consumption comprising the BMR in the adult human body. The large energy consumption of the brain is surprising; the brain of a five-year-old child may account for up to 50% of its BMR. 300 280 Male 260 240 220 200 Human metabolic rate 180 160 140 Female 120 100 0 1020304050607080 Age (years) FIGURE 17.6 Average BMR per unit area for humans vs. age. Basal metabolic rate 2 per unit surface area (kJ/m h)17.5 Metabolism 703 Table 17.3 Breakdown of the Contributions to the Basal Metabolic Rate of the Various Organs of the Adult Human Body Organ Mass (kg) % of Body Mass % of BMR Liver 1.5 2.14 27 Brain 1.4 2.00 20 Heart 0.3 0.43 10 Kidneys 0.3 0.43 8 Muscles 30.0 42.8 26 Remaining body tissue 36.5 52.2 9 Total 70.0 100.0 100.0 Source: Reprinted by permission of the publisher and the author from Margen, S. Energy metabolism. In: McCally, M. (Ed.), Hypodynamics and Hypogravics,1968 ed. Academic Press, New York. CRITICAL THINKING The average basal metabolic rates per unit surface area for male and female humans areshowninFigure17.6.Whydo you think the values for females are less than those for males over their life spans? Also, why do these curves level off at about age 20? Measuring an animal’s metabolic heat transfer directly is called direct calorimetry. The technique is very difficult to carry out because the animal’s conductive, convective, and radiation heat transport rates must all be measured directly. This is commonly done by putting the animal in a closed box that has water circulating through all six of its sides. If the outside of the box is well insulated, then an energy balance shows that all of the metabolic heat produced by the animal ends up in the circulating water. However, virtually all metabolic measurements done today use a method called indirect calorimetry, wherein the CO production and the O consumption are 2 2 measured instead. Generally, indirect calorimetric techniques are found to be as accurate as the direct techniques and are usually considerably easier and less expensive to use. The ratio of the number of moles of CO produced to the number of moles of O consumed during an indirect 2 2 calorimetry test is called the respiratory quotient (RQ), and its value depends on the type of food being metabo- lized. For example, in the metabolism of 1 mole of a typical carbohydrate, glucose, C H O +6ðO Þ 6ðCO Þ+6ðH OÞ 6 12 6 2 2 2 6molesofO and 6 moles of CO are involved. Therefore, the RQ of carbohydrate is 1.0. On the other hand, 2 2 the RQ of protein is 0.8 and that of fat is 0.7. An animal generally consumes a mixture of these substances, so consumed? Tests show that, under how do we know which value to use as the energy equivalent per liter of O 2 basal conditions, the RQ is approximately 0.82 (which is very nearly the average value for the RQs of carbohy- drate, protein, and fat), and it can be shown that this gives a mixture composition of these three substances that 3 corresponds to an energy equivalent of 20.2 MJ/m of O,or20.2kJ/LofO . Thus, if we measure the number 2 2 of liters of O consumed per unit time by an animal in the resting state and multiply this value by 20.2 kJ/L 2 O , we obtain the resting energy consumption rate (or basal metabolic rate) of the animal. In the case of a fast- 2 ing (starving) animal, which is living on the consumption of its own body fat and protein, the energy equivalent per liter of O consumed is 21.3 kJ/L O . 2 2 WHAT IS KLEIBER’S LAW? In 1932, Max Kleiber (1893–1976) published a paper, “Body Size and Metabolism,” which included a graph (Figure 17.7) 3/4 that showed that an animal’s metabolic rate scales to the three-quarter power of the animal’s mass, or BMR∝M . Kleiber’s law has been found to hold across 18 orders of size, from microbes to whales. (Continued)704 CHAPTER 17: Thermodynamics of Biological Systems WHAT IS KLEIBER’S LAW? Continued 3 10 Warm-blooded organisms 0 10 −3 10 Cold-blooded organisms −6 10 Unicellular organisms −9 10 −12 10 −12 −9 −6 −3 0 3 6 9 10 10 10 10 10 10 10 10 Mass (grams) FIGURE 17.7 Kleiber’s law graph. WHAT DO A BANANA, AN ORANGE, AND A PERSON HAVE IN COMMON? According to a new study, all living organisms share roughly the same resting metabolic rate when body size and tempera- ture are taken into account. The finding suggests that widely diverse species burn energy in predictable patterns. “The cor- rected basal metabolic rate of an apple or tree is remarkably similar to that of bacteria, which is remarkably similar to a fish or person,” says James Gillooly, at the University of New Mexico in Albuquerque. A comparison of the basal metabolic rates for a large number of warm-blooded animals from mice to elephants and birds produced the following empirical correlation, called Kleiber’s law:  3/4 BMR = 293 m (17.16) where BMR is the animal’s basal metabolic rate in kJ/d, and m is the animal’s body mass in kg. Thus, the basal metabolic rate per unit mass of the animal is 293 −1/4 BMR/m = 293ðm Þ = (17.17) 1/4 m and it clearly increases with decreasing body mass. EXAMPLE 17.3 Determine the basal metabolic rate (BMR) per unit mass of an 80.0 kg adult human and an 8.00 gram mouse. Since both are warm-blooded mammals, explain why there is a difference in these values. Solution The basal metabolic rate per unit mass of a warm-blooded animal is given by Eq. (17.17) as −0:25 BMR/m = 293ðm Þ Metabolic rate (kcal/h)17.6 Thermodynamics of Nutrition and Exercise 705 where BMR is in kJ/d and m is in kg. Then, the BMR per unit mass of an 80.0 kg human is kJ −0:25 ðBMR/mÞ = 293ð80:0 Þ = 98:0 human . kg d and the BMR per unit mass of an 8.00 gram mouse is kJ −0:25 ðBMR/mÞ = 293ð0:00800 Þ = 980: mouse . kg d This calculation shows that the basal metabolic rate per unit mass of a mouse is ten times that of a human. This large differ- ence is primarily due to the difference in surface area to volume ratio of these mammals. Since heat loss from the body is primarily by convection heat transfer, which is proportional to surface area, and internal heat generation inside the body is proportional to its volume, then as the ratio of surface area to volume increases the internal heat generation rate must also increase if a mammal is to maintain its body temperature. To produce higher internal heat generation rates, small animals must feed very often if they are not to starve. It is a fact that the smaller any object becomes, the larger its surface area to volume ratio becomes. This is easiest to under- 2 3 stand with spherical objects. The surface area of a sphere is 4πR whereas its volume is (4/3)πR . Therefore, its surface area to volume ratio is  2 Surfacearea 4πR 3 ¼ ¼ 4 Volume 3 R sphere πR 3 and this ratio decreases inversely with increasing R. Thus, there is a lower limit to the size of warm-blooded animals. The shrew and the hummingbird are the smallest known animals of this kind. The body temperature of insects and cold-blooded animals is approximately equal to the temperature of their surroundings. Consequently, there is no thermodynamic lower limit to their size. Exercises 7. Determine the basal metabolic rate (BMR) and the basal metabolic rate per unit mass (BMR/m) of a 1300. kg elephant. Answers: BMR = 63,400 kJ/d and (BMR/m) = 48.8 kJ/kg·d. elephant elephant 8. If a 0.500 gram house fly had to maintain the body temperature of warm-blooded mammal using the same internal heat generation rate mechanisms, what would be its basal metabolic rate (BMR) and its basal metabolic rate per unit mass (BMR/m)? Answers: BMR = 0.980 kJ/d and (BMR/m) = 1960 kJ/kg·d. fly fly 9. Since whales are aquatic mammals, determine the basal metabolic rate (BMR) and basal metabolic rate per unit mass of a 136,000 kg (150. ton) great blue whale. Answer: BMR = 2,080,000 kJ/d and (BMR/m) = 15.3 kJ/kg·d. whale whale HOW DOES TEMPERATURE AFFECT METABOLISM? Temperature governs metabolism through its effects on rates of biochemical reactions. Reaction kinetics vary with tempera- −E/kT ture according to Boltzmann’s factor e , where T is the absolute temperature, E is the activation energy, and k is Boltz- mann’s constant. The combined effects of body mass (M) and body temperature (T) on the basal metabolic rate can then 3/4 −E/kT e , where E is the average activation energy for the enzyme-catalyzed biochemical reactions of be written as BMR∝M metabolism. 17.6 THERMODYNAMICS OF NUTRITION AND EXERCISE The molecular form of the food we eat can be broken down into the following three categories: 1. Carbohydrates. Carbohydrates always contain hydrogen and oxygen atoms in a 2 to 1 ratio, as in water; and they can have very large macromolecules built up from the glucose (C H O ) monomer, with molecular 6 12 6 6 masses as high as 2 × 10 (as in the case of plant starch and glycogen). 2. Proteins. Proteins are very large molecules containing carbon, hydrogen, oxygen, and often nitrogen. For example, a single molecule of human hemoglobin (C H O N S Fe ) contains a total of 9512 atoms 3032 4816 872 780 8 4 and has a molecular mass of 66,552 kg/kgmole. 3. Fats (glycerol and fatty acids). Fatty acids are much smaller molecules, with typically 16 or 18 carbon atoms per molecule plus attached hydrogen atoms and a carboxyl group (—COOH) at one end. An example of a saturated (with hydrogen atoms) fatty acid is shown in Figure 17.8. An example of the same acid unsaturated is shown in Figure 17.9.706 CHAPTER 17: Thermodynamics of Biological Systems H H H H H H H H H C C C C COOH HC C C C COOH H H H H H FIGURE 17.8 FIGURE 17.9 A saturated fatty acid hydrocarbon chain. An unsaturated fatty acid hydrocarbon chain. Mixer 110 V motor Thermometer Fuse wire Test sample Bomb (closed rigid vessel) Insulation Water FIGURE 17.10 A schematic of a typical bomb calorimeter. The energy value of different foods is normally determined by direct calorimetry in a device called a bomb calori- meter (see Figure 17.10). In this device, a sample of known mass is ignited in a pressurized atmosphere of excess pure oxygen. The liberated heat of combustion is transferred to water surrounding the combustion chamber, and it can easily be calculated from an energy balance on the calorimeter. The end product of this type of combus- tion is always CO and H O (and nitrogen products when the sample contains bound nitrogen). Since this is 2 2 exactly the same end state that occurs in the body as a result of enzyme decomposition of food molecules, the same amount of energy must be released in each case. Thus, bomb calorimeter energy measurements represent the total energy available in the sample that can be converted into heat or another form of energy. Bomb calorimeter studies on dry (water-free) foods give the following averaged results for the specific energies of the basic food components: ) Carbohydrate: 18:0MJ/kg Protein: 22:2MJ/kg Total energy contentðwater freeÞ Fat: 39:8MJ/kg When these same substances are metabolized in the human body they produce the following specific energy releases: ) Carbohydrate: 17:2MJ/kg Protein: 17:2MJ/kg Metabolizable energy contentðwater freeÞ Fat: 38:9MJ/kg Using these two sets of values we can compute the food energy conversion efficiency of the human body as 17:2 η = ×100 = 95:5% carbohydrate 18:0 17:2 η = ×100 = 77:5% protein 22:2 38:9 η = ×100 = 97:7% fat 39:8 Thus, 22.5% of the energy in the protein we eat passes through the body unused. The low protein energy con- version efficiency supports the theory that humans were not always meat-eating animals.17.6 Thermodynamics of Nutrition and Exercise 707 These energy content values were for dry or water-free foods. Most foods, especially carbohydrates, contain a large amount of functional water (the mass of the human body, for example, is about 72% water). The energy content of natural or wet food is lower that that of dry food due to the dilution effect of the energetically inert water. The average natural state metabolizable specific energy content of the three basic food components is ) Carbohydrate: 4:20MJ/kg Protein: 8:40MJ/kg Metabolizable energy content natural state foods Fat: 33:1MJ/kg Note the extraordinarily large specific energy content of natural state fat. EXAMPLE 17.4 If the average meal consumed by an adult consists of about 45.0% carbohydrate, 15.0% protein, and 40.0% fat, determine (a) the specific energy content of an average meal with natural state foods and (b) the total mass of an average meal needed to provide 10.5 MJ per day. Solution a. The average energy content of natural (or wet) food components is Carbohydrate: 4:20MJ/kg Protein: 8:40MJ/kg Fat: 33:1MJ/kg Therefore, the specific energy content of the average meal is e = 0:450ðÞ 4:20 +0:150ðÞ 8:40 +0:400ðÞ 33:1 = 16:4MJ/kgmeal avg:meal b. A person who requires a daily food energy intake of 10.5 MJ must then consume 10:5 m _ = = 0:640kg of average meal/day = 1:4lbm of average meal/day avg:meal 16:4 Exercises 10. Suppose the person in Example 17.4 who needed 10.5 MJ of food energy per day goes on a diet that requires a food energy intake of only 5.0 MJ per day. What total mass of average meal food should this person consume per day? _ Answer: m = 0:300kgavg:meal/d = 0:670: avg:meal 11. If the amount of fat described in the average meal in Example 17.4 is reduced from 40.0% to 30.0% and the carbohydrate and protein increase to 50.0% and 20.0%, respectively, determine the new specific energy content of this meal and the total mass of this meal required to produce 10.5 MJ of food energy per day. Answer: _ m = 0:770kgavg:meal/d = 1:69: avg:meal 12. People who live in very cold climates usually have diets that have a very high fat content. Suppose their average meal consisted of 20% carbohydrate, 20% protein, and 60% fat. Determine the specific energy content of this meal and the m _ = total mass of this meal required to produce 10.5 MJ of food energy per day. Answer: avg:meal 0:470kgavg:meal/d = 1:03lbmavg:meal/d: Overweight conditions place a greatly increased load on the heart and other organs. For example, each kilogram of body tissue contains 0.885 km of tiny blood vessels. If an individual is 10. kg (22 lbm) overweight, the heart must pump blood through an extra 8.9 km (5.5 miles) of small blood vessels. EXAMPLE 17.5 People living in affluent societies generally know very little about starvation. Most feel that death is imminent if food is withheld for only a week. However, we know that 1.00 kg of human body fat contains about 33.1 MJ of metabolizable energy, and it would be useful to know: a. The mass of body fat consumed per day if one uses 10.5 MJ of energy in normal activities. b. How many days of total fasting are required to lose 10.0 kg of body fat. Solution a. A fasting person requiring 10.5 MJ of metabolizable energy per day consumes about 10:5MJ=d _ m = ¼ 0:317kgof bodyfat=d fat 33:1MJ=kgbodyfat (Continued)708 CHAPTER 17: Thermodynamics of Biological Systems EXAMPLE 17.5 (Continued) b. The number of fasting days required to lose (consume) 10.0 kg of body fat is 10:0kgof bodyfat m fat t = = ¼ 31:3d m _ 0:317kgof bodyfat=d fat Exercises 13. If the person in Example 17.5 requires only 8.40 MJ per day instead of 10.5 MJ per day, how much fat would be lost (consumed) by fasting just one day? Answer: m = 0.250 kg fat. fat 14. You are stranded in the wilderness without any food (but you have plenty of water to drink). If you have 15.0 kg of excess body fat, how long can you wait to be rescued before your body fat is consumed by your metabolism of 9.00 MJ/d? Answer: t = 55.2 d. 15. Suppose you have absolutely no excess body fat, and you are stranded without food. Your body then begins to consume your muscle protein, which has a metabolizable value of 8.4 MJ/kg. How much protein would you lose (consume) after being stranded for 30.0 days with a metabolism of 10.5 MJ/da? Answer:m = 12.5 kg protein. protein Example 17.5 shows that, if you have 10. kg (220 lbm) of excess body fat, you can theoretically fast for 31.3 days just living on that body fat alone. This also gives you some idea why weight loss by dieting is such a slow process. Fasting for long periods is not a medically sound method of weight loss since the body soon begins to consume its own protein, and this can seriously affect the functioning of the body’sorgansystems (especially the heart). No one should ever willingly attempt a total fasting diet without consulting a qualified physician. Whereas most adult humans can survive long periods of fasting, they cannot withstand long periods without water intake. Since the body continually loses water through the skin and lungs, it must be replaced or the body soon becomes dehydrated and death quickly follows. Healthy adults have been known to fast for over 100 days, but no human can survive for more than 10 to 20 days without water. Tables 17.4 and 17.5 present the metabolizable energy content values for various common foods and the aver- age energy expenditure requirements for various human exercises. Common nutritional tables today have food energy content and exercise energy expenditure levels listed in Calories. The capitalization of the word Calorie indicates what nutritionists call a large calorie, that is, a kilocalorie: 1 Calorie = 1000 calories = 1 kilocalorie. This is confusing notation, since only the capital C tells you that it is not the normal calorie energy unit, a subtle point often overlooked by the publishers of nutrition tables. Table 17.4 Approximate Energy Content of Some Common Foods Metabolizable Energy Content Food Calories MJ Btu Fast foods (average values) Hamburger 275 1.15 1090 Cheeseburger 325 1.36 1490 Quarter pound hamburger 450 1.88 1790 With cheese 550 2.30 2180 With cheese and bacon 650 2.72 2580 Fish sandwich 450 1.88 1790 With cheese 500 2.09 1980 Hot dog 300 1.26 1190 With chili or cheese 350 1.47 1390 Regular fries 250 1.05 992 Regular onion rings 350 1.47 1390 Baked potato 250 1.05 992 With sour cream and chives 450 1.88 1790 With chili and cheese 500 2.09 1980 With broccoli and cheese 500 2.09 1980 With bacon and cheese 550 2.30 2180 With cheese 550 2.30 218017.6 Thermodynamics of Nutrition and Exercise 709 Table 17.4 Approximate Energy Content of Some Common Foods continued Metabolizable Energy Content Food Calories MJ Btu Pizza (per slice, 8 slices per 13-in. pizza) With cheese 350 1.47 1390 With cheese and pepperoni 500 2.09 1980 Salads (1 cup each) Lettuce with French dressing 150 0.63 595 Potato with mayonnaise 375 1.57 1490 Chicken and mayonnaise 550 2.30 2180 Egg and mayonnaise 400 1.67 1590 Tuna fish and mayonnaise 500 2.09 1980 Drinks Shakes (all flavors, 10 fluid ounces) 350 1.47 1390 Milk (skim, per pint) 180 0.75 714 Cola (all flavors, 10 fluid ounces) 130 0.54 516 Diet cola 0 0.0 0 Beer (per fluid ounce) 8 0.033 30 Whiskey (per fluid ounce) 38 0.16 150 Desserts Ice cream (per pint, 10% fat) 600 2.51 2380 Pie (per slice, 8 slices per 9-in. pie) 300 1.26 1190 Chocolate candy (milk, per ounce) 150 0.63 595 Marshmallows (1 large) 25 0.10 99 Source: Jacobsen, M., Fritschner, S. The Fast-Food Guide. Workman Publishing, 1992, New York. Copyright © The Center for Science in the Public Interest. Reprinted by permission of Workman Publishing. Table 17.5 Approximate Adult Human Energy Expenditure in Exercise Energy Required during Exercise Exercise Calories/h MJ/h Btu/h Fast running 910 3.8 3610 Cross-country skiing 910 3.8 3610 Fast swimming 860 3.6 3410 Wrestling 810 3.4 3210 Boxing 690 2.9 2740 Hard cycling 600 2.5 2380 Jogging 600 2.5 2380 Football 600 2.5 2380 Fast dancing 600 2.5 2380 Basketball 550 2.3 2180 Handball 550 2.3 2180 Sawing wood 500 2.1 1980 Shoveling 500 2.1 1980 Tennis 480 2.0 1900 Climbing stairs normally 410 1.7 1630 Baseball 360 1.5 1430 Volleyball 360 1.5 1430 Fast walking 310 1.3 1230 Sexual intercourse 270 1.1 1070 Golf 240 1.0 952 Hoeing 190 0.8 754 Driving a car 140 0.6 556 Card playing 96 0.4 381 Watching TV 72 0.3 286 Basal metabolism 72 0.3 286710 CHAPTER 17: Thermodynamics of Biological Systems IS IT A Calorie OR A calorie? When the word calorie is capitalized, it indicates what nutritionalists call a large calorie, or a kilocalorie. That is, in nutrition jargon, 1 Calorie = 1 kilocalorie, yet it takes 1000 calories to equal 1 kilocalorie. Only the capital C distinguishes the two. So, when a nutrition table indicates that your caloric intake should be 2500 Calories per day, it really means 2500 kilocal- ories per day. Now, since 1 kilocalorie = 4.186 kilojoules, then 2500 Calories/d = 2500 kcal/d = (2500 kcal/d)(4.186 kJ/ kcal) = 10,465 kJ/d = 10.456 MJ/d. From Table 17.5, we find that the energy expenditure required to jog, play football, or fast dance is about 600 Calories per hour, and Table 17.4 tells us that the energy content of milk chocolate is about 150 Calories per ounce. So, if you want to exercise off the energy content of one 1.5-oz milk chocolate candy bar you would have to jog, play football, or fast dance continuously for ð1:5ozÞð150Calories/ozÞ = 0:375hours 600Calories/h This is a lot of hard exercise for one small candy bar. EXAMPLE 17.6 Suppose you want to exercise off the energy added to your body as a result of eating one pint of ice cream by lifting weights. The external work done by the body equals the change in potential energy of the weights as they are lifted (there is no significant energy recovery within the body, however, when the weights are lowered again). Suppose you are lifting 490. N (110. lbm) a vertical distance of 1.00 m, and you can make one lift in 1.00 second. Approximately how many lifts are required and how long will it take to work off the energy content of the ice cream? Solution Each lift requires that an amount of energy be put into the weights of . ðÞ mgZ/g =ðÞ 490:NðÞ 1:00m/1ðÞ = 490:N m = 490:J c weights If we take the human body as the thermodynamic system and apply the energy rate balance and ignore all mass flow energy movements into or out of the system during the exercise period (thus, we are ignoring perspiration energy losses and all O 2 and CO exchanges), then we can write 2    2 mgZ dU d mV d _ _ Q−W = + + dt dt 2g dt g c c body Since the kinetic and potential energies of the human body do not change significantly during the exercise, we can set    2 mgZ d mV d + = 0 dt 2g dt g c c body and the energy rate balance becomes  dU _ _ _ Q−W = = U body dt body Now, the external work rate that must be done by the system is ðÞ mgZ/g c weights 490:J _ W = − = = 490:J=s Δt 1:00s It has been shown experimentally that the energy conversion efficiency of animal muscular contraction defined by Eq. (17.12) is about 25%, or _ W ðÞ η = = 0:250 T muscle _ U body17.7 Limits to Biological Growth 711 Then, the rate of total internal energy expenditure within the body is _ −490J/s −W _ U = = = −1960J/s body ðÞ η 0:250 T muscle _ _ _ Therefore, Q = U +W = −1960+490 = −1470J/s: Consequently, the time, τ, required to produce a change in the total internal energy of the system that equals the energy content of one pint of ice cream (see Table 17.4) is  −ð1pintÞð2:51MJ/pintÞ ΔU τ = = = 1280s = 21:3min −3 _ U −1:96×10 MJ/s body Hence, the 490. N weight in this example must be lifted continuously at a rate of one lift per second until a total of 1280 lifts have been made. This is clearly a great deal of physical labor just to overcome the enjoyment of a pint of ice cream. Note that only 25% of the energy in the ice cream gets converted into external work while 75% of its energy is utilized else- where within the body to keep the circulatory, respiratory, and other subsystems operating and is ultimately converted into heat inside the body due to the internal irreversibilities of these processes. Exercises 16. If the muscle efficiency in Example 17.6 is 30.0% instead of 25.0%, how many lifts would be required to work off the energy content of the ice cream? Assume all the other variables remain unchanged. Answer: 1540 lifts. 17. If the 490 N weight in Example 17.6 is lifted only 0.50 m instead of 1.00 m, how long would it take to work off the energy content of the ice cream? Assume all the other variables remain unchanged. Answer:τ = 2560 s = 42.7 min. 18. Suppose the person in Example 17.6 consumes a cheeseburger instead of a pint of ice cream. Assuming all the other variables remain unchanged, how long would it take to work of the energy content of the cheeseburger? Answer:τ = 694 s = 11.6 min. Physiologically, it is very hard to lose weight by exercising alone. Most of the weight loss that appears after exercising is really water loss due to perspiration. Perspiration is a convection-evaporation heat transfer mechan- ism that removes the heat generated within the body due to the biological irreversibilities of exercise. Its func- tion is to help maintain a constant body temperature. This type of water loss is quickly replaced in the meals following the exercise and should never be considered as part of a permanent weight loss. 17.7 LIMITS TO BIOLOGICAL GROWTH For purposes of simplification, consider living systems to have a characteristic length L such that their surface 2 3 and cross-sectional areas are proportional to L and their volumes are proportional to L . The most obvious effect of size on animal evolution is the ability of an animal’s skeleton to support its body weight. The ability of a leg bone to withstand direct compression loading is proportional to its yield modulus and to the cross- 2 sectional area of the bone. Hence, the strength of a leg varies with L . However, the body weight of the ani- 3 malisproportionaltoitsvolume,whichvarieswith L .Theratioofbodyweighttolegloadingthen increases with the animal’s size, L. Clearly there exists an upper limit (dictated by the elastic properties of bone) to an animal’s growth, where its legs can no longer support its weight. The giant dinosaurs of 100 mil- lion years ago apparently evolved up to this critical size. Some aquatic dinosaurs were too large to leave the water because without the buoyant supporting force of the water their skeletons could not support their body weight. Even more crucial to mobile land animals are the bending stresses developed in their bones during walking and running. Small animals can run with very nimble and flexible legs while heavy animals like elephants must walk stiff legged to minimize leg bone bending stresses. The internal heat generated by biochemical irreversibilities in animals is proportional to the amount of tissue 3 present, and consequently, it varies with L . The rate of heat loss by an animal depends on the convective and radiative heat transfer mechanisms, whichinturndependdirectlyontheanimal’s surface area and, conse- 2 3 2 quently, vary with L . Therefore, the ratio of heat generation to heat loss is proportional to L /L = L, and if an animal’s size were to increase indefinitely, a point would eventually be reached where the animal would over- heat and die. Thus, at least two mechanisms provide an upper limit to the size of animals: the strength of their supporting tissue and their ability to maintain a moderate body temperature. The rate at which oxygen and food reach the body’s cells depends on the volume of blood in the circulatory system and the pumping capacity of the heart. The volume of blood delivered to the heart is proportional 2 to the cross-sectional area of the aorta and, consequently, varies with L , whereas the volume of the heart712 CHAPTER 17: Thermodynamics of Biological Systems WHAT ABOUT BIRDS? The pulse to breathing rate ratio is about 9.0 for all birds (regardless of their size) because birds have a continuous flow of air in only one direction through their lungs, in contrast to the two-way in-out breathing of mammals. The unidirectional air flow in birds is also countercurrent (in the opposite direction to) the blood flow in the lungs, thus improving the effi- ciency of gas exchange. 3 2 3 −1 itself is proportional to L . Therefore, the ratio of blood flow rate to heart volume varies with L /L = L ,and –1 consequently, the heart pulse rate also varies with L . For mammals, the heart pulse rate has been correlated with body mass according to 241 −0:25 Heart pulse rate ðin beats per minuteÞ = 241ðm Þ = (17.18) 1/4 m where the body mass m is in kilograms. The same argument can be made for the respiratory system. The ratio of the gas transport rate through the lung –1 –1 wall to the lung volume also varies with L , and the breathing rate is also proportional to L . Experimentally, we find that the ratio of pulse rate to breathing rate is constant at about 4.5 in all mammals regardless of their size. The breathing rate for mammals has been correlated with body mass as 54:0 −0:25 BreathingrateðinbreathsperminuteÞ = 54:0ðm Þ = (17.19) 1/4 m where the body mass m is in kilograms. EXAMPLE 17.7 As a famous biomedical engineer, you are challenged on your medical exam to determine the heart rate and respiratory rate of a 0.0300 kg mouse, a 70.0 kg human, and a 4000. kg elephant. Solution The heartbeat rate for mammals in beats per minute is given by Eq. (17.18) as −0:25 Heartbeat rate = 241ðm ÞBeats/min so the heartbeat rates of the mouse, human, and elephant are −0:25 ðÞ Heartbeat rate = 241ð0:0300 Þ = 579:Beats/min mouse −0:25 ðÞ Heartbeat rate = 241ð70:0 Þ = 83:3Beats/min human −0:25 ðÞ Heartbeat rate = 241ð4000: Þ = 30:3Beats/min elephant The respiratory rate of mammals in breaths per minute is given by Eq. (17.19) as −0:25 RespiratoryðbreathingÞrate = 54:0ðm ÞBreaths/min So the breathing rates of the mouse, human, and elephant are −0:25 ðÞ Breathing rate = 54:0ð0:0300 Þ = 130:Breaths/min mouse −0:25 ðÞ Breathing rate = 54:0ð70:0 Þ = 18:7Breaths/min human −0:25 ðÞ Breathing rate = 54:0ð4000: Þ = 6:79Breaths/min elephant Exercises 19. Determine the heatbeat rate of a 2.80 kg house cat. Answer: 186 beats/min. 20. Determine the breathing rate of a 700. kg racehorse. Answer: 10.5 breaths/min. 21. Since whales are mammals, determinethe heartrateofa 136,000 kg(150. ton)bluewhale.Answer: 12.6 beats/min. Because plants lack mobility, their size criteria are generally simpler than those for animals. The main strength concerns in plants center on the buckling of their central trunk and excessive deflections of their cantilevered limbs. Consider a circular cylinder of height h and diameter d. Then, for slender cylinders ðh/d25Þ the critical height for a cylinder buckling under its own weight can be shown to be  1/3 E 2/3 h = 0:85 d (17.20) critical γ

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