1

Antoine Lavoisier and Pierre Simon de Laplace used this calorimeter to make the firstdirect measurements of animal heat, and relate it to respiration. A mouse or guinea pigwas placed in the inner basket, which was surrounded by ice, and the quantity of heatwas determined by measuring the weight of ice melted. From Lavoisier, TraitéÉlémentaire de Chemie (1789).

AIMS

1. To determine your metabolic rate by measuring the carbon dioxide content of yourexhaled breath, your breath volume, and your breathing rate.

2. To determine the food calories in corn oil by burning it in a calorimeter, and to relatethe result to your metabolic rate.

INTRODUCTION

Professor Germain Hess never knew the pleasures of french fries and hamburgers.After all, fast food drive-ins were hard to find near the University of St. Petersburg in1840. But today's consumers of those oleaginous snacks, and the gurus who write thediet books, owe Hess a great debt: his law of heat summation eventually gave other

CALORIES: HAMBURGERS,FRENCH FRIES,

AND HESS'S LAW

2

people a way to measure the Calories that tag along with food. Now, thanks to Hess,numbers can be attached to our worries about overeating. To trace the connectionbetween St. Petersburg in 1840 and Burger king in 2000, we should start with a briefreminder of Hess's law.

Hess's Law

Germain Hess deduced his law of heat summation during three years ofcalorimetric measurements, from 1839 to 1842. In his words, "the heat developed in achemical change is constant, whether the change occurs directly, or indirectly in severalstages." That is, the total heat involved in a chemical change depends solely upon thebeginning reactants and the ending products, no matter how many reactions are involvedin getting from the beginning to the end. The enthalpy change is independent of the pathtaken from reactants to products. It follows that the enthalpy changes for chemicalreactions must be additive, just as the chemical reactions themselves are additive.

For instance, charcoal (carbon) burns in oxygen in a one-step reaction whichproduces carbon dioxide and releases heat:

2C(s) + 2O2(g) → 2CO2(g) ∆ H1 = -787 kJ (1)

However, carbon can also react with oxygen in two steps, with carbon monoxide as anintermediate compound:

2C(s) + O2(g) → 2CO(g) ∆ H2 = -221 kJ (2)

2CO(g) + O2(g) → 2CO2(g) ∆ H3 = -566 kJ (3)

Reaction (1) is the sum of reactions (2) and (3):

2C(s) + O2(g) → 2CO(g) (2)2CO(g) + 2CO2(g) (3)

______________________ 2C(s) + 2O2(g) → 2CO2(g)

and the heat evolved in reaction 1 is the sum of the heats of reactions (2) and (3):

∆ H1 = ∆ H2 + ∆ H3 = -221 kJ -566 kJ = -787 kJ

Whether CO2 is formed in one step, two steps, or dozens of steps, the heat evolved ingoing from carbon and oxygen to CO2 is the same. We need to know only that the initialreactants are carbon and oxygen, and the final product is carbon dioxide.

3

Conservation of energy

Although Hess did not realize it at the time, his law is a special case of the law ofconservation of energy, otherwise known as the first law of thermodynamics. Theoversight is excusable, for a clear general statement of the conversation of energy wasalmost a decade away. According to the first law, there can be no net production ordestruction of energy when it is converted from one form to another, say from heat tomechanical work, or from chemical energy to electrical work. Energy, though it can existin many forms, is indestructible. Energy cannot be created out of nothing, nor can itvanish into nothingness.

It is easy to understand why the first law of thermodynamics was a longer timecoming than the other conservation law which is so important to chemists, theconservation of mass. Weight is far less obscure than energy, and to determine that themasses of products and reactants are identical they simply need to be weighed. Butenergy is an extreme abstraction, and for much of the 19th century the notions of force,momentum, and kinetic energy were hopelessly entangled. Arriving at a conservationlaw is not easy if you cannot define what is being conserved.

The names Mayer, Joule, and Helmholtz are inseparable from the first law ofthermodynamics. In 1842, Julius Mayer published the first attempt to clarify theconfusion surrounding the various forces. His paper was a statement of the conservationof energy, but it was too early and written too obscurely. In addition, there was littleexperimental evidence for his ideas, and he provided none himself. The paper generatedso little interest that the editor of the scientific journal asked him not to send any more!Poor Mayer became depressed, dropped from sight, tried to kill himself, spent some timein a mental institution, and died in obscurity. Acceptance of the first law ofthermodynamics was easier later, after more experimental results had accumulated.

Needed evidence was provided, beginning in the decade of the 1840's, by JamesPrescott Joule, a wealthy brewery owner, superb experimentalist, and eventuallypresident of the British Association for the Advancement of Science. He demonstratedhow mechanical, electrical, and chemical energy are interrelated, but his best-knownexperiments were focused upon measuring the mechanical equivalent of heat (the amountof heat produced by a given amount of work) by measuring the temperature rise producedin stirred liquids. His first papers fell on deaf ears - his contemporaries did notunderstand what he was doing - but he persisted until his work was recognizeduniversally. Finally, it was Herman von Helmholtz1 in his epochal memoir The 1 No scientist contributed more widely than Helmholtz. Trained as a physician, he began as a surgeon inthe Prussian army, then became professor of physiology in Königsberg, then professor of physiology inHeidelberg, then professor of physics in Berlin. The first law of thermodynamics was far from his onlyscientific achievement. In mathematics and physics, he made major contributions to non-Euclideangeometry, electrodynamics, optics, and acoustics. In physiology, he was first to measure the speed of nerveimpulses and the heat produced by muscles, explained the function of the cochlea in hearing, and showedhow eyes accommodate. He explained the role of overtones in music, invented the opthalmoscope,established the theory of three primary colors, wrote the landmark book on the theory of vision, and on andon..!

4

Conservation of Force (1847), who set forth the first general account of the conservationof energy. Even then, Helmholtz found it difficult to get the paper published in ascientific journal, and eventually printed it in pamphlet form.

Perpetuum mobile

Another statement of the first law could be a real time-saver for the crackpotfringe: A machine which works by producing more energy than it consumes is impossible.Such a fantastic device, which is called a perpetual motion machine of the first kind,offers the ultimate something for nothing: the production of unlimited amounts of powerfrom sheer emptiness. The earliest-known reference to perpetual motion appeared in thetwelfth century (1159), and medieval Europeans, with their interest in mechanisms,attempted to design machines to achieve it. A good many descriptions have survived;two well-known ones, by Pierre de Maricourt and Villard de Honnecourt, are shown inFigure 1. Pierre de Maricourt was one of the few experimentalists of his time, and a verygood one, too. His great treatise on magnetism, Epistolae de Magnete (1269), includedthe diagram shown here of a magnetic perpetuum mobile machine, which may or may nothave been built. Villard, who was an architect, is famous mostly because his sketchbookhas survived nearly intact. The architectural and mechanical devices that appear thereinclude a perpetual motion wheel (circa 1240) which may have been the first of its type inEurope. The quest for perpetual motion continues even today, 150 years after Helmholtzlaid the idea to rest, but it is now the province of ignoramuses and cranks.

Figure 1. Two medieval perpetual motion machines. On the left is Pierre de Maricourt'smagnetic device of 1269. Nobody has figured out how it was supposed to work. Theright one is Villard de Honnecourt's machine of 1240. Unbalanced hammers aresupposed to supply the motive force. From J. Gimpel, The Medieval Machine.

5

If Germain Hess had been wrong, and enthalpies were not additive, perpetualmotion devices would be on sale at Walmart every day. For instance, reconsiderreactions (1), (2), and (3), and imagine the ∆ H2 and ∆ H3 are unchanged (∆ H2 = -221 kJand ∆ H3 = -566 kJ), but, ∆ H1 is -887 kJ instead of -787 kJ:

2C(s) + 2O2(g) → 2CO2(g) ∆ H1= -887 kJ (1)

2C(s) + O2(g) → 2CO(g) ∆ H2 = -221 kJ (2)

2CO(g) + O2(g) → 2CO2(g) ∆ H3 = -566 kJ (3)

Hess's law is now violated, since ∆ H1 ≠ ∆ H2 + ∆ H3. A perpetual motion cycle can beimagined, which starts with CO2(g), and reverses reactions (2) and (3), which thereforebecome endothermic (∆ H > O). By heating the CO2, carbon and oxygen could beobtained in two steps:

2CO2(g) → 2CO(g) + O2(g) ∆ H4 = +566 kJ (4)

2CO(g) → 2C(s) + O2(g) ∆ H2 = +221 kJ (5) ________________________

2CO2(g) → 2C(s) + 2O2(g) (6)

The heat absorbed in this process is ∆ H6 = ∆ H4 + ∆ H5 = +787 kJ. Next, burn thecarbon in a single step to produce CO2 again and release heat:

2C(s) + 2O2(g) → 2CO2(g) ∆ H1 = -887 kJ

The net result is a cycle, beginning and ending with CO2, which produces 100 kJ moreheat than it consumes each time the CO2 is decomposed and remade:

CO2(g) → CO2(g) ∆ H = ∆ H1 + ∆ H2 + ∆ H3 = -100 kJ

Energy is produced from nowhere. Furthermore, the same result is obtained no matterwhat scheme is proposed: to violate Hess' law is to violate the first law ofthermodynamics.

Lest you feel that Hess' law is perfectly obvious, examine Figure 2 and thinkagain. The scheme shown there is simply another chemical perpetual motion machine,where the inventor proposes to decompose water into hydrogen and oxygen by heating it,then reverse the decomposition by reacting the two gases to make water and release heat.Somehow, enough gas would be left over to be "conducted by branch pipes to anyapartment where light is wanted, and will readily produce a beautiful intense light"2.

2 Wrong. Besides violating Hess' law, it would produce an intense explosion. Hydrogen and oxygen makeone of the most explosive mixtures known. Lighting it would blow the furnace, apartment, and lamplighterto atoms.

6

Notice the gullible editor's enthusiasm and the date 1848, only one year after Helmholtz'smemoir on the first law. We know better now, and would write

2H2O(g) → O2(g) + 2H2(g) ∆ H1 = 483.6 kJ/mol

O2(g) + 2H2(g) → 2H2O(g) ∆ H2 = -483.6 kJ/mol _______________________

2H2O(g) → 2H2O(g) ∆ H = ∆ H1 + ∆ H2 = O

Figure 2. A perpetual motion device proposed at the time of Helmholtz's publication ofthe conservation of energy. When you read it closely, you will see that the proposalviolates Hess' law, and hence violates the first law of thermodynamics. This bootstrapcontraption nicely illustrates the confusion abroad at the time. The title of the journalconjures up a different America, where the rural former - mechanic tradition stilldominated.

Hess's law and energy metabolism

The first law of thermodynamics is obeyed, of course, by both animate andinanimate nature. Otherwise, people would be perpetual motion machines, which they

7

are not. For instance, carbohydrates are oxidized to CO2 and H2O whether they areburned in a single step in a fire, or metabolized within a creature by a multitude ofenzymatic reactions. Since the initial reactants and the final products are the same byeither path, Hess's law insists that the heat released is precisely the same by either path.For instance, one gram of glucose (blood sugar, C6 H12O6) releases 3.73 cal/g (∆ Ho =-2815 kJ/mol) when combusted to CO2 and H2O, but the heat does no useful work. Itmerely increases the temperature of the room. When the gram of sugar is metabolized, itis oxidized slowly in many intermediate steps, and some of the energy released by theoxidation is diverted to power biochemical reactions and muscles. However, eventuallyall of it is converted to CO2 and H2O, and all 3.73 cal of heat appears in the environment.The upshot is that the dietary calories in a sugar or other digestible carbohydrate can befound by combusting it in a calorimeter.

Fats, like carbohydrates, are also oxidized completely to CO2 and H2O, whetherthey are combusted or metabolized. Again, the initial reactants and the final products arethe same by both paths, and, according to Hess's law, the enthalpies are also the same.For a typical fat (C55H106O6), the heat of combustion is about 9.5 cal/g.

Combustion is not a good estimate of the heat released by metabolized proteins,because the two different routes produce different end products. Proteins contain about16 percent nitrogen, which is excreted as urea by mammals and as uric acid by birds andreptiles. The energy biologically available from a protein must be determined by feedingit to an animal and measuring the heat that the animal produces. For most proteins, this isabout 4.2 cal per gram metabolized.

The calories in food

The calories listed on your cereal box or potato chip bag measure the heat thatyour body would produce by metabolizing the food. A dietary calorie is simply a unit ofenergy. One Calorie3 equals 4.18 kilojoules, which is the heat required to raise thetemperature of 1 kg of water from 15.0oC to 16.0oC. This is a formidable amount ofenergy. A modest food energy intake of 2000 Cal/day produces enough body heat dailyto warm 7.5 gallons of water (about 60 pounds) from body temperature to the boilingpoint, and all of it must be dissipated!

A calorie is not, as many people apparently suppose, a mysterious invisible obeseparticle designed to make you fat. The problem of weight gain is simply a matter of thefirst law of thermodynamics: if the enthalpy of combustion of the food that you eat isgreater than the heat that you release in daily activity, the excess energy cannot vanish. Itis conserved in the most efficient way, by synthesizing fat, which has a large ∆ H ofcombustion. The solution is inescapable: lower the enthalpy of combustion of the food(eat less) and increase the dissipation of energy (exercise more).

3 The dietary calorie is sometimes written Calorie, to distinguish it from the smaller chemist's calorie: 1 Cal(dietary) = 1 kcal (chemical). In this chapter, a calorie is always defined as 4.18 kJ.

8

Some carbohydrates, such as starch and most sugars, are completely digested,absorbed from the gut into the bloodstream, and metabolized. For these, the heat releasedby metabolic reactions is the same as the heat of combustion of the food eaten. Fats, too,are almost completely (~95%) digested, absorbed, and metabolized. But some foodscontain substances, such as fiber (mostly cellulose), which furnish no energy at all to thebody because they are indigestible. When burned, cellulose releases even more heat pergram than sugar does, but when eaten by people it produces no heat at all because thehuman digestive system lacks the enzymes needed to break it down to glucose so that itcan be absorbed4.

Today we know which foods are or are not digested, absorbed from the gut, andmetabolized, but that information was obtained by measuring the heat produced bypeople and other animals. To do that, a person, cow, or other creature lives for days in acalorimeter having a specially designed chamber lined with a water jacket. Heatproduction is determined by measuring the temperature increase in the water, and arecord is kept of O2 consumption, CO2 production, food and water intake, waterevaporation, and urine and feces production. The enthalpy of combustion is measured forsamples of the food, feces, and even the hair which grows during the time ofconfinement, and a detailed energy balance sheet is written. The method is called directcalorimetry. When a human being in a calorimeter is given a gram of glucose, the heatproduced is the same as the heat of combustion (3.73 cal/g), but when cellulose is givenno heat is produced. If only a fraction of the food is digested and metabolized, its caloricvalue is known from the heat evolved. The heroic efforts expended on direct calorimetrypeaked in the late 19th and early 20th centuries5, but today's tables of nutritionalinformation trace directly back to those earlier experiments. The commonly-acceptedaverage caloric contents of nutrients are listed in Table 1.

Table 1. The energy content of nutrients. The numbers are average values obtained bydirect calorimetry, and take into account digestibility and absorption of the nutrients

Nutrient Cal/gCarbohydrate 4.0Fat 9.0Protein 4.0

Although these numbers originated from direct calorimetry many years ago, thatmethod is no longer needed to determine the calories in food. To obtain the caloriccontent of Banana Creamies (Figure 3), the Tasty Baking Company does not combust theCreamies, or seal employees in a calorimeter chamber while force-feeding them with the

4 Cattle can live on grass because they process cellulose. They enzymes are provided by organisms livingin the rumen, which is essentially a fermentation vat.5 The Armsby calorimeter at the Pennsylvania State University is one of the most impressive examples. Itwas designed for farm animals. The copper-lined calorimeter chamber and auxiliary equipment occupy anentire building. The Armsby is no longer used, but it stands as a memorial to infinite patience andpainstaking toil.

9

company's products. Instead, those nutritious morsels are chemically analyzed forcarbohydrate, fat, and protein, and the results are multiplied by 4, 9, and 4 cal/g.

Figure 3. Nutritional label from a box of Tastykake Banana Creamies. Notice thestandards listed at the end: fat 9 cal/g, carbohydrate 4 cal/g, protein 4 cal/g. The totalcalories in a serving are 7(9) + 25(4) + 1(4) = 167 cal, which is rounded off to 170 cal.The calories from fat are 7(9) = 63, rounded off to 60.

Gas exchange and metabolic rate

The metabolic rate is the rate of energy consumption (heat production) by anorganism. Common units are cal/hr or cal/day. When we speak of a food which provides2000 cal, we mean that the food, if metabolized, would produce 2000 cal of heat. Thanksto the scientific pioneers who painstakingly used direct calorimetry to measure gasexchange and heat production in people, physiologists now have enough confidence tocalculate metabolic rates simply by measuring gas exchange during respiration.

Needed are estimates of the calories of heat released per liter of CO2 gasproduced. The numbers for carbohydrates and fats can be estimated from the combustionof glucose (a carbohydrate):

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆ Ho = -673 cal/mol (7)

and the combustion of a typical fat:

C55H106O6(s) + 78.502(g) → 55CO2(g) + 53H2O(l) ∆ Ho = -8177 cal/mol (8)

According to eq.(7), the combustion of 1 mole of glucose (180.2g) produces 6 moles ofcarbon dioxide. Since one mole of gas at standard temperature and pressure (STP, 0oCand 1 atm) occupies 22.41 L, the volume of the CO2 is (6 mol) (22.41 L/mol) = 134.5 L,

10

and the energy released is 673 cal/ 134.5 L = 5.00 cal/L CO2 produced at STP. Similarly,according to eq.(8), one mole of fat (863.4g) produces (55 mols) (22.41 L/mol) = 1233 Lof CO2 at STP, and the energy released is 8177 cal/1233 L = 6.63 cal/L CO2. The valuefor proteins, which was established by direct calorimetry of humans, is about 5.6 cal/L ofCO2 produced at STP. Commonly-accepted values are listed in Table 2.

Table 2. The energy content of nutrients, in terms of mass metabolized and volume ofCO2 produced at 0oC and 1 atm pressure. The numbers are average values obtained bydirect calorimetry.

Nutrient cal/g cal/LCO2 at STP g/LCO2 at STPCarbohydrate 4.0 5.0 1.25Fat 9.0 6.6 0.73Protein 4.0 5.6 1.40

This table suggests that a person's metabolic rate, and the mass of nutrientsmetabolized, can be calculated by measuring the CO2 content in exhaled air. Thecomplication is that all three nutrients are metabolized at the same time, and the cal/L aredifferent for each one. The metabolic rate cannot be calculated precisely unless theseparate contributions by carbohydrate, fat, and protein are known. These amounts canbe determined accurately, and an unequivocal metabolic rate can be calculated, bysimultaneously measuring O2 consumption, CO2 production, and urinary nitrogenexcretion. However, we will measure only CO2 production, and in order to calculate themetabolic rate we must assume the contribution that each of the three nutrients makes tothe overall metabolic energy. That is, we must guess the percentages of carbohydrate,fat, and protein being simultaneously metabolized. Given such assumptions, numberssuch as those in Table 3 can then be calculated from the data given in Tables 1 and 2.

Table 3. Energy produced by metabolizing carbohydrates, fats, and proteins in variousproportions. The volumes of CO2 are corrected to 0oC and 1 atm pressure (STP).

% carb % fat % protein cal/g cal/LCO2at STP60 30 10 5.5 5.750 30 20 5.5 5.850 40 10 6.0 5.940 40 20 6.0 5.930 50 20 6.5 6.1

To clarify the use of this table, suppose that a person produces 21.7L/hr of CO2 atSTP. If the nutrients being metabolized consist of 60% carbohydrate, 30% fat, and 10%protein, the metabolic rate is (21.7L/hr) (5.7cal/L) = 120cal/hr. The total weight ofnutrients consumed is (120cal/hr)/(5.5cal/g) = 22g, consisting of (0.60) (22g) = 13g/hrcarbohydrate, (0.30) (22g) = 6.6g/hr fat, and (0.10) (22g) = 2.2g/hr of protein. Note thatthe calories listed in Table 3 are not strongly dependent upon the proportions of threenutrients being metabolized; using an average value of 5.9cal/L introduces less than 4percent error.

11

OVERVIEW OF THE EXPERIMENTS

The work is divided into two parts. In Part I, you will determine your metabolicrate in cal/hr by measuring the partial pressure of carbon dioxide gas in your breath, thevolume of your breath, and your breathing rate. In Part II, you will determine thebiologically-available energy in corn oil by burning it in a calorimeter. By combining thedata from both parts, you will then be able to calculate the weight of corn oil that wouldbe needed to supply enough energy to maintain your metabolic rate.

PROCEDURE

PART I. MEASURING METABOLISM

We have already seen that metabolic rates can be determined roughly bymeasuring the rate of production of carbon dioxide gas during respiration. In this part,you will estimate the L/hour of CO2 (corrected to STP) that you produce during normalbreathing, and, by multiplying the volume by the conversion factor discussed earlier(Table 3) you will be able to calculate your metabolic rate in cal/hour.

Three separate measurements are needed: the partial pressure of CO2 in yourbreath, the rate of breathing in cycles/hour, and the volume of each breath.

Chemicals and special equipment

CO2 - measuring apparatus (Figure 4)Soda straws6M NaOH (sodium hydroxide) solutionDye (dropper bottle)Spirometer and mouthpieces (Figure 5)

A. Partial pressure of carbon dioxide

The pressure of a particular gas in a mixture with other gases acts independently ofthe other gases in the mixture. That is, each gas exerts its own pressure, as if the othergases were not even present. If this were not true, PV = nRT would not be true for allgases. The separate pressure contributed by each gas in a mixture is called a partialpressure. This independent behavior of gases is summarized by Dalton's law of partial

Safety Precaution6M NaOH is very caustic, and can burn youreyes and skin severely. If you contact it, wash

immediately with plenty of water.

12

pressures: the total pressure exerted by any mixture of gases is the sum of the partialpressures of the individual gases in the mixture. Think of the partial pressure of a gas ina mixture as the pressure that it would exert if all the other gases were removed withoutchanging the volume of the container. For instance, imagine a closed flask containingnitrogen at a partial pressure of 0.74 atm (562 mmHg) and carbon dioxide at a partialpressure of 0.39 atm (297 mmHg). The total pressure is the sum of the partial pressures:0.74 atm + 0.39 atm = 1.13 atm. If the CO2 were removed, the N2 remaining in the flaskwould exert a pressure of 0.74 atm, and if the N2 were removed, the CO2 would exert apressure of 0.39 atm. Similarly, if the CO2 were removed, the decrease in pressurecaused by its removal would be 0.39 atm, so that the pressure decrease equals the partialpressure of CO2 initially present. This principle is the basis of our measurement of thepartial pressure of CO2 in respired air.

A breath sample will be trapped in a closed flask, then the CO2 will be removed.The disappearance of the CO2 from the flask decreases the pressure inside, and bymeasuring the pressure decrease the initial partial pressure of CO2 is found. The CO2 isremoved from the respired air by absorbing it into a solution of sodium hydroxide:

CO2(g) + 2NaOH(aq) → Na2CO3(aq) + H2O(l) (10)

The sodium carbonate remains dissolved in the solution, thus sequestering all the CO2.

The pressure drop in the flask is measured with a manometer, which in this case isa small-bore glass capillary tube held vertically, its bottom dipping into a beaker of water(Figure 4). Its top connects to the flask. The partial vacuum created by absorbing theCO2 causes the water to rise in the tube, and the height of the rise measures the pressuredifference between the outside atmosphere and the inside of the flask. Because thevolume of the tube is much smaller than the flask volume, the volume change in the tubeis negligible and the entire system can be considered to have a fixed volume.

Assembling the apparatus

The apparatus is shown in Figure 4. The 500-mL flask and 50-mL beaker are inyour locker, but the additional equipment will be issued. Inspect the fully-assembled set-up at the back of the room before assembling the apparatus. Put about 40mL of water inthe beaker, add a few drops of dye to make the water more visible, and immerse thebottom of the capillary in the colored water. Caution: the rubber tube connecting the topof the manometer to the plastic T-connector must be short enough to bring the manometertube into contact with the T. The rubber tube will bend over and close if it is too long.

13

Figure 4. Apparatus for measuring CO2 in respired air. The entire system must be madeleakless.

Preparing for the measurement

After assembling the apparatus, tie a piece of thread to the plastic vial as shown inFigure 4. The thread should be long enough to allow you to lower the vial to the bottomof the upright flask. Place about 10mL of 6M NaOH in the vial (careful!). Next place aslotted cap on the vial, with the thread passing through the slot. Do not screw down thecap. The purpose of the cap is to cover the NaOH solution until it is needed, so that theCO2 is not prematurely absorbed. Carefully lower the vial so that it rests upright on thebottom of the flask. If the vial tips and spills its contents, or if its lid is dislodged, rinsethe flask and vial thoroughly and start over. Finally, drop the thread into the flask andfirmly seal the rubber stopper in place.

Close the Hoffman clamp next to the T-connector (the tube must be centered inthe clamp), and gently blow out any water droplets trapped inside the capillary tube.Then check the system for leaks by gently sucking on the straw until the water is abouthalfway up the manometer tube. Close the clamp next to the straw, and watch the liquidlevel in the tube. It should remain stationary; if it falls, there is a leak which must becorrected. Likely sources of leaks are:

a) Leaky stopper. Press it down firmly. A little water smeared around the top ofthe flask may help seal the joint.

b) Hoffman clamps. The rubber tubing must be centered in the jaws, which mustbe screwed down tightly.

14

c) Rubber tubing. Squeeze it, and if cracks appear replace it.

d) Wrong rubber tubing size. Use smaller rubber tubing if the fit with the glasstube is loose.

Determining respired carbon dioxide

1. When the apparatus is leak-free, open both clamps and exhale eight or ten timesthrough the straw to completely flush the flask with respired air. Although breathingis restricted, try to breathe as normally as possible. Do not hold your breath or exhalemore than you usually do.

2. Firmly close both clamps, and record the distance in millimeters between the surfaceof the water in the beaker and the top of the water in the capillary tube. They will notbe identical even though the pressures are identical inside and outside; the surfacetension of the water pulls the water up the tube just as water rises up blotter paper.

3. Release the NaOH by tipping the flask to topple the vial and dislodge the lid. The lidmust come off the vial. Swirl the flask gently and slowly to spread the solutionevenly over the bottom, and watch the rise of the water in the capillary tube. Itshould go well over halfway. Hold the flask by the neck in a paper towel tominimize heating it with your hand; expansion of the heated air would affect themanometer reading. If the column does not rise very far, or if rises and then drops,there is a leak; check the apparatus and start over. When the water stops rising andthe level is stable, record the distance in millimeters between the top of the columnand the surface of the water in the beaker. At this point, you should no longer betouching the flask.

4. Remove the stopper, and record the temperature in the flask.

How to calculate the partial pressure of CO2

Before the NaOH is spilled, the gases in the flask consist mostly of CO2, N2, andO2. The total pressure, which equals atmospheric pressure, is the sum of the partialpressures of these individual gases. When the NaOH is spilled, the CO2 is absorbed andits contribution to the total pressure vanishes. As a result, the pressure inside the flaskdrops below the outside atmospheric pressure, which therefore pushes the water up themanometer tube. The height of the water column, corrected for capillary rise, measuresthe difference between the outside and inside pressures. Since this pressure difference iscaused by removing CO2, it is also the partial pressure of the CO2 originally in the flask.

Begin the calculations by finding the pressure difference in mm of water:

∆ P (mm H2O) = (hgt before absorbing CO2) - (hgt after absorbing CO2)

15

Next, convert ∆ P to units of mm Hg by dividing the ratio of the density of mercury to thedensity of water, which is 13.6g/cm3. The partial pressure of CO2 expressed in mm Hg istherefore

Pp= ∆P(mmHg) =∆P(mmH2O)

13.6 (11)

Once the partial pressure is known, the rate of production of CO2 can be determined if thevolume of each breath and the rate of breathing is measured.

B. Tidal volume of the lungs

The volume of air inspired or expired in a single normal breath is called the tidalvolume. It is much less than the total volume of the lungs, which can expand andcontract well beyond the tidal volume. You can easily convince yourself of this fact byinhaling and exhaling more than you usually do. Taking a deep breath inflates bringsbeyond normal breathing, and it is easy to forcefully exhale more air than usual.

Tidal volume is measured with a spirometer. Our modest version is constructedof sewer pipe and a soda bottle (Figure 5), but it works quite well.

Figure 5. The spirometer in perspective (left) and in cross section (right). Breathing inthe tube traps the expired air in the bottle, which floats on water and is free to move upand down.

16

The plastic soda bottle floats on water, and moves vertically within the four guiderods. Exhaled air is delivered to it by the central pipe, and causes it to float higher. Thevolume is read directly from 100-mL graduations on the bottle. The pressure inside isatmospheric because the mass of the bottle is very small.

Use a new sterile mouthpiece, and return it to the designated container whenfinished (the wastebasket is not the designated container). It is difficult to exhale anormal breath into the spirometer, especially when you become conscious of your ownbreathing. One way is to breath through your nose repeatedly until you are convincedthat your breathing is routine, then, after inhaling that last breath, pinch your nose andexhale normally through the spirometer tube. Seal your lips tightly around themouthpiece. Repeat the measurement 5-10 times, and average the volumes. Tominimize bias, avoid looking at the float when making a measurement. Ask someone toobserve and record, while you do the breathing. Tidial volumes vary, of course, fromperson to person, but 300 mL is quite low and 700 mL is quite high.

C. Breathing rate

Measuring breathing rate is difficult if you are aware of doing it, which you will be.It is recorded in cycles per minute (a cycle is one breath in, one breath out), and should bemeasured for at least one minute. There is no special technique, but something isprobably being done improperly if the rate measures lower than 10 breaths/min or greaterthan 20.

D. Calculating metabolic rate

Since the calories per liter of exhaled CO2 are known (Table 3), the metabolic rate(cal/hr) can be determined by calculating the volume of CO2 produced. The steps are:

1. Calculate the volume of CO2 exhaled

The volume of CO2 exhaled per hour (L/hr) is

V = (cycles/min) (min/hr) (tidal volume)

but its pressure is the partial pressure Pp defined by eq(11), and it is at roomtemperature.

2. Correct the volume to STP (00C and 760 mm Hg).

Use the well-known expression

P1V1

T1

=P2V2

T2

17

The volume of CO2 exhaled per hour (L/h) at STP is therefore

VSTP = P1V1T2

P2T1

(12)

Where P2 and T2 refer to standard conditions (0oC and 760 mm Hg). Temperature iseq(12) must be in kelvins.

3. Calculate the metabolic rate

The metabolic rate in cal per hour is given by

rate, cal/hr = (L/hr)(cal/L) = (VSTP)(cal/Lr) (13)

At this point, it is necessary to assume the percentages of carbohydrate, fat, and proteinbeing metabolized. However, as pointed out earlier, the cal/L CO2 (Table 3) are notstrongly affected by the composition of the metabolites. If the average of the values inTable 3 is chosen, the metabolic rate is

rate (cal/hr) = 5.9 VSTP

PART II. THE CALORIES IN CORN OIL (work in pairs)

As already discussed at some length, the enthalpy of combustion of a fat equalsthe calories that it provides to the body when it is metabolized. Therefore, by measuringthe heat produced per gram of fat burned in a calorimeter (cal/g), and combining it withyour metabolic rate measured in Part I, you can calculate the grams of fat that would berequired to maintain your metabolic rate if fat were the only nutrient metabolized. Thiscaveat is artificial, of course. In reality, a mix of carbohydrate, fat, and protein aremetabolized, and the separate contribution of each one cannot be determined unlessoxygen consumption, carbon dioxide production, and urinary nitrogen are all measured.Notwithstanding that fact,

Chemicals and special equipment

Corn oilCalorimeter and accessories

Safety PrecautionsEating corn oil makes you fat

18

Description of the calorimeter

The heat of combustion is obtained by burning the corn oil, and measuring thetemperature increase in a sample of water heated by the flame. The water is contained inan aluminum can, which is painted black to increase the absorption of radiant heat(Figure 6).

Figure 6. The calorimeter. Left: A black aluminum can, containing water, slips inside aplastic cylinder and rests on stops. At the bottom of the cylinder are slots to admit air forthe oil burner, which is in the cylinder below the can. An outside shield blocks drafts.Middle: The oil burner, consisting of an aluminum pan containing the oil, a fiberglasswick, and a wick holder. It burns like a candle. Right: Cross-section, showing theposition of the can, the stops and guides, and the location of the burner.

Corn oil consumption is measured by weighing the burner before and after heating thewater, and the temperature rise is calculated per mg of oil consumed. To calibrate thecalorimeter, the same procedure is carried out by burning paraffin wax, which has aknown heat of combustion.

Measuring the heat of combustion of corn oil

Obtain a calorimeter, draft shield, oil burner, and paraffin calibration candle. Fillthe burner with corn oil to 2-3mm above the top of the wick support. Before setting upthe calorimeter, light the burner and the candle and examine the flames. They should beabout the same size, 1/4 inch high, and not smoky. The oil flame tends to be smokierthan the candle flame. It may be necessary to lower the empty cylinder over the flame toshield it from drafts. The paraffin candle flame can be reduced, if absolutely necessary,by carefully pinching back the wick a little. Remember, if the wick is shortened too

19

much it cannot be lengthened again. Adjust the size of the oil flame by adding orremoving oil from the pan.

Blow out the flames, and wipe the outside of the oil burner (side and bottom) toremove errant oil. To ensure reproducible positioning, place the oil burner or candle on asheet of a paper and trace around it, then center the plastic cylinder over the candle andtrace around it. Nearly fill the calorimeter can with water, empty it, drain it inverted on apaper towel for a few minutes, and fill it with exactly 200 mL of water at roomtemperature (graduated cylinder). Be careful; the black paint is fragile, and scratcheseasily. Do not handle the can with bare hands; finger grease ruins the surface. Carefully(to avoid scratching the paint) lower the can into the cylinder, and make sure that it isproperly seated. Clamp thermometer to a ringstand, positioned with the bulb wellimmersed in the water but not touching the bottom of the can.

Weight the oil burner on the analytical balance, place it in position on the markedpaper, light it, and immediately place the cylinder in position on the marked paper. Stirthe water constantly, and record temperature and time at one-minute intervals. When thetemperature has risen by about 10oC, remove the cylinder and immediately blow out theflame. Timing starts when the flame is lit, and stops when it is extinguished. Finally,reweigh the burner.

Calibrating the calorimeter

Empty the can, drain it inverted on a paper towel for a few minutes, and refill itwith exactly 200mL of water. Repeat the procedure used for the oil, but use the candle,weighing it at the beginning and end of the run.

Calculating the heat of combustion

To calculate the enthalpy of combustion of the oil (cal/g), we assume, quitereasonably, that the rate of temperature rise of the water is directly proportional to theenthalpy of combustion and the mass of fuel consumed.

For the oil and for the calibration candle, plot temperature against time inminutes, and draw the best-fit straight line. Determine the slope ∆ T/∆ t in deg/min.Also calculate the rate of consumption of oil in g/min:

∆m

∆t=

mo − m

to − t(14)

when mo - m is the mass of oil consumed, and to -t is the time between lighting andextinguishing the flame. Dividing the slope by eq(14) gives the temperature rise pergram of oil consumed:

R =∆T

∆m (15)

20

Since R is proportional to the heat of combustion (cal/g), and the heat capacity of thecalorimeter is the same for both runs, we can write

Ro

Rp

= ∆Ho

∆Hp

∆Ho =Ro

Rp

∆Hp( )

where ∆ Ho, Ro refer to oil and ∆ Hp, Rp refer to paraffin. The heat of combustion ofparaffin is 10.6cal/g.

PRE-LABORATORY QUESTIONS

1. A student's breathing rate is 14 breaths/min, and his tidal volume is 540mL. When theCO2 content in his breath is measured by the method described, the water rise in thecapillary manometer is 42.4cm. Room temperature is 22.3oC, and the barometricpressure is 746mmHg. Use the average value of 5.9cal/L CO2 at STP (Table 3) tocalculate the metabolic rate in cal/hour. Ans: 102 Cal/hr.

2. Olive oil and paraffin are burned separately in two calorimeter runs. Both use exactly200mL of water. For the olive oil, the following data were obtained:

t(min) T(oC) t(min) T(oC)1 22.8 6 31.22 24.3 7 32.53 26.1 8 34.34 27.9 9 36.05 29.5

The mass of oil consumed was 0.4980g, and the burning time was 10.00 minutes. Acalibration run was made with paraffin, and R was found to be 37.3deg/g. Calculatethe heat of combustion of the olive oil. Ans: 9.46cal/g

3. Calculate the weight of olive oil that the student would need to consume per hour inorder to maintain his metabolic rate, if his sole source of energy were the oil.