In the Classroom edited by
Resources for Student Assessment
John Alexander University of Cincinnati Cincinnati, OH 45221
Three Forms of Energy Sigthór Pétursson Department of Natural Resource Science, University of Akureyri, 600 Akureyri, Iceland;
[email protected] Thermodynamics is of fundamental importance to chemistry. Over one hundred papers in this Journal show the appreciation of this importance by the chemical community in the last five years. Reference to only a few of these papers is given here (1–9). The importance of reactions that are primarily carried out for the production of energy and the principles used to work out the quantity of energy released are well known. Of no less importance are the fundamental principles relating to entropy and the free energy of reactions to determine the spontaneity of reactions and even the equilibrium constant (10, 11). The use of the common thermodynamic equations, as presented in general chemistry courses, is relatively straightforward, even if some of the underlying concepts, covered further in more advanced physical chemistry courses, are difficult for the beginner. The modern students and practitioners of chemistry have at their disposal a vast quantity of thermodynamic data; it is therefore essential that the practical aspects of the subject are not obscured by too much theory. The authors of contemporary books on thermodynamics are aware of this as noted in recent reviews (12, 13). Calculations It is probably true that chemists are more familiar with heat energy than mechanical energy. To appreciate fully the important and common transformation of chemical energy into mechanical energy, for example in the internal combustion engine or in our bodies, it is helpful to compare the energy involved in familiar events. Three different transformations that everybody should be able to envisage are examined. The three forms of energy are described and illustrated in Figures 1–3. We may have our own feelings on which of these three events involves the greatest amount of energy, but let us work it out.
Heat What is the energy needed to heat 200 g of water from 7.0 ⬚C to 37.0 ⬚C? specific heat capacity of H2O (c) = 4.184 J ∆t = 37.0 ⬚C − 7.0 ⬚C = 30.0 ⬚C mass of H2O (m) = 200 g
What is the energy (heat) involved in warming 200 g of water from 7.0 °C to 37.0 °C? Remember that every time you drink a glass of cold water your body expends this quantity of heat to warm the water up to your body temperature. glass of water
200 g of water heated from 7.0 oC to 37.0 oC
Figure 1. Heat.
Elevation of a body against Earth’s gravitational force. What is the energy (work) needed to lift a 50.0 kg sack of cement to a height of 10.0 m? This is roughly equal to carrying a sack of cement to the third floor of a building.
50.0 kg elevated by 10.0 m against gravity, g = 9.80 m s-2 cement, 50.0 kg
Figure 2. Mechanical energy.
What is the work performed in expanding a cylinder by 90.0 L against an external pressure of 1.0 atm? external pressure P = 1.00 atm
g᎑1 ⬚C᎑1
area = A
The heat energy, q, involved is therefore: q = (m)(∆t)(c) = (200 g) × (30.0 °C) × (4.184 J g᎑1 °C᎑1) = 25104 J = 25.1 kJ 776
piston extracted against external pressure
∆V ∆ = 90.0 L Figure 3. Expansion.
Journal of Chemical Education • Vol. 80 No. 7 July 2003 • JChemEd.chem.wisc.edu
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In the Classroom
Mechanical Energy What is the energy needed to lift a 50.0 kg sack of cement to a height of 10.0 m? The gravitational acceleration, g, is 9.80 m s᎑2. The force acting on the sack is therefore: f = (m)(g) = (50.0 kg) × (9.80 m s᎑2) = 490 kg m s᎑2 = 490 N Mechanical energy or work is by definition the product of force and displacement. The work, w, performed by displacing the sack of cement by 10.0 m against gravity is therefore: w = ( f )(l) = (490 N) × (10.0 m) = 4900 N m = 4.90 kJ
Expanding Cylinder What is the work, w, performed to expand a cylinder by 90.0 L against a pressure of 1.00 atm? w = ( f )(d) = ( f )(l); where d is the displacement P = f 兾A; pressure is the force per unit area, A rearranging f = P × A, thus w = (P)(A)(l) Since (A)(l) is the change of volume, ∆V, w = (P)(∆V) w =(1.00 atm) × (90.0 L) = 90.0 atm L The unit atm L must be equivalent to energy (work). We can confirm this and convert the unit to joule by representing the pressure in SI units 1 atm = 1.013 × 105 Pa or in fundamental SI units 1 atm = 1.013 × 105 kg m᎑1 s᎑2 Thus atm L = 1.013 × 105 kg m᎑1 s᎑2 L Since L = dm3 = 10᎑3 m3, then atm L = (1.013 × 105 kg m᎑1 s᎑2) × (10᎑3 m3) = 101.3 kg m2 s᎑2 = 101.3 joule
as kg m2 s᎑2 is equivalent to joule. Therefore the work done on the surroundings is: w = (90.0 atm L) × [101.3 J兾(atm L )] = 9117 J = 9.12 kJ Discussion It is interesting to compare these results, especially the energy involved in heating the water and elevating the sack
of cement. It takes a reasonably fit person to carry a mass of 50.0 kg of cement to the third floor of a house. This activity could be considered a good exercise. Do that a few times every day and you would be justified in feeling that you were getting rid of a few excess Calories (capital C is used to denote nutritional calories, equal to 1000 heat calories)! The energy expended in doing this activity is, however, only 19.5% of the energy expenditure in drinking a glass of cold water. Even if we add the body weight of the person, let us say 75 kg, the energy expenditure is still only about 49% of that involved in drinking the water. We can take a more extreme case regarding the heat energy. Consider the example of a person who ingests, perhaps a bit excessively, 1 L of hot (67 ⬚C) liquid daily (coffee, tea, etc.). If that person changed to drinking 1 L of cold (7 ⬚C) water he would be expending 251 kJ per day, since he is switching from heat gain in cooling 1 L at 67 ⬚C to 37 ⬚C to heat expenditure in heating 1 L at 7 ⬚C to 37 ⬚C. Let us look at the expanding cylinder. The thermodynamic importance of the expansion or compression of gases, when reactions take place at constant pressure, is familiar. Keeping in mind Torricelli’s barometer and remembering that mercury has a density of 13.5 g兾cm3, this is equivalent to lifting [(90.0 dm3) × (1000 cm3兾dm3) × (13.5 g兾cm3)] = 1.22 × 106 g = 1.22 × 103 kg or 1.22 metric tons to a height of 760 mm or 0.760 m. This is best appreciated if we think in terms of extracting a piston with a radius of 19.42 cm or 1.942 dm, by 0.760 m or 7.60 dm. The pressure is 1 atm or 760 mm Hg and the volume swept by the process is (1.942 dm)2 × π × 7.60 dm = 90.0 dm3 or 90.0 L. The weight of the mercury is about 24 times the weight of the sack of cement and a much larger force is needed to perform this task, but the displacement is also much smaller. Nutritional Energy Comparison Since two of the examples above led us into comparisons relevant to nutrition we can develop that theme very briefly. First it must be stressed that these examples are not illustrated to trivialize the need for healthful physical activity because we need to exercise our muscles and drinking cold water is not the best way to accomplish that. However the comparison does draw attention to a worthwhile and economical nutritional fact. Consider that the basal metabolic rate—the energy expenditure when lying at complete rest— of an average middle-aged man (180-cm high, 75 kg) is about 7100 kJ (about 1700 kcal or Calories) per 24 hours. The total energy use of the same man doing light physical work during the day is about 8400 kJ (about 2000 Calories) per 24 hours. Thus the energy requirement for physical activity for the man is about 1300 kJ or 310 Calories (14). The extra 251 kJ of energy the man would utilize by switching from drinking a liter of hot drinks to drinking a liter of cold water is about 20% of the energy needed for light physical activity. Literature Cited 1. Mills, Pamela; Sweeney, William V.; Cieniewicz, Waldemar. J. Chem. Educ. 2001, 78, 1360–1361. 2. Weiss, Hilton M. J. Chem. Educ. 2001, 78, 1362–1364. 3. Bartell, Lawrence S. J. Chem. Educ. 2001, 78, 1059–1067.
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In the Classroom 4. Bartell, Lawrence S. J. Chem. Educ. 2001, 78, 1067–1069. 5. Wadsö, Lars; Smith, Allan L.; Shirazi, Hamid; Mulligan, S. Rose; Hofelich, Thomas. J. Chem. Educ. 2001, 78, 1080– 1086. 6. Jacobson, Nathan. J. Chem. Educ. 2001, 78, 814–819. 7. Howard, Irmgard K. J. Chem. Educ. 2001, 78, 505–508. 8. Jansen, Michael P. J. Chem. Educ. 2000, 77, 1578–1579. 9. Jensen, William B. J. Chem. Educ. 2000, 77, 713–717. 10. Masterton, W. L.; Hurley, C. N. Chemistry, Principles and
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11. 12. 13. 14.
Reactions, 4th ed.; Harcourt College Publishers: Orlando, FL, 2001; Chapters 8, 17. Noggle, J. H. Physical Chemistry, 3rd ed.; Harper Collins: New York, 1996; Chapters 2, 3. Gislason, Eric A. J. Chem. Educ. 2001, 78, 1186. Minderhout, Vicky. J. Chem. Educ. 2001, 78, 457. Passmore, R.; Eastwood, M. A. Davidson and Passmore Human Nutrition and Dietetics; Churchill Livingstone: Edinburgh, Scotland, UK, 1986; Chapter 3.
Journal of Chemical Education • Vol. 80 No. 7 July 2003 • JChemEd.chem.wisc.edu
