1 4 When sweat evaporates, it cools the skin by absorbing heat from the body. Energy and Matter CAPTER OUTLINE 4.1 Energy 4.2 The Three States of Matter 4.3 Intermolecular Forces 4.4 Boiling Point and Melting Point 4.5 Specific eat 4.6 Energy and Phase Changes 4.7 eating and Cooling Curves CAPTER GOALS In this chapter you will learn how to: ➊ Define energy and become familiar with the units of energy ➋ Identify the characteristics of the three states of matter ➌ Determine what types of intermolecular forces a compound possesses ➍ Relate the strength of intermolecular forces to a compound s boiling point and melting point ➎ Define specific heat and use specific heat to determine the amount of heat gained or lost by a substance ➏ Describe the energy changes that accompany changes of state ➐ Interpret the changes depicted in heating and cooling curves 114
2 4.1 Energy 115 Why does water ( 2 O) have a much higher boiling point than methane (C 4 ), the main component of natural gas? Why does chloroethane, a local anesthetic, numb an injury when it is sprayed on a wound? To answer these questions, we turn our attention in Chapter 4 to energy and how energy changes are related to the forces of attraction that exist between molecules. 4.1 Energy Energy is the capacity to do work. Whenever you throw a ball, ride a bike, or read a newspaper, you use energy to do work. There are two types of energy. Potential energy is stored energy. Kinetic energy is the energy of motion. The water at the top of a waterfall has potential energy because of its position. This potential energy becomes kinetic energy as the water falls. A ball at the top of a hill or the water in a reservoir behind a dam are examples of potential energy. When the ball rolls down the hill or the water flows over the dam, the stored potential energy is converted to the kinetic energy of motion. Although energy can be converted from one form to another, one rule, the law of conservation of energy, governs the process. The total energy in a system does not change. Energy cannot be created or destroyed. Throwing a ball into the air illustrates the interplay of potential and kinetic energy (Figure 4.1). Figure 4.1 Potential and Kinetic Energy All of the ball s kinetic energy is converted to potential energy. Kinetic energy is converted to potential energy as the ball rises. Potential energy is converted to kinetic energy as the ball falls. Kinetic energy is transferred from the thrower s hand to the baseball. Some of the kinetic energy of the ball is transferred to the ground. When a ball is thrown into the air it possesses kinetic energy, which is converted to potential energy as it reaches its maximum height. When the ball descends, the potential energy is converted back to kinetic energy, until the ball rests motionless on the ground.
3 116 Chapter 4 Energy and Matter The joule, named after the nineteenthcentury English physicist James Prescott Joule, is pronounced jewel. 4.1A The Units of Energy Energy can be measured using two different units, calories (cal) and joules (J). A calorie is the amount of energy needed to raise the temperature of 1 g of water 1 C. Joules and calories are related in the following way. 1 cal = J Since both the calorie and the joule are small units of measurement, more often energies in reactions are reported with kilocalories (kcal) and kilojoules (kj). Recall from Table 1.2 that the prefix kilo means 1, kcal = 1,000 cal 1 kj = 1,000 J 1 kcal = kj To convert a quantity from one unit of measurement to another, set up conversion factors and use the method first shown in Section 1.7B and illustrated in Sample Problem 4.1. SAMPLE PROBLEM 4.1 A reaction releases 421 kj of energy. ow many kilocalories does this correspond to? Analysis and Solution  Identify the original quantity and the desired quantity. 421 kj? kcal original quantity desired quantity  Write out the conversion factors. Choose the conversion factor that places the unwanted unit, kilojoules, in the denominator so that the units cancel. kj kcal conversion factors kj 1 kcal or 1 kcal kj Choose this conversion factor to cancel kj.  Set up and solve the problem. Multiply the original quantity by the conversion factor to obtain the desired quantity. 421 kj 1 kcal = kcal, rounded to 101 kcal kj Kilojoules cancel. Answer PROBLEM 4.1 Carry out each of the following conversions. a. 42 J to cal b kcal to cal c. 326 kcal to kj d kcal to J PROBLEM 4.2 Combustion of 1 g of gasoline releases 11.5 kcal of energy. ow many kilojoules of energy is released? ow many joules does this correspond to? 4.1B FOCUS ON TE UMAN BODY Energy and Nutrition When we eat food, the protein, carbohydrates, and fat (lipid) in the food are metabolized to form small molecules that in turn are used to prepare new molecules that cells need for maintenance
4 4.1 Energy 117 and growth. This process also generates the energy needed for the organs to function, allowing the heart to beat, the lungs to breathe, and the brain to think. The amount of stored energy in food is measured using nutritional Calories (upper case C), where 1 Cal = 1,000 cal. Since 1,000 cal = 1 kcal, the following relationships exist. Nutritional Calorie 1 Cal = 1 Cal = 1 kcal 1,000 cal Upon metabolism, proteins, carbohydrates, and fat each release a predictable amount of energy, the caloric value of the substance. For example, one gram of protein or one gram of carbohydrate typically releases about 4 Cal/g, while fat releases 9 Cal/g (Table 4.1). If we know the amount of each of these substances contained in a food product, we can make a first approximation of the number of Calories it contains by using caloric values as conversion factors, as illustrated in Sample Problem 4.2. Table 4.1 Caloric Value for Three Classes of Compounds Protein Carbohydrate Fat Cal/g cal/g 4,000 4,000 9,000 One nutritional Calorie (1 Cal) = 1,000 cal = 1 kcal. CONSUMER NOTE Knowing the number of grams of protein, carbohydrates, and fat allows you to estimate how many Calories a food product contains. A quarter-pound burger with cheese with 29 g of protein, 40 g of carbohydrates, and 26 g of fat contains 510 Calories. When an individual eats more Calories than are needed for normal bodily maintenance, the body stores the excess as fat. The average body fat content for men and women is about 20% and 25%, respectively. This stored fat can fill the body s energy needs for two or three months. Frequent ingestion of a large excess of Calories results in a great deal of stored fat, causing an individual to be overweight. SAMPLE PROBLEM 4.2 If a baked potato contains 3 g of protein, a trace of fat, and 23 g of carbohydrates, estimate its number of Calories. Analysis Use the caloric value (Cal/g) of each class of molecule to form a conversion factor to convert the number of grams to Calories and add up the results. Solution  Identify the original quantity and the desired quantity. 3 g protein 23 g carbohydrates? Cal original quantities desired quantity  Write out the conversion factors. Write out conversion factors that relate the number of grams to the number of Calories for each substance. Each conversion factor must place the unwanted unit, grams, in the denominator so that the units cancel. Cal g conversion factor for protein 4 Cal 1 g protein Cal g conversion factor for carbohydrates 4 Cal 1 g carbohydrate  Set up and solve the problem. Multiply the original quantity by the conversion factor for both protein and carbohydrates and add up the results to obtain the desired quantity. Calories due to protein Calories due to carbohydrate 4 Cal 4 Cal Total Calories = 3 g + 23 g carbohydrate 1 g protein 1 g carbohydrate Grams cancel. Grams cancel. Total Calories = = 12 Cal + 92 Cal 104 Cal, rounded to 100 Cal Answer
5 118 Chapter 4 Energy and Matter PROBLEM 4.3 ow many Calories are contained in one tablespoon of olive oil, which has 14 g of fat? PROBLEM 4.4 One serving (36 crackers) of wheat crackers contains 6 g of fat, 20 g of carbohydrates, and 2 g of protein. Estimate the number of calories. 4.2 The Three States of Matter As we first learned in Section 1.2, matter exists in three common states gas, liquid, and solid. A gas consists of particles that are far apart and move rapidly and independently from each other. A liquid consists of particles that are much closer together but are still somewhat disorganized since they can move about. The particles in a liquid are close enough that they exert a force of attraction on each other. A solid consists of particles atoms, molecules, or ions that are close to each other and are often highly organized. The particles in a solid have little freedom of motion and are held together by attractive forces. As shown in Figure 4.2, air is composed largely of N 2 and O 2 molecules, along with small amounts of argon (Ar), carbon dioxide (CO 2 ), and water molecules that move about rapidly. Liquid water is composed of 2 O molecules that have no particular organization. Sand is a solid composed of SiO 2, which contains a network of covalent silicon oxygen bonds. Figure 4.2 The Three States of Matter Solid, Liquid, and Gas O 2 O Si N 2 solid SiO 2 liquid 2 O N 2 and O 2 gases Most sand is composed of silicon dioxide (SiO 2 ), which forms a three-dimensional network of covalent bonds. Liquid water is composed of 2 O molecules, which can move past each other but are held close together by a force of attraction (Section 4.3). Air contains primarily N 2 and O 2 molecules that move rapidly with no force of attraction for each other.
6 4.3 Intermolecular Forces 119 Whether a substance exists as a gas, liquid, or solid depends on the balance between the kinetic energy of its particles and the strength of the interactions between the particles. In a gas, the kinetic energy of motion is high and the particles are far apart from each other. As a result, the attractive forces between the molecules are negligible and gas molecules move freely. In a liquid, attractive forces hold the molecules much more closely together, so the distance between molecules and the kinetic energy is much less than the gas. In a solid, the attractive forces between molecules are even stronger, so the distance between individual particles is small and there is little freedom of motion. The properties of gases, liquids, and solids are summarized in Table 4.2. PROBLEM 4.5 ow do gaseous, liquid, and solid methanol (C 4 O) compare in each of the following features: (a) density; (b) the space between the molecules; (c) the attractive force between the molecules? PROBLEM 4.6 Why is a gas much more easily compressed into a smaller volume than a liquid or solid? Table 4.2 Properties of Gases, Liquids, and Solids Property Gas Liquid Solid Shape and volume Expands to fill its container A fixed volume that takes the shape of the container it occupies A definite shape and volume Arrangement of particles Randomly arranged, disorganized, and far apart Randomly arranged but close Fixed arrangement of very close particles Density Low (< 0.01 g/ml) igh (~1 g/ml) a igh (1 10 g/ml) Particle movement Interaction between particles Very fast Moderate Slow None Strong Very strong a The symbol ~ means approximately. 4.3 Intermolecular Forces To understand many of the properties of solids, liquids, and gases, we must learn about the forces of attraction that exist between particles atoms, molecules, or ions. Ionic compounds are composed of extensive arrays of oppositely charged ions that are held together by strong electrostatic interactions. These ionic interactions are much stronger than the forces between covalent molecules, so it takes a great deal of energy to separate ions from each other (Section 3.5). In covalent compounds, the nature and strength of the attraction between individual molecules depend on the identity of the atoms. Intermolecular forces are the attractive forces that exist between molecules.
7 120 Chapter 4 Energy and Matter There are three different types of intermolecular forces in covalent molecules, presented in order of increasing strength: London dispersion forces Dipole dipole interactions ydrogen bonding Thus, a compound that exhibits hydrogen bonding has stronger intermolecular forces than a compound of similar size that has dipole dipole interactions. Likewise, a compound that has dipole dipole interactions has stronger intermolecular forces than a compound of similar size that has only London dispersion forces. The strength of the intermolecular forces determines whether a compound has a high or low melting point and boiling point, and thus if the compound is a solid, liquid, or gas at a given temperature. 4.3A London Dispersion Forces London dispersion forces can also be called van der Waals forces. London dispersion forces are very weak interactions due to the momentary changes in electron density in a molecule. For example, although a nonpolar methane molecule (C 4 ) has no net dipole, at any one instant its electron density may not be completely symmetrical. If more electron density is present in one region of the molecule, less electron density must be present some place else, and this creates a temporary dipole. A temporary dipole in one C 4 molecule induces a temporary dipole in another C 4 molecule, with the partial positive and negative charges arranged close to each other. The weak interaction between these temporary dipoles constitutes London dispersion forces. London dispersion force between two C 4 molecules δ δ δ δ δ + δ δ + δ δ + δ + δ + δ δ δ + δ + δ + Although any single interaction is weak, a large number of London dispersion forces creates a strong force. For example, geckos stick to walls and ceilings by London dispersion forces between the surfaces and the 500,000 tiny hairs on each foot. More electron density in one region creates a partial negative charge (δ ). Less electron density in one region creates a partial positive charge (δ + ). All covalent compounds exhibit London dispersion forces. These intermolecular forces are the only intermolecular forces present in nonpolar compounds. The strength of these forces is related to the size of the molecule. The larger the molecule, the larger the attractive force between two molecules, and the stronger the intermolecular forces. Atoms like the noble gases helium and argon exhibit London dispersion forces, too. The attractive forces between argon atoms are stronger than the attractive forces between helium atoms because argon atoms are considerably larger in size.
8 4.3 Intermolecular Forces 121 PROBLEM 4.7 Which of the following compounds exhibit London dispersion forces: (a) N 3 ; (b) 2 O; (c) Cl; (d) ethane (C 2 6 )? PROBLEM 4.8 Which species in each pair exhibits the stronger London dispersion forces? a. C or C C C b. e atoms or Ne atoms 4.3B Dipole Dipole Interactions ow to determine whether a molecule is polar is shown in Section Dipole dipole interactions are the attractive forces between the permanent dipoles of two polar molecules. For example, the carbon oxygen bond in formaldehyde, 2 C O, is polar because oxygen is more electronegative than carbon. This polar bond gives formaldehyde a permanent dipole, making it a polar molecule. The dipoles in adjacent formaldehyde molecules can align so that the partial positive and partial negative charges are close to each other. These attractive forces due to permanent dipoles are much stronger than London dispersion forces. dipole dipole interactions C O = δ + δ δ + δ δ + δ formaldehyde PROBLEM 4.9 Draw the individual dipoles of two dipole dipole interaction. Cl molecules and show how the dipoles are aligned in a 4.3C ydrogen Bonding ydrogen bonding occurs when a hydrogen atom bonded to O, N, or F, is electrostatically attracted to an O, N, or F atom in another molecule. = O O = hydrogen bond between two 2 O molecules covalent O bond ydrogen bonding can occur only when molecules contain a hydrogen atom bonded to a very electronegative atom that is, oxygen, nitrogen, or fluorine. For example, two 2 O molecules can hydrogen bond to each other: a hydrogen atom is covalently bonded to oxygen in one water
9 P P P P P P P P 122 Chapter 4 Energy and Matter molecule, and hydrogen bonded to an oxygen atom in another water molecule. ydrogen bonds are the strongest of the three types of intermolecular forces. Table 4.3 summarizes the three types of intermolecular forces. Table 4.3 Summary of the Types of Intermolecular Forces Type of Force Relative Strength Exhibited by Example London dispersion Weak All molecules C 4, 2 CO, 2 O Dipole dipole Moderate Molecules with a net dipole 2 CO, 2 O ydrogen bonding Strong Molecules with an O, N, or F bond 2 O ydrogen bonding is important in many biological molecules, including proteins and DNA. DNA, which is contained in the chromosomes of the nucleus of a cell, is responsible for the storage of all genetic information. DNA is composed of two long strands of atoms that are held together by hydrogen bonding as shown in Figure 4.3. A detailed discussion of DNA appears in Chapter 17. Figure 4.3 ydrogen Bonding and DNA DNA double helix ydrogen bonding interactions are shown as dashed red lines. N N N O N N N N O N P P P P C 3 N O N N N N N N P P O DNA is composed of two long strands of atoms that wind around each other in an arrangement called a double helix. The two strands are held together by an extensive network of hydrogen bonds. In each hydrogen bond, a hydrogen atom of an N bond on one chain is intermolecularly hydrogen bonded to an oxygen or nitrogen atom on an adjacent chain. Five hydrogen bonds are indicated.
10 4.4 Boiling Point and Melting Point 123 SAMPLE PROBLEM 4.3 What types of intermolecular forces are present in each compound: (a) Cl; (b) C 2 6 (ethane); (c) N 3? Analysis London dispersion forces are present in all covalent compounds. Dipole dipole interactions are present only in polar compounds with a permanent dipole. ydrogen bonding occurs only in compounds that contain an O, N, or F bond. Solution Cl has London forces like all covalent compounds. a. polar bond δ + δ Cl Cl has a polar bond, so it exhibits dipole dipole interactions. Cl has no atom on an O, N, or F, so it has no intermolecular hydrogen bonding. b. C C C 2 6 is a nonpolar molecule since it has only nonpolar C C bonds. Thus, it exhibits only London forces. C and nonpolar molecule c. δ N δ + δ + δ + net dipole PROBLEM 4.10 N 3 has London forces like all covalent compounds. N 3 has a net dipole from its three polar bonds (Section 3.12), so it exhibits dipole dipole interactions. N 3 has a atom bonded to N, so it exhibits intermolecular hydrogen bonding. What types of intermolecular forces are present in each molecule? a. Cl 2 b. CN c. F d. C 3 Cl e. 2 PROBLEM 4.11 Which of the compounds in each pair has stronger intermolecular forces? a. CO 2 or 2 O b. CO 2 or Br c. Br or 2 O d. C 4 or C Boiling Point and Melting Point The boiling point (bp) of a compound is the temperature at which a liquid is converted to the gas phase, while the melting point (mp) is the temperature at which a solid is converted to the liquid phase. The strength of the intermolecular forces determines the boiling point and melting point of compounds. The stronger the intermolecular forces, the higher the boiling point and melting point. In boiling, energy must be supplied to overcome the attractive forces of the liquid state and separate the molecules to the gas phase. Similarly, in melting, energy must be supplied to overcome the highly ordered solid state and convert it to the less ordered liquid phase. A stronger force of attraction between molecules means that more energy must be supplied to overcome those intermolecular forces, increasing the boiling point and melting point.
11 124 Chapter 4 Energy and Matter In comparing compounds of similar size, the following trend is observed: Compounds with London dispersion forces only Compounds with dipole dipole interactions Compounds that can hydrogen bond Increasing strength of intermolecular forces Increasing boiling point Increasing melting point Methane (C 4 ) and water ( 2 O) are both small molecules with hydrogen atoms bonded to a second-row element, so you might expect them to have similar melting points and boiling points. Methane, however, is a nonpolar molecule that exhibits only London dispersion forces, whereas water is a polar molecule that can form intermolecular hydrogen bonds. As a result, the melting point and boiling point of water are much higher than those of methane. In fact, the hydrogen bonds in water are so strong that it is a liquid at room temperature, whereas methane is a gas. C methane London forces only bp = 162 C mp = 183 C O water hydrogen bonding bp = 100 C mp = 0 C stronger forces higher bp and mp Methane, the main component of natural gas, is a gas at room temperature because the C 4 molecules have weak forces of attraction for each other. In comparing two compounds with similar types of intermolecular forces, the larger compound generally has more surface area and therefore a larger force of attraction, giving it the higher boiling point and melting point. Thus, propane (C 3 8 ) and butane (C 4 10 ) have only nonpolar bonds and London forces, but butane is larger and therefore has the higher boiling point and melting point. C C C C C C C propane butane larger molecule stronger forces higher bp and mp bp = 42 C mp = 190 C bp = 0.5 C mp = 138 C SAMPLE PROBLEM 4.4 (a) Which compound, A or B, has the higher boiling point? (b) Which compound, C or D, has the higher melting point? N 3 C 4 C O C Cl ammonia methane A B methanol chloromethane C D Analysis Determine the types of intermolecular forces in each compound. The compound with the stronger forces has the higher boiling point or melting point.
12 4.5 Specific eat 125 Solution a. N 3 (A) has an N bond, so it exhibits intermolecular hydrogen bonding. C 4 (B) has only London forces since it has only nonpolar C bonds. N 3 has stronger forces and the higher boiling point. b. Methanol (C) has an O bond, so it can intermolecularly hydrogen bond. Chloromethane (D) has a polar C Cl bond, so it has dipole dipole interactions, but it cannot hydrogen bond. C has stronger forces, so C has the higher melting point. PROBLEM 4.12 Which compound in each pair has the higher boiling point? Which compound in each pair has the higher melting point? a. C 4 or C 2 6 b. C 2 6 or C 3 O c. Br or Cl d. C 2 6 or C 3 Br PROBLEM 4.13 Explain why CO 2 is a gas at room temperature but 2 O is a liquid. 4.5 Specific eat In addition to boiling point and melting point, there are many other physical properties that characterize a substance. For example, specific heat (S) is a physical property that reflects the ability of a substance to absorb heat. Specific heat relates energy, mass, and temperature change ( T). The specific heat is the amount of heat energy (in calories or joules) needed to raise the temperature of 1 g of a substance by 1 C. specific heat = heat mass ΔT = cal (or J) g C The specific heat of water is 1.00 cal/(g C), meaning that 1.00 cal of heat must be added to increase the temperature of 1.00 g of water by 1.00 C. The amount of heat that must be added for a particular temperature increase depends on the amount of sample present. To increase the temperature of 2.00 g of water by 1 C requires 2.00 cal of heat. Table 4.4 lists the specific heat values for a variety of substances. The larger the specific heat of a substance, the less its temperature will change when it absorbs a particular amount of heat energy. Table 4.4 Specific eats of Some Substances Substance cal/(g C) J/(g C) Substance cal/(g C) J/(g C) Aluminum Propanol Carbon (graphite) Rock Copper Sand Ethanol Silver Gold Water(l) Iron Water(g) Mercury Water(s)
13 126 Chapter 4 Energy and Matter The specific heat of graphite is about twice that of copper. Adding 1.00 cal of heat will raise the temperature of 1.00 g of graphite by 5.9 C, but raise the temperature of 1.00 g of copper by 11.1 C. Because metals like copper have low specific heats, they absorb and transfer heat readily. We use copper, iron, or aluminum cookware because the metals readily transfer heat from the stove to the food in the pan. The specific heat of water is very high compared to other liquids. As a result, water absorbs a large amount of heat with only a small change in temperature. Water has a high specific heat because of its strong intermolecular forces. Since water molecules are held together with an extensive array of hydrogen bonds, it takes a great deal of heat energy to break these bonds and increase the disorder of the water molecules. Moreover, since the amount of heat absorbed for a given temperature increase equals the amount of heat released upon cooling, water releases a great deal of energy when its temperature drops even a few degrees. This explains why the climate in a coastal city is often more moderate than that of a city located 100 miles inland. A large body of water acts as a reservoir to absorb or release heat as temperature increases or decreases, so it moderates the climate of the land nearby. SAMPLE PROBLEM 4.5 Consider the elements aluminum, copper, gold, and iron in Table 4.4. (a) If 10 kcal of heat is added to 10 g of each element, which element will have the highest temperature? (b) Which element would require the largest amount of heat to raise the temperature of a 5-g sample by 5 C? Analysis The larger the specific heat, the less the temperature of a substance will change when it absorbs heat energy. The larger the specific heat, the more heat that must be added to increase the temperature of a substance a given number of degrees. Solution The specific heats of the metals in Table 4.4 increase in the following order: gold < copper < iron < aluminum. In part (a), gold has the lowest specific heat, so its temperature will be the highest if the same amount of heat is added to the same mass of all four elements. In part (b), aluminum has the largest specific heat, so it will require the largest amount of heat to raise its temperature the same number of degrees as the same mass of the other elements. PROBLEM 4.14 A student has two containers one with 10 g of sand and one with 10 g of ethanol. (a) Which substance has the higher temperature after 10.0 cal of heat is added to each container? (b) Which substance requires the larger amount of heat to raise its temperature by 10 C? PROBLEM 4.15 The human body is composed of about 70% water. ow does this help the body to maintain a steady internal temperature? We can use the specific heat as a conversion factor to calculate how much heat is absorbed or lost from a substance as long as its mass and change in temperature are known, using the equation: heat absorbed or released heat = mass temperature change specific heat ΔT cal = g C cal g C
14 4.5 Specific eat 127 Sample Problem 4.6 shows how to use specific heat to calculate the amount of heat absorbed by a substance when the mass and temperature change are known. SAMPLE PROBLEM 4.6 ow many calories are needed to heat a pot of 1,600 g of water from 25 C to 100. C? Analysis Use specific heat as a conversion factor to calculate the amount of heat absorbed given the known mass and temperature change. Solution  Identify the known quantities and the desired quantity. mass = 1,600 g T 1 = 25 C T 2 = 100. C? calories known quantities desired quantity Subtract the initial temperature (T 1 ) from the final temperature (T 2 ) to determine the temperature change: T 2 T 1 = T = = 75 C. The specific heat of water is 1.00 cal/(g C).  Write the equation. The specific heat is a conversion factor that relates the heat absorbed to the temperature change ( T) and mass. heat = mass T specific heat cal = g C cal g C  Solve the equation. Substitute the known quantities into the equation and solve for heat in calories. cal = 1600 g 75 C 1.00 cal 1 g 1 C = cal Answer PROBLEM 4.16 ow much energy is required to heat 28.0 g of iron from 19 C to 150. C? Report your answer in calories and joules. PROBLEM 4.17 ow much energy is released when 200. g of water is cooled from 55 C to 12 C? Report your answer in calories and kilocalories. Building on what you have learned in Sample Problem 4.6, Sample Problem 4.7 shows how to determine the temperature change of a given mass of a substance when the amount of heat absorbed is known. SAMPLE PROBLEM 4.7 If 400. cal of heat is added to 25.0 g of 2-propanol at 21 C, what is the final temperature? Analysis Use specific heat as a conversion factor to determine the temperature change ( T) given the known mass of the substance and the amount of heat absorbed. Add the temperature change to the initial temperature (T 1 ) to obtain the final temperature (T 2 ).
15 128 Chapter 4 Energy and Matter Solution  Identify the known quantities and the desired quantity. mass = 25.0 g heat added = 400. cal T 1 = 21 C T 2 =? known quantities desired quantity According to Table 4.4, the specific heat of 2-propanol is cal/(g C).  Write the equation and rearrange it to isolate ΔT on one side. Divide both sides of the equation by mass (in g) and specific heat [in cal/(g C)] to place T (in C) on one side. heat = mass T specific heat heat mass specific heat cal g cal/(g C) = T = T  Solve the equation to determine the change in temperature. T = cal g cal/(g C) = 400. cal 25.0 g cal/(g C) = 26.1 C temperature change  Add the change in temperature (ΔT) to T 1 to obtain the final temperature T 2. PROBLEM 4.18 T 2 = = 47.1 C rounded to 47 C Answer If 20. cal of heat is added to 10.0 g each of copper and mercury at 15 C, what is the final temperature of each element? PROBLEM 4.19 If the initial temperature of 120. g of ethanol is 20. C, what will be the final temperature after 950. cal of heat is added? 4.6 Energy and Phase Changes In Section 4.4 we learned how the strength of intermolecular forces in a liquid and solid affect a compound s boiling point and melting point. Let s now look in more detail at the energy changes that occur during phase changes. When energy is absorbed, a process is said to be endothermic. When energy is released, a process is said to be exothermic. In a phase change, the physical state of a substance is altered without changing its composition. 4.6A Converting a Solid to a Liquid Converting a solid to a liquid is called melting. Melting is a phase change because the highly organized water molecules in the solid phase become more disorganized in the liquid phase, but the chemical bonds do not change. Each water molecule is composed of two O bonds in both the solid and the liquid phases.
16 4.6 Energy and Phase Changes 129 Melting is an endothermic process. Energy must be absorbed to overcome some of the attractive intermolecular forces that hold the organized solid molecules together to form the more random liquid phase. The amount of energy needed to melt 1 g of a substance is called its heat of fusion. endothermic melting freezing 2 O exothermic ice liquid water Freezing is the opposite of melting; that is, freezing converts a liquid to a solid. Freezing is an exothermic process because energy is released as the faster moving liquid molecules form an organized solid in which particles have little freedom of motion. For a given mass of a particular substance, the amount of energy released in freezing is the same as the amount of energy absorbed during melting. eats of fusion are reported in calories per gram (cal/g). A heat of fusion can be used as a conversion factor to determine how much energy is absorbed when a particular amount of a substance melts, as shown in Sample Problem 4.8. SAMPLE PROBLEM 4.8 ow much energy in calories is absorbed when 50.0 g of ice cubes melt? The heat of fusion of 2 O is 79.7 cal/g. Analysis Use the heat of fusion as a conversion factor to determine the amount of energy absorbed in melting. Solution  Identify the original quantity and the desired quantity g? calories original quantity desired quantity  Write out the conversion factors. Use the heat of fusion as a conversion factor to convert grams to calories. g cal conversion factors 1 g 79.7 cal or 79.7 cal 1 g Choose this conversion factor to cancel the unwanted unit, g. When an ice cube is added to a liquid at room temperature, the ice cube melts. The energy needed for melting is pulled from the warmer liquid molecules and the liquid cools down.  Solve the problem g 79.7 cal 1 g Grams cancel. = 3,985 cal rounded to 3,990 cal Answer PROBLEM 4.20 Use the heat of fusion of water from Sample Problem 4.8 to answer each question. a. ow much energy in calories is released when 50.0 g of water freezes? b. ow much energy in calories is absorbed when 35.0 g of water melts? c. ow much energy in kilocalories is absorbed when 35.0 g of water melts?
17 130 Chapter 4 Energy and Matter 4.6B Converting a Liquid to a Gas Converting a liquid to a gas is called vaporization. Vaporization is an endothermic process. Energy must be absorbed to overcome the attractive intermolecular forces of the liquid phase to form gas molecules. The amount of energy needed to vaporize 1 g of a substance is called its heat of vaporization. endothermic vaporization condensation 2 O exothermic liquid water steam EALT NOTE Chloroethane (C 3 C 2 Cl), commonly called ethyl chloride, is a local anesthetic. When chloroethane is sprayed on a wound it quickly evaporates, causing a cooling sensation that numbs the site of an injury. Condensation is the opposite of vaporization; that is, condensation converts a gas to a liquid. Condensation is an exothermic process because energy is released as the faster moving gas molecules form the more organized liquid phase. For a given mass of a particular substance, the amount of energy released in condensation equals the amount of energy absorbed during vaporization. eats of vaporization are reported in calories per gram (cal/g). A high heat of vaporization means that a substance absorbs a great deal of energy as it is converted from a liquid to a gas. Water has a high heat of vaporization. As a result, the evaporation of sweat from the skin is a very effective cooling mechanism for the body. The heat of vaporization can be used as a conversion factor to determine how much energy is absorbed when a particular amount of a substance vaporizes, as shown in Sample Problem 4.9. SAMPLE PROBLEM 4.9 ow much heat in kilocalories is absorbed when 22.0 g of 2-propanol, rubbing alcohol, evaporates after being rubbed on the skin? The heat of vaporization of 2-propanol is 159 cal/g. Analysis Use the heat of vaporization to convert grams to an energy unit, calories. Calories must also be converted to kilocalories using a cal kcal conversion factor. Solution  Identify the original quantity and the desired quantity g? kilocalories original quantity desired quantity  Write out the conversion factors. We have no conversion factor that directly relates grams and kilocalories. We do know, however, how to relate grams to calories using the heat of vaporization, and calories to kilocalories. g cal conversion factors cal kcal conversion factors 1 g 159 cal or 159 cal 1 g 1000 cal 1 kcal or 1 kcal 1000 cal Choose the conversion factors with the unwanted units g and cal in the denominator.
18 4.6 Energy and Phase Changes 131  Solve the problem g 159 cal 1 g 1 kcal 1000 cal = 3.50 kcal Grams cancel. Calories cancel. Answer PROBLEM 4.21 Answer the following questions about water, which has a heat of vaporization of 540 cal/g. a. ow much energy in calories is absorbed when 42 g of water is vaporized? b. ow much energy in calories is released when 42 g of water is condensed? 4.6C Converting a Solid to a Gas CONSUMER NOTE Occasionally a solid phase forms a gas phase without passing through the liquid state. This process is called sublimation. The reverse process, conversion of a gas directly to a solid, is called deposition. Carbon dioxide is called dry ice because solid carbon dioxide (CO 2 ) sublimes to gaseous CO 2 without forming liquid CO 2. endothermic sublimation deposition Freeze-drying removes water from foods by the process of sublimation. These products can be stored almost indefinitely, since bacteria cannot grow in them without water. CO 2 (s) exothermic CO 2 (g) Carbon dioxide is a good example of a solid that undergoes this process at atmospheric pressure. At reduced pressure other substances sublime. For example, freeze-dried foods are prepared by subliming water from a food product at low pressure. PROBLEM 4.22 Label each process as endothermic or exothermic and explain your reasoning: (a) sublimation; (b) deposition. Sample Problem 4.10 illustrates how molecular art can be used to depict and identify phase changes. SAMPLE PROBLEM 4.10 What phase change is shown in the accompanying molecular art? Is the process endothermic or exothermic? A B Analysis Identify the phase by the distance between the spheres and their level of organization. A solid has closely packed spheres that are well organized; a liquid has closely packed but randomly arranged spheres;
19 132 Chapter 4 Energy and Matter a gas has randomly arranged spheres that are far apart. Then, classify the transformation as melting, freezing, vaporization, condensation, sublimation, or deposition, depending on the phases depicted. Solution A represents a solid and B represents a liquid, so the molecular art represents melting. Melting is an endothermic process because energy must be absorbed to convert the more ordered solid state to the less ordered liquid state. PROBLEM 4.23 What phase change is shown in the accompanying molecular art? Is the process endothermic or exothermic? 4.7 eating and Cooling Curves The changes of state described in Section 4.6 can be illustrated on a single graph called a heating curve when heat is added and a cooling curve when heat is removed. 4.7A eating Curves A heating curve shows how the temperature of a substance (plotted on the vertical axis) changes as heat is added. A general heating curve is shown in Figure 4.4. Figure 4.4 eating Curve Temperature C boiling point melting point B melting C D liquid boiling E gas A solid eat added A heating curve shows how the temperature of a substance changes as heat is added. The plateau B C occurs at the melting point, while the plateau D E occurs at the boiling point.
20 4.7 eating and Cooling Curves 133 A solid is present at point A. As the solid is heated it increases in temperature until its melting point is reached at B. More heat causes the solid to melt to a liquid, without increasing its temperature (the plateau from B C). Added heat increases the temperature of the liquid until its boiling point is reached at D. More heat causes the liquid to boil to form a gas, without increasing its temperature (the plateau from D E). Additional heat then increases the temperature of the gas. Each diagonal line corresponds to the presence of a single phase solid, liquid, or gas while horizontal lines correspond to phase changes solid to liquid or liquid to gas. PROBLEM 4.24 Answer the following questions about the graph D E Temperature C B C A eat added a. What is the melting point of the substance? b. What is the boiling point of the substance? c. What phase(s) are present at plateau B C? d. What phase(s) are present along the diagonal C D? PROBLEM 4.25 If the substance shown in the heating curve in Figure 4.4 has a melting point of 50 C and a boiling point of 75 C, what state or states of matter are present at each temperature? a. 85 C b. 50 C c. 65 C d. 10 C e. 75 C
21 134 Chapter 4 Energy and Matter 4.7B Cooling Curves A cooling curve illustrates how the temperature of a substance (plotted on the vertical axis) changes as heat is removed. A cooling curve for water is shown in Figure 4.5. Figure 4.5 Cooling Curve for Water V Temperature C W gas condensation X Y liquid freezing Z solid eat removed The cooling curve shows how the temperature of water changes as heat is removed. The plateau W X occurs at the boiling point, while the plateau Y Z occurs at the freezing point. Gaseous water is present at point V. As the gas is cooled it decreases in temperature until its boiling point is reached at W. Condensation at 100 C forms liquid water, represented by the plateau from W X. Further cooling of the liquid water takes place until its freezing point (melting point) is reached at Y. Freezing water forms ice at 0 C, represented by the plateau from Y Z. Cooling the ice further decreases its temperature below its freezing point. PROBLEM 4.26 If the cooling curve in Figure 4.5 represented a substance with a melting point of 40 C and a boiling point of 85 C, what state or states of matter would be present at each temperature? a. 75 C b. 50 C c. 65 C d. 10 C e. 85 C 4.7C Combining Energy Calculations We have now performed two different types of energy calculations. In Section 4.5 we used specific heat to calculate the amount of energy needed to raise the temperature of a substance in a single phase such as heating a liquid from a lower to a higher temperature. In Section 4.6 we learned how to calculate energy changes during changes of state using heats of fusion or vaporization. Sometimes these calculations must be combined together to determine the total energy change when both a temperature change and a change of state occur. Sample Problem 4.11 illustrates how to calculate the total energy needed to convert a given mass of liquid water to steam at its boiling point.
22 4.7 eating and Cooling Curves 135 SAMPLE PROBLEM 4.11 ow much energy is required to heat 25.0 g of water from 25 C to a gas at its boiling point of 100. C? The specific heat of water is 1.00 cal/(g C), and the heat of vaporization of water is 540 cal/g. Analysis Use specific heat to calculate how much energy is required to heat the given mass of water to its boiling point. Then use the heat of vaporization to calculate how much energy is required to convert liquid water to a gas. Solution  Identify the original quantities and the desired quantity. mass = 25.0 g T 1 = 25 C T 2 = 100. C known quantities? calories desired quantity Subtract the initial temperature (T 1 ) from the final temperature (T 2 ) to determine the temperature change: T 2 T 1 = T = = 75 C.  Write out the conversion factors. Conversion factors are needed for both the specific heat and the heat of vaporization. specific heat conversion factors heat of vaporization conversion factors 1.00 cal 1 g 1 C or 1 g 1 C 1.00 cal 540 cal 1 g or 1 g 540 cal Choose the conversion factors that place the unwanted units (g C) and g in the denominator.  Solve the problem. Calculate the heat needed to change the temperature of water 75 C using specific heat. heat = mass T specific heat 1.00 cal cal = 25.0 g 75.0 C = 1,875 cal rounded to 1,900 cal 1 g 1 C Calculate the heat needed for the phase change (liquid water heat of vaporization. gaseous water) using the cal 540 cal = 25.0 g = 13,500 cal rounded to 14,000 cal 1 g Add the two values together to obtain the total energy required. Total energy = 1,900 cal + 14,000 cal = 15,900 cal rounded to 16,000 cal Answer PROBLEM 4.27 ow much energy (in calories) is released when 50.0 g of water is cooled from 25 C to solid ice at 0.0 C? The specific heat of water is 1.00 cal/(g C), and the heat of fusion of water is 79.7 cal/g. PROBLEM 4.28 ow much energy (in calories) is required to melt 25.0 g of ice to water at 0.0 C, heat the liquid water to 100. C, and vaporize the water to steam at 100. C? The specific heat of water is 1.00 cal/(g C), the heat of fusion of water is 79.7 cal/g, and the heat of vaporization of water is 540 cal/g.
23 136 Chapter 4 Energy and Matter KEY TERMS Boiling point (bp, 4.4) Calorie (4.1) Condensation (4.6) Cooling curve (4.7) Deposition (4.6) Dipole dipole interactions (4.3) Endothermic (4.6) Energy (4.1) Exothermic (4.6) Freezing (4.6) eating curve (4.7) eat of fusion (4.6) eat of vaporization (4.6) ydrogen bonding (4.3) Intermolecular forces (4.3) Joule (4.1) Kinetic energy (4.1) Law of conservation of energy (4.1) London dispersion forces (4.3) Melting (4.6) Melting point (mp, 4.4) Potential energy (4.1) Specific heat (S, 4.5) Sublimation (4.6) Vaporization (4.6) KEY CONCEPTS ❶ What is energy and what units are used to measure energy? (4.1) Energy is the capacity to do work. Kinetic energy is the energy of motion, whereas potential energy is stored energy. Energy is measured in calories (cal) or joules (J), where 1 cal = J. One nutritional calorie (Cal) = 1 kcal = 1,000 cal. ❷ What are the characteristics of the three states of matter? (4.2) A gas consists of randomly arranged, disorganized particles that are far apart and move very fast. A liquid consists of randomly arranged particles that are much closer and held together by attractive interactions. A solid consists of highly organized, very close particles held together by strong attractive forces. ❸ What types of intermolecular forces exist? (4.3) Intermolecular forces are the forces of attraction between molecules. Three types of intermolecular forces exist in covalent compounds. London dispersion forces are due to momentary changes in electron density in a molecule. Dipole dipole interactions are due to permanent dipoles. ydrogen bonding, the strongest intermolecular force, results when a atom bonded to an O, N, or F, is attracted to an O, N, or F atom in another molecule. ❹ ow are intermolecular forces related to a compound s boiling point and melting point? (4.4) The stronger the intermolecular forces, the higher the boiling point and melting point of a compound. ❺ What is specific heat? (4.5) Specific heat is the amount of energy needed to raise the temperature of 1 g of a substance by 1 C. Specific heat is used as a conversion factor to calculate how much heat a known mass of a substance absorbs or how much its temperature changes. ❻ Describe the energy changes that accompany changes of state. (4.6) A phase change converts one state to another. Energy is absorbed when a more organized state is converted to a less organized state. Thus, energy is absorbed when a solid melts to form a liquid, or when a liquid vaporizes to form a gas. Energy is released when a less organized state is converted to a more organized state. Thus, energy is released when a gas condenses to form a liquid, or a liquid freezes to form a solid. The heat of fusion is the energy needed to melt 1 g of a substance, while the heat of vaporization is the energy needed to vaporize 1 g of a substance. ❼ What changes are depicted on heating and cooling curves? (4.7) A heating curve shows how the temperature of a substance changes as heat is added. Diagonal lines show the temperature increase of a single phase. orizontal lines correspond to phase changes solid to liquid or liquid to gas. A cooling curve shows how the temperature of a substance changes as heat is removed. Diagonal lines show the temperature decrease of a single phase. orizontal lines correspond to phase changes gas to liquid or liquid to solid.
24 Understanding Key Concepts 137 UNDERSTANDING KEY CONCEPTS Selected in-chapter and odd-numbered end-of-chapter problems have brief answers at the end of each chapter. The Student Study Guide and Solutions Manual contains detailed solutions to all in-chapter and odd-numbered end-of-chapter problems, as well as additional worked examples and a chapter self-test. a. Which line segment corresponds to the following changes of state?  4.29 What phase change is shown in the accompanying molecular art? Is energy absorbed or released during the process?  4.30 What phase change is shown in the accompanying molecular art? Is energy absorbed or released during the process? b. What is the melting point of the substance? c. What is the boiling point of the substance? 4.32 Which line segments on the cooling curve in Problem 4.31 correspond to each of the following physical states? 4.31 Consider the cooling curve drawn below. V A B Temperature C W X Y Z A B C eat removed 4.33 Riding a bicycle at miles per hour uses 563 Calories in an hour. Convert this value to (a) calories; (b) kilocalories; (c) joules; (d) kilojoules Estimate the number of Calories in two tablespoons of peanut butter, which contain 16 g of protein, 7 g of carbohydrates, and 16 g of fat.
25 138 Chapter 4 Energy and Matter 4.35 What types of intermolecular forces are exhibited by each compound? Acetaldehyde is formed when ethanol, the alcohol in alcoholic beverages, is metabolized, and acetic acid gives vinegar its biting odor and taste Consider the two beakers containing the same mass of X and Y that were at the same initial temperature. Which compound, X or Y, has the higher specific heat if the same amount of heat was added to both substances? a. b. acetaldehyde acetic acid 4.36 Ethanol and dimethyl ether have the same molecular formula. 75 C 63 C X Y ethanol dimethyl ether a. What types of intermolecular forces are present in each compound? b. Which compound has the higher boiling point? 4.38 Consider the following three containers (A C) drawn below. (a) If the same amount of heat was added to all three samples, which sample has the highest temperature? (b) Which sample has the lowest temperature? ethanol ethanol 2-propanol A B C ADDITIONAL PROBLEMS Energy 4.39 Carry out each of the following conversions. a. 50 cal to kcal c kj to cal b. 56 cal to kj d. 4,230 kj to cal 4.40 Carry out each of the following conversions. a. 5 kcal to cal c kj to cal b. 2,560 cal to kj d. 4,230 J to kcal 4.41 Running at a rate of 6 mi/h uses 704 Calories in an hour. Convert this value to (a) calories; (b) kilocalories; (c) joules; (d) kilojoules Estimate the number of Calories in a serving of oatmeal that has 4 g of protein, 19 g of carbohydrates, and 2 g of fat A can of soda contains 120 Calories, and no protein or fat. ow many grams of carbohydrates are present in each can? 4.44 Alcohol releases 29.7 kj/g when it burns. Convert this value to the number of Calories per gram Which food has more Calories: 3 oz of salmon, which contains 17 g of protein and 5 g of fat, or 3 oz of chicken, which contains 20 g of protein and 3 g of fat? 4.46 Which food has more Calories: one egg, which contains 6 g of protein and 6 g of fat, or 1 cup of nonfat milk, which contains 9 g of protein and 12 g of carbohydrates?
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The first scheduled quiz will be given next Tuesday during Lecture. It will last 5 minutes. Bring pencil, calculator, and your book. The coverage will be pp 364-44, i.e. Sections 0.0 through.4. 0.7 Theory
CHAPTER 6 Chemical Bonding SECTION 1 Introduction to Chemical Bonding OBJECTIVES 1. Define Chemical bond. 2. Explain why most atoms form chemical bonds. 3. Describe ionic and covalent bonding.. 4. Explain
Balancing chemical reaction equations (stoichiometry) This worksheet and all related files are licensed under the Creative Commons Attribution License, version 1.0. To view a copy of this license, visit
Assessment Chapter Test A States of Matter MULTIPLE CHOICE Write the letter of the correct answer in the space provided. 1. Boyle s law explains the relationship between volume and pressure for a fixed
Name: Period: Chemical Formulas, Equations, and Reactions Test Pre-AP Write all answers on your answer document. 1. Which of the following is a NOT a physical property of hydrogen? A. It is gas C. It is
Alkanes Chapter 1.1 Organic Chemistry The study of carbon-containing compounds and their properties What s so special about carbon? Carbon has 4 bonding electrons. Thus, it can form 4 strong covalent bonds
Test 1 General Chemistry CH116 Summer, 2012 University of Massachusetts, Boston Name ESSAY. Write your answer in the space provided or on a separate sheet of paper. 1) Sodium hydride reacts with excess
.1.1 Measure the motion of objects to understand.1.1 Develop graphical, the relationships among distance, velocity and mathematical, and pictorial acceleration. Develop deeper understanding through representations
Chapter 5 Chemical Quantities and Reactions 5.1 The Mole Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans 1
Chapter 4 Lecture Notes Chapter 4 Educational Goals 1. Given the formula of a molecule, the student will be able to draw the line-bond (Lewis) structure. 2. Understand and construct condensed structural
Chapter 10: Temperature and Heat 1. The temperature of a substance is A. proportional to the average kinetic energy of the molecules in a substance. B. equal to the kinetic energy of the fastest moving
1. The elements on the Periodic Table are arranged in order of increasing A) atomic mass B) atomic number C) molar mass D) oxidation number 2. Which list of elements consists of a metal, a metalloid, and
Key Concepts 6.1 Systems Chapter 6 Thermodynamics: The First Law Systems, States, and Energy (Sections 6.1 6.8) thermodynamics, statistical thermodynamics, system, surroundings, open system, closed system,
Lake County, Lakeview, 9 th grade, Physical Science, Brent Starr Standard: H1P1 Explain how atomic structure is related to the properties of elements and their position in the Periodic Table. Explain how
Chapter 18 Temperature, Heat, and the First Law of Thermodynamics Problems: 8, 11, 13, 17, 21, 27, 29, 37, 39, 41, 47, 51, 57 Thermodynamics study and application of thermal energy temperature quantity
Week lectures--tentative 0.7 Kinetic-Molecular Theory 40 Application to the Gas Laws 0.8 Molecular Effusion and Diffusion 43 Graham's Law of Effusion Diffusion and Mean Free Path 0.9 Real Gases: Deviations
Modern Construction Materials Prof. Ravindra Gettu Department of Civil Engineering Indian Institute of Technology, Madras Module - 2 Lecture - 2 Part 2 of 2 Review of Atomic Bonding II We will continue
All of the chemical changes you observed in the last Investigation were the result of chemical reactions. A chemical reaction involves a rearrangement of atoms in one or more reactants to form one or more
EXAMPLE EXERCISE 3.1 Metric Basic Units and Prefixes Give the symbol for each of the following metric units and state the quantity measured by each unit: (a) gigameter (b) kilogram (c) centiliter (d) microsecond