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1 Energy and States of Matter 5 Answers and Solutions to Text Problems 5.1 At the top of the hill, all of the energy of the car is in the form of potential energy. As it descends down the hill, potential energy is being converted into kinetic energy. When the car reaches the bottom, all of its energy is in the form of motion (kinetic energy). 5.2 As the elevator (lift) moves to the top of the ramp, the skier s potential energy is increasing. As the skier descends down the ramp (ski jump), potential energy is being converted into kinetic energy. 5.3 a. potential b. kinetic c. potential d. potential 5.4 a. potential b. potential c. kinetic d. kinetic 5.5 a. Using a hair dryer converts electrical energy into heat energy, (the air is warmed). b. Using a fan converts electrical energy into kinetic energy. c. Burning gasoline converts chemical energy into kinetic energy as the car is propelled down the road, and heat energy as the car s components warm up. d. Radiant energy is converted into heat energy by the solar water heater. 5.6 a. Electrical energy becomes radiant and heat energies when the filament glows. b. Burning gas converts chemical energy into radiant and heat energies. c. Electrical energy is converted into heat energy. d. The chemical energy of the log is converted into radiant and heat energies. 5.7 Copper has the lowest specific heat of the samples and will reach the highest temperature. 5.8 The specific heat of A must be less than the specific heat of B because the amount of temperature change depends on the amount of energy supplied (or removed, if cooling), the amount of material present, and its specific heat. If the same amount of energy is applied to each substance and equal masses of A and B are used, the only variable that remains is the specific heat for each material. A substance with a low specific heat will produce a larger temperature change that a substance with a high specific heat. 5.9 a cal = 3.5 kcal b. 28 cal 4.18 J = 120 J 1 cal c. 425 J 1 cal = 102 cal 4.18 J d. 4.5 kj 1000 J 1 cal = 1100 cal 1 kj 4.18 J

2 Chapter 5 Answers and Solutions 5.10 a. 8. = 8100 cal b. 325 J 1 kj = kj 1000 J c cal 4.18 J 1 kj = 9.41 kj 1 cal 1000 J d kcal 4.18 J = 1.05 x 10 4 J 1 cal 5.11 a. T = 25 C 15 C = 10. C 25 g 1.00 cal 10. C = 250 cal b. 150 g 4.18 J 75 C = J c g 0.39 J 250. C = 980 J d. 150 g 4.18 J 62 C 1 kj = 39 kj 1000 J e g Ag 222 C 0.24 J = 530 J 5.12 a. T = 45 C 25 C = 20. C 85 g 1.00 cal 20. C = 1700 cal b kg 1000 g 4.18 J 20. C = J or 4.2 x 10 4 J 1 kg c g 0.19 cal 113 C = 5.4 kcal d. 25 g 4.18 J 25 C = 2600 J or J e. 5.0 kg 1000 g 1.00 cal 6 C = 30 kcal = kg 5.13 a. Rearrange the equation cal = g x _T x specific heat for mass (g) g = cal = 450 cal = 1200 g _T spec. heat (12 C)(0.031 cal/) b. g = cal = 2500 cal = 89 g _T spec. heat (28 C)(1.00 cal/) c. Rearrange the equation cal = g _T specific heat for specific heat spec. heat = cal = 320 cal = 0.13 cal/ g _T (25 g)(102 C) d. spec. heat = cal = 1500 cal = 0.19 cal/ g _T (36 g)(217 C)

3 Energy and States of Matter 5.14 Heat = g x T x sp. ht. Mass (g) = heat/( T x sp. ht.) a. T = 26 C 14 C = 12 C g = 180 cal = 260 g (2 SF) 12 C cal/g C b. g = 3200 cal = 80. g (2 SF) 40. C x 1.0 cal/g C c. spec. heat = heat/(g x T) = 260 cal = cal/ 28 g x 212 C d. sp. ht. = 2100 cal = 0.25 cal/ 45 g x 184 C 5.15 a g 1.00 cal 10. C 1 Cal = 5.0 Cal b g 1.00 cal 42 C 1 Cal = 210 Cal c g 1.00 cal 25 C 1 Cal = 25 Cal 5.16 To calculate heat, we need the mass of the water, its specific heat, and the temperature change. The only additional task is to divide the heat energy absorbed by the water, (which is equivalent to the energy released by the combustion of fuel), by the grams of octane burned, as follows: 1200 g 4.18 J 4 C 1 kj 1 = 40 kj/g octane 1000 J 0.50 g octane 5.17 Recall that a Calorie is equivalent to a kilocalorie! a. Because the orange juice contains both carbohydrate and protein, two calculations will be needed. 26 g carbohydrate 4 Cal = 100 Cal g carbohydrate 2 g protein 4 Cal = 8 Cal g protein Total: 100 Cal + 8 Cal = 110 Cal b. With only carbohydrate present, a single calculation is all that is required. 72 kcal 1 g carbohydrate = 18 g carbohydrate 4 kcal c. With only fat present, a single calculation is all that is required. 14 g fat 9 Cal = 130 Cal g fat d. Three calculations are needed: 30. g carbohydrate 4 kcal 1 Cal = 120 Cal g carbohydrate 15 g fat 9 kcal 1 Cal = 140 Cal g fat 5 g protein 4 kcal 1 Cal = 20 Cal g protein Total: 120 Cal Cal Cal = 280 Cal

4 Chapter 5 Answers and Solutions 5.18 Recall that a Calorie is equivalent to a kilocalorie. (All values are rounded to the tens place) a. Three calculations are needed: 6 g carbohydrate 4 kcal = 20 kcal g carbohydrate 16 g fat 9 kcal = 140 kcal g fat 7 g protein 4 kcal = 30 kcal g protein Total: 20 kcal kcal + 30 kcal = 190 kcal b. First, we need to calculate how many Calories are due to fat and carbohydrate: 7 g fat 9 Calories = 60 Calories g fat 9 g carbohydrate 4 Calories = 40 Calories g carbohydrate Total: 60 Calories + 40 Calories = 100 Calories Subtracting the Calories due to fat and carbohydrate from 110 Calories yields the Calories due to protein. From this information, we can calculate the mass of protein present in the soup, as follows: 110 Calories 100 Calories = 10 Calories 10 Calories 1 g protein = 3 g protein 4 Calories c. With just carbohydrate present, only a single calculation is required. 140 Calories 1 g carbohydrate = 35 g carbohydrate (sugar) 4 Calories d. First, we need to calculate how many Calories are due to carbohydrate and protein: 13 g carbohydrate 4 Calories = 50 Calories g carbohydrate 5 g protein 9 Calories = 50 Calories g protein Total: 50 Calories + 50 Calories = 100 Calories Subtracting the Calories due to carbohydrate and protein from the total of 405 Calories yields the Calories due to fat. From this information, we can calculate the mass of fat present in the avocado, as follows: 405 Calories 100 Calories = 305 (or 310) Calories 310 Calories 1 g fat = 34 g fat 9 Calories 5.19 Three calculations are needed: 9 g protein 4 kcal = 40 kcal g protein 12 g fat 9 kcal = 110 kcal g fat 16 g carbohydrate 4 kcal = 64 kcal g carbohydrate Total: 40 kcal kcal + 64 kcal = 210 kcal Then 210 kcal 4.18 kj = 880 kj

5 5.20 Three calculations are needed: 70. g carbohydrate 4 kcal = 280 kcal g carbohydrate 150 g protein 4 kcal = 600 kcal (5.0 x 10 2 kcal) g protein 5.0 g fat 9 kcal = 45 kcal g fat Total: 280 kcal kcal + 45 kcal = 930 kcal Energy and States of Matter Then 280 kcal 4.18 kj = 1200 kj 600 kcal 4.18 kj = 2500 kj 45 kcal 4.18 kj = 190 kj Total: 1200 kj kj kj = 3900 kj 5.21 a. gas b. gas c. solid 5.22 a. liquid b. gas c. gas 5.23 a. dipole-dipole b. ionic c. dispersion d. hydrogen bond e. dispersion 5.24 a. dipole-dipole b. ionic c. dipole-dipole d. dispersion forces e. hydrogen bond 5.25 a. HF; hydrogen bonds are stronger than dipole-dipole interactions of HBr b. NaF; ionic bonds are stronger than the hydrogen bonds in HF c. MgBr 2 ; ionic bonds are stronger than the dipoles-dipole interactions in PBr 3 d. C 4 H 10 has more electrons and therefore more dispersion forces than CH a. MgCl 2 b. H 2 O c. NH 3 d. HF 5.27 Both b (liquid water freezes) and d (solid butter melts) describe processes that involve a change of state Both a (solid water sublimes) and c (solid1 nitrogen melts) describe processes that involve a change of state a. melting b. melting c. sublimation d. freezing 5.30 a. sublimation b. melting c. freezing d. freezing 5.31 a. 65 g ice 80. cal = 5200 cal absorbed 1 g ice b g ice 80. cal 4.18 J = 5700 J absorbed 1 g ice 1 cal c. 225 g water 80. cal = 18 kcal released 1 g water d g water 80. cal 4.18 J 1 kj = 17 kj released 1 g water 1 cal 1000 J

6 Chapter 5 Answers and Solutions 5.32 a. 35 g water 80. cal = 2800 cal (released) 1 g water b. 250 g water 80. cal 4.18 J = J or 8.4 x 10 4 J (released) 1 g water 1 cal c. 140 g ice 80. cal = 1 (absorbed) 1 g ice d. 5.0 kg ice 1000 g 80. cal 4.18 J 1 kj = 1700 kj (absorbed) 1 kg 1 g ice 1 cal 1000 J 5.33 a. condensation b. evaporation c. boiling d. condensation 5.34 a. boiling b. condensation c. evaporation d. boiling 5.35 a. The liquid water in perspiration absorbs heat and changes to vapor. The heat needed for the change is removed from the skin. b. On a hot day, there are more liquid water molecules in the damp clothing that have sufficient energy to become water vapor. Thus, water evaporates from the clothes more readily on a hot day. c. Some water molecules evaporate but they cannot escape from the sealed bag. The high humidity in the bag allows some of the gaseous water to condense back to liquid, so the clothes will not dry a. The liquid absorbs heat from the skin to evaporate, thus the skin is chilled (numbed). b. In a wide dish, there are more molecules at the surface from which evaporation must occur. The number of molecules with sufficient energy to evaporate and in position to evaporate is greater in the wide dish, so the water evaporates faster than it can in a tall glass. c. The sandwich on a plate dries out faster. Any water that evaporates from the sandwich sealed in plastic wrap cannot escape, and the sandwich does not dry out a g water 540 cal = 5400 cal absorbed 1 g water b g water 540 cal 4.18 J = 110,000 J absorbed 1 g water 1 cal c. 8.0 kg steam 1000 g 540 cal = 4300 kcal released 1 kg 1 g steam d. 170 g steam 540 cal 4.18 J 1 kj = 380 kj released 1 g steam 1 cal 1000 J 5.38 a g steam 540 cal = cal (released) 1 g steam b. 75. g steam 540 cal 4.18 J = J or 1.7 x J (released) 1 g steam 1 cal c. 44 g water 540 cal = 24 kcal (absorbed) 1 g water d. 5.0 kg water 1000 g 540 cal 4.18 J 1 kj = kj (1.1 x 10 4 kj absorbed) 1 kg 1 g water 1 cal 1000 J

7 Energy and States of Matter 5.39 T C 100 boiling point gas 50 liquid 0 melting point solid Heat added 5.40 gas T C 100 condensation point (boiling point) 50 liquid 0 freezing point solid Energy removed 5.41 a g 1.00 cal 57 C = 1100 cal b. Two calculations are needed: 50.0 g ice 80. cal = 4000 cal (2 sig figs) 1 g ice g 1.00 cal 65 C = 3300 cal Total: 4000 cal cal = 7300 cal c. Two calculations are needed: 15 g steam 540 cal 4.18 J 1 kj = 34 kj 1 g steam 1 cal 1000 J 15 g 4.18 J 100. C 1 kj = 6.3 kj 1000 J Total: 34 kj kj = 40. kj d. Three ca1culations are needed: 24 g ice 80. cal = 1.9 kcal 1 g ice 24 g 1.00 cal 100. C = 2.4 kcal 24 g water 540 cal = 13 kcal 1 g water Total: 1.9 kcal kcal + 13 kcal = 17 kcal

8 Chapter 5 Answers and Solutions 5.42 a. Two calculations are needed: 125 g steam 540 cal = cal or cal 1 g steam 125 g 1.00 cal 85 C = cal or 1.1 x 10 4 cal Total: cal + 1 = cal or cal b. Two calculations are needed: 525 g ice 80. cal 4.18 J = J or J 1 g ice 1 cal 525 g 4.18 J 15 C = J or J Total: J J = J or J c. Three ca1culations are needed: 85 g steam 540 cal = 46 kcal 1 g steam 85 g 1.00 cal 100. C = 8.5 kcal 85 g water 80. cal = 6.8 kcal 1 g water Total: 46 kcal kcal kcal = 6 d. Two calculations are needed: 55 ml water 1.0 g 1.00 cal 90. C = 5.0 kcal 1 ml water 55 ml water 1.0 g 540 cal = 30. kcal 1 ml water 1 g Total: 5.0 kcal kcal = 35 kcal 5.43 Two calculations are needed: 250 g ice 80. cal = 20. kcal 1 g ice 250 g 1.00 cal 21 C = 6.0 kcal Total: 20. kcal kcal = 26 kcal 26 kcal 4.18 kj = 110 kj

9 Energy and States of Matter 5.44 Three ca1culations are needed: 115 g steam 540 cal = 62 kcal 1 g steam 115 g 1.00 cal 100. C = 11.5 kcal 115 g water 80. cal = 9.2 kcal 1 g water Total: 62 kcal kcal kcal = 83 kcal 62 kcal 4.18 kj = 260 kj 11.5 kcal 4.18 kj = 48.1 kj 9.2 kcal 4.18 kj = 38 kj Total: 260 kj kj + 38 kj = 350 kj 5.45 From Table 5.1, we see that liquid water has a high specific heat (1.00 cal/g C), which means that a large amount of energy is required to cause a significant temperature change. Sand, on the other hand, has a low specific heat (0.19 cal/g C). Even a small amount of energy will cause a significant temperature change in the sand Because the iced tea is cold, the air surrounding the glass is also cooled as the air molecules collide with the glass. Any moisture (humidity) in the cooled air will also be cooled sufficiently to undergo a phase change from gas to liquid. The condensed water forms the droplets on the glass of iced tea Both water condensation (formation of rain) and deposition (formation of snow) from the gaseous moisture in the air are exothermic processes (heat is released). The heat released in either of these processes warms the surrounding air, and so the air temperature is in fact raised a. As the liquid water freezes, heat must be released into the air. The temperature in the orchard will not go below 0 C until all the water has frozen. b. Two calculations are required: 5.0 kg 1000 g 1.00 cal 15 C = 75 kcal 1 kg g C 5.0 kg 1000 g 80. cal = 400 kcal ( kcal) 1 kg 1 g For a total of: 75 kcal x 10 2 kcal = 480 kcal ,000 Cal = cal 1 Cal cal = g specific heat T T = cal = cal = 40 C g spec. heat 50,000 g 1.00 cal/g C Final T = 20 C + 40 C = 60 C

10 Chapter 5 Answers and Solutions 5.50 The data that 18.9 kj are released when 0.50 g of oil is burned start the calculation. The remainder of the calculation involves unit conversions kj 1000 J 1 cal = 9.0 kcal/g oil 0.50 g oil 1 kj 4.18 J 5.51 a. For 15% of one s total Calories (kcal) to be supplied by protein, a conversion factor of 15 kcal from protein/100 kcal total in the daily diet will be used in the calculation. Similar factors will be used for the carbohydrate (carbs) and fat calculations kcal (total) 15 kcal (protein) 1 g protein = 45 g protein 100 kcal (total) 4 kcal (protein) 1200 kcal (total) 45 kcal (carbs) 1 g carbs = 140 g carbohydrate 100 kcal (total) 4 kcal (carbs) 1200 kcal (total) 40. kcal (fat) 1 g fat = 53 g fat 100 kcal (total) 9 kcal (fat) b. The calculations for part b differ from part a only in the total kcal per day kcal (total) 15 kcal (protein) 1 g protein = 71 g protein 100 kcal (total) 4 kcal (protein) 1900 kcal (total) 45 kcal (carbs) 1 g carbs = 210 g carbohydrate 100 kcal (total) 4 kcal (carbs) 1900 kcal (total) 40. kcal (fat) 1 g fat = 84 g fat 100 kcal (total) 9 kcal (fat) c. The calculations for part c again differ only in the total kcal per day kcal (total) 15 kcal (protein) 1 g protein = 98 g protein 100 kcal (total) 4 kcal (protein) 2600 kcal (total) 45 kcal (carbs) 1 g carbs = 290 g carbohydrate 100 kcal (total) 4 kcal (carbs) 2600 kcal (total) 40. kcal (fat) 1 g fat = 120 g fat 100 kcal (total) 9 kcal (fat) 5.52 The meal contains a total of 45 g protein, (31 g + 3 g + 11 g), a total of 49 g fat, and a total of 120 g carbohydrate. The total caloric content of the meal must be determined to answer the question. 45 g protein 4 kcal = 180 kcal due to protein 1 g protein 49 g fat 9 kcal = 440 kcal due to fat 1 g fat 120 g carbohydrate 4 kcal = 480. kcal due to carbohydrate 1 g carbohydrate Total: 180 kcal kcal kcal = 1.10 x 10 3 kcal (1100 kcal) Using Table 5.5, 2.0 hours are needed to burn off the total caloric content of the meal kcal 1 hr = 2.0 hr 550 kcal

11 Energy and States of Matter 5.53 Because each gram of body fat contains 15% water, a person actually loses 85 grams of fat per hundred grams of body fat. (We considered 1 lb of fat as exactly 1 lb.) 1 lb body fat 454 g 85 g fat 9 kcal = 3500 kcal 1 lb 100 g body fat 1 g fat 3500 kcal 4.18 kj = 15,000 kj 5.54 This question requires three calculations to obtain the total kcal. 12 g carbohydrate 4 kcal = 50 kcal due to carbohydrate 1 g carbohydrate 9 g fat 9 kcal = 80 kcal due to fat 1 g fat 9 g protein 4 kcal = 40 kcal due to protein 1 g protein Total: 50 kcal + 80 kcal + 40 kcal = 170 kcal g 4.18 J 28 C 1 kj = 85 kj g C 1000 J gal 4 qt 946 ml 0.74 g 11.5 kcal = 4.8 x 10 4 kcal 1 gal 1 qt 1 ml 1 g 5.57 a. HF b. H 2 O c. KCl d. NH a. Dipole-dipole attractions occur between the positive end of one polar molecule and the negative end of another polar molecule. b. Hydrogen bonds are strong dipole-dipole bonds that occur between a partially positive hydrogen atom and one of the strongly electronegative atoms of F, O, or N. c. Dispersion forces occur between temporary dipoles that form within nonpolar molecules. Larger nonpolar compounds containing a greater number of electrons produce more temporary dipoles within the molecule In nonpolar compounds, there are no dipoles. However, they do interact by dispersion forces from temporary dipoles that result from a momentary shift of electrons. Because octane, C 8 H 18, has more electrons than ethane, C 2 H 6, octane can form larger temporary dipoles, which results in a higher melting point The hydrogen bonds between NH 3 molecules are stronger than the dipole-dipole bonds in PH 3 and require higher boiling point to break a. 3 b. 3 c. 4 d. 2 e. 3 f C 0 C: 325 g 25 C 4.18 J 1 kj = 34kJ g C 1000 J water 0 C ice 0 C: 325 g 80. cal 4.18 J 1 kj = 110 kj 1 g 1 cal 1000 J Total: 34 kj 110 kj = 140 kj

12 Chapter 5 Answers and Solutions kg 1000 g cal 300. C = cal available to melt ice 1 kg g C Since 80. cal are needed to melt 1 gram of ice, we can calculate the number of grams of ice that can melt cal 1 g = 320 g of ice will be melted 80. cal boiling gas T( C) liquid 20 0 solid melting a. liquid b. benzene melts c. liquid d. gas e. boiling temperature of 80.1 C 5.65 First, calculate heat required to melt the ice. Second, combine the masses of melted ice and liquid water, (both of which are at 0 C), and then continue with the heating and evaporation calculations g ice 80. cal = 4.0 kcal are required to melt the ice 1 g ice 50.0 g melted ice g liquid water = g liquid water at 0 C g 1.00 cal 100. C = 10.0 kcal (to warm the liquid) g C g water 540 cal = 54 kcal (to boil the liquid) 1 g water Total: 4.0 kcal kcal + 54 kcal = 68 kcal

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