PROPERTIES OF MATTER

Transcription

1 HAPTER 4 PROPERTIES OF MATTER SOLUTIONS TO REVIEW QUESTIONS 1. Gaseous state: 393 K ¼ ; boiling point of acetic acid is Liquid state: melting point of chlorine is and boiling point of chlorine is Small bubbles appear at each electrode, and a gas collects above each electrode. The system now contains water and two different gases. 4. A new substance is always formed during a change, but never formed during changes. 5. Potential energy is the energy of position. By the position of an object, it has the potential of movement to a lower energy state. Kinetic energy is the energy matter possesses due to its motion. 6. (a) 118:0 þ 273:15 ¼ 391:2K (b) ð118:0 Þð1:8Þþ32 ¼ 244:4 F 7. (a) ð16:7 Þð1:8Þþ32 ¼ 62:1 F (b) 16:7 þ 273:15 ¼ 289:9K 8. Boiling water is a change so the make-up of the water does not change. Therefore, the bubbles that you see in boiling water are... water vapor! 9. Dissolving salt in water is a change. If the salt-water solution is boiled the water will evaporate away leaving the salt behind. -24-

2 SOLUTIONS TO EXERISES 1. (a) (b) (c) 2. (a) (b) (c) (g) (h) (g) (h) 3. Although the appearance of the platinum wire changed during the heating, the original appearance was restored when the wire cooled. No change in the composition of the platinum could be detected. 4. A copper wire, like the platinum wire, changes to a glowing red color when heated ( change). Upon cooling, the original appearance of the copper wire is not restored, but a new substance, black copper(ii) oxide appears ( change). 5. Reactants: copper, oxygen Product: copper(ii) oxide 6. Reactant: water Product: hydrogen, oxygen 7. (a) (b) (c) 8. (a) (b) (c) 9. (a) kinetic energy (b) kinetic energy (c) potential energy 10. (a) kinetic energy (b) potential energy (c) potential energy potential energy kinetic energy kinetic energy kinetic energy 11. The kinetic energy is converted to thermal energy (heat), chiefly in the brake system, and eventually dissipated into the atmosphere. 12. The transformation of kinetic energy to thermal energy (heat) is responsible for the fiery reentry of a space vehicle. -25-

3 13. (a) þ (b) þ (c) 14. (a) þ (b) (c) þ þ 15. heat ¼ ðmþðspecific heatþðdtþ ¼ ð125 gþð0:900 J=g Þð95:5 19:0 Þ ¼ 8: J 16. heat ¼ ðmþðspecific heatþðdtþ ¼ ð65 gþð0:128 J=g Þð98:5 22:0 Þ ¼ 6: J 17. heat ¼ ðmþðspecific heatþðdtþ ¼ ð25:0gþð2:138 J=g Þð78:5 22:5 Þ ¼ 2: J 18. heat ¼ ðmþðspecific heatþðdtþ ¼ ð35:0gþð2:604 J=g Þð82:4 21:2 Þ ¼ 5: J 19. heat ¼ ðm Þðspecific heatþðdtþ; change kj to J specific heat ¼ heat mðdtþ ¼ 2: J ð135 gþð100:0 19:5 Þ ¼ 0:230 J=g 20. heat ¼ ðm Þðspecific heatþðdtþ; change kj to J specific heat ¼ heat mðdtþ ¼ 1: J ð275 gþð327:5 21:2 Þ ¼ 0:128 J=g 21. heat lost by copper ¼ heat gained by water x ¼ final temperature ð155:0gþð0:385 J=g Þð150:0 xþ ¼ ð250:0gþð4:184 J=g Þðx 19:8 Þ 8950 J 59:675x J= ¼ 1046 x J= J 8950 J þ J ¼ 1046 x J= þ 59:675x J= J ¼ 1106 x J= 26:8 ¼ x -26-

4 22. heat lost by copper ¼ heat gained by water x ¼ final temperature ð225:0gþð0:900 J=g Þð125:5 xþ ¼ ð500:0gþð4:184 J=g Þðx 22:5 Þ J 202:5x J= ¼ 2090 x J= J J þ J ¼ 2090 x J= þ 202:5x J= J ¼ 2290 x J= 31:7 ¼ x 23. (a) heat lost by metal ¼ heat gained by water ð110:0gþðspecific heatþð92:0 24:2 Þ ¼ ð75:0gþð4:184 J=g Þð24:2 21:0 Þ ðspecific heatþ 7458 g ¼ 1004:16 J specific heat ¼ 0:13 J=g (b) Iron has a specific heat of 0.47 J/g and lead has a specific heat of 0.13 J/g. The metal could possibly be lead, but further tests would be needed to determine this. The metal cannot be iron. 24. (a) heat lost by metal ¼ heat gained by water ð40:0gþðspecific heatþð62:0 21:0 Þ ¼ ð85:0gþð4:184 J=g Þð21:0 19:2 Þ specific heat 1640 g ¼ 640:152 J specific heat ¼ 0:39 J=g (b) Gold has a specific heat of 0.13 J/g. Therefore, the metal could not be pure gold. It is possible that the metal is copper, since the specific heat of copper is J/g. Further tests would be needed to determine the identity of the metal. 25. When wood burns carbon dioxide is released into the air. Therefore, the mass of the wood before burning will not equal the mass of the wood after burning. However, the law of conservation of mass still holds true. Some of the original mass of the wood is converted to gases that are released into the atmosphere. 26. The product is a compound. The tin reacted with oxygen from the air, producing an oxide of tin. 27. heat ¼ ðmþðspecific heatþðdtþ heat mass ¼ ðspecific heatþðdtþ ¼ 1: J ð4:184 J=g Þð100:0 23:0 Þ ¼ 52:5gH 2O -27-

5 28. heat ¼ ðmþðspecific heatþðdtþ heat mass ¼ ; convert grams to pounds ðspecific heatþðdtþ 3: J ¼ ð0:131 J=g Þ 1064:4 23:2 ¼ 238 g gold ð Þ ð238 gþ 1lb 453:6g ¼ 0:525 lb gold 29. heat ¼ ðmþðspecific heatþðdtþ change kj to J x ¼ final temperature 4: J ¼ ð500:0gþð4:184 J=g Þðx 10:0 Þ 4: J ¼ 2092x J= J 60,920 J ¼ 2092x J= 60,920 ¼ 2092x 29:1 ¼ x 30. heat lost by iron ¼ heat gained by water x ¼ mass of iron in grams xð0:473 J=g Þð125:0 25:6 Þ ð375 gþð4:184 J=g Þð25:6 19:8 Þ 47:0x J=g ¼ 9: J x ¼ 9:1 103 J 47:0J=g ¼ 190 g Fe 31. heat lost by copper ¼ heat gained by water x ¼ mass of copper in grams xð0:385 J=g Þð275:1 29:7 Þ ¼ ð272 gþð4:184 J=g Þð29:7 21:0 Þ 94:5x J=g ¼ 9: J x ¼ 9:9 103 J 94:5J=g ¼ 1:0102 gu 32. (a) heat ¼ ðmþðspecific heatþðdtþ ð100:0g Þð0:0921 cal=g Þð100:0 10:0 Þ ¼ 829 cal to heat u (b) let x ¼ temperature of Al after adding 829 cal 829 cal ¼ ð100:0 gþð0:215 cal=g Þðx 10:0 Þ 829 cal ¼ ð21:5 cal= Þx 215 cal x ¼ 48:6 ðfinal temperature for aluminumþ Therefore the copper gets hotter since it ended up at Note: You can figure this out without calculation if you consider the specific heats of the metals. Since the specific heat of copper is much less than aluminum the copper heats more easily. -28-

6 33. heat lost by iron ¼ heat gained by water x ¼ initial temperature of iron ðm ð500:0gþð0:473 J=g 237 J Þðspecific heatþðdtþ ¼ ðmþðspecific heatþðdtþ Þðx 90:0 Þ ¼ ð400:0gþð4:184 J=g Þð90:0 10:0 Þ x 2: J ¼ 1: J 237 J x ¼ 1: J x ¼ 1: J 237 J x ¼ heat lost by metal ¼ heat gained by water x ¼ specific heat of metal ð20:0gþðxþð203: 29:0 Þ ¼ ð100:0gþð4:184 J=g Þð29:0 25:0 Þ ð3480 g Þx ¼ 1674 J x ¼ 0:48 J 35. heat lost ¼ heat gained x ¼ final temperature ðspecific heats are the sameþ Let x ¼ final temperature J ð10:0gþ g 50:0 J ð xþ ¼ ð50:0gþ g ðx 10:0 Þ 500:g 10:0gx ¼ 50:0 gx 500 g 1: g ¼ 60:0 gx x ¼ 1: g ¼ 16:7 60:0g 36. Specific heats for the metals are Fe: J/g ; u: J/g ; Al: J/g. The metal with the lowest specific heat will warm most quickly, therefore, the copper pan heats fastest, and fries the egg fastest. 37. In order for the water to boil both the pan and water must reach Specific heat for copper is J/g. ð300:0gþ 0:385 J g ð100: 25 Þþð800:0gÞ 4:184 J g ð100: 25 Þ ¼ 8: J þ 2: J ¼ 2: J needed to heat the pan and water to 100 ð2: s JÞ ¼ 414 s ¼ 6:9 min ¼ 6 min þ 54 s 628 cal The water will boil at 6:06 and 54 s p.m. -29-

7 38. Heat is transferred from the molecules of water on the surface to the air above them. As you blow you move the warmed air molecules away from the surface replacing them with cooler ones which are warmed by the coffee and cool it in a repeating cycle. Inserting a spoon into hot coffee cools the coffee by heat transfer as well. Heat is transferred from the coffee to the spoon lowering the temperature of the coffee and raising the temperature of the spoon. 39. The potatoes will cook at the same rate whether the water boils vigorously or slowly. Once the boiling point is reached the water temperature remains constant. The energy available is the same so the cooking time should be equal. 40. ð250 mlþð0:04þ ¼10 ml fat ð10 mlþð0:8g=mlþ ¼8 g fat in a glass of milk 41. mercury þ sulfur! compound The mercury and sulfur react to form a compound since the properties of the product are different from the properties of either reactant. ð100:0mlþ 13:6 g ¼ 1: g mercury ml mercury þ sulfur! compound 1360 g þ 100:0 g 1460 g This supports the Law of onservation of Matter since the mass of the product is equal to the mass of the reactants. 42. According to the law of conservation of energy the amount of potential energy the ball has initially should equal the amount of kinetic energy it has at the bottom of the hill. If the hill that the ball rolled down was frictionless then the ball should role up the other side until it has reached the same level as where it started. However, no hill is really frictionless. Therefore, some of the ball s potential energy is converted to kinetic energy of motion and some is lost by way of heat friction between the ball and the surface of the hill. 43. A change has occurred. Hydrogen molecules and oxygen molecules have combined to form water molecules. 44. (a) As the water is heated, molecules that exist in the liquid state are changed into molecules in the gas gaseous state. (b) A change has occurred. No new substance was formed during the heating process. Only a change in state occurred. 45. Upon heating the substance, the appearance of the bright blue solid changes to black and a brownish orange gas is released. A change has occurred. -30-