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Daniel R. Barnes Init 3/12/2014

HEAT. HYPERINDEX BUTTON Jump to a place!. Daniel R. Barnes Init 3/12/2014. Based on Prentice Hall Chemistry, Chapter 17.

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Daniel R. Barnes Init 3/12/2014

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  1. HEAT HYPERINDEX BUTTON Jump to a place! Daniel R. Barnes Init 3/12/2014 Based on Prentice Hall Chemistry, Chapter 17 WARNING: This presentation includes images and other intellectual property taken from the Internet without the permission of the owners. It is meant to be viewed only by Mr. Barnes’ current chemistry students on a non-profit basis. Do not download, copy, post, or otherwise distribute this presentation. Do not store this presentation anywhere, including on any hard drive, flash drive, or other storage device. Do not upload or store any copy of it on the cloud or any other remote storage facility.

  2. 17.1 The Flow of Energy -- Heat and Work

  3. WORD OF ADVICE You will not have time in class to copy down every single word in this presentation. * Use abbreviations & symbols so you can write quickly. * Write only the most important ideas, but leave lots of space in your notes so you can write down other stuff you remember when you study your notes at home tonight. * Click through this ppt online on hhscougars.org – the media center has computers even if you don’t, and it’s usually open late.

  4. 221oC Look at Figure 13.3 on page 388. 100oC water 21oC water 50oC water 10oC water 1oC water Relative # of molecules Molecular speed

  5. 221oC Look at Figure 13.3 on page 388. 100oC water 21oC water 50oC water 10oC water 1oC water Relative # of molecules Molecular speed

  6. 221oC Look at Figure 13.3 on page 388. 21oC water 21oC water, larger amount THINK-PAIR-SHARE! Relative # of molecules Molecular speed

  7. 221oC Look at Figure 13.3 on page 388. 21oC water >21oC water, larger amount THINK-PAIR-SHARE! Relative # of molecules Molecular speed

  8. 221oC Look at Figure 13.3 on page 388. 21oC water <21oC water, smaller amount THINK-PAIR-SHARE! Relative # of molecules Molecular speed

  9. SWBAT . . . . . . explain the difference between temperature and heat. Temperature = average molecular kinetic energy T = KEave Kinetic energy = the energy of motion KE = ½ mv2 m = mass v = speed (In physics, you learn that velocity is speed in a certain direction.) The numerical value of temperature is proportional to average molecular kinetic energy when expressed in Kelvins, but not when expressed in oC or oF. If all you care about is how hot something is, use any units you want. If the energy of each particle matters, use only Kelvins.

  10. SWBAT . . . . . . explain the difference between temperature and heat. T = KEave KE = ½ mv2

  11. SWBAT . . . . . . explain the difference between temperature and heat. Temperature = average molecular kinetic energy Heat = q = energy transferred between two systems as a result of a temperature difference. Heat flows because of differences in temperature. “Thermal equilibrium” Heat flows from the hotter system to the colder system.

  12. SWBAT . . . . . . explain the difference between temperature and heat. T = KEave . WARNING: OVERSIMPLIFICATION ALERT!. H = KEtotal [$/person and total $ for group analogy]

  13. SWBAT . . . . . . explain the difference between endothermic and exothermic  q = positive Endothermic = absorbing heat  q = negative Exothermic = releasing heat Do some Q-squats! H = enthalpy = heat energy content q = DH = change in heat energy content = energy absorbed q & DH, being energy, are measured in calories or joules

  14. Cha-CHING! The bank has $3.7 million in deposits. That’s like . . . h KE q T ? I have $20,000 in my little bank account. That’s like . . . h KE q T ? The average bank account has $13,237 in it. That’s like . . . h KE q T ? On Sunday, all the customers combined made a total of $40,821 in deposits That’s like . . . h KE q T ?

  15. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems DT = Tf - Ti DT = change in temperature Tf = final temperature = temperature at the end of the story Ti = initial temperature = temperature at the beginning of the story Do #1 from Mr. Barnes’ Heat Math Worksheet (First Exposure) Do #2 from Mr. Barnes’ Heat Math Worksheet (First Exposure)

  16. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems C = specific heat capacity = “specific heat” C = how hard it is to change the temperature of a particular material Look at Table 17.1 on page 508 in your textbook Which material has the highest C? Which material has the lowest in the table? Notice anything about the trend in the metals in the table? The bigger the metal’s atomic mass is . . . the easier it is to DT it.

  17. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems C = how hard it is to DT a material C = how much energy a material abs/rel when it DT’s

  18. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems q = m DT Cp (Cp = C at constant pressure = when the material changes temperature in an open container – don’t stress about the “p”) q = m DT Cp q = energy absorbed m = mass DT = change in temperature = Tf - Ti * The more massive something is, the more energy it takes to heat it up. * Bigger temperature increases require more energy to happen. * The higher the specific heat of a material, the more energy it takes to heat it up.

  19. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems q = m DT Cp This is the form of the formula on your CST reference sheet. As written, it’s really set up to solve for . . . q What if you want to solve for Cp? q = m DT Cp q Cp = m DT m DT m DT Remember that DT = Tf – Ti Other than that you can just plug in the #’s & solve.

  20. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems Let’s try one as a class! This is #7 from Mr. Barnes’ “Heat Math Worksheet (First Exposure)”. 7. What is the specific heat of an oily liquid if it takes 2000 joules to heat up 7 grams of the liquid from 20oC to 720oC? Cp = ? J/(g oC) q = 2000 J m = 7 g Ti = 20oC Tf = 720oC DT = Tf - Ti = 720oC – 20oC = 700oC q = m DT Cp q q = m DT Cp Cp = m DT m DT m DT 2000 J 2000 J/goC Cp = = = 0.408 J/goC = Cp (7 g)(700oC) 4900

  21. C = ? Q = ? T = ? m = ?

  22. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems Take about two minutes to get started on #8 . . . TPS . . . SKIP TO ANSWER

  23. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for Cp) . . . solve q = m DT Cp problems

  24. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for q) . . . solve q = m DT Cp problems Let’s try one where q is the unknown. This is easier in a way because the formula on the reference sheet is already set up to solve for q. Take about two minutes to get started on #3 . . . TPS . . . 3. The specific heat of water is 1 cal/goC. How much heat energy is required to heat 2 grams of water up from 23 degrees Celsius to 93 degrees Celsius?

  25. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for q) . . . solve q = m DT Cp problems 3. The specific heat of water is 1 cal/goC. How much heat energy is required to heat 2 grams of water up from 23 degrees Celsius to 93 degrees Celsius? Cp = 1 cal/goC q = ? calories m = 2 g Ti = 23oC Tf = 93oC DT = Tf - Ti = 93oC – 23oC = 70oC q = m DT Cp = (2 g)(70oC)(1 cal/goC) = 140 cal = q SKIP TO ANSWER

  26. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for q) . . . solve q = m DT Cp problems 3. The specific heat of water is 1 cal/goC. How much heat energy is required to heat 2 grams of water up from 23 degrees Celsius to 93 degrees Celsius? Cp = 1 cal/goC q = ? calories m = 2 g Ti = 23oC Tf = 93oC DTf = Tf - Ti = 93oC – 23oC = 70oC q = m DT Cp = (2 g)(70oC)(1 cal/goC) = 140 cal = q

  27. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for q) . . . solve q = m DT Cp problems If you’re an honors section, do #4 for homework. If you’re a normal section, do it now. You have two minutes.

  28. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for T) . . . solve q = m DT Cp problems And now, to solve for temperature . . . Unfortunately, there isn’t one on the worksheet, so we’ll have to pull one out of my question bank . . . Let’s make it #17 on the ws. YOU MUST BECOME A PROBLEM-SOLVER!

  29. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for T) . . . solve q = m DT Cp problems 17. A corn cob is heated up in boiling water to a temperature of 100 oC. The corn cob has a mass of 400 g and absorbed 16,000 cal of heat from the boiling water. What was the original temperature of the corn cob? Assume that the specific heat of the corn cob is 1 cal/goC. Tf = 100 oC m = 400 g q = 16,000 cal Ti = ? oC Cp = 1 cal/goC DT = Tf – Ti = ? Ti = Tf – DT q = m DT Cp q = m DT Cp 16,000 cal DT = = 40 oC = DT m Cp m Cp (400 g) (1 cal/goC) SKIP TO ANSWER Ti = Tf – DT = 100 oC – 40 oC = 60 oC = Ti

  30. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for T) . . . solve q = m DT Cp problems 17. A corn cob is heated up in boiling water to a temperature of 100 oC. The corn cob has a mass of 400 g and absorbed 16,000 cal of heat from the boiling water. What was the original temperature of the corn cob? Assume that the specific heat of the corn cob is 1 cal/goC. Tf = 100 oC m = 400 g q = 16,000 cal Ti = ? oC Cp = 1 cal/goC DT = Tf – Ti = ? Ti = Tf – DT q = m DT Cp q = m DT Cp 16,000 cal DT = = 40 oC = DT m Cp m Cp (400 g) (1 cal/goC) Ti = Tf – DT = 100 oC – 40 oC = 60 oC = Ti

  31. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for m) . . . solve q = m DT Cp problems And now, let’s do a q = m DT Cp problem where m = ? . . . 6. How much unobtainium can be heated from 40 oC to 440 oC if only 40,000 joules of energy are available to heat it up? Assume that the specific heat of unobtanium is 0.4 J/goC. m = ? g Ti = 40 oC Tf = 440 oC q = 40,000 J = DT Cp = 0.4 J/goC DT = Tf - Ti = 440 oC - 40 oC = 400 oC q = m DT Cp q = m DT Cp 40,000 J m = DT Cp DT Cp (0.4 J/goC) (400 oC) 40,000 g 40,000 g = 250 g = m m = = (400)(0.4) 160 SKIP TO ANSWER

  32. C = ? Q = ? T = ? m = ? SWBAT . . . (Solving for m) . . . solve q = m DT Cp problems And now, let’s do a q = m DT Cp problem where m = ? . . . 6. How much unobtainium can be heated from 40 oC to 440 oC if only 40,000 joules of energy are available to heat it up? Assume that the specific heat of unobtanium is 0.4 J/goC. m = ? g Ti = 40 oC Tf = 440 oC q = 40,000 J = DT Cp = 0.4 J/goC DT = Tf - Ti = 440 oC - 40 oC = 400 oC q = m DT Cp q = m DT Cp 40,000 J m = DT Cp DT Cp (0.4 J/goC) (400 oC) 40,000 g 40,000 g = 250 g = m m = = (400)(0.4) 160

  33. Q Q Q Q = m DT Cp Cp = DT = m = DT Cp m DT m Cp

  34. Q Q Q = m DT Cp Tf - Ti = DT = m Cp m Cp DT = Tf - Ti Q + Ti Tf = m Cp Tf = DT + Ti Q Ti = Tf - m Cp Ti = Tf - DT

  35. SWBAT . . . . . . solve q = m DT Cp problems Try to do problems 1, 2, 4, & 5 on the Heat Math Worksheet (First Exposure).

  36. Okay, it’s time for the 17.1 Post-Quiz This might or might not count toward your grade. Barnes?

  37. 17.2 Measuring and Expressing Enthalpy Changes

  38. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Calorimetry = “the precise measurement of the heat flow into or out of a system for chemical and physical processes” (Prentice Hall Chemistry, 2005) System = “a part of the universe upon which you focus your attention” (Prentice Hall Chemistry, 2005) surroundings A freshly-boiled egg is a single object, but it’s made of parts (yolk, white, shell) system The hot egg is surrounded by cool water.

  39. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters You could use this apparatus to measure the energy given off by a boiling-hot (100 oC) egg as it cools to room temperature in one liter of cool water. Tf = 28 oC You would have to measure the temperature of the water before the egg was dropped into it . . . Ti = 20 oC surroundings As the egg cools off, it gives energy to the water, causing the water to heat up. system Eventually, the water and the egg reach the same temperature. Ti = 100 oC Tf = 28 oC

  40. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters The energy the egg gives off is exactly equal to the energy the water absorbs . . . and keeps IF the calorimeter is well-insulated. Tf = 28 oC CONSERVATION of ENERGY Ti = 20 oC surroundings system styrofoam

  41. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters This is a close-up of a tiny bubble, but it’s not a normal bubble. polystyrene Instead of the skin being made of water or soapy water, the skin of this bubble is made of polystyrene plastic.

  42. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters The inside of the bubble is filled with air. polystyrene air

  43. air air air air air air air air air air air air air SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters A bunch of these bubbles stuck together makes a “foam”. The air in each bubble is a “dead air space” because the air in one bubble can’t mix with air in other bubbles.

  44. air SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters air air air The air in each bubble is a “dead air space” because the air in one bubble can’t mix with air in other bubbles. air air air air air air air air air

  45. air SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Dead air spaces are almost as good as vacuums at preventing heat flow. air air Each bubble in a “closed cell” foam is a tiny dead air space. air air air air Dead air spaces are excellent thermal insulators. Each bubble in a “closed cell” foam is a tiny dead air space. air air air air air air

  46. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters It’s not the polystyrene plastic in the styrofoam that makes styrofoam such a good insulator . . . . . . it’s the air trapped in the bubbles that makes it such a good insulator. Dead air spaces, trapped air masses that don’t mix with other air, are excellent insulators.

  47. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Q 100 oC 50 oC 100 oC 99 oC Q metal styrofoam 50 oC 0 oC 0 oC 1 oC

  48. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters Metals are excellent thermal conductors Styrofoam is an excellent thermal insulator Q 50 oC 99 oC Q metal styrofoam 1 oC 50 oC

  49. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters The energy the egg gives off is exactly equal to the energy the water absorbs . . . and keeps IF the calorimeter is well-insulated. The egg is exothermic and the water is endothermic. Tf = 28 oC CONSERVATION of ENERGY q egg = - q water Ti = 20 oC q water = m DT Cp surroundings q = (1000 g)(8 oC)(1 cal/goC) system q water = 8000 cal q egg = - 8000 cal styrofoam

  50. SWBAT . . . . . . Compare and contrast constant-pressure calorimeters and bomb calorimeters The egg may have given off more heat than that, but some heat may have leaked into the air. This calorimeter would be more accurate if it had a styrofoam lid. The lid provides even more insulation. A styrofoam lid on a styrofoam cup is not typically an airtight seal. styrofoam surroundings Therefore, the air pressure inside the calorimeter is always equal to the air pressure outside. system styrofoam

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