OBJECTIVE: Introduction to Thermochemistry, energy units + energy unit conversion math, review of Table B with descriptions, review of Table I and descriptions, and the inclusion of heat emergy to the Stoichiometry of balanced chemical equations. 

Get out reference tables,

and a calculator

1. Thermo = heat, the energy changes that happen in a chemical reaction.

2. Sometimes heat is given off which is exothermic

3. Sometimes energy is absorbed to make the reaction happen, this is endothermic.   4. There are 4 Energy Units. Some you “know”, some will sound odd, that’s okay.

5A. Calories are “food” calories. That’s a loose word that really means 1000 calories (with a small “c”)

5B. 1 Calorie = 1000 calories

5C. A Calorie is a kilo-calorie or 1000 calories.  

6. In our class we will use CAPITAL C for the food or kilo calories.   

With a lower case “c”, calories are the “science” calories, a smaller amount of energy. There are 1000 calories in 1 Calorie.

 

7. A calorie is the amount of energy required to raise the temperature of one gram of pure water by exactly 1 Kelvin (or 1.0°C) 

1000 cal = 1 C = 1 kcal 

8. A Hershey bar has 230 Calories, how many calories is that? This is a one step unit conversion problem (do it now…)

 

230 C1

x 1000 cal1 C

= 230,000 cal

8. A Hershey bar has 230 Calories, how many calories is that? This is a one step unit conversion problem (do it now…)

 

9. Another unit of energy is the JOULE, named after J.P. Joule, the English physicist who studied energy & work. His ideas led to the development of the theory (now a law) of conservation of energy in a chemical reactions. (wowza!)   Joules and calories are related with this equality: 

4.18 Joules = 1.0 calories 

10. The last unit is kilo-Joule or kJ 11. 1000 Joules = 1 kilo-Joule  12. 1.0 C = 1000 cal = 4180 j = 4.180 kJ    Converting back and forth between units will

take some practice, but not much.

An idea that sometimes escapes students is this: Food is the equivalent of energy. You eat food, which has its energy content measured in Calories. Too much food eaten = too many Calories of energy eaten = too much Charlie Eat less food means taking in less energy. You get tired quicker, and begin to use the stored energy in your body (fat becomes energy, you lose mass) That’s called going on a diet. Food is stored by the body (this is basic biology) as fat. Food (energy) is stored to be used when the body needs energy. If you never need more energy because you live now, and there are supermarkets and lunchrooms, you don’t need to worry about your hunting skills, or the weather, then the season of less food never comes. Your body stores the excess energy you eat (in the form of cupcakes or salmon, food is energy), and you grow.  Food is energy. Calories turn into usable energy, or stored energy. The more food = the more energy 

Table B (let’s take a lookie-look at what gifts we have there!)

14. The title of table B is…

15. Three Physical Constants of Water are on table B. These 3 constants are for water. Every substance will have these 3 constants, although their numbers will be different than water’s numbers.

 

Table B  

16. The Heat of Fusion symbol is: HF

17. The heat of fusion for water means the amount of energy required to be added to one gram of ice at the melting point, to melt it into one gram of

water, without changing the temperature.

18. This means, 273 K solid to 273 K liquid, NO TEMPERATURE CHANGE This is only phase change

 

It’s the cold phase change energy for water.

 19. It also means the reverse: how much energy needs to be removed from one gram of liquid water at the freezing point to turn it into a solid.  

20. Heat of fusion for water is 334 Joules/gram 

This is ONE GRAM of ICE (CUBE)

= 1 gram of H2O(S)

To MELT this gram of ice it takes a CERTAIN AMOUNT of energy added in, called the HEAT OF FUSION for water.

21. To turn 1 gram of ICE into water it requires

334 Joules of energy be added

This is ONE GRAM of WATER.

= 1 gram of H2O(L)

To FREEZE this gram of ice it takes a CERTAIN AMOUNT of energy removed, called the HEAT OF FUSION for water.

22. To turn 1 gram of waterinto an ICE CUBE it requires

334 Joules of energy be removed

23. The Heat of Vaporization symbol is HV

24. The heat of vaporization for water means the amount of energy required to be added to one gram of water at the boiling point, to

vaporize it into one gram of steam gas without changing the temperature. This means, 373 K liquid to 373 K gas.

It’s the hot phase change energy for water.

 

25. It also means the reverse: how much energy needs to be removed from one gram of steam at the condensing point to turn it into a liquid.

 

26. The Heat of vaporization for water is 2260 J/g-----------------------------------------------

For water… HF = 334 J/g HV = 2260 J/g 

27. The last constant for water here is the specific heat capacity who’s symbol is C

28. The Specific Heat Capacity is the amount of energy required to change the temperature of one gram of water by one Kelvin (or one degree centigrade). This change can be heating, or cooling. Energy added will heat the water, energy removed will cool the water.

29. This constant can only be used when H2O is liquid phase, and

there is a change in temperature.

30. The specific heat capacity constant for water is C = 4.18 Joules/gram x Kelvin (which we write as…) 

31. C = 4.18 J/g·K The “dot” means multiply

-----------------------------------------------

DO this NOW: Go to table B. Under it, in small letters, add:1 C = 1000 cal 4.18 J = 1 cal 1000 J = 1 kJ 

Objective Thermo Class #2: Heats of Reaction, ΔH

32. Table I shows us the HEATS OF REACTION for 25 different balanced chemical reactions.

33. This means how much energy is released (exo) or absorbed (endo) by these reactions when they occur at room temperature and normal pressure. (this is not STP)

Let’s look over this first reaction on Table I Heats of Reaction the combustion of methane gas.

CH4 + 2O2 ---> CO2 + 2H2O

we can read this as: 1 mole CH4 combusts with 2 moles O2

forming into 1 mole CO2 + 2 moles H2O

34. The ΔH for this reaction is -890.4 kJ/mole35. This means: that when one mole of methane combusts, it releases 890.4 kJ of energy.

36. (the - sign only means exothermic!)

37. There’s NO SUCH THING as negative energy. All energy is energy.

38. So, this becomes...

CH4 + 2O2 ---> CO2 + 2H2O + 890.4 kJ

we can read this as:

1 mole CH4 combusts with 2 moles oxygen,

into 1 mole CO2 and 2 moles water + 890.4 kJ energy 

CH4 + 2O2 ---> CO2 + 2H2O + 890.4 kJ 

You should be able to do this stoichiometry problem:If 23.4 moles of methane combust, how many molecules of water form?

39. Well, now I’m telling you

that ENERGY is also in the mole ratio as well.

40. The mole ratio for the combustion of methane is 1:2:1:2:890.4 kJ

 

41. If 23.4 moles of methane combust, how much energy is released?

 

If I were feeling devilish today, I could ask this…

If 23.4 moles of methane combust, how many Calories of energy are released (or joules, or calories)?

Wait until tomorrow!

MRCH4

energy1

890.4 kJ23.4X kJ

X = (890.4)(23.4) = 20,835.36 kJ

X = 20,800 kJ with 3 SF

OBJECTIVE: to practice and master energy unit conversions, and to learn how to include energy into Stoichiometry.  

Calculators and reference tables now

42. You eat about 1450 Calories per day. Convert that amount of energy into calories, joules, and kilojoules. 

1450 C1

X 1000 cal1 C

= 1,450,000 caloriesKeep the 3 SF

42. You eat about 1450 Calories per day. Convert that amount of energy into calories, joules, and kilojoules. 

1,450,000 calories

1

x 4.18 Joules1 calorie

= 6,061,000 J

= 6,060,000 J with 3 SF

6,060,000 J1

X 1 kJ1000 J

= 6,060 kJ keeping 3 SF

43. A teaspoon of sugar (sucrose, C12H22O11) has 41 Calories. How many calories, joules and kilojoules is in 2 teaspoons of sugar? A normal amount in a cup of tea or coffee?

Show work for all three problems. 

82 C1

X 1000 cal1 C

= 82,000 cal with 2 SF

82,000 cal1

X 4.18 J1 cal

= 342,760 J = 340,000 Joules with 2 SF

340,000 J1

X 1 kJ1000 J

= 340 kJ with 2 SF

43. A teaspoon of sugar (sucrose, C12H22O11) has 41 Calories. How many calories, joules and kilojoules is in 2 teaspoons of sugar?

44. Propane gas, C3H8 combusts according to Table I How much energy (in kilo joules) is released when 5.75 moles C3H8 combusts?

Hint: this is a simple mole to mole type stoich problem

 

44. Propane gas, C3H8 combusts according to Table I How much energy (in kilo joules) is released when 5.75 moles C3H8 combusts?

C3H8 + 5 O2 ---> 3CO2 + 4 H2O ΔH = -2219.2 kJ

 

You need to get right now, that when 1 mole C3H8 combusts that 2219.2 kJ of energy is released –ΔH indicates exothermic. You also need to grasp that energy is part of the actual mole ratio. So,

  

MR C3H8

energy

12219.2 kJ

5.75X kJ X = 12760.4 kJ

X = 12,800 kJ with 3 SF

45. How much energy is absorbed by the reaction of 99.0 moles of hydrogen mono-iodide gas forming? (hint, look on table I)

H2(G) + I2(G) ---> 2HI (G) ΔH = +53.0 kJ

Which means, when 2 moles HI gas forms, then 53.0 kJ of energy is absorbed. Use the mole ratio to do this problem for 99.0 moles HI gas.

MRHI

energy2

53.0 kJ99.0X kJ

2X = 5247 kJ

X = 2623.5 kJ

X = 2620 kJ with 3 SF

45.

OBJECTIVE: Using table B constants with energy changes in real life problems

(get a calculator)

You have 35.0 g of ice at 273 K. If you put it into your mouth to melt it, how much energy will you impart onto that ice to convert it into liquid water without temperature change?

To do this we need a formula from table T, the heat of fusion formula. In this case we need to “un-fuse” ice into liquid water.

46. Write the formula, then under it, fill in the blanks, with units!

q = mHF

47. q is the amount of heat in joules m is the mass in grams of H2O HF is the specific heat capacity constant = 334 J/g

which is how much energy will it take to melt 1 g of ice with no change in temperature

48. You have 35.0 g of ice at 273 K. How much energy to melt it without raising the temp?

q = mHF

q = (35.0 g) x (334 J/g)

q = 11,690 J = 11,700 J with 3 SF

48. You have 35.0 g of ice at 273 K. How much energy to melt it without raising the temp?

q = mHF

49. Calculate how much energy,in kilojoules, it takes to freeze

1550 mL of H2O.

q = mHF

q = mHF

q = (1550 g) x (334 J/g)

q = 517,700 joules

q = 518,000 with 3 SF

49. Calculate how much energy,in kilojoules, it takes to freeze 1550 mL of H2O.

50. How many joules of energy will it take to vaporize 550. grams of water at the boiling point into gas at the same temperature?

Hint: this time we’ll use the heat of vaporization formula with

the HV constant.

This is not the fusing or unfusing of water at 273 K,

This is the HOT phase change at 373 Kelvin.50. First , write the formula

q = mHV

50. How many joules of energy will it take to vaporize 550. grams of water at the boiling point into gas at the same temp?

q = mHV

q = (550. g) x (2260 J/g)

q = 1,243,000 Joules = 1,240,000 J with 3 SF

51. When you jump into the bath water it will cool down as you absorb some energy and warm up and some energy is lost to the air. If you bath contains 625 liters of water, and it is at a comfortable 318 Kelvin, and after 15 minutes the water is cooled down to 301 Kelvin, how much energy has this water lost? Calculate joules and Kilojoules

Hint: this is a change in temperature formula problem, so you cannot use the

HF or HV formula, it’s time for the q = mCΔT

51. When you jump into the bath water it will cool down as you absorb some energy and warm up and some energy is lost to the air. If you bath contains 625 liters of water, and it is at a comfortable 318 Kelvin, and after 15 minutes the water is cooled down to 301 Kelvin, how much energy has this water lost?

q = mCΔT

51. When you jump into the bath water it will cool down as you absorb some energy and warm up and some energy is lost to the air. If you bath contains 625 liters of water, and it is at a comfortable 318 Kelvin, and after 15 minutes the water is cooled down to 301 Kelvin, how much energy has this water lost?

q = mCΔT

q = (625,000 g)(4.18 J/g.K)(17.0 K)

q = 44,412,500 J = 44,400,000 Joules with 3 SF

44,400,000 J1

X 1 kJ1000J

= 44,400 kJ

You know how when you go out to eat and order an appetizer and a meal and adessert, and each part has it’s own price on the check?

And then you order a soda too. You also have to pay tax on top of all this. You don’t get a little billfor each of these parts, the server adds it all up for you and you pay one price.

You’re smart enough (and suave enough)to realize how simple that makes things.

The same thing happens in thermochem (really).

52. What if you had say 155 g of ice at 273 K and let it sit in a room for four hours. The water reaches 294 K. What is the total number of joules that the H2O absorb?

This problem is like eating out, there are smaller energy events that must be added to a whole energy total (no tax though).

155 g of ice at 273; the water reaches 294 K.

52. First, MELT the ice…

q = mHF

Then, warm up the water

q = mCΔT

Finally, add them together.

52. What if you had say 155 g of ice at 273 K and let it sit in a room for four hours. Then the water has reached 294 K. How many joules of energy did this H2O absorb to do this?

q = mHF q = (155 g)(334 J/g) = 51770 J = 51,800 J

q = mCΔT q = (155 g)(4.18 J/g.K)(21.0 K) = 13605.9 J = 13,600 J

ANSWER = 51,800 J + 13,600 J = 65,400 J with 3 SF

53. Can you now convert that energy into Calories?

ANSWER = 51,800 J + 13,600 J = 65,400 J with 3 SF

53. Convert that energy into Calories?

65,400 J1

x 1 cal4.18 J

= 15,645.933 cal

= 15,600 cal

15,600 cal1

x1 Calorie1000 cal

= 15.6 C

54. On a particularly bad day, 16.3 grams of steam at exactly 373 K condenses onto your finger, then cools quickly to 305 Kelvin. How much energy have youabsorbed from this event? (you will be crying)

Hint: this first must condense through the hot phase change (heat of vaporization) then cool with a change in temperature formula as well.

q = mHV

q = mCΔT

54. On a particularly bad day, 16.3 grams of steam at exactly 373 K condenses onto your finger, then cools quickly to 305 Kelvin. How much energy have youabsorbed from this event?

q = mHV q = (16.3 g)(2260 J/g) = 36,838 J = 36,800 J with 3 SF

q = mCΔT q = (16.3 g)(4.18 J/g.K)(68.0 K) = 4633.112 J = 4630 J

ANSWER: 36,800 J + 4630 J = 41430 J =

41,400 J with 3 SF

ThermoChem Class #4 – Calorimetry

OB: how is energy in food measured? (It’s the “bomb”!)

You will need a calculator and need to put your thinking hat now.

By Law, all food sold in the USA has to have a food Nutrition Label, with one of the “facts” being the Caloric Content, how much energy are you eating, put onto the package.

It takes a special machine, and careful measurements, and fancy mathematics and conversions, to calculate this data.

In our Dorito’s Lab tomorrow, we will attempt to measure the energy content in the chips, and compare it to the nutrition label on the package.

55. Let’s Convert 450. Calories (3 SF) into Joules now.

55. Let’s Convert 450. Calories (3 SF) into Joules now.

450. C1

X 1000 cal1C

= 450,000 cal

450,000 cal1

X 4.18 Joules1 cal

= 1,880,000 J

An person emits about 100 Joules of energy/SECOND.You might imagine this: a WATT of energy is one Joule/second. A 100 Watt bulb uses 100 Joules/second.

56. There is no easy way to directly measure the energy that is in food. An indirect way has been well figured out, using a machine called a calorimeter.

The food is burned up in a sealed box containing oxygen, and the heat that it gives off, the energy that is inside the food, is used to heat pure water. This water is carefully measured, and the energy gained by the water changes its temperature. This temperature change is pushed through our q = mC∆T formula, and you can calculate the number of joules it took to heat up the water, and then convert the joules into cal, then into calories.

It seems harder than it is, but it is easy enough to do tomorrow.

In our Dorito’s Lab we will use a five cent calorimter machine. Real calorimeters are much more complex, and insulated. They measure all of the energy gain, we will “do” the same thing, but MOST of our energy will get lost to the air.

We’ll do the same process, just get poor results. Terrible measuring, great learning!

57. Let’s draw and label a calorimeter now. Picture on next slide.

Food goes into the cup. It’s sealed in the “bomb” with oxygen. The electricity leads are attached. This is placed into a certain mass of pure water, at measured temperature. The stirrer moves the water around and makes sure it’s all at the same temperature. At some point you spark and burn the food. Heat is released into the water and the temperature rises.

Using q = mC∆T, calculate the number joules required to make that temperature change, convert to cal, then Calories.

58. Let’s assume that there is exactly 2120. mL of water in our bomb calorimeter and it’s at exactly 295.0 Kelvin. After burning up our food sample, the temperature of the water rises to 354.5 Kelvin. How many Calories of energy are in this food sample?

To do this, start with the q = mC∆T formula. Solve for q.

58. Let’s assume that there is exactly 2120. mL of water in our bomb calorimeter and it’s at exactly 295.0 Kelvin. After burning up our food sample, the temperature of the water rises to 354.5 Kelvin. How many Calories of energy are in this food sample?

To do this, start with the q = mC∆T formula. Solve for q.

q = mC∆Tq = (2120. g)(4.18 J/g·K)(59.50 K)q = 527,265.2 J

527,265.2 J1 X 1 cal

4.18 J= 126,140 cal

126,140 cal = 126.1 Calories (4 SF)

59. In a different calorimeter you measure some food energy content. You have exactly 4005 mL of pure water at 274.2 Kelvin and it warms up to 365.4 Kelvin when you burn up 50.0 grams of chocolate delight dessert. What is the Calorie content of the dessert?

59. In a different calorimeter you measure some food energy content. You have exactly 4005 mL of pure water at 274.2 Kelvin and it warms up to 365.4 Kelvin when you burn up 50.0 grams of chocolate delight dessert. What is the Calorie content of the dessert?

q = mC∆Tq = (4005. g)(4.18 J/g·K)(91.20 K)q = 1,526,770.08 Joules

1,526,770.08 J1

X 1 cal4.18 J

= 365,256 cal

365,256 cal = 365.3 Calories (4 SF)

OBJECTIVE: Students will develop mastery of all thermochem math problems

Calculators, reference tables, and thinking caps (again).

60. Calculate the specific heat capacity constant for aluminum if your hunk of pure aluminum has mass of 147.2 g and the temperature changes from 279 K to 365 K when you impart 11,355 Joules of energy onto it.

60. Calculate the specific heat capacity constant for aluminum if your hunk of pure aluminum has mass of 147.2 g and the temperature changes from 279 K to 365 K when you impart 11,393 Joules of energy onto it.

q = mCΔT

11,355 J = (147.2g)(C)(86.0K)

11,355 J (147.2g)(86.0K)

0.897 J/g·K = C

= C

NOTE: metals have a fairly low specific heat capacity, water’s is 4.18 J/g·K

Water is a common enough substance in your life that the regents will not remind you what phase changes it might go through in any problem. You will have to determine that from the temperatures in the problem.

Ice can be colder than 273 K (0°C) and can be made colder or warmer while staying a solid. Water phase changes at 273 K.

When ever there is a temperature change (ice can change temp) we use the basic heat formula: q = mCΔT

If water is liquid, we use the 4.18 J/g·K constant.

When the H2O is solid ice, it has a different constant,

61. CICE = 2.10 J/g·K

When you are given information like this, you must use it in your problems, like the next slide will do. Write the ice constant near table Bin your reference table now.

62. For fun, you obtain a block of ice (2550 g) at -5.00°C and sit on it, in a tub until it melts to body temperature of 36.0°C. How much energy does that take? (do math on next slide)

q=mCΔT

C = 2.10 J/g·K

q = mHF

q=mCΔT

C = 4.18 J/g·K

268 K 309 K

0°C-5.00°C36.0°C

62. For fun, you obtain a block of ice (2550 g) at -5.00°C and sit on it in a tub until it melts to body temperature of 36.0°C. How much energy does that take?This is a 3 step thermochem problem, ice warms, ice melts, water warms.

Make note to use the proper “C” value for your H2O in the proper place.

Add up the three sets of Joules for one total answer.

62. For fun, you obtain a block of ice (2550 g) at -5.00°C and sit on it in a tub until it melts to 36.0°C

q=mCΔT = (2550g)(2.10J/g·K)(5.00 K) = 26,775 J

q = mHF = (2550g)(334J/g) = 851,700 J

q=mCΔT = (2550g)(4.18J/g·K)(36.0K) = 383,724 J

1,262,199 J

1,260,000 J with 3 SF

Thermochem is like eating out ina fancy restaurant, multi courses

with ONE total bill.

Add up the parts - sum.for total

63. Compare how much energy it takes to condense 15.0 g steam into water - to the amount of energy it takes to vaporize 15.0 g water into steam.

Vaporize 15.0 g waterq = mHV

Condense 15.0 g waterq = mHV

63. Compare how much energy it takes to condense15.0 g steam into water with the amount of energy it takes to vaporize 15.0 g water into steam.

vaporizeq = mHV

q = (15.0g)(2260 J/g)

q = 33,900 J

condenseq = mHV

q = (15.0g)(2260 J/g)

q = 33,900 J

64. When 56.0 grams of carbon and sufficient hydrogen synthesize completely into ethane (C2H6) gas, how much energy is released (or absorbed) in JOULES?

NOTE: ΔH is a negative number, that means energy is a product, which we can write this way…

2C + 3H2 C2H6 + 84.0 kJ

56.0 g C1

X 1 mole C

12 g C= 4.67 moles C

MR carbonenergy

184.0

4.67x

2X = 392.28 kJ X = 196 kJ

64. When 56.0 grams of carbon and sufficient hydrogen synthesize completely into ethane (C2H6) gas, how much energy is released (or absorbed) in JOULES?

The Cooling curve for Chromium metal

A

B

C

D

E

F

Heat removed at a constant rate (time)

KELVIN

1

2

65.

A. What temps are 1 + 2?

B. What’s PE doing BC and CD?

C. What’s KE doing AB and DE

D. Why is BC longer than DE?

E. Which thermochem formula do you use for BC?

F. How about for EF?

A

B

C

D

E

Heat removed at a constant rate (time)

KELVIN

1

2

Cooling curve for Chromium metal

F

66.

Why is BC longer than DE?

BC represents the condensation phasechange for chromium gas to liquid, which is a “bigger” energy event that the freezing of chromium.

A

B

C

D

E

F

Heat removed at a constant rate (time)

KELVIN

2944

2180

Cooling curve for Chromium metal

PE

PE

KE

KE

q = mHV

q = mCΔT

66.

Converting balanced chemical equations into balanced thermochemical equations.

67. Look at table I, choose the second equation: propane combusts. Write the balanced chemical equation with the ΔH

68. Write a balanced thermochemical equation

69. Find the most endothermic equation on Table I, write the balanced chemical equation with the ΔH

70. Write the balanced thermochemical equation for this reaction as well.

Converting balanced chemical equations into balanced thermochemical equations.

Look at table I, choose the second equation, propane combusts. Write the balanced chemical equation with the ΔH, then, write a balanced thermochemical equation below it (properly).

67. C3H8 + 5O2 3CO2 + 4H20 ΔH= -2219.2 kJ

68. C3H8 + 5O2 3CO2 + 4H20 + 2219.2 kJ

Energy is a PRODUCT in an exothermic reaction

Then, write the most endothermic equation of all, with the ΔH, then below that one, the balanced thermochemical equation properly.

69. 2C + H2 C2H2 ΔH = +227.4 kJ

70. 2C + H2 + 227.4 kJ C2H2

Energy is a REACTANT in an endothermic reaction

OBJECTIVE: A review of all types of thermochemistry problems

Celebration of Knowledge in

ThermochemistryIs on

____________________________

71. You decide to warm up some water for oatmeal. How much energy does it take to warm up 354 mL of water from 24.5°C to the boiling point and to let 7.50 grams vaporize making the teapot whistle. Be careful and do good calculations.

71. You decide to warm up some water for oatmeal. How much energy does it take to warm up 354 mL of water from 24.5°C to the boiling point and to let 7.50 grams vaporize making the teapot whistle. Be careful and do good calculations.

q = mCΔT = (354g)(4.18J/g·K)(75.5 K) = 111,718.86 J

q = mHV = (354g)(2260J/g) = 16950 J

111,718.86 J + 16960J = 128,668.86 J

129,000 Joules with 3SF

72. Write the balanced thermochemical equation for the synthesis of aluminum oxide, then calculate how much energy is released (or absorbed) when you form 21.0 moles of Al2O3.

72. Write the balanced thermochemical equation for the synthesis of aluminum oxide, then calculate how much energy is released (or absorbed) when you form 21.0 moles of Al2O3.

4Al + 3O2 → 2Al2O3 + 3351 kJ

MR EnergyAl2O3

3351 kJ2

X kJ21.0

2x = 70371 x = 35,185.5 kJ 35,200 kJ with 3 SF

73. Draw the heating curve for phosphorous. Indicate the actual MP and FP. Label graph points ABCDEF left to right.

What formula would you use

For BC

For CD

For DE

What happens to PE + KE at BC and CD?

73. Heating Curve for Phosphorous

A

B

C

D

E

F

317 K

554 K

BC use: q = mHF

CD use: q = mCΔT

DE use: q = mHV

BC: KE steady, PE increasesCD: KE increases, PE steady

Heat added at a constant rate →

74. 45.0 g of an unknown metal absorbs 1.51 kiloJoules of heat. The temperature changes from 268 K to 345 K. What is the specific heat capacity constant for this metal?

74. 45.0 g of an unknown metal absorbs 1.50 kiloJoules of heat. The temperature changes from 268 K to 345 K. What is the specific heat capacity constant for this metal?

q = mCΔT1510 J = (45.0g)(C)(77.0K)

1510J(45.0g)(77.0K) = C = 0.436

J/g·K3465g·K

75. Which takes more energy?

Melting 50.0 g of ice into water or

vaporizing 50.0 g of water into steam?

75. Which takes more energy?

Melting 50.0 g of ice into water or

vaporizing 50.0 g of water into steam?

No math needed, just think…

HF constant for water is 334 J/g

HV constant for water is 2260 J/g

It will take more than 7X the energy to vaporize than melt

76. Which takes more energy?

Heating 23.0 mL of water from 274 K to 299 K, or,

23.0 mL of water from 299 K to 323 K?

76. Which takes more energy?

Heating 23.0 mL of water from 274 K to 299 K, or,

23.0 mL of water from 299 K to 323 K?

Think. The ΔT for the first one is 25.0 K

The ΔT for the second one is only 24.0 K

NO PHASE CHANGES to think about, so the first one requires slightly more energy (4.18 J/g because it’s one more temperature unit of ΔT)

77. Which takes more energy?

Vaporizing 21.0 g water from 373 K liquid to gas

or,

changing the temp of 21.0 g H2O by 97.0 K?

77. Which takes more energy? [DO THE MATH THIS TIME]

Vaporizing 21.0 g water from 373 K liquid to gas, or,

100 g water from 274 K to 371 K?

q = mHV = (21.0g)(2260J/g) = 47,460 J = 47,500 J 3SF

q = mCΔT = (21.0g)(4.18J/g·K)(97.0K) = 8514.66 J = 8520 J 3SF

The first one takes more energy!

78. Which has the LOWEST AVERAGE KINETIC ENERGY?

100 mL water at 51.0°C

100 mL water at 50.0°C

175 mL water at 49.0°C

15.0 mL water at 23.0°C

78. Which has the LOWEST AVERAGE KINETIC ENERGY?

100 mL water at 50.0°C

100 mL water at 50.0°C

175 mL water at 49.0°C

15.0 mL water at 23.0°C

Lowest temp = lowest average KE always.

Same in reverse: Highest temp = Highest Average KE

This horse has a boo boo.

That’s an ICE PACK on his leg.

79. Describe thethermochem:

A. Heat flows from ice pack → leg C. Heat flows from leg → ice packB. Cold flows from ice pack → leg D. Cold flows from leg → ice pack

Energy can be transferred. Heat is energy, cold is THE LACK OF ENERGY.

Cold cannot flow.

What happens to you when you jump into a cold pool?

Heat leaves your body and you get chilly.

The COLD DOES NOT FLOW INTO YOU.

79. C. Heat flows from leg → ice pack

Now time to run through the Thermochem BASICS

Then, let’s look over the Thermochem Practice tests

2nd Q ends January 23, next Friday.

ALL BACK WORK DUE by 3:00 PM 1-23-15

Midterm is NEXT THURSDAY. Counts 2x celebrations.