1. Specific heat and phase changes
ENERGY
Specific heat and phase changes

2. DEFINITION OF CHEMISTRY
Reminder: At the beginning of the year we defined chemistry as the study of matter and energy.
So far:
We have looked at many properties of matter (structure of atoms, molecular shapes, physical/chemical properties, etc)
Now:
We look at the properties of energy in matter

3. WHAT IS ENERGY? Energy: the ability to do work or produce heat
Two types of energy
Potential Energy: energy due to the composition or position of an object (stored energy)
Kinetic Energy: energy of motion

4. ENERGY Chemical systems contain both kinetic and potential energy.
Potential Energy: the energy stored in the chemical bonds of a molecule
Kinetic Energy: the vibration or movement of molecules in a substance
This is related to the temperature of an object

5. IMPORTANT CONCEPT
An important concept in science is the law that describes energy
Law of conservation of energy: In any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created or destroyed.
This concept will become more important when we discuss chemical reactions

6. TOPICS IN ENERGY
We will be discussing two aspects of energy in this unit
Specific Heat
Changes in States of Matter

7. WHAT IS HEAT?
For the next minute, discuss in your groups what you think is the definition of heat.

8. DEFINITION OF HEAT
Heat: energy that is in the process of flowing from a warmer object to a cooler object
Therefore, heat is a relative term
95° is cool when it is summer in Vegas
50° is warm when it is winter in Wisconsin
Heat is different from energy and temperature. Again, temperature is a measure of the movement of molecules

9. UNIT FOR HEAT AND ENERGY
The unit for both heat and energy are the same. We will be discussing 2 units:
Calorie (cal)
On your food lables, the calorie actually represents 1000 cal or 1 kilocalorie (1 kcal)
Joule (J)

10. CALORIE
Calorie (cal): defined as the amount of energy needed to raise the temperature of one gram of water 1°C
Joule (J): this is the SI (metric) unit of energy
1 cal = 4.184J

11. DIMENSIONAL ANALYSIS
Again, we must use dimensional analysis to convert between units
REMEMBER: 1 cal = 4.184J
This is your conversion
Always write down the unit you start with and cancel out the units

12. EXAMPLE
You release 250 calories of energy from a chemical sample. How much energy is this in the unit of Joules?

13. ANSWER 250 cal | 4.184 J = 1046J 1 cal REMINDER: Sig figs.
You start with 2 sig figs. (250)
You need 2 sig figs. in your answer
Therefore: 1.0x103 J

14. TRY THESE
A food item says it contains 2.35x103J of energy. What is this in calories (cal)?
You start with 399 cal of energy. What is this in Joules (J)?
A yogurt contains 170 calories according to the label. What is this in Joules (J)? (REMINDER: Food label calories are really kilocalories)

15. WHAT ARE SOME USES OF ENERGY?
Have you ever noticed that when you boil water, the pot heats up much faster than the water in the pot?
Why does the pot heat up faster than the water?
Discuss in your groups for 1 minute why you think this is true.

16. SPECIFIC HEAT
The reason is that water needs to absorb more heat than the pot to increase in temperature
This property of matter is called specific heat
SPECIFIC HEAT: the amount of heat required to raise the temperature of one gram of a substance by 1°C

17. EXAMPLE The specific heat of water is: 4.184 J/(g • °C)
We’ll discuss the unit in a minute
The specific heat of concrete is J/(g • °C)
Therefore the specific heat of water is about 5 times bigger than concrete
What does this mean?

18. ANSWER On a hot day, concrete requires less energy to heat
Therefore, concrete heats up much faster than water
This is why you don’t walk on concrete barefoot, but you can still go in the water to cool off
Water doesn’t heat up very fast so it stays cooler

19. HOW HOT DOES SOMETHING GET?
We defined earlier what specific heat was
We also showed the units involved in specific heat:
J/(g • °C)
We will now discuss how to calculate how hot something gets when exposed to a certain amount of heat or energy

20. q = c x m x δT EQUATION FOR HEAT
To explain temperature change, we have to explain the equation for calculating heat:
q = c x m x δT

21. DEFINING THE TERMS
q: the heat absorbed or released from a substance in Joules (J) or calories (cal)
c: the specific heat of the substance
m: the mass of the substance in grams (g)
δT: the change in temperature in °C (or Tfinal – Tinitial)

22. EXAMPLE
Aluminum has a specific heat of J/(g • °C). If you add 250 cal of energy to 150g of aluminum, how much does the temperature increase?

23. STEP 1
Write down your variables and make sure they are in the proper units:
q = 250 cal
c = J/(g • °C)
m = 150g
δT = ?
NOTE: Heat is in calories and specific heat is in Joules, we need to convert

24. STEP 1 q = 250 cal 250 cal | 4.184J = 1.0x103 J 1 cal
c = J/(g • °C)
m = 150g
δT = ?

25. STEP 2 Plug your values into the formula: q = c x m x δT
1.0x103J = J/(g • °C) x 150g x δT

26. STEP 3 Perform your calculations as you cancel out units:
1.0x103J = J/(g • °C) x 150g x δT
1.0x103J = 135 J/°C x δT
135 J/°C J/°C
7.4 °C = δT

27. TRY THESE
Water has a specific heat of J/(g • °C). If you add 125J of energy to 22.2g of water, how much does the temperature increase?
Granite has a specific heat of J/(g • °C). If you add 125J of energy to 22.2g of granite, how much does the temperature increase?

28. ANSWER
1.35°C
7.01°C
Therefore granite heats up about 5times faster than water

29. OTHER CALCULATIONS WITH SPECIFIC HEAT
In addition for solving for temperature change, you can solve for specific heat, amount of heat or the mass of the object.
Let’s explore some examples

30. EXAMPLE 1
An object absorbs 1.33x104J of energy. If it has a mass of 15.7g and the temperature increased 25°C, what is the specific heat of the object?
REMINDER: Make sure you have the correct units in the answer.

31. ANSWER 1
34 J/g•°C

32. EXAMPLE 2
An object has a mass of 35.0g and the temperature increased 15°C. If the object has a specific heat of J/g•°C, how much heat was absorbed?

33. ANSWER 2
340 J

34. EXAMPLE 3
A piece of lead has an initial temperature of 25°C. The specific heat of lead is J/g•°C. If 55.5g of the lead absorbs 250J of energy, what is the final temperature?
REMINDER: δT = (Tfinal – Tinitial)

35. ANSWER 3 δT = 35°C Tinitial = 25°C δT = Tfinal – Tinitial
35°C = Tfinal - 25°C
Tfinal = 60°C

36. EXAMPLE 4
An object has a mass of 12.5g and the temperature DECREASED 5.7°C. If the object has a specific heat of J/g•°C, how much heat was released?
NOTE: If temperature decreases, you will have a negative “-” δT

37. ANSWER 4
-46J

38. HOW DO WE MEASURE HEAT?
So far we have had a lot of example problems, solving for heat (q), change of temperature (δT), mass (m) and specific heat (c)
What method do scientists have to measure “heat”?

39. CALORIMETER Scientists measure heat using a calorimeter.
Calorimeter: an insulated device that is used to measure the amount of heat released or absorbed during a physical or chemical process

41. SUMMARY
Almost all of the energy from the reaction vessel is transferred to the water
The temperature change of the water is observed (no energy is lost because it is insulated)
You calculate the energy gained by the water (you know the specific heat of water, the mass of water, and can read the δT)
This is the amount of heat from the reaction in the chamber

42. EXAMPLE
You heat up 3.243kg of an unknown metal and place it in the water of a calorimeter. As the metal cools by 8.2°C, it raises the temperature of the water 15°C and then remains constant. What is the specific heat of the metal if the metal was placed in 100g of water?
KNOWN: Specific heat of water: J/g•°C

43. STEP 1 Calculate the amount of heat absorbed by the water
q = c x m x δT (for water)
q = (4.184 J/g•°C)(100g)(15°C)
qwater = 6276J
qwater = qmetal
Therefore, qmetal = 6276J

44. STEP 2 Calculate the specific heat of the metal
q = 6276J
c = ?
m = 3.243kg = 3243g
δT = 15°C
q = c x m x δT (for water)
6276J = c (3243g)(8.2°C)
c = J/g•°C

45. STEP 3 Use the specific heat to figure out the unknown metal
Look on page 520 of the text at your desk
Which metal was put in the calorimeter?

46. ANSWER
SILVER

47. TRY THE FOLLOWING
You heat up 1.617kg of an unknown metal and place it in the water of a calorimeter. As the metal cools 5.5°C, it raises the temperature of the water 5.5°C and then remains constant. What is the metal, if the metal was placed in 250g of water?
KNOWN: Specific heat of water: J/g•°C

48. ANSWER
CALCIUM

49. REVIEW Earlier in the year, we discussed phase change diagrams
In your groups, create a phase change diagram:
Solid  Liquid;
Liquid  Gases

50. PHASE CHANGE DIAGRAM Stays in gaseous state. Gas gets hotter
Liquid is heating up. Liquid molecules speed up
Substance is going from liquid to gas. The bonds holding the liquid together break apart
Solid is heating up. Solid increases in temperature
Substance is going from a solid to liquid. The bonds holding the solid molecules together break apart

51. SPECIFIC HEAT
We have already discussed the energy needs for one aspect of the phase change diagram
The energy needed to increase the temperature of a substance is dependent on the specific heat
SPECIAL NOTE: the specific heat of a substance depending on what state of matter it is in

52. Liquid: 4.184 J/g•°C Solid: 2.03 J/g•°C Gas: 2.01 J/g•°C EXAMPLE
Specific heats for water:
Liquid: J/g•°C
Solid: J/g•°C
Gas: 2.01 J/g•°C

53. CALCULATING HEAT NEEDS
Knowing the specific heats of substances allows us to determine how much energy is needed to change the temperature of a substance
What happens to energy needs when there is no change in temperature?
EXAMPLE: Phase changes

54. SOLID  LIQUID
Before we describe the energy needs for going from solid to liquid, we need to describe how solids and liquids behave at the molecular level

55. SOLID  LIQUID WHAT WE KNOW ABOUT SOLIDS: Solids are close together
They are not free moving (bound tightly together)
The bonds vibrate, but are fixed
Solids tend to be more dense than liquid (this is why solids tend to sink in the liquid of the same matter)
EXCEPTION: Water

56. SOLID  LIQUID WHAT WE KNOW ABOUT LIQUIDS:
Liquids are still close together
They are free moving
The bonds that held them together as a solid have been broken apart
Liquids are free to move around, but cannot expand further because of intermolecular forces (δ+ and δ-)
EXAMPLE: Water  Hydrogen bonds

57. SOLID  LIQUID
Therefore, to go from a solid to a liquid, you need enough energy to break the bonds holding the solid together
This is called the heat of fusion
Heat of Fusion: the energy needed to melt one gram of a solid substance

58. SOLID  LIQUID Since there is no temperature change for heat of fusion
All the energy is being used to break the bonds holding the solid together
The formula is:
q = m x Hfus

59. EXAMPLE
You have 250g of ice. The heat of fusion for water is 334 J/g. How much energy is need to melt the ice?
q = ?
m = 250g
Hfus = 334 J/g
q = (250g)(334 J/g) = 8.4x104 J

60. TRY THESE
You have 125g of solid acetic acid. The heat of fusion is 195 J/g. What is the heat needed to melt the acetic acid?
You have g of solid ethanol. The heat of fusion is 107 J/g. What is the heat needed to melt the ethanol?

61. ANSWERS
2.44X104 J
0.368J

62. LIQUID  GAS WHAT WE KNOW ABOUT GASES: Gases are very far apart
They are free moving and rapid
The bonds that held them together as a liquid have been broken apart
Gases do not have any force of attraction holding them together

63. LIQUID  GAS
Therefore, to go from a liquid to a gas, you need enough energy to break apart the intermolecular forces holding the liquid together
This is called the heat of vaporization
Heat of Vaporization: the energy needed to vaporize (or boil) one gram of a liquid substance

64. LIQUID  GAS
Since there is no temperature change for heat of vaporization
All the energy is being used to break the intermolecular forces holding the liquid together
The formula is:
q = m x Hvap

65. EXAMPLE
You have 250g of liquid water. The heat of vaporization for water is J/g. How much energy is need to boil the water?
q = ?
m = 250g
Hvap = 2262 J/g
q = (250g)(2262 J/g) = 5.7x105 J

66. TRY THESE
You have 125g of liquid acetic acid. The heat of vaporization is 390 J/g. What is the heat needed to boil the acetic acid?
You have g of liquid ethanol. The heat of vaporization is 836 J/g. What is the heat needed to boil the ethanol?

67. ANSWERS
4.88X104 J
28.8 J

68. PUTTING IT ALL TOGETHER
Now it is possible to determine the total energy used in a system
The things to remember
There is a different specific heat for each state of matter (solid, liquid, gas)
Each substance has a heat of fusion for melting or freezing
Each substance has a heat of vaporization for boiling or condensing

69. EXAMPLE
You have 3500g of solid water (ice) at -15°C. How much energy is needed to convert all of this water to a gas with a temperature of 110°C?
Hfus = 334 J/g
Hvap = 2262 J/g
csolid = 2.03 J/g•°C
cliquid = J/g•°C
cgas = 2.01 J/g•°C

70. STEP 1 Calculate the heat for increasing the temperature of the solid
Make sure to use the correct specific heat
q1 = c x m x δT
q1 = (2.03 J/g•°C) x (3500g) x (15°C)
q1 = 1.1x105 J

71. STEP 2 Calculate the heat needed to convert from a solid to a liquid
q2 = m x Hfus
q2 = (3500g) x (334 J/g)
q2 = 1.2x106 J

72. STEP 3 Calculate the heat for increasing the temperature of the liquid
Make sure to use the correct specific heat
q3 = c x m x δT
q3 = (4.184 J/g•°C) x (3500g) x (100°C)
q3 = 1.5x106 J

73. STEP 4 Calculate the heat needed to convert from a solid to a liquid
q4 = m x Hvap
q4 = (3500g) x (2262 J/g)
q4 = 7.9x106 J

74. STEP 5 Calculate the heat for increasing the temperature of the gas
Make sure to use the correct specific heat
q5 = c x m x δT
q5 = (2.01 J/g•°C) x (3500g) x (10°C)
q5 = 7.0x104 J

75. STEP 6 Add all of the heats together to get a total
q = q1 + q2 + q3 + q4 + q5
q = 1.1x105J + 1.2x106J + 1.5x106J + 7.9x106J + 7.0x104J
q = 1.1 x107 J

76. SPECIAL NOTE You will not always have to do all the steps
Only make the conversions you need
Example: Taking 20g of water from 105°C to 75°C
You will only have to do the following:
Water gas from 105°C  100°C
Phase change from gas  liquid
Water liquid from 100°C  75°C

77. TRY THE FOLLOWING Copper Melting point: 1084°C
Boiling point: 2562°C
Hvap: 4688 J/g
Hfus: 202 J/g
Csolid: J/g•°C
Cliquid: 3.45 J/g•°C
Cgas: J/g•°C
To the left is the information for copper. If you have 1.25g of copper that starts at 1000°C, what is the energy needed to get it to 2500°C?

78. ANSWER FIRST 2 STEPS: Heat up the solid: q1 = c x m x δT q1 = 40.4 J
q1 = (0.385 J/g•°C)(1.25g)(1084°C-1000°C)
q1 = 40.4 J
Convert from solid to liquid
q2 = m x Hfus
q2 = (1.25g)(202J/g)
q2 = 253 J

79. ANSWER – NOTE WE DID NOT NEED ALL STEPS
LAST 2 STEPS
Heat up the liquid
q3 = c x m x δT
q3 = (3.45 J/g•°C)(1.25g)(2500°C-1084°C)
q3 = 6.11x103 J
Add all the heats
q = q1 + q2 + q3
q = 40.4J + 253J x103 J
q = 6.40x103 J