1. Heat, Temperature Phase Changes
And
Phase Changes
Pisgah High School Chemistry
Mr. Jones
Last rev

2. Part One
Heat and Temperature

3. What temperature does the thermometer indicate?

4. What might be going on that would cause this temperature?

5. This is the view out the window, past the thermometer.

6. Yep.
It is snowing.

7. Water freezes and ice melts.
Why would snow cause the air temperature to be at precisely 0C?
What occurs at 0C?
Water freezes and ice melts.

8. Ice, in the form of snow, falls through the slightly warmer air.
The snow melts and absorbs heat from the air, causing the air to cool.
Ice melts at 0 C, so the air cools to that temperature.

9. The temperature hovers at zero Celsius as the snow melts.

10. So why is there snow on the ground if it is melting?
Yep. That’s what allows the snow to accumulate.

11. As the snow melts, it absorbs heat and cools the ground, the car, and the grill.

12. This allows more snow to lay
This allows more snow to lay. It doesn’t melt because the ground is now at 0C.

13. What does it mean to have a temperature of 0 C?
What is temperature?
Is temperature the same thing as heat?

14. Temperature is a measure of how “hot” or “cold” something is.
Temperature is measured in arbitrary units, like Fahrenheit or Celsius.

15. Temperature is proportional to the average kinetic energy of the molecules of the substance.
T µ KE
KE = ½ mv2

16. Temperature is therefore proportional to the speed of the molecules of a substance.
Velocity or speed
T µ KE
KE = ½ mv2
T µ v

17. The higher the temperature, the greater the average speed of the molecules.
Velocity or speed
T µ KE
KE = ½ mv2
T µ v

18. Heat is the thermal energy transferred from a hot object to a cold object.
Heat is measured in energy units -- Joules or calories.

19. The heat transferred is proportional to the mass of the object, the specific heat capacity of the object and the temperature change the object undergoes.

20. Heat has the symbol q and is calculated using …
q = mcDT

21. q = mcDT mass specific heat capacity Quantity of heat
temperature change

22. The specific heat capacity of water is 4.18 J/gC
Quantity of heat
q = mcDT
The specific heat capacity of water is 4.18 J/gC

23. q = m c DT q = (25.6g)(4.18J/gC)(30.0C) q = 3210 J
How much heat is needed to raise the temperature of 25.6 grams of water from 20.0 C to 50.0 C?
q = m c DT
q = (25.6g)(4.18J/gC)(30.0C)
q = 3210 J

24. q = m c DT = q DT m c Answer: 6.20 C
What is the final temperature of 27.0 grams of liquid water, initially at 0C, after it absorbs J of energy?
q = m c DT
Hint: start by solving for DT. = q DT
m c
Answer:
6.20 C

25. Calorimetry and Specific Heat Capacity
Part Two
Calorimetry and Specific Heat Capacity

26. Calorimetry is a collection of laboratory procedures used to investigate the transfer of heat.
In calorimetry experiments, one might be looking for a final temperature or a specific heat capacity.

27. Investigate:
Suppose two different masses of water at different temperatures are mixed. Can you predict the final temperature?

28. Investigate:
Will the final temperature be cooler than the cool water, or will it be warmer than the warm water?
Or will the final temperature be somewhere in between?

29. Investigate:
Develop a procedure where you mix a known mass of cool water with a different mass of water at an elevated temperature and measure the final (equilibrium) temperature.
What equipment will you need?

30. Investigate:
Develop a procedure where you mix a known mass of cool water with a different mass of water at an elevated temperature and measure the final (equilibrium) temperature.
You could use a balance, a thermometer, a coffee cup calorimeter, and a hot plate.

31. Investigate: What do we need to record in a data table?
Mass of calorimeter cup _________
Mass of cool water and cup _________
Mass of cool water _________
Initial temperature of cool water _________
Initial temperature of hot water _________
Final temperature after mixing _________
Mass of mixed water and cup _________
Mass of hot water _________

32. Investigate:
Whenever we design an experiment we make some assumptions. Here are a couple, can you add any more?
The calorimeter cup is a perfect insulator and no heat is exchanged with the surroundings.
Warning: Hot plates and boiling water can cause severe burns.

33. Investigate: You might need a hint about how to calculate the results.
What is the law of conservation of energy?
Energy is neither created nor destroyed, only changed in form.

34. Investigate: You might need a hint about how to calculate the results.
The law of conservation of energy suggests that the heat lost by the hot water as it cools is equal to the heat gained by the cool water as it warms up.

35. qlost = -qgained = mhcDTh = -mccDTc Investigate:
To put it mathematically:
qlost = -qgained
Heat lost by the hot water
Heat gained by the cold water =
And since q = mcDT then
mhcDTh = -mccDTc

36. mhc(Tf -Th) = -mcc(Tf -Tc)
Investigate:
The convention for DT is final temperature minus initial temperature or Tfinal – Tinitial
mhcDTh = -mccDTc
becomes
mhc(Tf -Th) = -mcc(Tf -Tc)
Use your algebra skills, to solve for Tf , the final temperature.

37. How did our predicted Tf compare to the one we observed?

38. In the next investigation you will …
develop a method to find the specific heat capacity of a metal.

39. Specific heat capacity …
…varies from one substance to another.
…a measure of how much heat something can “hold”.
…the amount of heat needed to raise one gram of a substance by one Celsius degree.

40. Specific heat capacity lab suggestions:
Heat a metal to a known temp.
Transfer the metal to a known quantity of water at a known temperature.
Measure the equilibrium temperature.
Use qlost = -qgained to compute the specific heat of the metal.

41. Get the initial temperature of the metal.
The temperature of boiling water.
metal
hotplate

42. Get initial temp of water in calorimeter cup.

43. Transfer the metal to the calorimeter.

44. Continue stirring until thermal equilibrium is reached.

45. Data: Mass of metal Initial temp of metal Mass of water
Initial temp of water
Final temp of water and metal

46. qlost = -qgained
mmcmDTm = -mwcwDTw
-mwcwDTw
cm =
mmDTm

47. Mass of metal 40.0 g
Initial T of metal 98.0 C
Mass of water in calorimeter 60.0 g
Initial T of water 20.0 C
Final T of water and metal 22.9 C
Calculate the specific heat capacity of the metal.

48. Table of selected specific heats.
What is the unknown metal?

49. Calorimetry and Phase Changes
Part Three
Calorimetry and Phase Changes

50. Is heat absorbed or released during a phase change?
How could you measure the heat absorbed or released as substances change phase?

51. Consider ice melting in water.
1. What is the temperature of a mixture of ice and water?
0 C
2. Does the temperature of the water change? No
3. Is the water absorbing or releasing heat?
Releasing heat

52. Consider ice melting in water.
4. Does ice absorb heat or release heat as it melts?
Absorbs heat
5. What is the temperature of the water from the melting ice?
0 C
6. When will the temperature of the water change?
When all ice melts

53. Consider ice melting in water.
The word fusion means “melting”.
How could you design an experiment to measure the heat of fusion of ice?

54. Consider ice melting in water.
You could measure the heat lost by some water as it cools.
Ice
That should equal the heat gained by the ice as it melts.

55. Consider ice melting in water.
Suppose some ice at 0C is placed into g of water at 25.5 C.
Ice
Copy down this information and the data that follows.

56. Consider ice melting in water.
When the system reaches equilibrium at 0C, grams of the ice has melted.
Ice

57. Consider ice melting in water.
Knowing that the heat lost by the water as it cools to 0C is equal to the heat gained by the ice as it melts at 0C …
Ice

58. Consider ice melting in water.
…we should be able to compute the heat of fusion of ice, Hf.
Ice
qlost by water= -qgained by ice
mwatercDT = -miceHf

59. Consider ice melting in water.
Go ahead and calculate the value of Hf.
Ice
qlost by water= -qgained by ice
mwatercDT = -miceHf

60. We now know that heat is either absorbed or released during a phase change.
Heat is absorbed as solids melt, or liquids vaporize.

61. We now know that heat is either absorbed or released during a phase change.
Heat is released as liquids freeze, or vapors condense.

62. Heat is absorbed by the ice.
And melts.

63. Heat is absorbed by the ice.
One gram of ice at 0C absorbs 334 J as it melts to form water at 0C.
… making liquid water

64. Heat is released by the water as it freezes.
334 joules is released when one gram of water freezes at 0C.
Ice
water

65. Ice Water releases 334 J per gram as it freezes at 0C
Ice absorbs 334 J per gram as it melts at 0C
Ice
Water releases 334 J per gram as it freezes at 0C

66. Heat is absorbed by the water as it vaporizes.
Hotplate

67. Heat is absorbed by the water as it vaporizes.
2260 joules is absorbed by one gram of water as it boils at 100C.
Hotplate

68. Steam releases 2260 J/g as it condenses at 100 C
Water absorbs 2260 J/g as it boils at 100 C
Hotplate

69. Heat is released by water vapor as it condenses.

70. Heat is released by water vapor as it condenses.
The heat released by condensing water vapor is a major factor in weather phenomena like thunderstorms and hurricanes.

71. Thunderhead
40,000 + feet
The heat released by condensing water vapor causes convection and updrafts in thunderstorms.

72. Phase changes occur at a constant temperature as heat is absorbed or released.

73. Question for discussion:
If phase changes occur at a constant temperature, then what happens to the heat when water boils?

74. Heat energy is converted to matter (E=mc2) making the water heavier.
Question for discussion – possible answers:
Heat energy is converted to matter (E=mc2) making the water heavier.
The heat increases the speed of the water molecules.
The heat energy breaks the intermolecular bonds which keep the water in the liquid phase.
The temperature really does change, you just missed it.

75. Heat energy is converted to matter (E=mc2) making the water heavier.
Question for discussion – possible answers:
Heat energy is converted to matter (E=mc2) making the water heavier.
The heat increases the speed of the water molecules.
The heat energy breaks the intermolecular bonds which keep the water in the liquid phase.
The temperature really does change, you just missed it.

76. The heat gained or lost in phase changes can be calculated using …
Heat of fusion (melting)
q = mHf
q = mHv
Heat of vaporization

77. Hf = 334 J/g Hv=2260 J/g The values for water are …
Heat of fusion (melting)
Hf = 334 J/g
Hv=2260 J/g
Heat of vaporization

78. q = m Hf q = (150.0 g)(334 J/g) q = 50,100 J or 50.1 kJ
How much heat is absorbed by g of ice as it melts at 0C?
q = m Hf
q = (150.0 g)(334 J/g)
q = 50,100 J
or 50.1 kJ

79. q = m Hv q = (20.0 g)(2260 J/g) q = 45,200 J or 45.2 kJ
How much heat is released by 20.0 grams of steam as it condenses at 100C?
q = m Hv
q = (20.0 g)(2260 J/g)
q = 45,200 J
or 45.2 kJ

80. Part Four
Sublimation
and
Phase Diagrams

81. Sublimation is an unusual phase change.
Sublimation occurs when a solid changes directly into a gas without going through the liquid phase.
Heat is absorbed when sublimation occurs.

82. Solid iodine, I2, undergoes sublimation when heated.
Iodine vapor fills the beaker.
Watchglass
Beaker with iodine
Hotplate

83. Solid iodine crystallizes on the bottom of the watchglass.
The color of the vapor fades as the iodine deposits on the watchglass
Watchglass
Beaker with iodine
Hotplate

84. A solid forming directly from the vapor is called deposition.
The color of the vapor fades as the iodine deposits on the watchglass
Watchglass
Beaker with iodine
Hotplate

85. Here we see the sublimation and deposition of iodine.

86. Heat is absorbed as the iodine undergoes sublimation.

87. Heat is released as the iodine undergoes deposition.

88. Dry ice is solid carbon dioxide, CO2.
At room temperature and normal atmospheric pressures dry ice undergoes sublimation.
It goes directly from the solid state to the vapor state.
Dry Ice
CO2 vapor

89. Dry ice is solid carbon dioxide, CO2.
CO2 vapor
Dry ice is solid carbon dioxide, CO2.
CO2 vapor
At room temperature and normal atmospheric pressures dry ice undergoes sublimation.
CO2 vapor
CO2 vapor
It goes directly from the solid state to the vapor state.
Dry Ice
CO2 vapor

90. A phase diagram can help explain why dry ice undergoes sublimation.
The phase diagram has three distinct regions.
Temperature
Pressure 2 1 3

91. 2 1 3 Which phase is in each region?
The phase diagram has three distinct regions.
Temperature
Pressure 2 1 3

92. 2 1 3 Which phase is in each region? 1= ??? 2= ??? 3= ???
Hint: What happens to ice as temperature increases?
1= ???
Temperature
Pressure 2
2= ??? 1
3= ??? 3

93. 2 1 3 Which phase is in each region? 1= Solid 2= ??? 3= ???
Hint: What happens to ice as temperature increases?
1= Solid
Temperature
Pressure 2
2= ??? 1
3= ??? 3

94. 2 1 3 Which phase is in each region? 1= Solid 2= Liquid 3= ???
Hint: What happens to ice as temperature increases?
1= Solid
Temperature
Pressure 2
2= Liquid 1
3= ??? 3

95. 2 1 3 Which phase is in each region? 1= Solid 2= Liquid 3= Gas
Hint: What happens to ice as temperature increases?
1= Solid
Temperature
Pressure 2
2= Liquid 1
3= Gas 3

96. The point where all three phases exist in equilibrium is called the
triple point.
triple point. L S
Pressure G
Temperature

97. At a pressure of 1 atm, most substances go through all three phases, as the temperature increases,
Solids melt to form liquids, which vaporize to form gases.
1 atm G
Temp.

98. At a pressure of 1 atm, most substances go through all three phases, as the temperature increases,
Notice the melting point and boiling point.
1 atm G MP BP
Temp.

99. L S G But the phase diagram for CO2 is a little different.
Notice that the triple point is above 1 atm. L S
5 atm
1 atm G
Temperature

100. At 1 atm CO2 goes directly from solid to vapor as the temperature increases.

101. At 1 atm CO2 goes directly from solid to vapor as the temperature increases.
The sublimation point is –78.5 C
1 atm
-78.5 G
Temperature

102. Carbon dioxide is a liquid at the bottom of the ocean where the pressure is well above 5 atmospheres.

103. For more common substances we see a phase diagram like this.
The line for one atmosphere of pressure tells us that all three phases can exist.
Pressure
1 atm
Temperature

104. An arrow will appear in the following phase diagrams.
Tell what phase change the arrow indicates.
Pressure
Temperature

105. L S G What phase change is occurring? Melting (fusion) Pressure
Temperature

106. L S G What phase change is occurring? Vaporization Pressure
Temperature

107. L S G What phase change is occurring? Condensation Pressure
Temperature

108. What phase change is occurring?L G
Sublimation
Pressure
Temperature

109. L S G What phase change is occurring?
Liquefying a gas by increasing the pressure. S L G
Pressure
Temperature

110. What do the lines between the different regions represent?
Each line
represents an equilibrium between two phases.

111. L S G Equilibria occur at the boundaries between the regions.
The equilibrium between the solid and liquid phases. S L G
Pressure
Temperature

112. L S G Equilibria occur at the boundaries between the regions.
The equilibrium between the liquid and gaseous phases. S L G
Pressure
Temperature

113. L S G Equilibria occur at the boundaries between the regions.
The equilibrium between the solid and gaseous phases. S L G
Pressure
Temperature

114. L S G Consider the equilibrium between two phases.
Ice and water are in an insulated container.
Pressure
Temperature

115. Ice and water are in an insulated container.
Acme Digital Thermometer
0.0 C
Ice and water are in an insulated container.

116. Some ice melts and forms liquid water.
Acme Digital Thermometer
0.0 C
Some ice melts and forms liquid water.

117. Some water freezes and forms ice.
Acme Digital Thermometer
0.0 C
Some water freezes and forms ice.

118. Acme Digital Thermometer
When the rates at which the ice melts and the water freezes are equal …

119. equilibrium is established.
Acme Digital Thermometer
0.0 C
… an
equilibrium is established.

120. The amounts of ice and water will remain constant…
Acme Digital Thermometer
0.0 C
The amounts of ice and water will remain constant…

121. …and the mixture of ice and water will remain at a constant 0C.
Acme Digital Thermometer
0.0 C
…and the mixture of ice and water will remain at a constant 0C.

122. Acme Digital Thermometer
A mixture of ice and water can be used to calibrate a thermometer at 0C.

123. That’s because phase changes occur at a single temperature.
Water freezes and ice melts at 0C.
At sea level, water boils and steam condenses at 100C.

124. Therefore, it can be seen that when ice and water are placed into a perfectly insulated container …
… the mixture will stay at a constant zero degrees Celsius by establishing an equilibrium.

125. The amount of ice and water will never change.
An ice/water equilibrium occurs when the rate at which water freezes is equal to the rate at which ice melts.
The amount of ice and water will never change.
If the container is completely insulated.

126. Heating and Cooling Curves
Part Five
Heating and Cooling Curves

127. A process that absorbs heat is called endothermic.
A process that gives off heat is called exothermic.

128. Endothermic: Exothermic: Melting (fusion) Vaporization Sublimation
Heat is absorbed.
Freezing
Condensation
Deposition
Exothermic:
Heat is released.

129. Investigate:
Heat the mixture of water and ice on a hotplate and record the temperature as a function of time.

130. Investigate:
The following are suggested procedures you could use to record the temperature of water at regular intervals.
Note: Hot plates and boiling water can cause severe burns.

131. Investigate:
Clamp a thermometer with the bulb in a mixture of ice and water in a beaker on a hot plate. (The hot plate is off.)
Allow the temperature to equilibrate.
Turn on the hot plate and continue to record temperature at regular intervals until some of the water boils away.
Plot temperature as a function of time.

132. Thermometer
Temperature
Time
Graph paper
Stirring hotplate

133. Thermometer
Time
Temperature
Stirring hotplate

134. 0.0 C Temperature probe CBL, LabPro, or computer Stirring hotplate
Time
CBL, LabPro, or computer
Stirring hotplate

135. Consider the following heating curve for water.
100
Time
Temp

136. Consider the following heating curve for water.
100
Time
Temp
Ice at –30C absorbs heat. Temperature rises to 0C.

137. Consider the following heating curve for water.
100
Time
Temp
Ice at 0C absorbs heat and melts at constant 0C to make water at 0C.

138. Consider the following heating curve for water.
When all ice melts, water at 0C absorbs heat and temperature rises to 100C.
100
Time
Temp

139. Consider the following heating curve for water.
100
Time
Temp
Water absorbs heat and boils at a constant temperature of 100C.

140. Consider the following heating curve for water.
100
Time
Temp
Temperature of steam rises as it absorbs heat after all of the water boils.

141. What is happening at each segment of the heating curve?
100
Time
Temp

142. Look at the different regions of the heating curve for water.
100
Time
Temp
Water and
steam
Steam
Water
Ice and
water
Phase changes?
Ice

143. The temperature is constant during a phase change.
100
Time
Temp
Water and
steam
Steam
Water
Ice and
water
Phase changes
Ice

144. Calculating heat at each segment of the heating curve.
100
Time
Temp
The temperature of the ice is increasing. The specific heat for ice is 2.05 J/gC.
q1=mciDT

145. Calculating heat at each segment of the heating curve.
100
Time
Temp
A phase change occurs at a constant temperature. Use the heat of fusion since ice is melting.
q2=mHf
q1=mciDT

146. Calculating heat at each segment of the heating curve.
100
Time
Temp
q3=mcwDT
The temperature of the water is increasing. The specific heat of water is 4.18 J/gC.
q2=mHf
q1=mciDT

147. Calculating heat at each segment of the heating curve.
100
Time
Temp
q4=mHv
q3=mcwDT
A phase change occurs at a constant temperature. Use the heat of vaporization since water is boiling.
q2=mHf
q1=mciDT

148. Calculating heat at each segment of the heating curve.
q5=mcsDT
100
Time
Temp
q4=mHv
q3=mcwDT
The temperature of the steam is increasing. The specific heat of steam is 2.02 J/gC.
q2=mHf
q1=mciDT

149. Calculating heat at each segment of the heating curve.
q5=mcsDT
100
Time
Temp
q4=mHv
q3=mcwDT
Use q=mcDT when there is a temperature change.
q2=mHf
q1=mciDT

150. Calculating heat at each segment of the heating curve.
q5=mcsDT
100
Time
Temp
q4=mHv
q3=mcwDT
Use q=mHf or q=mHv when there is a phase change.
q2=mHf
q1=mciDT

151. Calculating heat at each segment of the heating curve.
q5=mcsDT
100
Time
Temp
q4=mHv
q3=mcwDT
The total amount of heat absorbed is the sum:
q2=mHf
qtot= q1+q2+q3+q4+q5
q1=mciDT

152. What would the cooling curve of steam look like?
100
Time
Temp

153. What would the cooling curve of steam look like?
100
Time
Temp
Water and
steam
Water
Steam
Ice and
water
Ice

154. What would the cooling curve of steam look like?
Heat energy is released at each step.
100
Time
Temp
Water and
steam
Water
Steam
Ice and
water
Ice

155. Why is a steam burn worse than one from boiling water, even if both are at 100C?
100
Time
Temp
Water and
steam
Water
Steam
Ice and
water
Ice

156. Part Six
Fractional
Distillation

157. Initial Observations
A clear, colorless liquid has a strong, odor. When placed on a watch glass and ignited, it burns, but not completely. Some nonflammable liquid remains.
Is the liquid a pure substance?
Is the liquid a mixture?
Heterogeneous or homogeneous?

158. How might you separate a mixture of two clear, colorless liquids?
What if the liquids have different boiling points?

159. Heating curve for two liquids, A and B
time
temperature
Boiling point of liquid B
Boiling point of liquid A

160. Fractional distillation can be used to separate the mixture into its various “fractions”.
Isolate each fraction at each of the different boiling points.

161. Collecting the first fractionB
time
temperature 3 A
Collect in the first test tube
what comes off in region 1. 2 1

162. Collecting the second fractionB
time
temperature 3 A
Collect in the second test tube
what comes off in region 2. 2 1

163. Collecting the third fractionB
time
temperature 3 A
Collect in the third test tube
what comes off in region 3. 2 1

164. Predict what is in each tube
Test tube 1
time
temperature A B 1 2 3
May contain only A
Test tube 2
May contain both A and B
Test tube 3
May contain only B

165. Equipment setup for doing fractional distillation.
Ring stand and finger clamp
thermometer or temperature probe
sidearm
rubber tubing
Jones condenser
boiling flask
beaker or test tube
hot plate

166. Change the beaker or test tube to collect each fraction.
Ring stand and finger clamp
thermometer or temperature probe
sidearm
rubber tubing
Jones condenser
boiling flask
beaker or test tube
hot plate

167. Test the contents of each test tube for …
Odor – does it have an odor? Is it strong or weak?
Flammability – does it burn? A lot or a little?
Place a small amount on a watch glass and ignite it with a match.

168. Record your observations
TT#
Start
Temp
End
Odor
Flammability 1 2 3

169. Questions?

170. Questions
1. Ice and water are placed in an insulated container. What will be the equilibrium temperature?
2. A substance freezes at -80.0C. At what temperature does it melt?

171. Questions
3. A liquid gradually turns solid at a constant temperature. Is heat being added, or removed?
4. How does melting snow affect the air temperature?

172. Questions
5. When water vapor condenses to form liquid water, is heat released or absorbed?
6. What is the connection between condensing water vapor and updrafts in thunderstorms?

173. Questions 7. Explain how sweating cools your body.
8. Explain how liquid water evaporating from a roadway can cause black ice to form on the road.

174. Questions 9. Explain why dry ice doesn’t form a puddle of liquid CO2.
10. What phase change is occuring?
Temperature
Pressure

175. Questions
11. How many joules of heat are released when 50.0 grams of water cools from 80.0C to 20.0C?
12. How many joules of heat are needed to melt 15.0 grams of ice at 0C?

176. Questions
13. How many joules of heat are needed to take 15.0 grams of ice at –20.0C to steam at 150.0C?
14. How many grams of steam must cool from 125.0C to 80.0C to release 2.00 x 106 J of energy?
Hint: there are three changes that take place.

177. Questions
15. Based on your experience in doing fractional distillation, how practical would it be to separate a mixture of benzene and ethanol? Benzene has a boiling point of 80.C. Be sure to fully explain your reasoning.

178. Heat
Deposition
Melting
Temperature
Phase change
Equilibrium
Joule
Phase diagram
Vaporization
Heating curve
Calorie
Condensation
Sublimation
Freezing
Boiling
Celsius