1. Heat (q)
Heat: the transfer of energy between objects due to a temperature difference
Flows from higher-temperature object to lower-temperature object
If T1 > T2
q system = -
exothermic
System
(T1)
Heat
Surroundings
(T2)
If T1 < T2
q system = +
endothermic
System
(T1)
Heat
Surroundings
(T2)

2. There are NO temp changes during a phase change.
Calorimetry: the measurement of heat flow
device used is called a...
calorimeter
specific heat capacity (C): amt. of heat needed to raise
temp. of 1 g of a substance 1oC (1 K)
Only useable within a state of matter (i.e. s, l, or g)
For energy changes involving…
heat of fusion (ΔHfus):
heat of vaporization (ΔHvap):
melting/freezing
boiling/condensing
There are NO temp changes during a phase change.

3. Various Specific Heat Capacities
heat capacity
(J/K g)
Substance
Gold
Silver
Copper
Iron
Aluminum
H2O(l)
H2O(s)
H2O(g)
0.129
0.235
0.385
0.449
0.897
4.184
2.03
1.998
Metals do not generally require much energy to heat them up
(i.e. they heat up easily)
Water requires much more energy to heat up

4. states of matter or during a phase change)
We can find the heat a substance loses or gains using:
where q = heat (J)
q = m C DT
m = mass of substance (g)
(used within a given
state of matter)
C = specific heat (J/goC)
DT = temperature change (oC)
AND
DH = heat of vap/fus (J/g)
q = m ΔH
Heating Curve
HEAT
Temp. s
s/l l
l/g g
(used between two
states of matter or during a phase change) + Cg
ΔHvap Cl –
ΔHfus Cs
D = final – initial

5. Using heat capacities…
q = m  C  ΔT
q (J) = mass (g)  C (J/goC)  ΔT (oC)
q = joules (J)
Mnemonic device: q = m “CAT”

6. Heating Curve of water Gas Boiling Point Liquid Melting Point Solid
140
120
l ↔ g
100 80
Liquid
Boiling Point 60 40
Temperature (oC) 20
s ↔ l
Melting Point
-20
Solid
-40
-60
-80
-100
Energy Added

7. Temperature (oC) Energy Added 140 120 100 80 60 40 20 -20 -40 -60 -80
-20
-40
-60
-80
-100
Energy Added

8. Heating Curve of Water Temperature (oC) Energy Added 140 120 100 80 60
-20
-40
-60
-80
-100
Energy Added

9. Heating Curves Temperature Change within phase
change in KE (molecular motion)
depends on heat capacity of phase
C H2O (l) = J/goC
C H2O (s) = J/goC
C H2O (g) = J/goC
Phase Changes (s ↔ l ↔ g)
change in PE (molecular arrangement)
temperature remains constant
overcoming intermolecular forces
(requires the most heat)
(requires the least heat)
ΔHfus = 333 J/g
(s ↔ l)
ΔHvap = 2256 J/g
(l ↔ g)
Why is this so much larger?

10. Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE 
140
Gas - KE 
120
100
Boiling - PE  80 60 40
Liquid - KE 
Temperature (oC) 20
Melting - PE 
-20
-40
Solid - KE 
-60
-80
-100
Energy Added

11. From Ice to Steam in Five Easy Steps
Heating Curve of Water
From Ice to Steam in Five Easy Steps q4 q5
q1: Heat the ice to 0°C
q2: Melt the ice into a liquid at 0°C
q3: Heat the water from 0°C to 100°C
q4: Boil the liquid into a gas at 100°C
q5: Heat the gas above 100°C
q1 = m Cs ΔT
q2 = m ΔHfus
q3 = m Cl ΔT
q4 = m ΔHvap
q5 = m Cg ΔT q3 q2 q1
Heat
Heat
qtot= q1 + q2 + q3 + q4 + q5

12. Heating Curve Practice
1. How much energy (J) is required to heat 12.5 g of ice at –10.0 oC to water at 0.0 oC? 5 4 3 2 1
Notice that your q values are positive because heat is added…
q1: Heat the ice from -10 to 0°C
q2: Melt the ice at 0°C to liquid at 0 oC
q1 = 12.5 g (2.077 J/g oC)( oC) =
q2 = 12.5 g (333 J/g) = J J
qtot = q1 + q2
= J + 4,162.5 J =
4,420 J

13. Heating Curve Practice
2. How much energy (J) is required to heat 25.0 g of ice at –25.0 oC to water at 95.0 oC?
Notice that your q values are positive because heat is added… 5 4 3 2 1
q1: Heat the ice from -25 to 0°C
q2: Melt the ice at 0°C to liquid at 0 oC
q3: Heat the water from 0°C to 95 °C
q1 = 25.0 g (2.077 J/g oC)( oC) =
q2 = 25.0 g (333 J/g) = J
8325 J
q3 = 25.0 g (4.184 J/g oC)(95.0 – 0.0oC) =
9937 J
qtot = q1 + q2 + q3
= J + 8,325 J J =
19,560 J

14. Heating Curve Practice
3. How much energy (J) is removed to cool 50.0 g of steam at oC to ice at -5.0 oC?
Notice that your q values are negative because heat is removed… 5 4 3 2 1
q5: Cool the steam from to 100°C
q4: Condense the steam into liquid at 100°C
q3: Cool the water from 100°C to 0 °C
q2: Freeze the water into ice at 0 °C
q2 = 50.0 g (- 333 J/g) =
q1: Cool the ice from 0°C to – 5.0 °C
q5 = 50.0 g (2.042 J/g oC)( oC) =
q4 = g ( J/g) = J -
-112,800 J
q3 = 50.0 g (4.184 J/g oC)(0.0 – 100.0oC) = J J
q1 = 50.0 g (2.077 J/g oC)(- 5.0 – 0.0oC) = J
qtot = q1 + q2 + q3+ q4 + q5
= J ,800 J J + -16,650 J J =
-152,000 J

15. Food and Energy Caloric Values 1 calorie = 4.184 joules
Food joules/gram calories/gram “Calories”/gram
Protein , ,
Fat , ,
Carbohydrates 17, ,
1 calorie = joules
1000 calories = 1 “Calorie”
"science" "food"
or… 1 Kcal = 1 “Calorie”
Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51

16. Does water have negative calories?
How many Calories (nutritional) will you burn by drinking 1.0 L of water, initially at 36.5 oF (standard refrigeration temperature)? Assume that the body must expend energy to heat the water to body temperature at 98.6 oF.
37 oC
1 L = 1000 mL
2.5 oC
1 mL = 1 g
1 calorie = joules
1000 calories = 1 “Calorie”
q = 1.0 x 103 g (4.184 J/g oC)(37 oC oC) = 144,348 J J
1 cal
1 “Cal” =
35 Cal
4.184 J
1000 cal

17. C H2O (l) = 4.184 J/goC C H2O (s) = 2.03 J/goC C H2O (g) = 1.998 J/goC
Heating Curve of H2O Constants and Graph
C H2O (l) = J/goC C H2O (s) = 2.03 J/goC C H2O (g) = J/goC
ΔHfusion = 6.02 kJ/mol ΔHvap = 40.7 kJ/mol
140
120
100 80 60 40
Temperature (oC) 20
-20
-40
-60
-80
-100
Energy Added 17

18. What will happen over time?
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

19. Let’s take a closer look…
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

20. Eventually, the temperatures will equalize
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291

21. (i.e. atmospheric pressure)
Thermometer
Styrofoam
cover
cups
Stirrer
Much calorimetry is carried out using a coffee-cup calorimeter, under constant pressure
(i.e. atmospheric pressure)
If we assume that no heat is
lost to the surroundings, then the
energy absorbed inside the
calorimeter must be equal to the
energy released inside the calorimeter.
i.e., q absorbed = – q released qx =
– qy

22. Heat Transfer Experiments
1. A 75.0 g piece of lead (specific heat = J/goC),
initially at 435oC, is set into g of water, initially
at 23.0oC. What is the final temperature of the mixture? Pb
125 g
23.0 °C
What is the final temperature, Tf, of the mixture?
75.0 g
435.0 °C
C = J/°C g
qwater = –qPb
q = m x C x ΔT for both cases, although specific values differ
Plug in known information for each side
Solve for Tf ...

23. A 75.0 g piece of lead (specific heat = 0.130 J/goC),
initially at 435oC, is set into g of water, initially
at 23.0oC. What is the final temperature of the mixture?
q = m x C x ΔT for both cases, although specific values differ
Plug in known information for each side
qwater = –qPb
mwater Cwater DTwater =
–mPb CPb DTPb
125 (4.18) (Tf – 23) =
–75 (0.13) (Tf – 435)
522.5 Tf – =
–9.75 Tf
+9.75 Tf
+9.75 Tf Tf =
Tf = 30.5oC

24. 2. A 97.0 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15.0oC. If gold has a specific heat of J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter.
T = 785oC Au
mass = 97.0 g
T = 15.0 oC
mass = 323 g
LOSE heat = GAIN heat -
- [(C Au) (mass) (DT)] = (C H2O) (mass) (DT)
- [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)]
- [(12.5) (Tf - 785oC)] = (1.35 x 103) (Tf - 15oC)]
-12.5 Tf x = x 103 Tf x 104
3 x 104 = x 103 Tf
Tf = oC

25. HW #2. If 59. 0 g of water at 13. 0 oC are mixed with 87
HW #2. If 59.0 g of water at 13.0 oC are mixed with 87.0 g of water at 72.0 oC, find the final temperature of the system.
T = 13.0 oC
mass = 59.0 g
T = 72.0 oC
mass = 87.0 g
LOSE heat = GAIN heat -
- [ (mass) (C H2O) (DT)] = (mass) (C H2O) (DT)
- [ (59 g) (4.184 J/goC) (Tf - 13oC)] = (87 g) (4.184 J/goC) (Tf - 72oC)]
- [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)]
Tf = Tf
= Tf
Tf = oC

26. HW #4. 240. g of water (initially at 20
HW # g of water (initially at 20.0oC) are mixed with an unknown mass of iron initially at 500.0oC (CFe = J/goC). When thermal equilibrium is reached, the mixture has a temperature of 42.0oC. Find the mass of the iron.
T = 500oC Fe
mass = ? grams
T = 20oC
mass = 240 g -
LOSE heat = GAIN heat
-q1 = q2
- [ (mass) (CFe ) (DT)] = (mass) (CH2O) (DT)
- [ (X g) ( J/goC) (42oC - 500oC)] = (240 g) (4.184 J/goC) (42oC - 20oC)]
- [ (X) (0.4495) (-458)] = (240 g) (4.184) (22)
205.9 X =
X = 107 g Fe

27. KEY: Assume that the ice melts and the final product is a liquid.
A 23.6 g ice cube at –31.0oC is dropped into 98.2 g of
water at 84.7oC. Find the equilibrium temperature.
KEY: Assume that the ice melts and the final product is a liquid.
qice = –qwater
qwater = –98.2 (4.18) (Tf – 84.7)
= – Tf
qice = 23.6 (2.077) (0 – –31) (333) (4.18) (Tf – 0)
= Tf
= Tf Tf =
Tf = 49.9oC

28. Heating Curve Challenge Problems
Temperature (oC) 40 20
-20
-40
-60
-80
-100
120
100 80 60
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, solid
1. A sample of ice at -25oC is placed into 75 g of water initally at 85oC. If the final temperature of the mixture is 15oC, what was the mass of the ice?
52.8 g ice
A 38 g sample of ice at -5oC is placed into 250 g of water at 65oC. Find the final temperature of the mixture assuming that the ice sample completely melts.
A 35 g sample of steam at 116oC are bubbled into 300 g water at 10oC. Find the final temperature of the system, assuming that the steam condenses into liquid water.
45.6 oC
76.6 oC

29. Heating Curve for Water (Phase Diagram)
140
120
100 80 60 40 20
-20
-40
-60
-80
-100 F
q4 = m DHvap
DHvap = +/ J/g 5 BP
q2 = m DHfus D E
DHfus = +/- 333 J/g 4
q5 = m C DT
C g = J/goC 3
q3 = m C DT
Temperature (oC)
Cl = J/goC MP B C 2 1
A  B warm ice
B  C melt ice (s  l)
C  D warm water
D  E boil water (l  g)
E  D condense steam (g  l)
E  F superheat steam
q1 = m C DT
Cs = J/goC A
Heat

30. Calculating Energy Changes - Heating Curve for Water
140
DH = mol x DHvap
120
DH = mol x DHfus
100 80 60
Heat = mass x Dt x Cp, gas 40
Temperature (oC) 20
Heat = mass x Dt x Cp, liquid
-20
-40
-60
Heat = mass x Dt x Cp, solid
-80
-100
Time