Heat (q)
• Heat: the transfer of energy between objects due to a
temperature difference
•
Flows from higher-temperature object to lower-temperature
object
System
(T1)
System
(T1)
Heat
Heat
Surroundings
(T2)
If T1 > T2
q system = exothermic
Surroundings
(T2)
If T1 < T2
q system = +
endothermic
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): melting/freezing
heat of vaporization (ΔHvap): boiling/condensing
There are NO temp changes during a phase change.
Various Specific Heat Capacities
Substance
Specific
heat capacity
(J/K g)
Gold
0.129
Silver
0.235
Copper
0.385
Iron
0.449
Aluminum
0.897
H2O(l)
4.184
H2O(s)
2.03
H2O(g)
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
We can find the heat a substance loses or gains using:
q = m C DT
where q = heat (J)
m = mass of substance (g)
(used within a given
state of matter)
C = specific heat (J/goC)
DT = temperature change (oC)
DH = heat of vap/fus (J/g)
Heating Curve
Temp.
+
s/l
ΔHfus
s
g
l
Cl
l/g
ΔHvap
–
Cs
HEAT
Cg
AND
q = m ΔH
(used between two
states of matter or
during a phase change)
D = final – initial
Using heat capacities…
q = m  C  ΔT
q (J) = mass (g)  C (J/goC)  ΔT (oC)
q = joules (J)
Mnemonic device: q = m “CAT”
Heating Curves
• Temperature Change within phase
• change in KE (molecular motion)
• depends on heat capacity of phase
C H2O (l) = 4.184 J/goC
C H2O (s) = 2.077 J/goC
C H2O (g) = 2.042 J/goC
(requires the most heat)
(requires the least heat)
• Phase Changes (s ↔ l ↔ g)
• change in PE (molecular arrangement)
• temperature remains constant
• overcoming intermolecular forces
(s ↔ l)
ΔHfus = 333 J/g
ΔHvap = 2256 J/g (l ↔ g) Why is this so much larger?
Heating Curve of Water
From Ice to Steam in Five Easy Steps
q4
q5
q1: Heat the ice to 0°C
q1 = m Cs ΔT
q3
q2
q2: Melt the ice into a liquid at 0°C
q2 = m ΔHfus
q3: Heat the water from 0°C to 100°C
q3 = m Cl ΔT
q1
Heat
q4: Boil the liquid into a gas at 100°C
Heat q4 = m ΔHvap
q5: Heat the gas above 100°C
q5 = m Cg ΔT
qtot= q1 + q2 + q3 + q4 + q5
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?
4
3
5
Notice that your q values are
positive because heat is added…
2
1
q1: Heat the ice from -10 to 0°C
q1 = 12.5 g (2.077 J/g oC)(0.0 - -10.0 oC) = 259.63 J
q2: Melt the ice at 0°C to liquid at 0 oC
q2 = 12.5 g (333 J/g) = 4162.5 J
qtot = q1 + q2 = 259.63 J + 4,162.5 J = 4,420 J
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?
4
3
2
1
5
Notice that your q values are
positive because heat is added…
q1: Heat the ice from -25 to 0°C
q1 = 25.0 g (2.077 J/g oC)(0.0 - -25.0 oC) = 1298.1 J
q2: Melt the ice at 0°C to liquid at 0 oC
q2 = 25.0 g (333 J/g) = 8325 J
q3: Heat the water from 0°C to 95 °C
q3 = 25.0 g (4.184 J/g oC)(95.0 – 0.0oC) = 9937 J
qtot = q1 + q2 + q3 = 1298.1 J + 8,325 J + 9937 J = 19,560 J
Heating Curve Practice
3. How much energy (J) is removed to cool 50.0 g of steam
at 115.0 oC to ice at -5.0 oC?
4
3
2
1
5
Notice that your q values are
negative because heat is removed…
q5: Cool the steam from 115.0 to 100°C
q5 = 50.0 g (2.042 J/g oC)(100.0 - 115.0 oC) = -1531.5 J
q4: Condense the steam into liquid at 100°C
q4 = 50.0 g ( - 2256 J/g) = -112,800 J
q3: Cool the water from 100°C to 0 °C
q3 = 50.0 g (4.184 J/g oC)(0.0 – 100.0oC) = -20920 J
q2: Freeze the water into ice at 0 °C
q2 = 50.0 g (- 333 J/g) = -16650 J
q1: Cool the ice from 0°C to – 5.0 °C
q1 = 50.0 g (2.077 J/g oC)(- 5.0 – 0.0oC) = -519.25 J
qtot = q1 + q2 + q3+ q4 + q5 = -1531.5 J + -112,800 J + -20920 J + -16,650 J + -519.25 J =
-152,000 J
Food and Energy
Caloric Values
Food
joules/gram
calories/gram
“Calories”/gram
Protein
17,000
4,000
4
Fat
38,000
9,000
9
Carbohydrates 17,000
4,000
4
1 calorie = 4.184 joules
1000 calories = 1 “Calorie”
"science"
"food"
or… 1 Kcal = 1 “Calorie”
Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51
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
5
C  F  32 
1 mL = 1 g
9
1 calorie = 4.184 joules
q  mCDT 1000 calories = 1 “Calorie”
q = 1.0 x 103 g (4.184 J/g oC)(37 oC - 2.5 oC) = 144,348 J
144348 J
1 cal
1 “Cal”
4.184 J 1000 cal
= 35 Cal
What will happen over time?
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291
Let’s take a closer look…
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291
Eventually, the temperatures will equalize
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 291
Thermometer
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
Styrofoam
cover
Styrofoam
cups
Stirrer
Heat Transfer Experiments
1. A 75.0 g piece of lead (specific heat = 0.130 J/goC),
initially at 435oC, is set into 125.0 g of water, initially
at 23.0oC. What is the final temperature of the mixture?
Pb
75.0 g
125 g
435.0 °C
23.0 °C
C = 0.130 J/°C g
What is the final
temperature, Tf, of
the mixture?
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 ...
A 75.0 g piece of lead (specific heat = 0.130 J/goC),
initially at 435oC, is set into 125.0 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 – 12017.5 =
+9.75 Tf +12017.5
532.25 Tf
–9.75 Tf + 4241.25
+9.75 Tf +12017.5
=
Tf = 30.5oC
16258.75
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 0.129 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 + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 104
3 x 104 = 1.36 x 103 Tf
Tf = 22.1oC
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 = 72.0 oC
mass = 87.0 g
T = 13.0 oC
mass = 59.0 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)]
-246.8 Tf + 3208 = 364 Tf - 26208
29416 = 610.8 Tf
Tf = 48.2oC
HW #4. 240. g of water (initially at 20.0oC) are mixed with an
unknown mass of iron initially at 500.0oC (CFe = 0.4495 J/goC).
When thermal equilibrium is reached, the mixture has a
temperature of 42.0oC. Find the mass of the iron.
Fe
T = 500oC
mass = ? grams
T = 20oC
mass = 240 g
- LOSE heat = GAIN heat
-q1 = q2
- [ (mass) (CFe ) (DT)] = (mass) (CH2O) (DT)
- [ (X g) (0.4495 J/goC) (42oC - 500oC)] = (240 g) (4.184 J/goC) (42oC - 20oC)]
- [ (X) (0.4495) (-458)] = (240 g) (4.184) (22)
205.9 X = 22091
X = 107 g Fe
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) = –410.48 Tf + 34767.32
qice = 23.6 (2.077) (0 – –31) + 23.6 (333) + 23.6 (4.18) (Tf – 0)
=
1519.53
= 9378.33 + 98.65 Tf
+ 7858.8 +
509.13 Tf = 25388.99
Tf = 49.9oC
98.65 Tf
Heating Curve Challenge Problems
1. A sample of ice at
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
Temperature (oC)
-25oC
140
120
100
80
60
40
20
0
-20
-40
-60
-80
-100
DH = mol x DHvap
DH = mol x DHfus
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, solid
Time
2. 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.
45.6 oC
3. 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.
76.6 oC
Heating Curve for Water
(Phase Diagram)
q4 = m DHvap
DHvap = +/- 2256 J/g
140
120
Temperature (oC)
100
q2 = m DHfus
DHfus = +/- 333 J/g
BP
80
60
0
B
2
-20
-40
q5 = m C D T
C g = 2.042 J/goC
C
AB
BC
CD
DE
ED
EF
1
q1 = m C D T
Cs = 2.077 J/goC
-60
-80
4
q3 = m C D T
Cl = 4.184 J/goC
20
MP
5
E
D
3
40
F
A
-100
Heat
warm ice
melt ice (s  l)
warm water
boil water (l  g)
condense steam (g  l)
superheat steam
Calculating Energy Changes Heating Curve for Water
140
120
DH = mol x DHfus
DH = mol x DHvap
Temperature (oC)
100
80
Heat = mass x Dt x Cp, gas
60
40
20
0
Heat = mass x Dt x Cp, liquid
-20
-40
-60
-80
Heat = mass x Dt x Cp, solid
-100
Time