Flows from highertemperature object to lowertemperature object System T 1 Surroundings T 2 Heat If T 1 gt T 2 q system exothermic System T 1 Surroundings T ID: 933720
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Slide1
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)
Surroundings(T2)
Heat
If T1 > T2q system = -exothermic
System(T1)
Surroundings(T2)
Heat
If T
1
< T
2
q
system
= +
endothermic
Slide2Calorimetry
: the measurement of heat flow
device used is called a...
calorimeterspecific 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)
heat of fusion (ΔH
fus):heat of vaporization (ΔHvap):For energy changes involving…melting/freezingboiling/condensingThere are NO
temp changes during a phase change.
Slide3Various
Specific Heat Capacities
SubstanceSpecificheat capacity
(J/K g)GoldSilverCopperIronAluminumH2O(l)H2O(s)H2
O(g)
0.1290.2350.3850.4490.8974.1842.031.998Metals do not generally require much energy to heat them up(i.e. they heat up easily)Water requires much more energy to heat up
Slide4We can find the heat a substance loses or gains using:
where q =
heat (J)
m = mass of substance (g) C = specific heat (J/goC)
DT = temperature
change (oC) q = m C DT(used within a givenstate of matter)AND
q = m ΔH
(used between twostates of matter or during a phase change)
HEAT
Temp.
s
s/
l
l
l
/g
g
Heating Curve
ΔH
fus
ΔH
vap
C
s
C
l
C
g
+
–
D
= final – initial
D
H
=
heat of
vap
/
fus
(
J/g)
Slide5q = m
C Δ
Tq (J) = mass (g) C (J/goC) ΔT (
oC) q = joules (J) Mnemonic device: q = m “CAT” Using heat capacities…
Slide6Heating Curve of water
Temperature (
o
C)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Energy Added
Melting Point
Boiling Point
Solid
Liquid
Gas
s ↔
l
l
↔ g
Slide7Temperature (
o
C)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Energy Added
Slide8Heating Curve of Water
Temperature (
o
C)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Energy Added
Slide9Heating 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/goCPhase Changes (s ↔ l ↔ g)change in PE (molecular arrangement)temperature remains constantovercoming intermolecular forces
(requires the most heat)(requires the least heat)ΔHfus = 333 J/gΔHvap = 2256 J/gWhy is this so much larger?
(l ↔ g)(s ↔
l)
Slide10Heating Curves
Temperature (
o
C)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Energy Added
Melting - PE
Solid - KE
Liquid - KE
Boiling - PE
Gas - KE
Slide11Heat
Heat
q
1: Heat the ice to 0°Cq2: Melt the ice into a liquid at 0°Cq3: Heat the water from 0°C to 100°Cq4: Boil the liquid into a gas at 100°Cq5: Heat the gas above 100°C
Heating Curve of WaterFrom Ice to Steam in Five Easy Steps
q1 = m Cs ΔT q2 = m ΔHfusq3 =
m Cl ΔT
q4 = m ΔHvapq5 = m Cg ΔT
q
1
q
2
q
3
q
4
q
5
q
tot
=
q
1
+ q
2
+
q
3
+ q
4
+
q
5
Slide125
4
3
2
1
Heating Curve Practice1. How much energy (J) is required to heat 12.5 g of ice at –10.0 oC to water at 0.0 oC?4,420 Jq1: Heat the ice from -10 to 0°Cq2: Melt the ice at 0°C to liquid at 0
oC
q1 = 12.5 g (2.077 J/g oC)(0.0 - -10.0 oC) = q2 = 12.5 g (333 J/g) =
qtot
=
q
1
+
q
2
259.63
J
=
259.63
J +
4,162.5
J =
4162.5
J
Notice that your q values are positive because heat is added…
Slide135
4
3
2
1
Heating Curve Practice2. How much energy (J) is required to heat 25.0 g of ice at –25.0 oC to water at 95.0 oC?19,560 Jq1: Heat the ice from -25 to 0°Cq2: Melt the ice at 0°C to liquid at 0 o
Cq3: Heat the water from 0°C to 95 °C
q1 = 25.0 g (2.077 J/g oC)(0.0 - -25.0 oC) = q2 = 25.0 g (333 J/g) =
q
tot
=
q
1
+
q
2
+
q
3
1298.1
J
8325 J
=
1298.1
J +
8,325
J + 9937 J =
q
3
= 25.0 g (4.184 J/g
o
C
)(95.0 – 0.0
o
C) =
9937 J
Notice that your q values are positive because heat is added…
Slide145
4
3
2
1
Heating Curve Practice3. How much energy (J) is removed to cool 50.0 g of steam at 115.0 oC to ice at -5.0 oC?-152,000 Jq5: Cool the steam from 115.0 to 100°Cq
4: 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)(100.0 - 115.0
o
C
) =
q
4
=
50.0 g
( 2256 J/g)
=
q
tot
=
q
1
+
q
2
+
q
3
+
q
4
+
q
5
-
1531.5
J
-112,800 J
= -
1531.5
J + -
112,800
J + -20920 J + -
16,650
J + -
519.25
J =
q
3
= 50.0 g (4.184 J/g
o
C
)(0.0 – 100.0
o
C) =
-20920 J-16650 J
q1
= 50.0 g (
2.077 J/g oC
)(- 5.0 – 0.0oC) =
-519.25
JNotice that your q values are
negative because heat is removed…-
Slide15Food and Energy
Caloric Values
Food joules/gram
calories/gram “Calories”/gramProtein 17,000 4,000 4Fat 38,000 9,000 9Carbohydrates 17,000 4,000 4Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 511000 calories = 1 “Calorie”
"science" "food"
1 calorie = 4.184 joulesor… 1 Kcal = 1 “Calorie”
Slide16Does 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
2.5 oC
1 L = 1000 mL1 mL = 1 gq = 1.0 x 103 g (4.184 J/g oC)(37 oC - 2.5
oC) = 144,348 J1000 calories = 1 “Calorie”
1 calorie = 4.184 joules 144348 J1 cal4.184 J=
35 Cal
1 “Cal”
1000 cal
Slide17C H
2O (l) = 4.184 J/goC C H2O (s) = 2.03 J/goCC H2O (g) = 1.998 J/g
oC
Temperature (
o
C)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Energy Added
Δ
H
fusion
= 6.02 kJ/mol
Δ
H
vap
= 40.7 kJ/mol
Heating Curve of H
2
O Constants and Graph
Slide18What will happen over time?
Zumdahl, Zumdahl, DeCoste,
World of Chemistry
2002, page 291
Slide19Let’s take a closer look…
Zumdahl, Zumdahl, DeCoste,
World of Chemistry
2002, page 291
Slide20Eventually, the temperatures will equalize
Zumdahl, Zumdahl, DeCoste,
World of Chemistry
2002, page 291
Slide21Thermometer
Styrofoam
cover
Styrofoam
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
q
x
=
–
q
y
Slide22125 g
23.0 °C
Heat Transfer Experiments qwater
= –qPb q = m x C x ΔT for both cases, although specific values differPlug in known information for each side Solve for Tf ...Pb
75.0 g
435.0 °C C = 0.130 J/°C g
What is the final temperature, Tf, of the mixture?
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 23.0
o
C. What is the final temperature of the mixture?
Slide23A 75.0 g piece of lead (specific heat = 0.130
J/
goC),initially at
435oC, is set into 125.0 g of water, initiallyat 23.0oC. What is the final temperature of the mixture?m
water Cwater D
Twater qwater = –qPb –mPb CPb DTPb
=125 (4.18) (Tf
– 23) =522.5 Tf – 12017.5 =–9.75 Tf + 4241.25+9.75 Tf
+9.75 Tf
+12017.5
+12017.5
532.25 T
f
=
16258.75
T
f
= 30.5
o
C
–75 (0.13) (
T
f
– 435)
q = m x C x
Δ
T for both cases, although specific values differ
Plug in known information for each side
Slide242. A
97.0 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15.0o
C. 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 = 785oCmass = 97.0 gT = 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 - 785
oC)] = (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 1043 x 10
4
= 1.36 x 10
3
T
f
T
f
= 22.1
o
C
Au
Slide25HW #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 oCmass = 59.0 gLOSE 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 TfTf = 48.2oC
T = 72.0 oC
mass = 87.0 g
Slide26HW #4.
240. g of water (initially at 20.0oC) are mixed with an unknown mass of iron initially at 500.0
oC (CFe = 0.4495 J/goC). When thermal equilibrium is reached, the mixture has a temperature of 42.0
oC. Find the mass of the iron.T = 500oCmass = ? gramsT = 20
oCmass = 240 g
LOSE heat = GAIN heat-- [ (mass) (CFe ) (DT)] = (mass) (CH2O) (DT)- [ (X g) (0.4495 J/goC) (42
oC - 500oC)] = (240 g) (4.184 J/goC) (42oC - 20o
C)] - [ (X) (0.4495) (-458)] = (240 g) (4.184) (22)205.9 X = 22091X = 107 g Fe-q1 = q2
Fe
Slide27A 23.6 g ice cube at –31.0
o
C is dropped into 98.2 g ofwater at 84.7oC. Find the equilibrium temperature.
KEY: Assume that the ice melts and the final product is a liquid.qice = –qwater q
ice = 23.6 (2.077) (0 – –31) + 23.6 (333) + 23.6 (4.18) (Tf
– 0)509.13 Tf=25388.99Tf = 49.9oC
qwater
= –98.2 (4.18) (Tf – 84.7)= 1519.53 + 7858.8 + 98.65 Tf = 9378.33 + 98.65 Tf
= –410.48 Tf + 34767.32
Slide28Heating Curve Challenge Problems
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?
Temperature (
o
C)
40
20
0
-20
-40
-60
-80
-100
120
100
80
60
140
Time
D
H =
mol x
D
H
fus
D
H =
mol x
D
H
vap
Heat = mass x
D
t x C
p, liquid
Heat = mass x
D
t x C
p, gas
Heat = mass x
D
t x C
p, solid
A 38 g sample of ice at -5
o
C is placed into 250 g of water at 65
o
C. Find the final temperature of the mixture assuming that the ice sample completely melts.
A 35 g sample of steam at 116
o
C are bubbled into 300 g water at 10
o
C. Find the final temperature of the system, assuming that the steam condenses into liquid water.
52.8 g ice
45.6
o
C
76.6
o
C
Slide29A
B warm iceB C melt ice (s
l)C D warm waterD E boil water (l g)
E D condense steam (g l)E F superheat steamHeating Curve for Water(Phase Diagram) 140
120 100
80 60 40 20 0 -20 -40 -60 -80-100Temperature (oC)
Heat
BPMP
A
B
C
D
E
F
q
2
= m
D
H
fus
D
H
fus
=
+/-
333 J/g
q
4
=
m
D
H
vap
D
H
vap
=
+/- 2256
J/g
q
3
= m
C
D
T
C
l
= 4.184 J/
g
o
C
q
1
=
m
C
DTC
s = 2.077 J/
goC
q5 = m
C D
T
C g = 2.042 J/g
oC
1
2
3
4
5
Slide30Calculating Energy Changes - Heating Curve for Water
Temperature (
o
C)
40
200-20-40
-60-80
-1001201008060140
Time
D
H =
mol x
D
H
fus
D
H =
mol x
D
H
vap
Heat = mass x
D
t x C
p, liquid
Heat = mass x
D
t x C
p, gas
Heat = mass x
D
t x C
p, solid