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Calorimetry
Heat Transfer
Al Al
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 20 gT = 20oC
20 g (40oC) 20 g (20oC) 30oC
Block “A” Block “B”Final
Temperature
Assume NO heat energy is “lost” to the surroundings from the system.
C30
g) 20 g (20C20g 20C40g 20 o
oo
What will be the final temperature of the system ?
a) 60oC b) 30oC c) 20oC d) ?
Heat Transfer
AlAl
m = 20 gT = 40oC
SYSTEM
Surroundings
m = 10 gT = 20oC
20 g (40oC) 20 g (20oC) 30.0oC
Block “A” Block “B”Final
Temperature
Assume NO heat energy is “lost” to the surroundings from the system.
20 g (40oC) 10 g (20oC) 33.3oC
C3.33
g) 10 g (20C20g 10C40g 20 o
oo
What will be the final temperature of the system ?
a) 60oC b) 30oC c) 20oC d) ?
?
Heat Transfer
AlAl
m = 20 gT = 20oC
SYSTEM
Surroundings
m = 10 gT = 40oC
20 g (40oC) 20 g (20oC) 30.0oC
Block “A” Block “B”Final
Temperature
Assume NO heat energy is “lost” to the surroundings from the system.
20 g (40oC) 10 g (20oC) 33.3oC
C7.26
g) 10 g (20C40g 10C20g 20 o
oo
20 g (20oC) 10 g (40oC) 26.7oC
Heat Transfer
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
20 g (40oC) 20 g (20oC) 30.0oC
Block “A” Block “B”Final
Temperature
20 g (40oC) 10 g (20oC) 33.3oC
C46
g) 30 g (75C100g 30C25g 75 o
oo
20 g (20oC) 10 g (40oC) 26.7oC
AgH2O
Real Final Temperature = 26.6oC
Why?
We’ve been assuming ALL materialstransfer heat equally well.
Specific Heat
• Water and silver do not transfer heat equally well. Water has a specific heat Cp = 4.184 J/goC Silver has a specific heat Cp = 0.235 J/goC
• What does that mean? It requires 4.184 Joules of energy to heat 1 gram of water 1oC and only 0.235 Joules of energy to heat 1 gram of silver 1oC.
• Law of Conservation of Energy… In our situation (silver is “hot” and water is “cold”)… this means water heats up slowly and requires a lot of energy whereas silver will cool off quickly and not release much energy.
• Lets look at the math!
“loses” heat
Calorimetry
C26.6 x
320.8x 8550
7845 313.8x x 05.7 705
algebra. the solve and units Drop
C25 -x g 75CgJ 184.4 C100 -x g 30CgJ 235.0
equation. into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
Tfinal = 26.6oC
Calorimetry
C26.6 x
8550 320.8x
7845 313.8x x 05.7 705
algebra. the solve and units Drop
C25 -x g 75CgJ 184.4 C100 -x g 30CgJ 235.0
equation. into values Substitute
TTmC TTmC
TmC TmC
q q
o
oooo
ifpinitialfinalp
pp
OHAg 2
m = 75 gT = 25oC
SYSTEM
Surroundings
m = 30 gT = 100oC
AgH2O
240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC). When thermal equilibrium is reached, the system has a temperature of 42oC. Find the mass of the iron.
Calorimetry Problems 2 question #5
FeT = 500oCmass = ? grams
T = 20oC
mass = 240 g LOSE heat = GAIN heat-
- [(Cp,Fe) (mass) (T)] = (Cp,H2O) (mass) (T)
- [(0.4495 J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)]
Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22)
205.9 X = 22091
X = 107.3 g Fe
A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15oC. 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.
Calorimetry Problems 2 question #8
AuT = 785oCmass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat-
- [(Cp,Au) (mass) (T)] = (Cp,H2O) (mass) (T)
- [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] Drop Units:
- [(12.5) (Tf - 785oC)] = (1.35x 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
If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature of the system.
Calorimetry Problems 2 question #9
T = 13oC
mass = 59 g
LOSE heat = GAIN heat-
- [(Cp,H2O) (mass) (T)] = (Cp,H2O) (mass) (T)
- [(4.184 J/goC) (59 g) (Tf - 13oC)] = (4.184 J/goC) (87 g) (Tf - 72oC)] Drop Units:
- [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)]
-246.8 Tf + 3208 = 364 Tf - 26208
29416 = 610.8 Tf
Tf = 48.2oC
T = 72oC
mass = 87 g
A 322 g sample of lead (specific heat = 0.138 J/goC) is placed into 264 g of water at 25oC.If the system's final temperature is 46oC, what was the initial temperature of the lead?
Calorimetry Problems 2 question #12
PbT = ? oCmass = 322 g
Ti = 25oC
mass = 264 g
LOSE heat = GAIN heat-
- [(Cp,Pb) (mass) (T)] = (Cp,H2O) (mass) (T)
- [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)] Drop Units:
- [(44.44) (46oC - Ti)] = (1104.6) (21oC)]
- 2044 + 44.44 Ti = 23197
44.44 Ti = 25241
Ti = 568oC
Pb
Tf = 46oC
25 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system.
Calorimetry Problems 2 question #11
- [(Cp,H2O) (mass) (T)] + (Cv,H2O) (mass) + (Cp,H2O) (mass) (T) = [(Cp,H2O) (mass) (T)]
- [ - 816.8 - 56400 + 104.5Tf - 10450] = 997Tf - 7972
- [qA + qB + qC] = qD
qA = [(Cp,H2O) (mass) (T)]
qA = [(2.042 J/goC) (25 g) (100o - 116oC)]
qA = - 816.8 J
qB = (Cv,H2O) (mass)
qA = (2256 J/g) (25 g)
qA = - 56400 J
qC = [(Cp,H2O) (mass) (T)]
qC = [(4.184 J/goC) (25 g) (Tf - 100oC)]
qA = 104.5Tf - 10450
qD = (4.184 J/goC) (238.4 g) (Tf - 8oC)
qD = - 997Tf - 7972
- [qA + qB + qC] = qD
816.8 + 56400 - 104.5Tf + 10450 = 997Tf - 7972
67667 - 104.5Tf = 997Tf - 7979
75646 = 1102Tf
1102 1102
Tf = 68.6oC
Tem
pera
ture
(o C
)
40200
-20-40-60-80
-100
120100
8060
140
Time
H = mol x Hfus
H = mol x Hvap
Heat = mass x t x Cp, liquid
Heat = mass x t x Cp, gas
Heat = mass x t x Cp, solid
A
B
C
D
(1000 g = 1 kg)
238.4 g
A Coffee Cup Calorimeter
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302
Thermometer
Styrofoamcover
Styrofoamcups
Stirrer
Thermometer
Glass stirrer
Cork stopper
Two Styrofoam ®cups nestedtogether containingreactants in solution
Bomb Calorimeter
thermometer
stirrer
full of water
ignition wire
steel “bomb”
sample
A Bomb Calorimeter