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Thermochemical Calculations. CA Standards. Units for Measuring Heat. The Joule is the SI system unit for measuring heat:. The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree. Specific Heat. - PowerPoint PPT Presentation
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Thermochemical Thermochemical CalculationsCalculations
CA StandardsCA Standards
Units for Measuring HeatUnits for Measuring HeatThe JouleJoule is the SI system unit for measuring heat:
The caloriecalorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree
2
2111
s
mkgmeternewtonJoule
Joulescalorie 18.41
Specific Specific HeatHeat
The amount of The amount of heat required to heat required to raise the raise the temperature of temperature of one gram of one gram of substance by substance by one degree one degree Celsius.Celsius.
1
2
3
45 6
7
8
9
1 10
2
3
45 6
7
8
9
11
Calculations Involving Specific Calculations Involving Specific HeatHeat
ccpp = Specific Heat
QQ = Heat lost or gained
TT = Temperature change
OROR
mm = Mass
Tm
Qcp
pcTmQ
Specific HeatSpecific HeatThe amount of heat required to raise the temperature of one gram of substance by one degree Celsius.SubstanceSubstance Specific Heat (J/g·K)Specific Heat (J/g·K)
Water (liquid) Water (liquid) 4.184.18
Ethanol (liquid) Ethanol (liquid) 2.442.44
Water (solid) Water (solid) 2.062.06
Water (vapor) Water (vapor) 1.871.87
Aluminum (solid) Aluminum (solid) 0.8970.897
Carbon (graphite,solid) Carbon (graphite,solid) 0.7090.709
Iron (solid) Iron (solid) 0.4490.449
Copper (solid) Copper (solid) 0.3850.385
Mercury (liquid) Mercury (liquid) 0.1400.140
Lead (solid)Lead (solid) 0.1290.129
Gold (solid) Gold (solid) 0.1290.129
Latent Heat of Phase ChangeLatent Heat of Phase ChangeMolar Heat of FusionMolar Heat of Fusion
The energy that must be The energy that must be absorbedabsorbed in order to convert one in order to convert one mole of mole of solid to liquidsolid to liquid at its at its melting pointmelting point..
The energy that must be The energy that must be removedremoved in order to convert one in order to convert one mole of mole of liquid to solidliquid to solid at its at its freezing pointfreezing point..
Molar Heat of Molar Heat of SolidificationSolidification
Latent Heat of Phase Change Latent Heat of Phase Change #2#2
Molar Heat of VaporizationMolar Heat of Vaporization
The energy that must be The energy that must be absorbedabsorbed in order to convert one in order to convert one mole of mole of liquid to gasliquid to gas at its at its boiling boiling pointpoint..
The energy that must be The energy that must be removedremoved in order to convert one in order to convert one mole of mole of gas to liquidgas to liquid at its at its condensation pointcondensation point..
Molar Heat of Molar Heat of CondensationCondensation
Latent HeatLatent Heat – Sample – Sample ProblemProblem
ProblemProblem: The molar heat of fusion of water is: The molar heat of fusion of water is
6.009 kJ/mol. How much energy is needed to convert 6.009 kJ/mol. How much energy is needed to convert 60 grams of ice at 060 grams of ice at 0C to liquid water at 0C to liquid water at 0C?C?
Massof ice
MolarMass ofwater
Heatof
fusion
kiloJoulesOHmol
kJ
OHg
OHmolOHg20
1
009.6
02.18
160
22
22
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 = ? oC
mass = 322 g
Ti = 25oCmass = 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
PbTf = 46oC
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 = 500oC
mass = ? grams
T = 20oCmass = 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 = 15oCmass = 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)] = (1.35 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
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 = 13oCmass = 59 g
LOSE heat = GAIN heat-
- [(Cp,H2O) (mass) (T)] = (Cp,H2O) (mass) (T)
- [(4.184 J/goC) (87 g) (Tf - 72oC)] = (4.184 J/goC) (59 g) (Tf - 13oC)Drop Units:
- [(364.0) (Tf - 72oC)] = (246.8) (Tf - 13oC)
-364 Tf + 26208 = 246.8 Tf - 3208
29416 = 610.8 Tf
Tf = 48.2oC
T = 72oCmass = 87 g
A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC. Find the system's final temperature.
Calorimetry Problems 2
question #10
iceT = -11oCmass = 38 g
T = 56oCmass = 214 g
LOSE heat = GAIN heat-
- [(Cp,H2O(l)) (mass) (T)] = (Cp,H2O(s)) (mass) (T) + (Cf) (mass) + (Cp,H2O(l)) (mass) (T)
-[(4.184 J/goC)(214g)(Tf-56oC)] = (2.077J/goC)(38g)(11oC) + (333J/g)(38g) + (4.184J/goC)(38g)(Tf-0oC) - [(895) (Tf - 56oC)] = 868 + 12654 + (159) (Tf)]
- 895 Tf + 50141 = 868 + 12654 + 159 Tf
- 895 Tf + 50141 = 13522 + 159 Tf
Tf = 34.7oC
36619 = 1054 Tf
Tem
pera
ture
(oC
)
40200
-20-40-60-80
-100
1201008060
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
warm icemelt ice
warm water
water cools
D BA C
Heat of SolutionHeat of SolutionThe Heat of SolutionHeat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.