thermochemical calculations

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.