1. Thermochemical Calculations

2. CA Standards
Students know energy is released when a material condenses or freezes and is absorbed when a material evaporates or melts.
Students know how to solve problems involving heat flow and temperature changes, using known values of specific heat and latent heat of phase change.

3. 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

4. Specific Heat
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.

5. Calculations Involving Specific HeatOR
cp = Specific Heat
Q = Heat lost or gained
T = Temperature change
m = Mass

6. Specific Heat
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.
Substance
Specific Heat (J/g·K)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (vapor)
1.87
Aluminum (solid)
0.897
Carbon (graphite,solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
Lead (solid)
0.129
Gold (solid)

7. Latent Heat of Phase Change
Molar Heat of Fusion
The energy that must be absorbed in order to convert one mole of solid to liquid at its melting point.
Molar Heat of Solidification
The energy that must be removed in order to convert one mole of liquid to solid at its freezing point.

8. Latent Heat of Phase Change #2
Molar Heat of Vaporization
The energy that must be absorbed in order to convert one mole of liquid to gas at its boiling point.
Molar Heat of Condensation
The energy that must be removed in order to convert one mole of gas to liquid at its condensation point.

9. Latent Heat – Sample Problem
Problem: The molar heat of fusion of water is
6.009 kJ/mol. How much energy is needed to convert 60 grams of ice at 0C to liquid water at 0C?
Mass
of ice
Molar
Mass of
water
Heat of
fusion

10. A 322 g sample of lead (specific heat = 0
A 322 g sample of lead (specific heat = 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?
T = ? oC Pb
mass = 322 g
Ti = 25oC
mass = 264 g Pb
Tf = 46oC -
LOSE heat = GAIN heat
- [(Cp,Pb) (mass) (DT)] = (Cp,H2O) (mass) (DT)
Drop Units:
- [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)]
- [(44.44) (46oC - Ti)] = (1104.6) (21oC)]
Ti =
44.44 Ti =
Ti = 568oC
Calorimetry Problems 2
question #12

11. Find the mass of the iron.
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. Fe
T = 500oC
mass = ? grams
T = 20oC
mass = 240 g -
LOSE heat = GAIN heat
- [(Cp,Fe) (mass) (DT)] = (Cp,H2O) (mass) (DT)
- [( 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 =
X = g Fe
Calorimetry Problems 2
question #5

12. 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 J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. Au
T = 785oC
mass = 97 g
T = 15oC
mass = 323 g
LOSE heat = GAIN heat -
- [(Cp,Au) (mass) (DT)] = (Cp,H2O) (mass) (DT)
Drop Units:
- [(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 x = x 103 Tf x 104
3 x 104 = x 103 Tf
Tf = oC
Calorimetry Problems 2
question #8

13. If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature of the system.
T = 13oC
mass = 59 g
T = 72oC
mass = 87 g
LOSE heat = GAIN heat -
- [(Cp,H2O) (mass) (DT)] = (Cp,H2O) (mass) (DT)
Drop Units:
- [(4.184 J/goC) (87 g) (Tf - 72oC)] = (4.184 J/goC) (59 g) (Tf - 13oC)
- [(364.0) (Tf - 72oC)] = (246.8) (Tf - 13oC)
-364 Tf = Tf
= Tf
Tf = oC
Calorimetry Problems 2
question #9

14. A 38 g sample of ice at -11oC is placed into 214 g of water at 56oC.
Find the system's final temperature.
Temperature (oC) 40 20
-20
-40
-60
-80
-100
120
100 80 60
140
Time
DH = mol x DHfus
DH = mol x DHvap
Heat = mass x Dt x Cp, liquid
Heat = mass x Dt x Cp, gas
Heat = mass x Dt x Cp, solid
ice
T = -11oC
mass = 38 g D
water cools
T = 56oC
mass = 214 g B
warm water A C
warm ice
melt ice
LOSE heat = GAIN heat - C D A B
- [(Cp,H2O(l)) (mass) (DT)] = (Cp,H2O(s)) (mass) (DT) + (Cf) (mass) + (Cp,H2O(l)) (mass) (DT)
-[(4.184 J/goC)(214g)(Tf-56oC)] = (2.077J/goC)(38g)(11oC) + (333J/g)(38g) + (4.184J/goC)(38g)(Tf-0oC)
- [(895) (Tf - 56oC)] = (159) (Tf)]
- 895 Tf = Tf
- 895 Tf = Tf
= Tf
Tf = oC
Calorimetry Problems 2
question #10

15. Heat of Solution
The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.
Substance
Heat of Solution
(kJ/mol)
NaOH
-44.51
NH4NO3
+25.69
KNO3
+34.89
HCl
-74.84