1. Calorimetry

2. Energy released to the surrounding as heat
Burning of a Match
System
Surroundings
D(PE)
(Reactants)
Potential energy
Energy released to the surrounding as heat
(Products)
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293

3. Conservation of Energy in a Chemical Reaction
In this example, the energy of the reactants and products increases,
while the energy of the surroundings decreases.
In every case, however, the total energy does not change.
Endothermic Reaction
Reactant + Energy Product
Surroundings
Surroundings
System
Energy
System
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41

4. Conservation of Energy in a Chemical Reaction
In this example, the energy of the reactants and products decreases,
while the energy of the surroundings increases.
In every case, however, the total energy does not change.
Exothermic Reaction
Reactant Product + Energy
Surroundings
Surroundings
System
System
Energy
Before
reaction
After
reaction
Myers, Oldham, Tocci, Chemistry, 2004, page 41

5. Direction of Heat Flow Surroundings ENDOthermic EXOthermic qsys > 0
System
ENDOthermic
qsys > 0
EXOthermic
qsys < 0
System
H2O(s) + heat  H2O(l)
melting
H2O(l)  H2O(s) + heat
freezing
Kotz, Purcell, Chemistry & Chemical Reactivity 1991, page 207

6. Caloric Values Food joules/grams calories/gram Calories/gram
Protein
Fat
Carbohydrates
1calories = joules
1000 calories = 1 Calorie
"science" "food"
Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51

7. Experimental Determination of Specific Heat of a Metal
Typical apparatus used in this activity include a boiler (such as large glass beaker), a heat source (Bunsen burner or hot plate), a stand or tripod for
the boiler, a calorimeter, thermometers, samples (typically samples of copper, aluminum, zinc, tin, or lead), tongs (or forceps or string) to handle samples, and a balance.

8. A Coffee Cup Calorimeter
Thermometer
Styrofoam
cover
cups
Stirrer
A Coffee Cup Calorimeter
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302

9. A Bomb Calorimeter

10. Calorimetry “loses” heat Surroundings SYSTEM Tfinal = 26.7oC H2O Ag
m = 75 g
T = 25oC
m = 30 g
T = 100oC

11. Calorimetry Surroundings SYSTEM H2O Ag m = 75 g T = 25oC m = 30 g

12. q = Cp . m . DT Cp(ice) = 2.077 J/g oC
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
Cp(ice) = J/g oC
It takes Joules to raise 1 gram ice 1oC.
X Joules to raise 10 gram ice 1oC.
(10 g)(2.077 J/g oC) = Joules
X Joules to raise 10 gram ice 10oC.
(10oC)(10 g)(2.077 J/g oC) = Joules
q = Cp . m . DT
Heat = (specific heat) (mass) (change in temperature)

13. 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
q = Cp . m . DT
Heat = (specific heat) (mass) (change in temperature)
Given Ti = -30oC
Tf = -20oC
q = Joules

14. 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

15. 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

16. 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

17. 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