Chapter 7: Hot & Cold Packs
Introductory Activity
How many things can you think of in everday life that either give off heat or absorb heat?
Which of these things are physical processes?
Which are chemical processes?
Hot & Cold Packs
This chapter will introduce the chemistry needed to understand how Hot & Cold Packs workSection 7.1: Endothermic & ExothermicSection 7.2: Calorimetry and heat capacitySection 7.3: Changes in StateSection 7.4: Heat of a Chemical ReactionSection 7.5: Hess’s Law
Hot/Cold Packs
Transfer of energy
Transfer of energy
Use
System & Surroundings
System & Surroundings
between
Materials ability to absorb energy
without noticeable temperature
change
Materials ability to absorb energy
without noticeable temperature
change
Effect on temperature depends on
Physical change
Physical change
Is determined with
CalorimetryCalorimetry
Can be done in
Chemical change
Chemical change
Section 7.1—Endothermic and Exothermic
Why are hot packs “hot” and cold packs feel “cold”
Endothermic & Exothermic
When the system absorbs energy from the surroundings, it’s an endothermic process
When the system releases energy to the surroundings, it’s an exothermic process
System & Surroundings
It’s very important to define the system & surroundings correctly to use the endo- and exothermic definitions!
Most people define the system too broadly They include everything in the beaker or
container as the system
However, the system is only the molecules undergoing the change
System & Surroundings
The system is only made of the molecules undergoing the change
The water molecules & the container…
Your hand and the air…
Even the thermometer…
They are all the surroundings
Note that water is made up of water molecules—not a solid chunk of water…but for this picture, it’s best to represent water as one thing since it’s the surroundings and focus on the molecules reacting as the system.
Exothermic & You
You touch the beaker and it feels hot
Energy is being transferred TO YOU
You are the surroundings
When energy moves from system to surroundings, it’s exothermic
Exothermic & the Thermometer
The temperature (measured by the thermometer) is related to the average kinetic energy of the molecules in the container
The majority of the molecules in a solution are water
If the temperature is increasing, the energy of the water molecules is increasing
Since water is the surrounding (it’s not actually reacting), energy is being transferred to the surroundings
Exothermic shows an increase in the temperature within the container
Endothermic
The opposite is also trueIf the container feels cold to you, energy is
being transferred FROM YOU (the surroundings) into the system—endothermic
If the thermometer goes down, energy is being transferred FROM the water molecules (surroundings) into the system--endothermic
Let’s Practice
Example:Identify the system and surroundings when you hold an ice cube while it
melts. Is this endo- or exothermic?
Let’s Practice
Example:Identify the system and surroundings when you hold an ice cube while it
melts. Is this endo- or exothermic?
System: Water molecules in the form of iceSurroundings: You and the air
It feels cold to you…so energy is leaving you (surroundings)
When energy goes from surroundings to system it’s endothermic
Section 7.2—Calorimetry & Heat Capacity
Why do some things get hot more quickly than others?
Temperature
Temperature – proportional to the average kinetic energy of the molecules
Energy due to motion(Related to how fast the molecules are moving)
As temperature increases
Molecules move faster
Heat & Enthalpy
Heat (q)– The flow of energy from higher temperature particles to lower temperature particles
Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy”
Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume
Energy Units
The most common energy units are Joules (J) and calories (cal)
4.18 J 1.00 cal
1000 J
1000 cal
1 kJ
1 Cal (food calorie)
=
=
=
Energy Equivalents
These equivalents can be used in dimensional analysis to convert units
Heat Capacity
Specific Heat Capacity (Cp) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C
Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C
Heat Capacity
High Heat Capacity Low Heat Capacity
Takes a large amount of energy to noticeably change temp
Small amount of energy can noticeably change temperature
Heats up slowly
Cools down slowly
Maintains temp better with small condition changes
Heats up quickly
Cools down quickly
Quickly readjusts to new conditions
A pool takes a long time to warm up and remains fairly warm over night.The air warms quickly on a sunny day, but cools quickly at night
A cast-iron pan stays hot for a long time after removing from oven.Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly
What things affect temperature change?
Heat Capacity of substanceThe higher the heat capacity, the slower the
temperature change
Mass of sampleThe larger the mass, the more molecules there are to
absorb energy, so the slower the temperature change
TCmH p
Energy added or removed
Mass of sample
Specific heat capacity of substance
Change in temperature
Positive & Negative T
Change in temperature (T) is always T2 – T1 (final temperature – initial temperature) If temperature increases, T will be positive
A substance goes from 15°C to 25°C. 25°C - 15°C = 10°CThis is an increase of 10°C
If temperature decreases, T will be negativeA substance goes from 50°C to 35°C35°C – 50°C = -15°CThis is a decrease of 15°C
Positive & Negative H
Energy must be put in for temperature to increaseA “+” T will have a “+” H
Energy must be removed for temperature to decreaseA “-” T will have a “-” H
Example
Example:If 285 J is added to 45 g of water at 25°C, what is the
final temperature? Cp water = 4.18 J/g°C
Let’s Practice #1
Example:How many joules must be
removed from 25 g of water at 75°C to drop the
temperature to 30°? Cp water = 4.18 J/g°C
Let’s Practice #2
Example:If the specific heat capacity of
aluminum is 0.900 J/g°C, what is the final temperature if 437 J is added to a 30.0 g
sample at 15°C
Calorimetry
1st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes
This is also referred to as the Law of Conservation of Energy
Conservation of Energy
If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa
Calorimetry
Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system
Hsurroundings = - Hsystem
Because of the Law of Conservation of Energy,The energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system.
(m×Cp×T)surroundings = - (m×Cp×T)system
Don’t forget the “-” sign on one sideMake sure to keep all information about surroundings together and all information about system together—you can’t mix and match!
Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature
Two objects at different temperatures
So you know that T2 for the system is the same as T2 for the surroundings!
An example of CalorimetryExample:
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water
at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of
the metal?
Metal:m = 23.8 gT1 = 100.0°CT2 = 32.5°CCp = ? Water:m = 50.0 gT1 = 24°CT2 = 32.5°CCp = 4.18 J/g°C Cp = 1.04 J/g°C
watermetal HH
waterpmetalp TCmTCm
CC
CgJgCCCg p
5.245.3218.40.500.1005.328.23
CCg
CCCg
JgC p
0.1005.328.23
5.245.3218.40.50
An example of CalorimetryExample:
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water
at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of
the metal?
Let’s Practice #3Example:
A 10.0 g of aluminum (specific heat capacity is 0.900 J/g°C) at 95.0°C is placed in a
container of 100.0 g of water (specific heat capacity is 4.18 J/g°C) at 25.0°. What’s the
final temperature?
Section 7.3—Changes in State
What’s happening when a frozen ice pack melts?
The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature)
Change in State
To melt or boil, intermolecular forces must be broken
Breaking intermolecular forces requires energy
A sample with solid & liquid will not rise above the melting point until all the solid is gone.
The same is true for a sample of liquid & gas
Melting
When going from a solid to a liquid, some of the intermolecular forces are broken
The Enthalpy of Fusion (Hfus) is the amount of energy needed to melt 1 gram of a substanceThe enthalpy of fusion of water is 80.87 cal/g or 334 J/g
All samples of a substance melt at the same temperature, but the more you have the longer it takes to melt (requires more energy).
fusHmH Energy needed to melt
Mass of the sample
Energy needed to melt 1 g
Example
Example:Find the enthalpy of
fusion of a substance if it takes 5175 J to melt 10.5 g of the
substance.
Vaporization
When going from a liquid to a gas, all of the rest of the intermolecular forces are broken
The Enthalpy of Vaporization (Hvap) is the amount of energy needed to boil 1 gram of a substanceThe Hvap of water is 547.2 cal/g or 2287 J/g
All samples of a substance boil at the same temperature, but the more you have the longer it takes to boil (requires more energy).
vapHmH Energy needed to boil
Mass of the sample
Energy needed to boil 1 g
Example
Example:If the enthalpy of
vaporization of water is 547.2 cal/g, how many calories are
needed to boil 25.0 g of water?
Solid
Liquid
Gas
Melting
Vaporizing or Evaporating
Condensing
Freezing
Incr
easi
ng m
olec
ular
mot
ion
(tem
pera
ture
)
Changes in State go in Both Directions
Going the other way
The energy needed to melt 1 gram (Hfus) is the same as the energy released when 1 gram freezes. If it takes 547 J to melt a sample, then 547 J would be
released when the sample freezes. H will = -547 J
The energy needed to boil 1 gram (Hvap) is the same as the energy released when 1 gram is condensed. If it takes 2798 J to boil a sample, then 2798 J will be
released when a sample is condensed. H will = -2798 J
Example
Example:How much energy is
released with 157.5 g of water is condensed?
Hvap water = 547.2 cal/g
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Heating Curves
Melting & Freezing
Point
Boiling & Condensing
Point
Heating curves show how the temperature changes as energy is added to the sample
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Going Up & Down
+H
-H
Moving up the curve requires energy, while moving down releases energy
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
States of Matter on the Curve
Gas OnlyEnergy added
increases temp
Liquid & gasEnergy added breaks remaining IMF’s
Liquid OnlyEnergy added
increases temp
Solid & LiquidEnergy added breaks IMF’s
Solid OnlyEnergy added
increases temp
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Different Heat Capacities
Gas OnlyCp = 0.48 cal/g°C
Liquid OnlyCp = 1.00 cal/g°C
Solid OnlyCp = 0.51 cal/g°C
The solid, liquid and gas states absorb water differently—use the correct Cp!
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Changing States
Liquid & gasHvap = 547.2 cal/g
Solid & LiquidHfus = 80.87 cal/g
Heating curve of water
-50
0
50
100
150
Energy input
Te
mp
era
ture
Adding steps together
If you want to heat ice at -25°C to water at 75°C, you’d have to first warm the ice to zero before it could melt.
Then you’d melt the ice
Then you’d warm that water from 0°C to your final 75°
You can calculate the enthalpy needed for each step and then add them together
Example:How many calories are
needed to change 15.0 g of ice at -12.0°C to steam at
137.0°C?
ExampleUseful information:Cp ice = 0.51 cal/g°C
Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C
Hfus = 80.87 cal/gHvap = 547.2 cal/g
TCpmH
fusHmH
vapHmH
Example:How many needed to
change 40.5 g of water at 25°C to steam at 142°C?
Let’s PracticeUseful information:Cp ice = 0.51 cal/g°C
Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C
Hfus = 80.87 cal/gHvap = 547.2 cal/g
TCpmH
fusHmH
vapHmH
Section 7.4—Energy of a Chemical Reaction
What’s happening in those hot/cold packs that contain chemical reactions?
Enthalpy of Reaction
Enthalpy of Reaction (Hrxn) – Net energy change during a chemical reaction
+Hrxn means energy is being added to the system—endothermic-Hrxn means energy is being released from the system—exothermic
Enthalpy of Formation
Enthalpy of Formation (Hf) – Energy change when 1 mole of a compound is formed from elemental states
Heat of formation equations: H2 (g) + ½ O2 (g) H2O (g) C (s) + O2 (g) CO2 (g)
A table with Enthalpy of Formation values can be found in the Appendix of your text
Be sure to look up the correct state of matter:H2O (g) and H2O (l) have different Hf values!
The overall enthalpy of reaction is the opposite of Hf for the reactants and the Hf for the products
Reactants are broken apart and Products are formed.
Breaking apart reactants is the opposite of Enthalpy of Formation.
Forming products is the Enthalpy of Formation.
reactHprodHH ffrxn
Hrxn = sum of Hf of all products – the sum of Hf reactants
Enthalpy of Formation & Enthalpy of Reaction
This is not the way a reaction occurs—reactants break apart and then rearrange…remember Collision Theory from Chpt 2! But for when discussing overall energy changes, this manner of thinking is acceptable.
Example
Example:Find the Hrxn for:
CH4 (g) + 2 O2 (g) 2 H2O (g) + CO2 (g)
Hf (kJ/mole)
CH4 (g) -74.81
O2 (g) 0
H2O (g) -241.8
CO2 (g) -393.5
Let’s Practice #1
Hf (kJ/mole)
CH3OH (l) -238.7
O2 (g) 0
H2O (l) -285.8
CO2 (g) -393.5
Example:Find the Hrxn for:
2 CH3OH (l) + 3 O2 (g) 2 CO2 (g) + 4 H2O (l)
Enthalpy & Stoichiometry
The Enthalpy of Reaction can be used along with the molar ratio in the balanced chemical equation
This allows Enthalpy of Reaction to be used in stoichiometry equalities
Example:If 1275 kJ is released, how many grams of B2O3 is
produced?B2H6 (g) + 3 O2 (g) B2O3 (s) + 2 H2O (g)
H = -2035 kJ
Example
Let’s Practice #2
If you need to produce 47.8 g B2O3, how many kilojoules will be released?
B2H6 (g) + 3 O2 (g) B2O3 (s) + 2 H2O (g) H = -2035 kJ
Section 7.5—Hess’s Law
How can we find the enthalpy of a reaction using step-wise reactions?
Hess’s Law
Hess’ Law – The sum of the energy changes during a series of reactions is equal to the sum of the reaction.
In other words…if you go from Reactant A to Product Z all in one step, you will have the same total energy change as someone that went from A to Z in 7 step—the energy from each of their 7 steps would add up to your 1 step energy change.
Example
1Label each step-wise equations with letters (“a”, “b”, “c”) if not already done for you.
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2(g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
For the first reactant in the overall reaction, find the step-wise reaction that has the same chemical and the same state of matter. It doesn’t have to be on the reactants side of the step-wise equation
If it is on the correct “side” write it as is. Write it’s label beside it, too (“a”, “b”)
If it’s on the wrong “side”, flip the equation. If you flip it, write it’s label as “-a” or “-b”.
2
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2(g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)
Example
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2(g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)
Repeat Step 2 for each reactant & product in the overall equation. If a reactant appears in more than one step-wise reaction, skip that reactant or product and move onto the next one.
3
Example
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2(g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)
Repeat Step 2 for each reactant & product in the overall equation. If a reactant appears in more than one step-wise reaction, skip that reactant or product and move onto the next one.
3
-b 2HNO3 (l) N2O5 (g) + H2O (l)
Example
-b 2 HNO3 (l) N2O5 (g) + H2O (l)
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2 (g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2 HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)
Use any un-used step-wise equations to get rid of unwanted things. Putting them on opposite sides will allow them to cancel.
4
-a 2 H2O (l) 2 H2 (g) + O2 (g)
Example
-b 2 HNO3 (l) N2O5 (g) + H2O (l)
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2 (g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2 HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)
-a 2 H2O (l) 2 H2 (g) + O2 (g)
Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation.
Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation.
5
2 2 6 2 4
Example
-b 2 HNO3 (l) N2O5 (g) + H2O (l)
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2 (g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2 HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)
-a 2 H2O (l) 2 H2 (g) + O2 (g)
Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation.
Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation.
5
2 2 6 2 42 4 2 2
Example
-b 2 HNO3 (l) N2O5 (g) + H2O (l)
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2 (g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2 HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)
-a 2 H2O (l) 2 H2 (g) + O2 (g)
Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation.
Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation.
5
2 2 6 2 42 4 2 2
5
2 N2 (g) + 5 O2 (g) 2 N2O5 (g)
Example
-b 2 HNO3 (l) N2O5 (g) + H2O (l)
Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?
Use: 2 H2 (g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2 HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ
abc
c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)
-a 2 H2O (l) 2 H2 (g) + O2 (g)
2 2 6 2 42 4 2 2
5
2 N2 (g) + 5 O2 (g) 2 N2O5 (g)
Use the step-wise “labels” as a math expression for solving for Hrxn.
6
2 × (-74.1 kJ)2 × -(76.6 kJ)1 × -(-571.6 kJ)
270.2 kJ
Example
What did you learn about hot/cold packs?
Hot/Cold Packs
Transfer of energy
Transfer of energy
Use
System & Surroundings
System & Surroundings
between
Materials ability to absorb energy
without noticeable temperature
change
Materials ability to absorb energy
without noticeable temperature
change
Effect on temperature depends on
Physical change
Physical change
Is determined with
CalorimetryCalorimetry
Can be done in
Chemical change
Chemical change