Energy and Temperature
I. The Nature of Energy
• The ability to do work
• 3 categories
-Radiant energy
-Kinetic energy
-Potential energy
Kinetic Energy
• Energy of motion
• Types
1. Mechanical - moving
parts of a machine
2. Thermal (heat) - energy
caused by the random
internal motion of
particles of matter
Potential Energy
• Stored energy
• Types
1. Electrical - ex. battery
2. Gravitational - ex. water
behind dam - used for
electricity
3. Chemical - ex. chemical
bonds in food
Law of Conservation of Energy
• In any process, energy is neither
created nor destroyed.
ex. 1 hitting a baseball transfers
kinetic energy from the bat to the
ball
ex. 2 igniting a match changes
chemical energy into heat and light
TEMPERATURE VS. HEAT
Temperature
Heat
• A measure of the average • Total amount of energy that flows
kinetic energy of molecules
between matter
in motion
• Flows from matter of higher
temperature to matter of lower
• Remember:
temperature
Kelvin = 273 + Celsius
• “Hot” molecules quickly move into
areas of slower moving “cold”
molecules
Chapter 10
Visual Concepts
Temperature and the
Temperature Scale
Heat
• The transfer of kinetic energy
from a hotter object to a colder
object.
• Symbol is q
-particles are always moving
-when you heat water  molecules
move faster
EXOTHERMIC REACTIONS
• Chemical reactions that release thermal energy
• Feels hot – temperature rises
• Examples: condensation, freezing
ENDOTHERMIC REACTIONS
• Chemical reactions that absorb thermal energy
• Feels cold – temperature drops
• Examples: boiling, evaporation, melting
Measuring Heat
• Kelvin Scale (K)
– 0 K - point at which there is no molecular
motion (absolute zero)
– All Kelvin temperatures are positive
• K = °C + 273
What is the boiling point of
water in Kelvin?
What is the freezing point
of water in Kelvin?
Measuring Heat
• Calorie = amount of heat needed to
raise the temperature of 1 gram of
water by 1 degree Celsius
(1 cal = 1 g x 1 C°)
• Energy stored in food = Calorie
(Cal)
1 Cal = 1000 cal = 1 kcal
Joule (J) : 1 cal = 4.184 J
Specific Heat
• Specific Heat
– Amount of heat
required to raise 1
gram of a substance
1°C.
– physical property
• Liquid water 4.184 J/g°C
• Fe 0.449 J/g°C
Specific Heat
• Water has high heat capacity:
– Absorbs a large quantity of heat with
only a small increase in temperature
– Gives up a large quantity of heat
with only a small decrease in
temperature
• Metals have low heat capacity
–Small amount of heat  large
temperature change
HEAT
q = cp• m • ΔT
Where
q = heat released (-) or heat absorbed (+)
cp = specific heat (value given on p. 343)
m = mass
ΔT (means change in Temperature)
= Final Temp – Initial Temp
HEAT
q = cp• m • ΔT
Where
q = Joules (J) or calories (cal)
cp = J/g•K or J/g•˚C or cal/g •˚C
m = grams
ΔT = K or °C
SAMPLE PROBLEM A
If 75 g of iron (cp = 0.449 J/g•K) is heated from
274 K to 314 K, what is the amount of heat
that is absorbed?
SAMPLE PROBLEM A
If 75 g of iron (cp = 0.449 J/g•K) is heated from 274
K to 314 K, what is the amount of heat that is
absorbed?
Step 1:
Outline what you know.
q=?
m = 75 g
cp = 0.449 J/g•K
ΔT = 314 - 274
SAMPLE PROBLEM A
If 75 g of iron (cp = 0.449 J/g•K) is heated from 274
K to 314 K, what is the amount of heat that is
absorbed?
Step 1:
Outline what you know.
q=?
m = 75 g
cp = 0.449 J/g•K
ΔT = 40 K
SAMPLE PROBLEM A
If 75 g of iron (cp = 0.449 J/g•K) is heated from 274
K to 314 K, what is the amount of heat that is
absorbed?
Step 2:
Plug-in and solve.
q = m • cp • ΔT
q = (75) (0.449) (40)
q = 1347 J
Endothermic
SAMPLE PROBLEM B
A 4.0 g sample of glass was heated from 274 K to
314 K and was found to have absorbed 32 J of
energy as heat. What is the specific heat of this
type of glass?
SAMPLE PROBLEM B
A 4.0 g sample of glass was heated from 274 K to
314 K and was found to have absorbed 32 J of
energy as heat. What is the specific heat of this
type of glass?
Step 1:
Outline what you know.
q = 32 J
m = 4.0 g
Cp = ?
ΔT = 314 – 274 = 40 K
SAMPLE PROBLEM B
A 4.0 g sample of glass was heated from 274
K to 314 K and was found to have
absorbed 32 J of energy as heat. What is
the specific heat of this type of glass?
Step 2:
Plug-in and solve.
q = m • cp • ΔT
32 = (4.0) (x) (40)
32 = 160 x
x = 0.20 J / g • K
Practice Problems
How much heat would be required to
raise the temperature of 20.0 g of
water from 50.0ºC to 100. ºC. (c of
H2O = 4.18J/gºC)
q = cm ΔT
= (4.18J/gºC)(20.0g)(100. ºC–50.0ºC)
q = 4,180 J
II.A. Calorimetry
• Calorimeter - an insulated
device to measure
temperature changes
• Can determine enthalpy
changes (heat) of reactions
Calorimetry
• Exothermic process 
releases heat  temperature
of surroundings increases
• Endothermic process 
absorbs heat  temperature
of surroundings decreases
Calorimeter
• qrxn = - qsur
Transferred Surroundings
Put hot iron ring into
cool water
Leave until the temp
Is the same for both
How does heat lost by
the iron compare to
heat gained by the
water?
Heat lost 
Iron
Heat gained
Water
LAB: Heat and Molecular Motion
Purpose: To illustrate what happens when a
substance is heated.
Procedure:
1. Fill 2 250 mL beakers ¾ full of tap water.
2. Heat 1 to near boiling on a hot plate.
3. Gently put 1 drop of food coloring in
each.
Quiz 12-9
A 10.0 g sample of metal X requires
25.0 J of heat to raise its temperature
from 17ºC to 67ºC. What is the
specific heat of the metal?
Quiz 12-8
How much heat is required to raise
the temperature of 30.0 g of H2O
from 18.0°C to 28.0°C?
c = 4.18 J/gºC
THERMOCHEMISTRY• Study of the changes in heat
in chemical reactions
• All chemical reactions
involve changes in energy.
• Heat energy is either
absorbed or released.
Enthalpy
• Enthalpy (H)-heat content of a
substance
• Depends on temperature, physical state,
and composition
• Enthalpy Change - the amount of heat
absorbed or released during a chemical
reaction; ΔH
ΔH = Hproducts – Hreactants
Exothermic Reactions
• Reactions that release heat to their
surroundings
• Combustion
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) + heat
- Heat is produced because the energy
released as new bonds are formed
(products) > energy required to break the
old bonds (reactants)
Thermite Reaction
Al(s) +Fe2O3(s) Al2O3(s) +Fe(l) +heat
Exothermic Reaction
• Products have
lower potential
energy than
reactants.
energy
released
Al(s) +Fe2O3(s) Al2O3(s) +Fe(l) +heat
• If heat is released,
• Hproducts < Hreactants
DH is negative (exothermic)
Endothermic Reactions
• Reactions that absorb heat from surroundings
• Ammonium nitrate in water
NH4NO3(s) + heat  NH4+(aq) + NO3-(aq)
• Energy released as new bonds are formed
(products) < energy required to break bonds
(reactants)
• This energy must be supplied by
surroundings and is stored in the chemical
bonds of the products
Endothermic Reaction
• Reactants have
lower potential
energy than
products
energy
absorbed
NH4NO3(s) + heat  NH4+(aq) + NO3-(aq)
• If heat is absorbed,
Hproducts > Hreactants
DH is positive (endothermic)
Calculating Heat of Reaction
How much heat will be released when 34.0g of
hydrogen peroxide decomposes?
2 H2O2(l)  2 H2O(l) + O2(g)
ΔH° = -190 kJ
-190 kJ
2 mol H2O2
Ratio of coefficients
34.0 g H2O2 1 mol H2O2
34.0 g H2O2
- 190 kJ
2 mol H2O2
= - 95.0 kJ
Draw an energy diagram for this reaction.
2. How much heat will be released when
184 g of NO2 is formed at STP?
N2(g) + 2 O2(g)  2 NO2(g) ΔH° = +68.0
kJ
184 g NO2 1 mol NO2
68.0 kJ
46 g NO2 2 mol NO2
= 136 kJ
Draw an energy diagram for this reaction.
1. What type of
reaction is this?
2. Would the value
of ΔH be positive
or negative?
1. What type of
reaction is this?
2. Would the value
of ΔH be positive
or negative?
Changes of State
•
•
•
•
•
•
Melting= s  l
Freezing= l  s
Condensation= g  l
Evaporation/Boiling= l  g
Deposition= g  s
Sublimation= s  g
ΔH=
ΔH=
ΔH=
ΔH=
ΔH=
ΔH=
Changes of State
• Heat of vaporization- The amount of heat required to
convert a unit mass of a liquid at its boiling point into vapor
without an increase in temperature.
• Heat of fusion- the amount of heat required to convert a
unit mass of a solid at its melting point into a liquid without
an increase in temperature.
Molar Heat of Fusion /
Vaporization
• How much heat does it take to melt /
evaporate something.
• Expressed in kJ / mol
Molar Heat =
Energy (kJ)
mol
Sample Problem A
How much energy is absorbed when 2.61
mol of ice melts? (Molar Heat of Fusion
for ice = 6.009 kJ / mol)
Step 1:
Outline what you know.
Molar Heat = 6.009 kJ / mol
Energy = ? kJ
mol = 2.61 mol
Sample Problem A
How much energy is absorbed when 2.61
mol of ice melts? (Molar Heat of Fusion
for ice = 6.009 kJ / mol)
Step 2:
Plug into the equation.
6.009 =
Energy (kJ)
2. 61 mol
Sample Problem A
How much energy is absorbed when 2.61
mol of ice melts? (Molar Heat of Fusion
for ice = 6.009 kJ / mol)
Step 3:
Solve.
(2.61 mol) 6.009 =
Energy (kJ) (2.61 mol)
2. 61 mol
Sample Problem A
How much energy is absorbed when 2.61
mol of ice melts? (Molar Heat of Fusion
for ice = 6.009 kJ / mol)
Step 3:
Solve.
Energy = 15.7 kJ
Changes of State
1. What is the heat of fusion of a metal if
28.0 g of the metal requires 1955 J to
melt?
(Express in J/g)
2. How much ice in moles can be melted
by 77.5 kilojoules of heat energy?
3. How much water(in grams) at its boiling
point can be vaporized by adding 1.50kJ
of heat?
4. How much heat is needed to vaporize
2.00 grams of copper if the enthalpy of
vaporization is 322 kJ/mol?
5. The molar enthalpy of fusion of
sodium metal is 2.602 kJ/mol. What is
the molar enthalpy of fusion of
sodium in J/g?
6. How much energy is absorbed by
500.0 g of water as it changes into
water vapor at its boiling point?
QUIZ 12-12
2 Fe + 1.5 O2 Fe2O3
∆H= -48.0kJ
How much heat is released
when 112 g of Fe reacts?
QUIZ 12-14
How much heat would be
produced by burning 44.8 L of
oxygen at STP?
C2H5OH (l)+3O2(g)2CO2 (g)+3 H2O(l)
∆H = -930 kJ