Transcript Section 10.1 Energy, Temperature, and Heat

Section 10.1
Energy, Temperature, and Heat
• Thermodynamics – study of energy
• First law of thermodynamics
– Energy of the universe is constant
– Energy can not be created or destroyed
Section 10.1
Energy, Temperature, and Heat
• Energy is the ability to do work or produce heat.
• 2 types of Energy
Potential energy
Energy of position
Kinetic energy
Energy of motion
E = ½ mv2
Section 10.1
Energy, Temperature, and Heat
• Law of conservation of energy
– Energy can not be created or destroyed but can be
converted from one form to another
– Ex chemical energy (gas) can be converted to
mechanical and thermal energy
Section 10.1
Energy, Temperature, and Heat
• Internal energy = E
“sum” of kinetic & potential energies of all “particles” in a system
Internal energy can be transferred by two types of energy flow:
• Heat (q)
• Work (w)
(q = quantity of heat)
E = q + w
Change of energy in a
system
Section 10.1
Energy, Temperature, and Heat
Measuring Energy Changes
• common energy units for heat are the calorie and the joule.
– Calorie – the amount of energy (heat) required to raise the
temperature of one gram of water 1oC.
– Joule – heat required to raise the temperature of 1 g of
water by 0.24 K
1 calorie = 4.184 joules
Section 10.1
Energy, Temperature, and Heat
• System – part of the universe on which we focus attention
• Surroundings – everything else in the universe
• Ex. Burning a match is the system releases heat to surroundings
Section 10.1
Energy, Temperature, and Heat
Exothermic – energy flows out of system
Endothermic – energy flows into system
Section 10.1
Energy, Temperature, and Heat
• C- Specific heat capacity
energy required to change the
mass of one gram of a
substance by one degree
Celsius. [joule/gram °C]
• Like energy storage bank
Section 10.1
Energy, Temperature, and Heat
• C- Specific heat capacity
energy required to change the
mass of one gram of a
substance by one degree
Celsius. [joule/gram °C]
• Like energy storage bank
• Water has a high specific heat
capacity that means it takes a lot
of heat (joules) to increase the
temperature of water compared
to metals (water better at storing
energy than metals
Section 10.1
Energy, Temperature, and Heat
B. Measuring Energy Changes
• To calculate the energy required for a reaction:
Q = C  m  t
Section 10.1
Energy, Temperature, and Heat
Q = C  m  t
How much heat is required to warm 145 g of water from 25 *C
to 65 *C
Section 10.1
Energy, Temperature, and Heat
Q = m x C  t
How much heat is required to warm 145 g of water from 25 *C
to 65 *C the specific heat of liquid water is 4.184 J/g *C
Q = m x C  t
Q=
145 g
x
4.184 J x (65 *C - 25 *C)
g *C
Section 10.1
Energy, Temperature, and Heat
Q = m x C  t
How much heat is required to warm 145 g of water from 25 *C
to 65 *C the specific heat of liquid water is 4.184 J/g *C
Q = C  m  t
Q=
145 g
x
4.184 J x (65 *C - 25 *C)
g *C
Q = 145 x 4.184 J x 40
Q = 24,267 joules
Section 10.1
Energy, Temperature, and Heat
A quantity of water is heated from 20.0 *C to 48.3 *C by
absorbing 623 Joules of heat energy, what is the mass of the
water? Specific heat of water is 4.184 J/g *C
Q = C  m  t
__Q__ = m
C x t
Section 10.1
Energy, Temperature, and Heat
A quantity of water is heated from 20.0 *C to 48.3 *C by
absorbing 623 Joules of heat energy, what is the mass of the
water? Specific heat of water is 4.184 J/g *C
Q = C  m  t
__Q__ = m
C x t
623 J
=
4.184 J /g *C x (48.3 *C - 20 *C)
m
Section 10.1
Energy, Temperature, and Heat
A quantity of water is heated from 20.0 *C to 48.3 *C by
absorbing 623 Joules of heat energy, what is the mass of the
water? Specific heat of water is 4.184 J/g *C
Q = C  m  t
__Q__ = m
C x t
623 J
= m
4.184 J /g *C x (48.3 *C - 20 *C)
623 J
4.184 J /g *C x (28.3 *C)
5.26 g = m
=
m
Section 10.1
Energy, Temperature, and Heat
In a household radiator a 25.5 g of steam at 100 *C condenses
(changes from gas to liquid) How much heat is released? Heat
of vaporization is 2260 J/g
Q = m  Hvap
Section 10.1
Energy, Temperature, and Heat
In a household radiator a 25.5 g of steam at 100 *C condenses
(changes from gas to liquid) How much heat is released? Heat
of vaporization is 2260 J/g
Q = m  Hvap
Q = 25.5 g x 2260 J/g
Q = 57,630 J
Section 10.1
Energy, Temperature, and Heat
If an endothermic reaction absorbs 482 J how many calories are
absorbed?
482 J x 1 calorie = 115 calories
4.184 J
Section 10.1
Energy, Temperature, and Heat
Q = m x C  t
How much heat is required to warm 275 g of water from 76 *C
to 87 *C the specific heat of liquid water is 4.184 J/g *C
Q = C  m  t
Q=
275 g
x
4.184 J x (87 *C - 76 *C)
g *C
Q = 275 x 4.184 J x 11
Q = 12,657 joules
Section 10.1
Energy, Temperature, and Heat
In a household radiator a 1000 g of steam at 100 *C
condenses (changes from gas to liquid) How much heat is
released? Heat of vaporization is 2260 J/g
Q = m  Hvap
Q = 1,000 g x 2260 J/g
Q = 2,260,000 J
Section 10.1
Energy, Temperature, and Heat
Objectives
1. To consider the heat (enthalpy) of chemical reactions
2. To understand Hess’s Law
Section 10.1
Energy, Temperature, and Heat
A. Thermochemistry (Enthalpy)
• Enthalpy, H – amount of heat content of a system absorbed
or released at constant pressure.
• H = heat = q
ΔH = q (valid for constant pressure ONLY!)
That is the Enthalpy ΔH (heat content) = q (quantity of heat)
The heat of formation is the enthalpy gain when a compound is CREATED from its elements at their conditions at some STP.
The heat of reaction involves CHEMICAL REACTIONS OF A SUBSTANCE WITH OTHER SUBSTANCES, and is the relative
enthalpy of the products compared to the reactants at some STP.
Section 10.1
Energy, Temperature, and Heat
• Enthalpy, H –
• Since H = q
• And q = m x C x ΔT then…
ΔH = mCΔT
ΔH = mΔHvap
ΔH = mΔHfus
Section 10.1
Energy, Temperature, and Heat
A. Thermochemistry
Calorimeter
• Enthalpy, H is measured
using a calorimeter.
Section 10.1
Energy, Temperature, and Heat
Temperature- average kinetic energy
Heat- flow of energy due to a temperature difference
Enthalpy- heat content of a system
Entropy- measure of disorder or randomness in a system
Section 10.1
Energy, Temperature, and Heat
Hess’s Law
• the change in enthalpy is the same whether the reaction takes
place in one step or a series of steps.
• Example:
N2(g) + 2O2(g)  2NO2(g)
H1 = 68 kJ
Since H = + 68 kJ then this is a endothermic reaction
Section 10.1
Energy, Temperature, and Heat
ΔHrxn = Heat of Reaction (represents change in enthalpy)
ΔHrxn = (ΔHproducts -ΔHreactants)
–ΔH = Exothermic
+ΔH = Endothermic
How much heat is required to raise the temperature of 445 g
of zinc from 25.0*C to 74.5*C if the specific heat of zinc is
0.390 J/g*C
How much heat is required to raise the temperature of 445 g
of zinc from 25.0*C to 74.5*C if the specific heat of zinc is
0.390 J/g*C
ΔH = mCΔT
How much heat is required to raise the temperature of 445 g
of zinc from 25.0*C to 74.5*C if the specific heat of zinc is
0.390 J/g*C
ΔH = mCΔT
ΔH = 445g x 0.390 J/g*C x 49.5*C
ΔH = 8,591 Joules
Determine the heat of reaction of the following equation
CaCO3  CO2 + CaO
Heat of formation for calcium carbonate is -1207 KJ/mol for 1 mole
Heat of formation for carbon dioxide is -394 KJ/mol for 1 mole
Heat of formation for calcium oxide is -635 KJ/mol for 1 mole
Hreaction = SUM (Hproducts) - SUM (Hreactants)
Determine the heat of reaction of the following equation
CaCO3  CO2 + CaO
Heat of formation for calcium carbonate is -1207 for 1 mole
Heat of formation for carbon dioxide is -394 for 1 mole
Heat of formation for calcium oxide is -635 for 1 mole
Hreaction = SUM (Hproducts) - SUM (Hreactants)
Hreaction = (-394 – 635) – (-1207)
Determine the heat of reaction of the following equation
CaCO3  CO2 + CaO
Heat of formation for calcium carbonate is -1207 for 1 mole
Heat of formation for carbon dioxide is -394 for 1 mole
Heat of formation for calcium oxide is -635 for 1 mole
Hreaction = SUM (Hproducts) - SUM (Hreactants)
Hreaction = (-394 – 635) – (-1207)
Hrxn = (-1029) – (-1207)
Hrxn = +178
Determine the heat of reaction of the following equation
CaCO3  CO2 + CaO
Heat of formation for calcium carbonate is -1207 for 1 mole
Heat of formation for carbon dioxide is -394 for 1 mole
Heat of formation for calcium oxide is -635 for 1 mole
Hreaction = SUM (Hproducts) - SUM (Hreactants)
Hreaction = (-394 – 635) – (-1207)
Hrxn = (-1029) – (-1207)
Hrxn = +178
ENDOTHERMIC
The Heat of Formation for ALL Diatomic molecules is zero
"All elements in their standard states (oxygen gas, solid carbon in the form of graphite, etc.)
have a standard enthalpy of formation of zero, as there is no change involved in their
formation."
The problem is that we can only measure CHANGES in the Enthalpy of the system, and have
no way to determine the ABSOLUTE Enthalpy. We have to define a convenient ZERO of
Enthalpy for all the chemical compounds that we can make. We do this by defining the
enthalpy of the Standard State of the Elements, this should cover all possible matter since all
matter is made of elements we only need about 100 descriptions of Standard Elements
Section 10.1
Energy, Temperature, and Heat
Objectives
1. To understand how the quality of energy changes as it is
used
2. To consider the energy resources of our world
3. To understand energy as a driving force for natural
processes
Section 10.1
Energy, Temperature, and Heat
A. Quality Versus Quantity of Energy
• When we use energy to do work we degrade its usefulness.
Section 10.1
Energy, Temperature, and Heat
A. Quality Versus Quantity of Energy
• Petroleum as energy
Section 10.1
Energy, Temperature, and Heat
B. Energy and Our World
• Fossil fuel – carbon based molecules from decomposing
plants and animals
– Energy source for United States
Section 10.1
Energy, Temperature, and Heat
B. Energy and Our World
• Petroleum – thick liquids composed of mainly hydrocarbons
– Hydrocarbon – compound composed of C and H
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Energy, Temperature, and Heat
B. Energy and Our World
• Natural gas – gas composed
of hydrocarbons
Section 10.1
Energy, Temperature, and Heat
B. Energy and Our World
• Coal – formed from the remains of plants under high
pressure and heat over time
Section 10.1
Energy, Temperature, and Heat
B. Energy and Our World
• Effects of carbon dioxide on climate
• Greenhouse effect
Section 10.1
Energy, Temperature, and Heat
B. Energy and Our World
• Effects of carbon dioxide on climate
• Atmospheric CO2
– Controlled by water cycle
– Could increase temperature by 10oC
Section 10.1
Energy, Temperature, and Heat
B. Energy and Our World
• New energy sources
– Solar
– Nuclear
– Biomass
– Wind
– Synthetic fuels
Section 10.1
Energy, Temperature, and Heat
C. Energy as a Driving Force
• Natural processes occur in the direction that leads to an
increase in the disorder of the universe.
• Example:
– Consider a gas trapped as shown
Section 10.1
Energy, Temperature, and Heat
C. Energy as a Driving Force
• What happens when the valve is opened?
Section 10.1
Energy, Temperature, and Heat
C. Energy as a Driving Force
• Two driving forces
– Energy spread
– Matter spread
Section 10.1
Energy, Temperature, and Heat
C. Energy as a Driving Force
• Energy spread
– In a given process concentrated energy is dispersed
widely.
– This happens in every exothermic process.
Section 10.1
Energy, Temperature, and Heat
C. Energy as a Driving Force
• Matter spread
– Molecules of a substance spread out to occupy a larger
volume.
– Processes are favored if they involve energy and matter
spread.
Section 10.1
Energy, Temperature, and Heat
C. Energy as a Driving Force
• Entropy, S – function which keeps track of the tendency for
the components of the universe to become disordered
Section 10.1
Energy, Temperature, and Heat
C. Energy as a Driving Force
• What happens to the disorder in the universe as energy and
matter spread?
Section 10.1
Energy, Temperature, and Heat
C. Energy as a Driving Force
• Second law of thermodynamics
– The entropy of the universe is always increasing.