Download Energy

1
Energy
SECTION 1: THERMOCHEMISTRY
2
Thermochemistry
Thermochemistry – thermo = energy/heat
Energy changes occur during reactions and phases changes
Many reactions
release energy making
the surrounding hotter.
Combustion gives off
energy as heat and light.
Some reactions
absorb energy. Their
surroundings feel colder.
This gas tube gets cold
as liquid under pressure
changes to gas.
Energy is the ability to do work
or transfer heat.
Energy used to cause an object that
has mass to move is called work.
Energy used to cause the
temperature of an object to rise is
called heat. (energy to make
molecules move)
Thermochemistry
Forms of Energy
Kinetic energy is energy an object
possesses by virtue of its motion:
1
Ek =  mv2
2
Notice energy is a
function of an
object’s velocity
(speed) and its mass
Kinetic Energy is gained by an object when it is accelerating
Thermochemistry
© 2012 Pearson Education, Inc.
Which has more kinetic energy:
A slow moving locomotive,
A fastball,
Or a bullet in flight?
Which would do more damage if it
ran into you?
That’s Kinetic energy!
Thermochemistry
© 2012 Pearson Education, Inc.
6
Forms of energy
Kinetic
Mechanical – motion of object
Electricity – electric current flow of electrons
Light (radiant) – light waves
Chemical Kinetic:
Heat – molecular motion
(speed energy – indicated by
temperature)
Potential Energy
Potential energy is energy
an object possesses by
virtue of its position or
chemical composition.
The energy always comes
from attraction (like gravity in
cartoon above)
Or repulsion (the coils of a
compressed spring)
Thermochemistry
8
Potential
Mechanical – ex:
compressed spring
Electric (“potential”)?
Stored in a battery, static
electricity
Chemical energy (potential)
when chemical bonds
and IM forces are formed
or broken.
9
Energy conversions - energy
is transferred and always
changing forms
Ex: In a car: gasoline’s
energy (chemical PE) is
converted into motion
(mechanical KE)
Light bulb: electrical energy
converted into light (and
heat – thermal) energy
Easy bake oven
heats with a light
bulb. (fire hazard!)
10
Law of conservation of energy
 Energy lost = Energy gained
Efficiency: most conversions
Lose some energy as waste heat
Typical Light bulb = 10% efficient
10% light
90% heat
Compact fluorescent = 60 to 100% efficient
LED lights: 5 x less energy
use than compact fluorescent
Light emitted in all directions makes them less efficient
11
Heat (kinetic energy) – two definitions
1) Total motion energy of all molecules in a sample
Depends on Temperature
(average motion energy of a molecule)
And Sample size (number of molecules present)
Which has more heat in it?
Lake in summer
Boiling water
(Hint: Which has more molecules?)
This type of heat energy can not be measured easily
2) Heat is the Energy transferred (lost or gained) when a
sample undergoes a change in form (PE)
Which way does it flow?
Ex: Ice is placed in
warm water.
Water: temp decreases,
KE is lost
Heat is transferred
from the water
to melt the ice
Ice: solid becomes liquid
PE is gained
Does water gain “Cold” from the ice?
Ice is a heat sucker!
12
13
1 scoop of NaOH
in 50 mL of H2O
Temp changes
o
From 100C to 200C
Δtemp = 10 C
1 scoop of NaOH
2 scoop of NaOH
in 100 mL of H2O
in 100 mL of H2O
Temp changes
o
From 100C to 150C
Temp changes
o
From 100C to 200C
Δtemp = 5 C
Δtemp = 10 C
Can you explain the difference?
Hint 1 : Potential energy is released when bonds form between molecules
Hint 2: Kinetic energy is absorbed by the water in the surrounding solution
How do the three trials compare in terms of heat?
Why did trial 2 cause half the temperature change as trial 1?
Why did trial 3 cause twice the temperature change as tial 2?
14
Energy flow Kinetic energy always flows “down hill”
(higher to lower KE)
Hotter substances lose heat to cooler substances
Energy flows from higher temp to lower temp – until both
samples at same final temperature.
Energy Transformations:
Potential energy
(change in form)
15
energy changes form:
 is converted into 
Kinetic energy
(change in Temp)
Endothermic changes – absorbing heat
MELTING ICE (COLD)
Phase change
Gains PE

heat flows in
WARM WATER
temp decreases
Loses KE
Endothermic Changes
- reactions or phase
changes, which
absorb heat
(KE  PE)

Evidence The
Surroundings
get colder
Energy Transformations:
16
energy changes form:
 is converted into 
Potential energy
(change in form)
Kinetic energy
(change in Temp)
Endothermic changes
Boiling (vaporizing)
Phase change
Gains PE
Ex: boiling/vaporizing
heat added from outside

heat flows in
frost
CO2 Tank
WARM metal
temp decreases
Loses KE

Expanding
vaporizing CO2
absorbs heat
from the
surroundings.
frost forms on
the cold metal
Energy Transformations:
Potential energy
(change in form)
17
energy changes form:
 is converted into 
Kinetic energy
(change in Temp)
Endothermic changes
Dissolving
Phase change
Gains PE
Ex: Also: Dissolving of
most salts (ex: cold pack)

heat flows in
WARM water
temp decreases
Loses KE
Salt crystals break
apart as they
dissolve - absorbing
heat from the
surrounding pack
making it cold.
Energy Transformations:
Potential energy
(change in form)
18
energy changes form:
 is converted into 
Kinetic energy
(change in Temp)
Exothermic changes – release heat
Ex:
BURNING (combustion)
Chemical change
  
Loses PE
heat flows out
Exothermic Changes reactions or phase
changes, which
release heat
(PE  while KE )
AIR
temp increases
gains KE
* Evidence –
surroundings get
warmer
Energy Transformations:
Potential energy
(change in form)
19
energy changes form:
 is converted into 
Kinetic energy
(change in Temp)
Exothermic changes – release heat
Ex:
Freezing
Physical change
Loses PE
Ex: Freezing
(solidifying) – heat
removed to freeze a
liquid
  
heat flows out
colder AIR
temp increases
gains KE
Solid ice has less
PE than liquid
water
Energy Transformations:
Potential energy
(change in form)
20
energy changes form:
 is converted into 
Kinetic energy
(change in Temp)
Exothermic changes – release heat
Ex:
Dissolving
Chemical change
Loses PE
  
heat flows out
Ex: Dissolving
NaOH – water
gains the heat
water
temp increases
gains KE
Water forms bonds
with the dissolved
salt releasing
energy
Water gets hot!
Energy Transformations:
Potential energy
(change in form)
21
energy changes form:
 is converted into 
Kinetic energy
(change in Temp)
Law of conservation of energy
– Energy lost by one substance = energy gained by other substance
22
GAS
(highest PE)
Energy
absorbed
LIQUID
Endo-thermic
Energy
Released
Exo-thermic
SOLID
(lowest PE)
23
Measuring Energy
Section 2A
Calorimetry
24
Measuring energy : Heat Capacity
Why does sand get hot while water stays cool?
Water has a greater heat capacity!
…than sand.
25
Measuring energy:
Heat Capacity Is the Energy required to raise the
temperature of a substance (make molecules speed up)
The same amount of heat is given to two different compounds
Ethanol’s
temperature
rises to 62oC
While
water’s
only
rises to
40oC
Which has a greater
heat capacity?
(ability to absorb and hold heat)
26
Measuring energy: Heat Capacity
Is the Energy required to raise the temperature of a
substance (make molecules speed up)
The same amount of heat is given to two different compounds
Water’s heat
capacity is
higher.
Ethanol’s heat
capacity is
lower.
It takes more
heat to raise
its
temperature
It takes less
heat to raise its
temperature
Which has a greater
heat capacity?
(ability to absorb and hold heat)
27
Measuring energy: Heat Capacity
Is the Energy required to raise the temperature of a
substance (make molecules speed up)
Its also the heat that must be removed to cool a substance.
Heat capacity depends on the Intermolecular (sticky) forces
Water has
stronger
hydrogen
bonding
Ethanol’s sticky
forces are weaker
28
Specific heat capacity (C) – is a physical constant
It’s the Heat required to raise 1.0 grams of a
substance by just 1o C
Water is commonly the standard
1 calorie = heat to raise 1 gram of water by 1oc
1000 calories = 1 kilocalorie = 1 “Calorie”
(notice the capitol C?)
1 Joule (J) – is the metric unit of energy
4.18 joules = 1 calorie
Water is the
Standard!
Example problem 1:
How many calories of heat
are needed to raise 10 grams of water
by 10 oc?
the heat capacity of water (specific heat) :
…is 1 calorie per gram per 0c
so 1 calorie x 10 g water x 10oc =
100 calories
Formula:
Q = C m ∆t
Heat = (specific heat capacity) X (mass) X ( change in temp oc)
29
30
Ex 2: What is the specific heat C
of a substance which absorbs 20. J of heat
to raise 5 grams of the substance by 2 oC?
Q = C m ∆t
Substitute: (20 joules) = C (5.0 g)(2.0 oC)
and solve:
20. J
(5.0 g)(2.0 oC)
=C
C = 2.0 joules / goC
Are the IM forces in this substance stronger or weaker than in liquid water?
31
Ex 2: What is the specific heat C
of a substance which absorbs 20. J of heat
to raise 5 grams of the substance by 2 oC?
Q = C m ∆t
Substitute: (20 joules) = C (5.0 g)(2.0 oC)
and solve:
20. J
(5.0 g)(2.0 oC)
=C
C = 2.0 joules / goC
Weaker, water’s specific heat is higher: 4.18 Joules/goC
32
Reference
Table
Formulas
Heat
q= mC∆t
C for water
Try one:
Ex 3: how many joules are absorbed when 300. grams of
water changes from 250C to 390C?
Write the formula:
Substitute values:
Solve:
Ans: 18000 Joules?
33
34
Ex 4: 100 grams of water at 250C loses 4200 joules of
energy. What is its final temperature?
Write the formula:
Q = C m ∆t
∆t = Q
Cm
Substitute values:
(4200 j )= (4.2 j/g0c) (100 g) ∆t
Solve:
∆t
Remember: equation
only allows you to
calculate ∆t
= (4200 j)
(4.2 j/g0c) (100 g)
∆t = 100c
Final temp = 250C – 100C = 150C
Original water temperature decreases by 10oC
35
Try these:
1. What quantity of heat is
absorbed by 35 grams of
water if its temperature
increases by 100C?
1463 Joules
2. What is the specific heat of a
substance which absorbs
2680 joules to change 85g
from 260C to 470C?
1.50 Joules per gram per o Celsius
3. What is the final temperature
when 35 grams of water at
150C absorbs 4630 j ?
Δtemp = 32oC, final temp = 47oC
36
Calorimetry: Calculating heat of a change
Suppose that a sample of wax is
melted in boiling water
How much heat is required to melt the wax?
This is called the heat of fusion (heat to melt)
Ex: 1 Heat of fusion of a solid (Hf)
How much heat is absorbed per gram, if 10 grams of
a solid melts in 300 grams of water, causing the
water temperature to decrease by 500C?
37
Calorimetry: Calculating heat of a change
Ex: 1 Heat of fusion of a solid (Hf)
How much heat is absorbed per gram, if 10 grams
of a solid melts in 300 grams of water, causing
the water temperature to decrease by 500C?
Step 1: how much heat did the water lose?
(heat gained by the solid to melt)
For the water heat lost: Q = C m ∆t
For the water heat lost: Q = (4.18Jg-oc-)(300 g)(50oC)
For the water heat lost: Q = 62700 Joules
38
Calorimetry: Calculating heat of a change
Ex: 1 Heat of fusion of a solid (Hf)
How much heat is absorbed per gram, if 10 grams
of a solid melts in 300 grams of water, causing
the water temperature to decrease by 500C?
Step 2: how much heat did the solid gain per gram?
(heat lost by the water)
Heat gained by the wax IS the heat lost from the water
Q = 62700 Joules lost by water = 62700 Joules gained by the wax
Heat gained per gram of wax is 62700 J of heat
10 grams of wax
39
Calorimetry: Calculating heat of a change
Energy released
during the system
(a reaction) is
absorbed by the
surrounding water
Find the heat
absorbed by the
water to find the
heat of reaction
Measuring energy (Calorimetry) – Water is used as the standard
> Heat lost or gained during a change will be exchanged with the water
> Calculate the heat exchanged by the water to find out about the change
40
Ex: 2 Calculate the heat of solution for KNO3 in
kJ/mole, if 10.1 grams of KNO3 is dissolved in
100g of water causing its temperature to
decrease by 8.3 0C?
Step one: how much heat did the water lose?
(heat gained by the chemical)
For the water heat lost: Q = C m ∆t
For the water heat lost: Q = (4.18Jg-oc-)(100 g)(8.3oC)
For the water heat lost: Q = 3500 Joules
For the water heat lost: Q = 3.5 kiloJoules
41
Step two: If the sample had a mass of 10.1 grams, and
Calculate the heat of solution (heat of dissolving)
of the KNO3 (kJ/mole)
Aka: How much heat (kJ) did each mole release?
(hint 2: KNO3 = 101 g/mole  39 + 14 + 3(16)
Q = 3.5 kiloJoules
0.100 moles
Q = 35 kJ/mole
42
Heat of Reaction (∆H)
– Reference Table I
Heat of reaction – heat lost or gained for
molar quantities indicated in balanced
equation
Ex: combustion of 2 moles of methanol
releases 1452 kilojoules of heat
Combustion reactions
synthesis reactions
Dissolving
(Heat of Solution)
43
Try one: Calculate Specific heat capacity of a metal
A 10 gram piece of hot metal at 1500C, is dropped into a
calorimeter containing 50. g of water? The water started at 250C,
and increased to 350C.
Step one: how much heat did the
water gain? (heat lost by the metal)
Hint: Q = C m ∆t for the water
Ans. 2090 Joules gained by the water
44
Try one: Calculate Specific heat capacity of a metal
A 10 gram piece of hot metal at 1500C, is dropped into a
calorimeter containing 50. g of water? The water started at 250C,
and increased to 350C.
Step two: If the metal had a mass of 10.
grams, and started at 150. 0C, Calculate
the specific heat (C) of the metal?
Hint: Q = 2090 Joules
Use Q = C m ∆t for the metal, find C
Ans: 1.8 J/gram ?
45
Section 2B
Heats of
Reaction
46
Heat of Reaction in Equations
Exothermic – reactions which release or lose energy
- ∆H is negative
Ex:
∆H =
CH4 + 2 O2  CO2 + H2O
– 890.4 kJ
(For 1 mole CH4 890.4 kJ is released)
In the equation:
CH4 + 2 O2  CO2 + H2O +
890.4 kJ
(energy is a product, written on right )
47
Heat of Reaction in Equations
Exothermic – reactions which release or lose energy
- ∆H is negative
Ex:
2CO + O2  2 CO2 + 566 kJ
∆H =
– 566.0 kJ
(Energy released for 2 moles CO)
48
Endothermic
Reactions which absorb or gain energy - ∆H is positive
Ex: dissolving of potassium nitrate:
KNO3(s)  K+(aq) + NO3–(aq)
KNO3(s) +
∆H = +34.89 kJ
34.89 kJ  K+(aq) + NO3–(aq)
(Energy is a reactant – must be absorbed, written on left )
Try a few:
1. According to reference table I, how much heat is released by the
combustion of 2.0 moles of propane C3H8 in a calorimeter?
2. Will the water in the calorimeter get hotter or colder? How do you
know?
3. When KNO3 is dissolved in water the temperature of the water
decreases. Is this change exothermic or endothermic? How could you
tell by looking at reference table I?
4. Using reference table I, write the thermochemical equation for the
synthesis of 2 moles of water vapor.
49
50
5. What will be the temperature change in 100.
grams of water if 0.100 moles of KNO3 are
dissolved in it?
Step one: Find heat absorbed by the 0.100 moles of chemical during dissolving
 This is kiloJoules per mole
Step two: Since this heat is absorbed from the water,
find the temp change in the water. (Recall: Q is the heat absorbed in step one!)
Q = C m ∆t
∆t = Q
Cm
∆t = 8.35 0C
6. 8.0 grams of methane undergoes combustion in a calorimeter containing
5000. grams of water. The reaction proceeds according to the equation
as shown on reference table I.
CH4 + O2  CO2 + H2O + 890.4 kJ
1. What will happen to the temperature of water in the calorimeter? Explain.
2. How much heat is released by the complete combustion of the 8.0 grams
of methane?
3. What will be the change in temperature of the water in the calorimeter?
51
52
53
54
Section 2C
Energy of Phase
Changes
Energy in Phase changes
Q=CmΔT
Energy Gained to melt
Energy Gained to boil
55
56
Heat of fusion (melting) / heat of vaporization (boiling)
Heat of fusion – Heat absorbed to melt a solid (Joules per gram)
Q = Hf m
Q = heat absorbed
Hf = heat of fusion (joules/g)
m= mass of sample
for water (Ice) Hf = 334 J/g (see ref. Tbl. B)
Ex1: How much heat is absorbed to melt 100 grams of ice?
Q = Hf m
Q = (334J/g) (100 g)
= 33400 Joules
=
33.4 kJ
57
Ex2: What is the heat of fusion of a solid if a 50 gram
sample absorbs 10 000 joules to melt?
Q = Hf m
Hf = 10 000 J
50 g
Hf = 200 J/g
and
Hf =
Q
m
Heat of vaporization
– Heat absorbed to boil water (or lost as a gas condenses)
For water Hv = 2260 Joules per gram
Ex3: What mass of water can be vaporized by the addition of 100 Kilojoules of
heat?
Q = Hv m
where
Q is the heat absorbed
Hv is the heat of vaporization (2260 J/gram)
m is the mass of the water in grams
Step 1: convert Kilojoules to joules
100 kJ x 1000 J
=
1 KJ
Step 2: setup and calculate
Q = Hv m
and
m=
m = 100 000 J
2260 J/g
m=
44.2 g
100 000 Joules
_Q_
Hv
58
Learning check: show work
1. How much heat is absorbed when 20. grams of ice melts in a calorimeter?
6680 joules
2. If the calorimeter contains 100. grams of water initially at 250C, what is the
final temperature of the water once all the ice is melted?
9 oC
3. What is the heat of vaporization for a liquid if it takes 5000 joules of energy
to vaporize 2.5 grams?
2000 J/g
4. Aluminum oxide decomposes according to the balanced equation
2 Al2O3  4 Al + 3 O2
According to reference table I, what is the ∆H for the decomposition of 1.0
mole of Al2O3?
+1680 kJ
59
60
61
62
63
64
In terms of energy flow, explain why the temperature of the water in the calorimeter used in the
investigation increases.
Show a correct numerical setup for calculating the number of joules of heat gained by the water
in the investigation.
In the investigation shown, the change in heat of the copper is greater than the change in heat of
the water. What error could account for this apparent violation of the law of conservation of
energy? [do not use human error as part of the answer]