Unit 4
Excerpt from The Web of Life by John H. Storer

“Air, rock, water, and sunlight – these are the four sources from which come all living things
and their environment. On the bare sands of the desert, the sun’s rays strike in tiny units of
energy moving with atomic speed. Some of them we can feel as heat or see as light. These
speeding units impart some of their energy to the dead sands, which temporarily store it in
the form of heat, but when the sun sinks, this newly acquired energy is radiated back into the
air and lost. The sand becomes as cold and dead as ever. But chlorophyll in the leaves of green
plants exists as an agent for garnering these units of solar energy. It makes of the green leaf a
laboratory in which nature creates food for living creatures and carries on unceasingly the
magic of building life.
 Like the sand, a field of grass absorbs the sun’s rays; but when night comes the grass does not
give back this newly gained energy. In its green laboratory, the chlorophyll blends the sun’s
captured radiance together with elements taken from the air, the water, and the soil, and
builds these dead materials into organized living form to make new blades of grass.
 This grass is cool and quiet, giving no hint of the sunlight stored within its framework. But
dry it out and touch a match to it. The blades of grass – these tiny bits of organized gas and
sunlight – blaze up with flame hot enough to kill a man. All of that fierce heat is merely a
release of the same energy that the cool, moist plants have been quietly gathering from the
sunlight and storing for later use.
 If the grass is not burned, the energy will remain stored within its substance. If it is eaten by
an animal, its life force is transferred with it into the body of the animal to sustain the spark
that we call life.”
THE CONCEPT OF ENERGY:
 All physical and chemical changes are accompanied by
changes in energy.
 Energy is defined as the ability to do work or to transfer
heat.
 Forms of Energy:
 Energy can be classified as potential or kinetic.
 The total energy an object possesses Etot = K.E. + P.E.
POTENTIAL ENERGY
 Potential energy is the energy that results from an object’s
position. It is stored energy that can be converted to
kinetic energy.
 EXAMPLES OF POTENTIAL ENERGY:
 Chemical potential energy: resulting from attractions among
electrons and atomic nuclei in molecules. This is the energy
associated with the forces within (Intramolecular) and
between (Intermolecular) molecules. Chemicals store energy
in bonds (Intramolecular forces).

The stronger the bond = more energy stored
 Gravitational potential energy: such as that of a ball held well
above the floor or that of water at the top of a waterfall.
Gravitational potential energy (GPE)
=
mgh
m = mass in kg; g = gravity, 9.801 m/s2; h = height in m
KINETIC ENERGY
 An object has kinetic energy because it is moving. Kinetic energy can be converted into potential
energy.
 EXAMPLES OF KINETIC ENERGY
 Thermal kinetic energy: atoms, molecules, or ions (atoms with a net charge) in motion at the
submicroscopic level—All matter has thermal kinetic energy because, according to the kineticmolecular theory, the submicroscopic particles of matter are in constant motion.
 Mechanical kinetic energy of macroscopic objects: moving baseball or automobile
 Electrical kinetic energy: electrons moving through a conductor
 Radiant kinetic energy: electrons transitioning between energy levels with the atoms producing light
 Sound kinetic energy: corresponds to the compression and expansion of the spaces between
molecules.
Kinetic energy
Where
=
mv2
2
m = mass in kg;
v = velocity in m/s
Let’s try some…..
 When gasoline is burned in a car engine, its chemical
potential energy is converted into other forms of energy.
 Name three.
 What kind of energy conversion takes place in a toaster?
 What kind of energy conversion takes place in a hair dryer?
UNITS OF ENERGY
 The SI unit for energy is Joule, J. This unit is named in honor of
James P. Joule, a British scientist who devoted much of his life to
investigating energy in all of its forms. He determined that work
(or energy) could be measured in terms of the heat it could
produce.
kg m2
J=
s2
 Another energy unit specifically used when talking about heat
energy is the calorie (cal).
1 calorie = 4.184 J
 One calorie represents a small quantity of heat, thus the
kilocalorie (kcal) is preferred. A dietary calorie, represented
Calorie (Cal), has the same value as kcal.
1000 cal = 1 kcal = 1 Cal
Pg 77 in your text: #50, 51, and 90
Definition of Energy Units
CONVERSION OF ENERGY AND ITS CONSERVATION:
 Law of Conservation of energy states that, “Energy cannot be
created nor destroyed, just transferred.”
 The total amount or energy in the universe is CONSTANT.
 Is mass and energy related?
E = mc2
E = Energy
m = mass
c = speed of light (2.998 x 108 m/s)
 This equation corresponds to when a body or system releases
energy, the body or system decreases in mass. Likewise, when a
body or system absorbs energy, the body or system increases in
mass.
 Usually, these changes in mass go undetected, unless we discuss
reactions that involve nuclei of atoms that occur in nuclear reactors
or an atomic bomb.
ENERGY AND CHEMICAL REACTIONS:
 Energy changes that accompany chemical changes are generally more noticeable
than those that take place during physical changes.
 The substances that we begin with before the chemical reaction are referred to as the
reactants.
 The substances that are produced as a result of a chemical change are referred to as the
products.
 If a reaction takes place and the temperature of the surroundings drops, does this
indicate the destruction of energy?
 NO, this indicates that energy was absorbed from the surroundings into the system
(reaction) and can is classified as an endothermic reaction.
 What if the temperature of the surroundings increases? Was energy created?
 NO, This means the system (reaction) released energy to the surroundings. This is an
exothermic reaction.
HOW CAN THIS BE EXPLAINED?????
 When two atoms chemically bond together, their bond has a specific
amount of energy due to its stability (strength). The energy required
to break this bond varies depending on the type of bond and the
atoms that are bonded together.
 During a chemical change, the reactants begin with a certain amount
of potential energy. In order to break these bonds a specific amount
of energy, called the activation energy, is needed. When the
products are yielded they may have a higher or lower potential
energy than the reactants due, once again, to the number of bonds
formed and the relative strengths of the bonds.
 If the reactants have more energy than the products, then heat must
be released from the reaction……. EXOTHERMIC.
 The products are more stable than the reactants.
 If the products require more energy than the reactants, then heat
must be absorbed from the surroundings……… ENDOTHERMIC.
 The reactants are more stable than the products.
 The heat of reaction is the measure of the energy flow; quantity of
heat released of absorbed during a reaction and is represented by q or
H. (q = H at constant pressure)
Bond
Bond
Energy
kJ/mol
F-F
C=O
O-H
Br-Br
C-Br
H-Br
Cl-Cl
C-O
H-H
C-C
C-Cl
C-H
C=C
H-F
H-Cl
C-F
I-I
C-I
CC
C-N
H-I
154
743
463
193
276
366
239
360
436
348
339
412
612
565
427
485
151
238
837
305
299
Potential Energy Diagrams
POTENTIAL ENERGY DIAGRAMS
Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
Potential energy
N2 + O2 + HEAT
2 NO
2 NO
Heat
N2 + O2
Direction

Every energy measurement has three parts.
1. A number (how many).
2. A unit (Joules or calories).
3. A sign to tell direction.


negative - exothermic
positive- endothermic
 The reaction:
2CO(g) + O2(g)  2CO2(g) is exothermic.
 What is the sign of H for this reaction?
 What is the sign of H for the reverse reaction?
 Is the heat content of the products greater or less than that of
the reactants?
 For the reaction:
CaCO3(s)  CaO(s) + CO2(g)
H = +176 KJ/mol
 Is this reaction exothermic or endothermic?
 b. What is the value of H for the reverse reaction?
 c. How does the heat content of the products compare to that
of the reactants?
HEAT ENERGY AND TEMPERATURE:
 We have all measured temperature
before using a thermometer, but how
does it work?
 When substances heat up, typically
they expand, which explains the
concept behind a thermometer. A
glass (or plastic) tube is attached to a
bulb. The bulb is filled with alcohol
or liquid mercury. Upon heating the
liquid expands and is forced from the
bulb into the tube. Upon cooling, the
liquid contracts and flows back into
the tube. The tube is then calibrated
to precisely record changes in
temperature.
TEMPERATURE SCALES
 The three temperature scales are
 Fahrenheit, F
 Celsius, C
 Kelvin, K.
 The Celsius temperature scale is the one most commonly used for
scientific work.
Comparison of the Three Temperature Scales
F
C
Water Boils
212.00
100.00
Body Temperature
98.6
37.0
Water Freezes
32.0
0.00
Absolute Zero
-459.67
-273.15
K
373.15
310.2
273.15
0.00
FAHRENHEIT:
 The Fahrenheit scale was created (1714,
Gabriel Fahrenheit) based on
temperatures he could recreate in the
lab.
 The lowest temperature he could achieve
was with a mixture of freezing ice, snow
salt, and water. This became 0F.
 The reference point for 100F is told to be
based on the body temperature of a cow,
around 98.6F, simply because Fahrenheit
liked cows!! Except he chose a sick cow
when he created the scale.
CELSIUS
 The Celsius scale was created in 1742 by
Anders Celsius. He designated the
freezing and boiling points of water to
be the 0C and 100C, respectively, and
then divided the difference into 100
graduations. This is why the Celsius
scale is sometimes referred to as
“Centigrade.”
KELVIN
 The Kelvin temperature scale (1848)
was derived from the observation of
gases.
 Theoretically, the lowest possible
temperature an object can obtain is
that of absolute zero. This
temperature has never been reached.
Scientists have come to within 1
billionth of a degree above absolute
zero.
 This value is the basis for Lord Kelvin’s
scale. At absolute zero, all motion
will cease.
TEMPERATURE CONVERSIONS
 The size of one degree is the same for both the Celsius and
Kelvin scales, however not for Fahrenheit. To convert among
these three scales:
F =
1.8 (C) + 32
C =
F – 32
K = C + 273.15
C = K – 273.15
1.8
Convert the following temperatures from one unit to another.
1.
2.
3.
4.
5.
263 K to oF
38 K to oF
13 oF to oC
1390 oC to K
3000. oC to oF
F – 32+ 32
F = C
1.8 =(K – 273.15)
KF= =C +
1.8273.15
(C) + 32
1.8
F
F==-391
14 F
F
F =
C
K =5432
-111663
CFK
How to Derive Equations
0º C = 32 ºF
Compare
Freezing Points
0ºC
32ºF
100 ºC = 212 ºF
Compare
Boiling Points
0ºC 100ºC
212ºF 32ºF
100ºC = 212ºF
0ºC = 32ºF
100ºC = 180ºF
0ºC 100ºC
Find the
distance
between the two
points on both
scales
212ºF 32ºF
100ºC = 212ºF
0ºC = 32ºF
100ºC = 180ºF
1ºC = (180/100)ºF
1ºC = 1.8ºF
0ºC 100ºC
A 1º change in
Celsius
corresponds to
212ºF
1.8ºF
32ºF
A 1º change in
Celsius
corresponds to
1.8ºF
Whatever data set
has a zero point,
assign that scale as
the x coordinate
m=
m=
y
x
1.8
y = mx + b
Lets set-up ordered
pairs to determine
the equation of the
line.
(0,32)
(100, 212)
m=
212 - 32
100 - 0
ºF = 1.8(ºC) + b
Find the y-intercept (b)
by substitution!!!
ºF = 1.8(ºC) + b
212 = 1.8(100) + b
32 = 1.8(0) + b
212 - 180 = b
32 = b
32 = b
ºF = 1.8(ºC) + 32
Water boils at 140 °X and freezes at 14 °X.
Derive the relationship between °X and °C.
°X
°C
oiling point
BBoiling
point
140
100
Freezing point
14
0
Difference
126
100
1ºC = (126/100)ºX
1ºC = 1.26ºX
Lets set-up ordered
pairs to determine
the equation of the
line.
A 1º change in Celsius
corresponds to 1.26ºX
(0,14)
(100, 140)
m=
y
x
m=
ºX = 1.26(ºC) + b
140 - 14
100 - 0
m = 1.26
y = mx + b
Find the y-intercept (b)
by substitution!!!
140 = 1.26(100) + b
14 = 1.26(0) + b
140 - 126= b
14 = b
14 = b
ºX = 1.26(ºC) + 14
HEAT AND ITS MEASUREMENT:
 We can now convert among the three temperature scales, but what
exactly is temperature?
 Temperature: the average kinetic energy of a substance’s particles.
 Are temperature and heat the same thing?
NO!!!
 Heat: form of energy that is often associated with the changing
temperature of an object. An objects temperature increases because
energy is transferred to it. The energy transfer, in the form of heat,
happens when two objects are brought into contact, as nature
attempts to equalize their temperatures, that is, to establish a
thermal equilibrium.
 A lit match burns at a temperature of 200C.
 A building on fire burns at a temperature of 200C.
 Which one is hotter?

Neither, they are the same temperature.
 Which one produces more heat?

The burning building.
 A hot cup of coffee is 85C.
 A Jacuzzi is 85C.
 A hot cup of coffee contains enough heat energy to probably melt one ice cube before
coming to room temperature.
 A Jacuzzi would melt an entire freezer full of ice before coming to room temperature.
Heat is affected by both temperature and mass.
THE KINETIC THEORY OF HEAT AND TEMPERATURE:
 Temperature can be used to determine the direction of flow of
energy. Energy spontaneously flows from a warmer object to a
cooler one. (HEAT DOES NOT RISE).
 Kinetic theory explains the flow of energy in terms of particle
collisions. Because the warm object is at a higher temperature than
the cool object, the average kinetic energy of its particles is greater
than that of the cool object. Because the objects are in contact, the
particles of the warm object collide with those of the cool object.
Gradually the particles in the warm object will transfer kinetic
energy to the particles in the cool object through collisions. As a
result, the particles in the cool object gain kinetic energy. Eventually
the average kinetic energy of the particles in both objects will be
equal and the two objects will have the same temperature.
 For example: You can feel that a bowl of hot soup warms the air
around it. Left undisturbed, the soup and bowl will eventually cool
to room temperature.
MEASURING CHANGES IN HEAT :
 In the early 1780’s Antoine Lavoisier
measured the heat given off by a guinea
pig. He kept the animal enclosed in a
container so that its body heat would melt
ice. From the amount of ice melted, he
calculated the body heat produced by the
animal. This was an example of an ice
calorimeter.
 A calorimeter is a device used to measure
the energy given off or absorbed during
chemical or physical changes.
 There are two main types of calorimeters, a
constant pressure calorimeter (coffeecup) and a constant volume calorimeter
(bomb calorimeter).
 In a coffee cup calorimeter, the reaction takes place in the water.
 A coffee cup calorimeter is great for measuring heat flow in a solution,
but it can't be used for reactions which involve gases, since they would
escape from the cup.
 The coffee cup calorimeter can't be used for high temperature reactions,
either, since these would melt the cup.
A bomb calorimeter works in the same manner as a coffee cup
calorimeter, with one big difference.
 A bomb calorimeter is used to measure heat flows for gases and high
temperature reactions.
 In a bomb calorimeter, the reaction takes place in a sealed metal container, which
is placed in the water in an insulated container. Heat flow from the reaction
crosses the walls of the sealed container to the water.
 The temperature difference of the water is measured, just as it was for a
coffee cup calorimeter. Analysis of the heat flow is a bit more complex than
it was for the coffee cup calorimeter because the heat flow into the metal
parts of the calorimeter must be taken into account:
Bomb Calorimeter
CALORIMETRY
 We will use the coffee-cup calorimeter in the lab. A bomb calorimeter is the
preferred calorimeter to measure the heat released in combustion reactions.
Both types of calorimeters contain water, which is used to measure the heat
absorbed or discharged.
 Some substances require little heat to cause a change in their temperature.
Other substances require a great deal of heat to cause a change in their
temperature.
 For example: One gram of liquid water requires 4.184 joules of heat to cause a
temperature change of 1C. It only takes 0.902 joules to raise the temperature
of one gram of aluminum 1C.
 The heat needed to raise the temperature of one gram of a substance by 1C is
called the specific heat (Cp) of a substance.
 Every substance has its own specific heat (intensive physical property).
 The heat required to raise the temperature of one gram of water 1C is 4.184 J or
1 cal.
The Specific Heat of water (Cp) is
4.184 J/gC or 1 cal/gC.
 Specific heats can be used in calculations involving the change in
temperature of a specific mass of a substance. Further, the amount of
energy transferred can be calculated from the relationship.
q = mCT




q = heat gained or lost by substance
m = mass in grams
C = specific heat
T = change in temperature (T2 – T1) or (Tfinal – Tinitial)
 According to the Law of Conservation of Energy, in a calorimeter
containing water, any heat gained by the substance is lost by the water.
Likewise, any heat lost by the substance is gained by the water. This
transfer of energy takes place between these two quantities of matter
that are at different temperatures until the two reach the same
(equilibrium) temperature.
 The heat released or absorbed by water is calculated using the
following equations:
q = mCT
q =
m
4.184 J
g C
T
or
q =
m
1 cal
T
g C
Note: Any heat lost by the substance (-q) equals the heat
gained by the water (+q) and vice versa.
 Notice that throughout our discussion of calorimetry, we have
assumed that the calorimeter itself does not absorb any heat. We
have also assumed that no heat escapes from the calorimeter.
Neither of these assumptions is completely true. However, we will
continue to make these assumptions in order to simplify our
calculations. The error from actual losses of heat should be
considered in any laboratory exercise using calorimeters.
EXAMPLE:
 Phosphorus trichloride, PCl3, is a compound used
in the manufacture of pesticides and gasoline
additives. How much heat is required to raise the
temperature of 96.7 g of PCl3 from 31.7°C to 69.2°C.
The specific heat of PCl3 is 0.874 J/g°C.
69.2 C
q = mCT
-69.2
31.7CC
q
???
m
96.7 g
96.7 g o.874 J
qC = 0.874 J/g
J/g°C
C
g C
T1
31.7°C
31.7
C
T2
q =
T
96.7
g Co.874 J
69.2°C
69.2
T2 – T1 g C
- 31.7
37.5CC
T
q = 3169.3425 J
37.5 C
q = 3170 J
Examples
 The specific heat of graphite is 0.71 J/g°C. Calculate the
energy needed to raise the temperature of 75 kg of graphite
from 294 K to 348 K.
348 K
q = mCT
q =
q =
75 kg
75 kg
1000
o.71 Jg
o.71
T J
g1 kg
C
g C
1000 g
o.71 J
1 kg
g C
T
54 C
q = 2.9 x 106 J
q = 2.9 x 103 kJ
-
294 K
54 K
EXAMPLE:
 An Oreo® is burned in a bomb calorimeter containing 5.0 L of
water originally at 25.0 °C. After igniting the Oreo®, the water
temperature increases to 30.6°C. How many Calories of energy
did the Oreo® contain?
Volume of water = 5.0 L
T1 = 25.o °C
T2 = 30.6 °C
How
What
does
happened
the heat
togained
the
Why?
bytemperature
the water
compare
of the to
the heat water?
lost by the Oreo?
Law of Conservation of Energy!!!
Volume of water = 5.0 L
If we calculate the
T1 = 25.o °C
heat gained by the
How can
we find
water,
this the
T2 = 30.6 °C
mass
of the water?
necessarily
equals
the heat lost by the
Density = mass / volume
cookie
CH2O= 1 cal/g °C
mass = Density * volume
mass =
1.0gg
1.0
mL
mL
5.0L L
5.0
mass =
1000 mL
1L
5.0 x 103 g
Putting it all together….
qH2O =
5.0 x 103 g
1 cal
5.6 °C
qH2O =
28 Calories
g °C
1 Cal
1000 cal
EXAMPLE:
 A reaction changes the temperature of a calorimeter
with 500. mL of water from 10.°C to 50.°C. What is the
ΔH for the reaction?
=
500. g
4.184 J
g °C
=
40. °C
???
m
500. g
C 4.184 J/g°C
84000 J
= - 84000 J
=
q
- 8.4 x 104 J
T1
10.°C
T22
50.°C
T
40. °C
Calorimetry Example
 A 12.5 g metal block was heated to 100.00°C and dropped
into a coffee-cup calorimeter containing 48.0 g of water at
24.37°C. After the block was dropped into the water, the
temperature of the water rose to 26.43°C. What is the
specific heat capacity of the metal?
m=
C=
T1 =
T2 =
ΔT = T2 – T1
METAL
12.5 g
100.00°C
26.43°C
-73.57 °C
WATER
48.0 g
4.184 J/g°C
24.37°C
26.43°C
2.06°C
Law of Conservation of Energy!!!
qH2O =
48.0 g
qH2O =
4.184 J
g °C
2.06°C
413.7139 J
- 413.7139 J
The unknown is Cmetal
- 413.7139 J
12.5 g
.4499 J/g °C
.450 J/g °C
-73.57 °C
The unknown is Cmetal
Cmetal =
Cmetal =
48.0 g
4.184 J
g °C
Cmetal =
2.06°C
- 12.5 g
.450
.4499J/g
J/g°C
°C
-73.57 °C
Examples
 A 46.2 g sample of copper is heated to
95.4°C and then placed in a calorimeter
containing 75.0 g of water at 19.6°C. The
final temperature of both the water and the
copper is 21.8°C. What is the specific heat of
copper?
COPPER
m=
46.2 g
C=
95.4°C
T1 =
21.8°C
T2 =
ΔT = T2 – T1
Cmetal =
75.0 g
-73.6°C
4.184 J
Cmetal
g °C =
WATER
75.0 g
4.184 J/g °C
19.6°
21.8°C
2.2°C
2.2°C
.20 J/g -°C
46.2 g
-73.6 °C
ENERGY AND CHANGES OF STATE:
 When energy is added to a solid substance, its temperature
increases. Before the melting point is reached, the added
energy increases the kinetic energy of the particles. Thus,
because the particles in a solid are held together in a fixed
position, the addition of heat causes the particles to vibrate
faster about a fixed position.
 At the melting point, the addition of more energy causes the
substance to melt. The temperature, however, remains the
same until all of the substance has melted. Melting is when the
added energy changes the positions of the particles and now
they can leave their original positions and move to other
locations in the substance. (This is why liquids take on the
shape of the container.)
 The potential energy changes as the physical state changes.
The energy required to melt one gram of a specific substance is
called the enthalpy of fusion, Hfus, of that substance.
 A similar phenomenon takes place at the boiling
point (condensation point). The addition of energy to
the liquid causes the particles to move about at a
faster rate. These liquid particles are free enough to
slip past one another and appear to vibrate about a
moving point.
 At the boiling point, the addition of more energy
causes the particles to spread apart even more, and
undergo a phase change into a gas. The energy
required to vaporize one gram of a substance at its
boiling point is called the enthalpy of vaporization,
Hvap of the substance. As a gas, these particles move
independent of each other in rapid, random
motion.
 When matter is heated further near temperatures of
5000C, the collisions between the particles are so
violent that electrons are knocked away from the
atoms. The action produces a state of matter
composed of electrons and positive ions, which is
called plasma.
CALCULATIONS:
 The enthalpy of vaporization of water is 2.25 kJ/g.
 The enthalpy of fusion of water is 0.333 kJ/g.
 Change in kinetic energy
q = mCT
 Increase in potential energy
H = m (Hvap)
at the boiling point
 Decrease in potential energy
H = m (-Hvap)
at the condensation point
 Increase in potential energy
H = m (Hfus)
at the melting point
 Decrease in potential energy
H = m (-Hfus)
at the crystallization point
 REMEMBER THAT q = H at constant pressure
Heating Curve for Water
120
Water
and
Steam
100
80
Steam
60
Water
40
20
0
Ice
Water
and Ice
-20
0
40
120
220
760
800
Heating Curve for Water
120
HeatWater
of
and
Vaporization
Steam
100
80
Steam
60
Water
40
20
0
Ice
Water
and Ice
-20
0
40
120
220
760
800
Heating Curve for Water
120
Water
and
Steam
100
80
Steam
60
Water
40
20
Heat ofWater
and Ice
Ice Fusion
0
-20
0
40
120
220
760
800
Heating Curve for Water
120
Both Water
Water
and
and Steam
Steam
100
80
Steam
60
Water
40
20
0
Ice
Water
and Ice
-20
0
40
120
220
760
800
Heating Curve for Water
120
Water
and
Steam
100
80
Steam
60
Water
40
20
Ice andWater
Ice Waterand Ice
0
-20
0
40
120
220
760
800
Examples
 How much heat does it take to heat 12 g of ice at -6C to
25C water?
When constructing the graph, note the
•Beginning and Ending temperatures
•T he temperature(s) of any phase change(s) within that
range.
For every segment of the
graph, you must calculate
the heat involved.
q1
q2
q3
 How much heat does it take to heat 12 g of ice at -6°C to
25°C water?
q1 = mCiceT
q2 = mHfusion
q3 = mCwater T
q1 = mCice T
q231 =
q1 = 149.76 J
q1
12 g
q2
q1 = 149.76 J
0.333
4.184
2.08 kJ
JJ
1000
25
6 °C
°CJ
q2 = 3996
J
g g°C
1 kJ
q3 = 1255.2 J
= J5400
= 5400.96
J q3q
qq2tot
= 3996
J
=tot
1255.2
q3
J
 How much heat does it take to heat 35 g of ice at 0 C
to steam at 150. C?
q3
q1 = mHfusion
q2 = mCwater T
q3 = mHvap
q2
q4 = mCsteam T
q1
q4
 How much heat does it take to heat 35 g of ice at 0 °C
to steam at 150.°C?
q1 = mHfusion
q1 =
35 g
0.333 kJ
1000 J
g
1 kJ
q1 = 11655 J
q2 = mCwater T
35 g
q2 =
q2= 14644 J
q3 = mHvap
q3= 79100 J
q4 = mCsteam T
q4= 3573.5 J
4.184 J
g °C
100. °C
35 g
q3 =
qTot= 108972.5 J
35 g
q4 =
qTot= 109000 J
2.26 kJ
g
2.042 J
g °C
1000 J
1 kJ
50. °C