PSC 151
Laboratory
Activity
7
Thermodynamic
Properties of
Water
Thermodynamic
Properties
of
Water
Heat of Fusion of Ice
What causes the temperatures of two objects placed in
thermal contact to change?
TH  TC
TC
TH
Something must move from the high temperature object to the low
temperature object.
Is it matter or energy?
If it is matter the mass of the high temperature object would
decrease while the mass of the low temperature object would
increase.
This is not observed.
It must be energy flowing between the two objects that causes a
change in their temperatures.
The energy flowing between the two objects must be on the
microscopic level because we can not see it.
Energy Review
Kinetic energy, KE-the macroscopic energy an object has due to
its motion, measured in Joules, J.
Gravitational potential energy energy, GPE-the macroscopic
energy an object has due to its position , measured in Joules, J.
Total mechanical energy, E-the sum of an object’s kinetic and
potential energies , measured in Joules, J.
Work, W-the process by which the total mechanical energy can be
changed , measured in Joules, J.
Work and the Related Changes in Macroscopic Energy
Work,W
Change in
Mechanical Energy
E
Change in Kinetic Energy
E
and/or
Change in Potential Energy
GPE
All macroscopic objects are composed of microscopic objects:
atoms and molecules
Macroscopic
Object
Atoms and
Molecules
These atoms and molecules are moving so they have a microscopic
kinetic energy.
These atoms and molecules are subject to conservative forces
(gravitational and electrical) so they have a microscopic potential
energy.
The sum of these microscopic kinetic and potential energies is
called Thermal Energy, U.
When two objects with different temperatures are placed in
thermal contact, thermal energy flows from the higher temperature
object to the lower temperature object until thermal equilibrium is
reached.
The thermal energy that flows between two objects because of a
difference in temperature is called heat, Q.
Since heat is a form of energy it is measured in Joules, J.
For historical reasons another unit of thermal energy or heat is
sometimes used: calorie, cal or kilocalorie, kcal.
Conversion Factor
1kcal = 4186J
When work is done on or by an object there is a change in the
object’s kinetic energy or its gravitational potential energy or both.
The change in kinetic energy is perceived as a change in object’s
velocity.
The change in gravitational potential energy is perceived as a
change in object’s position (height above the reference level).
When heat flows into or out of an object there is a change in the
object’s thermal energy.
How is the change in the thermal energy of an object perceived?
A change in thermal energy is perceived as a change in the
object’s temperature or phase (solid, liquid, or gas).
It has been observed that the change in temperature and the
change in phase never occur at the same time.
When the temperature is changing the phase remains constant
and when the phase is changing the temperature remains
constant.
Heat, Q
Change in Temperature
T
Phase Constant
Change in Thermal Energy
U
or
Change in Phase
T=0
What variables determine the magnitude of the change in
temperature?
Q
Q
Water
Water
More heat results in a larger change in temperature: change in
temperature is directly proportional to the amount of heat.
T  Q
Q
1kg
Water
Q
2kg
Water
When the same quantity of heat flows into (or out of) a larger
mass the change in temperature is less.The change in
temperature is inversely proportional to the mass.
T  1
m
Q
1kg
Water
Q
1kg
Alcohol
When the same quantity of heat flows into (or out of) equal
masses of different substances the change in temperature is
different.
The change in temperature depends on the specific heat, c of the
substance.
The specific heat of a substance is the amount of heat required to
change the temperature of 1kg of the substance by 1C°.
Units :
J or kcal
kg  C
kg  C
Q
1kg
Water
c
water
Q
 1 kcal
kg  C
1kg
Alcohol
c
alcohol
 0.6 kcal
kg  C
The substance with the higher specific heat experiences a smaller
change in temperature. The change in temperature is inversely
proportional to the specific heat.
1
T 
c
Combing All of the Proportions
1
1
T  Q, ,
m c
Q
T 
mc
Q  mcT
Heat flow that results in a phase change can not be described by
the equation above since during a phase change the temperature
remains constant…T=0.
It has been observed that phase changes can only occur at certain
temperatures which depend on the particular substance.
Temperature
increases
Q
Temperature remains
constant
0°C
0°C
-20°C
-20°C
Ice @
-20°C
Q
Ice @
0°C
Ice begins melting
0°C is the melting (or freezing) point of water.
As more heat is added the temperature will remain constant as
more ice melts.
Once all of the ice has melted the addition of more heat will result
in an increase in the temperature of the water.
100°C
Temperature
increases
Q
Temperature remains
constant
100°C
0°C
0°C
-20°C
-20°C
Water @
0°C
Q
Water @
100°C
Water begins boiling
100°C is the boiling point of water.
As more heat is added the temperature will remain constant as
more water converts to steam.
Once all of the water has converted to steam the addition of more
heat will result in an increase in the temperature of the steam.
What determines how much ice melts or water converts to steam?
Q
0°C
0°C
-20°C
-20°C
Ice @
0°C
Q
Ice @
0°C
More heat results in a larger mass of ice melting. The mass of ice
melted is directly proportional to the quantity of heat.
m Q
0°C
-114°C
Q
1kg ice @
0°C
Q
1kg solid
alcohol @
-114°C
When the same quantity of heat flows into (or out of) equal masses
of the solid phase of different substances, each at its melting point,
different masses will melt.
The mass that melts depends on the latent heat of fusion, Qf of the
substance.
The latent heat of fusion of a substance is the quantity of heat
required to change 1kg of the solid phase of the substance, at its
melting point, to 1kg of liquid at the same temperature.
J
kcal
Units :
or
kg
kg
Q f,water  80 kcal
kg
Q f,alcohol
0°C
Q
1kg ice @
0°C
Q
 -114°C
25 kcal
kg
1kg solid
alcohol @
-114°C
The substance with the greater heat of fusion experiences less
melting. The mass melted is inversely proportional to the heat of
fusion
m 1
Qf
Combing All of the Proportions
1
m  Q,
Qf
Q
m
Qf
Q  mQ f
solid  liquid
There is a similar equation describing the phase change between
liquid and gas.
Q  mQ v liquid  gas
Qv is the latent heat of vaporization, the quantity of heat required
to change 1kg of the liquid phase of the substance, at its boiling
point, to 1kg of gas at the same temperature.
J
kcal
Units :
or
kg
kg
Heat, Q
Change in Temperature
T
Phase Constant
Change in Thermal Energy
U
Q  mcT
Change in Phase
T=0
Q  mQ f solid  liquid
Q  mQ v liquid  gas
Summary of Thermodynamic Properties and Relationships
I) Phase Change Temperatures- °C, K, °F
A) Melting / Freezing Point, Tf
B) Boiling Point, Tb
II) Specific Heat -
J , kcal
kg  C kg C
A) Solid Phase, csolid
B) Liquid Phase, cliquid
C) Gas (vapor) Phase, cgas
1. cp , constant pressure
2. cv , constant volume
III) Latent Heat -
J , kcal
kg kg
A) Latent Heat of Fusion, Qf
B) Latent Heat of Vaporization, Qv
IV) Thermodynamic Relationships
A) Change in Temperature, Q = mcT
B) Change in Phase (solid  liquid), Q = mQf
C) Change in Phase (liquid  gas), Q = mQv
Example: Water
Tf = 0°C, 273K, 32°F
Tb = 100°C, 373K, 212°F
c ice  2093
J , 0.5 kcal
kg  C
kg  C
J , 1.0 kcal
kg C
kg  C
 2020 J , 0.483 kcal
kg C
kg  C
c water  4186
c p,steam
c v,steam  1520
J , 0.363 kcal
kg  C
kg C
Q f  3.34  10 5 J , 80.0 kcal
kg
kg
Q v  2.26  106 J , 540.0 kcal
kg
kg
Graphical Representation of Heat Flow, Temperature Change, and
Final
Tfinal
Phase Change
Boiling
Point
Q  mQv
Temperature
Tb
Q  mc vaporT
Temperature T, °C
Step #1 Raise the temperature to the melting point
Tf
Melting
Point
Tinitial
Step #2 Melt all of the solid
Step #3 Raise temperature of liquid to boiling point
Step #4 Convert the liquid to gas
Step #5 Raise the temperature of the gas to the final
temperature
Q  mc liquidT
Q  mQ f
Q  mcsolidT
Initial
Temperature
Heat flow w/ Temperature Change
Heat flow w/ Phase Change
Heat Flow Q, kcal
Calorimetry
When heat transfer between two systems occurs inside of an
insulated environment, no heat is lost to the external environment
and no heat enters the systems from the outside. In this case we
can say that the heat lost by the high temperature system equals
the heat gained by the low temperature system. Further, we can
recognize when the two systems have reached thermal
equilibrium by noting when the temperature of the combined
system remains constant.
Qlost  Qgained
Suppose we place a known mass of tap water (mw) into an insulated
calorimeter cup and measuring its initial temperature (Ti). Then a
small amount of ice (assumed to be at 0°C) is added to the cup, and
when thermal equilibrium is reached, the final temperature (Tf) is
measured. Heat will flow from the tap water into the ice.
Temperature, °C
The temperature of the tap water will decrease while the ice will
first melt and then the temperature of the ice water will increase
until a final common temperature is reached. The mass of ice (mi)
added can be determined by measuring the increase in the
amount of water in the cup.
water @ T
Ti
tap water cooling
to final temperature
Tf
0C
ice
melting
ice water warming
to final temperature
ice @ 0°C
Qgained
Qlost
Heat
i
Qlost (tap water) = mw c w Tw
Tw  Ti  Tf
Qgained (ice) = mi Qf
Qgained (ice water) = m ic w Tiw
Tiw  Tf  0  Tf
Qlost  Qgained
mw c w Tw  mi Qf  mic w Tiw
mi Qf  m w c w Tw  m ic w Tiw
m w c w Tw  m i c w Tiw
Qf 
mi
Procedure
Step 1. Measure the mass of the empty calorimeter cup…Record in
Data Table 1.
Step 2. Place about 150ml of tap water in the calorimeter cup and
measure the mass…Record in Data Table 1.
Step 3. Insert the temperature probe, and measure the initial
temperature (Ti) of the tap water …Record in Data Table 1.
Step 4. Remove the temperature probe. Place 3-4 pieces of ice in
the calorimeter cup and quickly replace the lid.
Step 5. Place the Styrofoam cover on the cup and insert the
temperature probe. Wait until thermal equilibrium is reached
(temperature reading remains constant). This will ensure that
all of the ice has melted and all of the water has reached the
common final temperature. Record the final equilibrium
temperature in Data Table 1.
Step 6. Remove the temperature probe and lid. Measure the final
mass and record in Data Table 1.
Data Table 1
Mass of Calorimeter Cup, g
Mass of Calorimeter Cup & Tap Water-m w, g
Mass of Tap Water-m w, g
Mass of Tap Water-mw, kg
Initial Temperature of Tap Water-T i , °C
Add ice and wait for thermal equilibrium.
Final Equilibrium Temperature-T f , °C
Final Mass of Calorimeter Cup & Water, g
Mass of Ice Added to Calorimeter Cup-m i , g
Mass of Ice Added to Calorimeter Cup-mi , kg
mw
Ti
Tf
mi
Supprt
Ring
Outer
Cup
Inner
Cup
Opening
for Stirrer
Outer
Cup
Inner
Cup
Opening for
Temperature
Probe
Support
Ring
Measure mass of
cup.
Place about 150ml of
water in the cup.
Measure mass of cup
and water, subtract
the mass of the cup to
determine the mass of
tap water - mw, kg.
Insert the CBL
temperature probe.
Measure initial
temperature of tap
water -Ti, °C.
23.8°C
Inner
Cup
Place the inner cup inside the outer cup.
Add 3-4 pieces of ice to the water in the cup.
Quickly place the top on the
calorimeter cup and reinsert the
temperature probe.
Wait for thermal equilibrium
to be reached.
Temperature Constant
Measure final equilibrium
temperature -Tf, °C
Measure final mass of
water, subtract initial
mass to determine mass
of ice added - mi, kg
18.5°C
Step 7. Calculate the heat of fusion of ice.
m w c w Tw  m i c w Tiw
Qf 
mi
Tw  Ti  Tf
Tiw  Tf  0  Tf
mw c w  Ti  Tf   mi c w Tf
Qf 
mi
mw c w  Ti  Tf   mi c w Tf
Qf 
mi
m i Qf  m w c w  Tf  Ti   m i c w Tf
mi Qf  mi c w Tf  m w c w  Tf  Ti 
mi  Qf  c wTf   mw c w  Tf  Ti 
m w c w  Tf  Ti 
mi 
 Q f  c w Tf 
mw c w  Ti  Tf   mi c w Tf
Qf 
mi
mi Qf  m w c wTi  mw c w Tf  mi c w Tf
mw c w Tf  mi c w Tf  m w c wTi  mi Qf
 mw  mi c wTf  mw c w Ti  mi Q f
m w c w Ti  m iQ f
Tf 
 m w  m i c w