Thermodynamic Properties of Water

PSC 151Laboratory Activity 7

Thermodynamic Properties of Water

Heat of Fusion of Ice

What causes the temperatures of two objects placed in thermal contact to change?

TH TC

TH > TC

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,WChange in

Mechanical EnergyΔE

Change in Kinetic EnergyΔΚE

Change in Potential EnergyΔGPE

and/or

Work and the Related Changes in Macroscopic Energy

All macroscopic objects are composed of microscopic objects: 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.

Macroscopic Object

Atoms and Molecules

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 Factor1kcal = 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 Thermal EnergyΔU

Change in PhaseΔ =0T

Change in TemperatureΔT Phase Constant

or

What variables determine the magnitude of the change in temperature?

Q QWaterWater

More heat results in a larger change in temperature: change in temperature is directly proportional to the amount of heat.

ΔT ∝Q

Water1kgQ

Water2kgQ

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.

ΔT ∝ 1m

Water1kgQ Q

Alcohol1kg

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 : Jkg⋅C° or

kcalkg⋅C°

Alcohol1kg

calcohol

=0.6 kcalkg⋅C°

QQWater1kg

cwater

=1 kcalkg⋅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.

ΔT ∝ 1c

Combing All of the Proportions

ΔT ∝Q, 1m , 1

c ΔT = Qm⋅c

Q =mcΔT

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…ΔT=0.

It has been observed that phase changes can only occur at certain temperatures which depend on the particular substance.

Q Q

Temperature remains constant

Ice begins melting

Ice @ -20°C

-20°C

0°C

Ice @ 0°C

-20°C

0°C

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.

Temperature increases

Water @ 0°C

-20°C

0°C

Q

100°C

Water @ 100°C

-20°C

0°C

100°C

Q

Temperature remains constant

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.

Temperature increases

What determines how much ice melts or water converts to steam?

Q Ice @ 0°C

-20°C

0°C

Ice @ 0°C

-20°C

0°C

QMore heat results in a larger mass of ice melting. The mass of ice melted is directly proportional to the quantity of heat.

m ∝Q

Q 1kg ice @ 0°C

0°C

Q

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.

1kg solid alcohol @

-114°C

-114°C

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.

Units :J

kgor kcal

kg

Q 1kg ice @ 0°C

0°C

Q 1kg solid alcohol @

-114°C

-114°C

Qf ,water =80 kcalkg

Qf ,alcohol =25 kcalkg

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

m ∝Q, 1Qf

m = QQf

Q =mQf

There is a similar equation describing the phase change between liquid and gas.

Q =mQv

solid ⇔ liquid

liquid ⇔ gasQv 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.

Units :J

kgor kcal

kg

Heat, Q

Change in Thermal EnergyΔU

Change in PhaseΔ =0T

Change in TemperatureΔT Phase Constant

Q =mcΔT

Q =mQf solid⇔ liquid

Q =mQv liquid⇔ gas

Summary of Thermodynamic Properties and RelationshipsI) Phase Change Temperatures- °C, K, °F

A) Melting / Freezing Point, Tf

B) Boiling Point, Tb

II) Specific Heat - Jkg⋅C° ,

kcalkg⋅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 - Jkg , kcal

kg

A) Latent Heat of Fusion, Qf

B) Latent Heat of Vaporization, Qv

IV) Thermodynamic Relationships

A) Change in Temperature, Q = mcΔT

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

cice =2093 Jkg⋅C° , 0.5

kcalkg⋅C°

cwater =4186 Jkg⋅C° , 1.0

kcalkg⋅C°

cp,steam =2020 Jkg⋅C° , 0.483

kcalkg⋅C°

cv,steam =1520 Jkg⋅C° , 0.363

kcalkg⋅C°

Qf =3.34×105 Jkg , 80.0kcalkg

Qv =2.26 ×106 Jkg , 540.0kcalkg

Graphical Representation of Heat Flow, Temperature Change, and Phase Change

Heat Flow Q, kcal

Heat flow w/ Temperature Change

Heat flow w/ Phase Change

Melting Point

Tf

Tb

Boiling Point

Initial Temperature

Tinitial

FinalTemperature

Tfinal

Step #1 Raise the temperature to the melting point

Step #2 Melt all of the solid

Step #3 Raise temperature of liquid to boiling point

Step #4 Convert the liquid to gasStep #5 Raise the temperature of the gas to the final temperature

Q =mcsolidΔT

Q =mQf

Q =mcliquidΔT

Q =mQv Q =mcvaporΔT

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.

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.

Heat

Ti

0°C

Tf

ice @ 0°C

water @ Ti

icemelting ice water warming

to final temperature

tap water coolingto final temperature

Qgained Qlost

miQf =mwcwΔTw−micwΔTiw

Qf =mwcwΔTw −micwΔTiw

mi

Qlost (tap water) = mwcwΔTw

Qgained (ice) = miQf

Qgained (ice water) = m icwΔTiw

Qlost =Qgained

mwcwΔTw =miQf +micwΔTiw

ΔTw = Ti − Tf

ΔTiw = Tf − 0 = Tf

ProcedureStep 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.

Mass of Calorimeter Cup, gMass of Calorimeter Cup & Tap Water-m w, gMass of Tap Water-m w, gMass of Tap Water-m w, kg mw

Initial Temperature of Tap Water-T i , °C Ti

Final Equilibrium Temperature-T f , °C Tf

Final Mass of Calorimeter Cup & Water, gMass of Ice Added to Calorimeter Cup-m i , gMass of Ice Added to Calorimeter Cup-m i , kg mi

Data Table 1

Add ice and wait for thermal equilibrium.

Outer Cup

Supprt Ring

Inner Cup

Outer Cup Support

Ring

Inner Cup

Opening for Temperature

ProbeOpening for Stirrer

Measure mass of cup and water, subtract the mass of the cup to determine the mass of tap water - mw, kg.

Measure initial temperature of tap

water -Ti, °C.Measure mass of

cup.

Place about 150ml of water in the cup.

Insert the CBL temperature probe.

23.8°C

Inner Cup

Measure final mass of water, subtract initial

mass to determine mass of ice added - mi, kg

Measure final equilibrium temperature -Tf, °C

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

Place the inner cup inside the outer cup.

18.5°C

Step 7. Calculate the heat of fusion of ice.

Qf =mwcwΔTw −micwΔTiw

mi

ΔTw = Ti − Tf ΔTiw = Tf − 0 = Tf

Qf =mwcw Ti −Tf( )−micwTf

mi

Qf =mwcw Ti −Tf( )−micwTf

mi

miQf =mwcw Tf −Ti( )−micwTf

miQf +micwTf =mwcw Tf −Ti( )

mi Qf + cwTf( ) =mwcw Tf −Ti( )

mi =mwcw Tf −Ti( )Qf +cwTf( )

Qf =mwcw Ti −Tf( )−micwTf

mi

miQf =mwcwTi −mwcwTf −micwTf

mwcwTf +micwTf =mwcwTi −miQf

mw +mi( )cwTf =mwcwTi −miQf

Tf =mwcwTi −miQf

mw +mi( )cw