ENTROPY
Phenomenological Interpretation of Entropy
For an ideal gas, the internal heat content of the gas is proportional to the temperature T of the gas. In particular
for a monatomic gas, U = 3/2 NkT.
If a quantity of heat Q is added to any substance, then one might expect the heat energy content of that
substance to increase and thus increase the substance's temperature. This does not always happen. For example,
during the melting of ice (or any phase change) the temperature remains constant while heat is being added.
If the added heat does not raise the temperature, what has happened to the energy added? For the melting of ice,
the added energy has gone into destroying the orderly arrangement of the atoms and turning the crystal-ice into
a less orderly arrangement of liquid-water.
How could we construct a quantity that would measure whether the energy added to a substance is going into
thermal energy or into destroying order? Conversely, how could we construct a quantity that would measure
whether the energy removed from a substance goes into lowering the substance's thermal energy content or into
creating a more ordered structure?
This quantity is defined as Entropy: S = Q/T.
What are the units of this quantity?
One such quantity that would - in my opinion - have been a better definition of entropy would be
This would make the units of entropy dimensionless and the change in entropy could be thought of as simply
being equal to the ratio of Heat Energy Absorbed over the Thermal Energy. If a substance absorbs heat and
doesn't change its temperature, then this ratio gets bigger positively and its entropy/disorder increases. If a
substance loses heat and doesn't change its temperature, then this ratio gets smaller negatively and its
entropy/disorder decreases.
Thermal Processes:
* When a substance changes from on state of equilibrium to another state of equilibrium, the steps or path
between the initial and final thermodynamic states is called the process.
Quasi-Equilibrium Processes:
* A process is call a quasi-equilibrium process if the intermediate steps in the process
are all close to equilibrium. In this way we can characterize the intermediate states of
the process using state variables (such as temperature, pressure, volume, entropy, etc)*
When a process is quasi-equilibrium we can plot the path of the process on say a
pressure vs. volume work diagram since all the variable used to characterize the substance's intermediate states
have well define values.
* Most of the process you will encounter will be quasi-equilibrium process and we will drop the "quasiequilibrium" when talking about a particular process.
Entropy Formal Definition:
Along any reversible path process
connecting states 1 and 2
Entropy is State Variable:
The Change in a system's Entropy is Independent
of the Path chosen
Ssys = 0
Ssys > 0
For any Reversible Process in which there is no heat flow,
Qsys = 0 (Both Adiabatic and Reversible)
For any Irreversible Process even if Qsys = 0
The Change in Entropy for an Irreversible Process can be determined by devising a reversible process
connecting the same two end states.
SOME IMPORTANT PROCESSES
Adiabatic and Reversible :
Isothermal:
Constant Pressure :
Ideal Gas
for any type of
Quasi-Equilibrium Processes:
(Isentropic Process)
(Phases Changes)
(For Solids and Liquids,
C is approximately constant)
Definition of Enthalpy
ENTHALPY



Chemist often use enthalpy because it lets them determine the heat flow in a constant pressure processes.
Mixing and heating chemicals open to the air is a constant pressure process.
The heat flow in or out during any process can be calculated using the constant pressure heat capacity
regardless of the type of process.
Like U the enthalpy H is also a state variable.
State Variable:
Examples of State Variables:
Temperature
Pressure
Volume
Entropy
Enthalpy
Internal
Energy
Mass
Density
* State Variables are Path Independent: meaning that the change in the value of the state variable will be the
same no matter what path you take between the two states. This is not true of either the work W or the heat Q.
* If a system is carried through a cycle that returns it to its original state, then a variable will only be a state
variable if variable returns to its original value.
* If X is a State Variable then:
* State Variables are only measurable when the system is in Equilibrium.
Reversible Processes:
* A process is reversible when the successive states of the process are Infinitesimally close to Equilibrium
States. i.e. the process is quasi-equilibrium.
* With a reversible process it is possible to restore the system to its original state without needing an external
agent or changing its surroundings.
* Reversible processes are an abstraction that aids the analysis of real processes.
* A reversible process is a standard of comparison for an actual system.
* Truly reversible thermal processes would require an infinite amount of time for completion.
Irreversible Processes:
* All Natural processes are Irreversible.
* The path of an irreversible process is indeterminate and cannot be drawn on a thermodynamic diagram. (We
use a hashed line to indicate the path because the intermediate states are in non-equilibrium.)
* The Entropy of the universe always increases during an irreversible process.
* It is always possible to restore an irreversible process to its original state by a reversible process, but the
Entropy of the universe can never be restored.
* An irreversible process always requires an external agent to restore it to its original state.
Examples of Irreversible Processes:
Friction
Heat Flow
Unrestrained
Expansion
Melting/Boiling
Mixing
Inelastic Deformation
Chemical
Reactions
Current Flow Your House Getting
Dirty
Equilibrium:
* A System is in Equilibrium if its Properties/Variables do not change with time.
Thermal Equilibrium
No Temperature or Pressure Gradients in the System.
Mechanical
Equilibrium
No Unbalanced Forces or Torques in the System.
Chemical Equilibrium No tendency of the System to undergo Chemical Reaction or Diffusion.
Electrical Equilibrium No Electrical Potential Gradients in the System.
LAWS OF THERMODYNAMICS
As related to Entropy
First Law : Energy cannot be created or destroyed.
Second Law : Entropy can be created but not spontaneously destroyed.
Third Law : It is the Entropy (not Energy) of a system that goes to
zero as the Absolute Temperature goes to zero.
A temperature entropy diagram, or T-s diagram, is used in thermodynamics to visualize changes to
temperature and specific entropy during a thermodynamic process or cycle. It is a useful and common tool,
particularly because it helps to visualize the heat transfer during a process. For reversible (ideal) processes, the
area under the T-s curve of a process is the heat transferred to the system during that process.
This example T-s diagram shows a thermodynamic cycle taking place between a hot reservoir at temperature TH
and a cold reservoir at temperature TC. For reversible processes, such as those found in the Carnot cycle, the
area in red QC is the amount of energy exchanged between the system and the cold reservoir. The area in white
W is the amount of work energy exchanged by the system with its surroundings. The amount of heat QH
exchanged with the hot reservoir is the sum of the two. The thermal efficiency of the cycle is the ratio of the
white area (work) divided by the sum of the white and red areas (total heat). If the cycle moves in a clockwise
sense then it is a heat engine that outputs work, if the cycle move in a count-clockwise sense it is a heat pump
the takes in work and moves heat QH from the cold reservoir to the hot reservoir.
An isentropic process is depicted as a vertical line on a T-s diagram, whereas an isothermal process is a
horizontal line.
THERMODYNAMIC PROCESSES FOR AN IDEAL GAS
PVn = Constant
Process
Isobaric
Isochoric
Isothermal
Adiabatic
Variable =>
Pressure
Volume
Temperature.
No Heat Flow
Quantity Constant
=>
P = 0
V = 0
T = 0
Q=0
n
0
1
 = Cp/Cv
First Law
U = Q - W
U = 0
Q=W
U = -W
Q=0
U = Q
W=0
Work
0
Heat Flow
0
Q
Heat
Capacity
0
Internal Energy
0
Enthalpy
0
Entropy
0*
Ideal Gas
Relations
* For Adiabatic Reversible Processes
n cp = m Cp
n cv = m C v
cp - c v = R
nR=Nk
 = Cp/Cv = c p/cv = Ratio of Specific Heats
Cp = Constant Pressure Specific Heat Capacity (J/kg/ oC)
Cv = Constant Volume Specific Heat Capacity (J/kg/ oC)
cp = Molar Constant Pressure Heat Capacity (J/mole/oC)
cv = Molar Constant Volume Heat Capacity (J/mole/oC)
DIESEL CYCLE
n moles of an ideal Diatomic gas undergoes the following 4-step cycle:
p
1 to 2: Adiabatic compression
2 to 3: at Constant pressure
3
2
3 to 4: Adiabatic expansion
4 to 1: constant volume.
4
p0
For each “corner” of the cycle calculate P, V, T.
For each process (A to B, etc.) and for calculate Q, W, ∆U, and ∆S.
Calculate total Q, W, ∆U, and ∆S for the entire cycle.
Calculate the efficiency of the heat engine running this particular
cycle.
Given: p0 = 105 N/m2
V0 = 10-3 m3
T0 = 300 K
Qin = _600_ J
V2 = 1/_15_ V0
CV = 5/2 R; CP = 7/2 R;  = 1.4
1
V
V0
NEATLY SHOW YOUR CALCULATIONS. Copy your answers into the table
At 1
At 2
At 3
At 4
(back at 1)
p
V
T
1 to 2
2 to 3
3 to 4
4 to 1
Q
W
∆U
∆S
Efficiency:
Total for the
cycle
IDEAL HEAT ENGINE GAS CYCLES
Otto Cycle - Internal Combustion Engine
Nicolaus August Otto the inventor of the four-stroke cycle was born on 14th June
1831 in Holzhausen (Germany). In 1862 he began first experiments with four-strokes
engines. Together with Eugen Langen he founded the first engine company "N.A.Otto & Cie". Then they improved the atmospheric gas engine and in 1867 they
won a gold medal at the Paris Exposition. One of the first four-stroke engines is
shown in Fig. 3. A working diagram of this engine is shown below. It corresponds to
the today's engines. He died on 26th January 1891 in Cologne.
Fig 3: four-stroke cycle engine 1876
Fig 2: atmospheric gas engine- about 1866/67 Fig 4: working diagram : 9th May 1876
(2) The Parts of the Engine
The figure at the right shows the main parts of a simple fourstroke cycle engine. These are: the Intake Valve (IV), the
Exhaust Valve (EV), the Piston (P), the Piston Rings (PR),
the Combustion Chamber (CC), the Connection Rod (CR),
the Crank Shaft (CS) and the Spark Plug (SP).
(3) The Cycle - Introduction
A 4-stroke cycle engine carried out four piston strokes during one combustion cycle - Intake Stroke,
Compression Stroke, Power Stroke and Exhaust Stroke. During the intake stroke the intake valve is opened
and the piston moves toward the crank shaft. The movement of the piston creates a negative pressure in the
combustion chamber. The air/fuel mixture is sucked into the chamber. If the Bottom Dead Center (BDC) is
reached the intake valve is closed and the piston moves upwards (compression stroke). The air/fuel mixture is
compressed. A short period before the Top Dead Center (TDC) is reached the spark plug ignites the air/fuel
mixture. Temperature and the pressure into the combustion chamber increased rapidly. The high pressure drives
the piston downward (power stroke). At the end of the power stroke - at the BDC - the exhaust valve is opened.
The piston is moved upward and the gases in the combustion chamber will be pushed through the exhaust valve
(exhaust stroke).
(4)The Cycle - The Four Strokes
If you have the ability to measure the pressure in the combustion chamber and its volume during all four strokes
you get a diagram like (p-V diagram). It represents the state of the gas in the chamber.
The single point shown in the p-V diagram represents the
common state. The color of the combustion chamber
indicates the temperature (red: hot; blue: cold).
Bild 6: p-V diagram
INTAKE STROKE
The piston moves down the
cylinder and the pressure
will drop (negative
pressure). The intake valve
is opend. Because of the
low pressure the air/fuel
mixtures is sucked into the
cylinder.
Fig 7: INTAKE STROKE
COMPRESSION STROKE
At Bottom Dead Center
(BDC) the cylinder is at its
maximum volume and the
intake valve is closed. Now
the piston moves backward
the Top Dead Center
(TDC) and compresses the
air/fuel mixtures. The
pressure is increased and
the volume is decreased.
The necessary work for the
compression increases the
internal energy of the
mixtures - the temperature COMPRESSION STROKE
is increased. Because of the
fast compression only a small part of the energy is transfered to the environment.
Near the end of the
compression stroke, the
ignition starts the
combustion and the
mixture burnes very
rapidly. The expanding gas
creates a high preasures
against the top of the
piston. The resulting force
drives the piston downward
in the cylinder.
Fig 9: IGNITION
POWER STROKE
The force drives the piston
downward to crank shaft
(the valves are closed). The
volume is increased and the
pressure is decreased. No
more energy is added and
because of this the internal
energy of the gas is
decreased as so as the
temperature.
Fig 10: POWER STROKE
EXHAUST STROKE
At BDC the exhaust valve is
opened and the piston
moves up the cylinder. The
pressure drops near the
pressure outside the cylinder
because of the opened
exhaust valve. Exhaust gas
leaves the cylinder. The
volume is decreased.
http://techni.tachemie.uni-leipzig.de/otto/index_e.html
http://www.animatedengines.com/otto.html
Standard four-stroke air cycle
Compression Stroke. Adiabatic compression of gas fuel mixture in the cylinder
Ignition of gas fuel mixture. Takes place rapidly at top of the compression stroke while the volume is
essentially constant.
Expansion Stroke. Adiabatic, isentropic expansion of gases in the cylinder after fuel mixture is ignited.
This the part of the cycle that does positive work.
Exhaust of the spent gases and the intake of a new fuel mixture into the cylinder. The volume is the
same at beginning and ending of the exhaust and intake stroke.
Diesel Cycle - Diesel Engine
Compression Stroke. Adiabatic compression of gas and diesel fuel mixture in the cylinder.
Ignition of gas fuel mixture. Fuel is ignited by high temperature due a large compression. Burning
takes places while the pressure is essentially constant.
Expansion Stroke. Adiabatic, isentropic expansion of gases. This is the part of the cycle that does
positive work.
Exhaust of the spent gases and the intake of a new fuel mixture into the cylinder. The volume is the
same at beginning and ending of the exhaust and intake stroke.
http://www.animatedengines.com/diesel.html
Stirling Cycle - Example of a Carnot Engine
Isothermal-compression compression of working gas.
Heat is absorbed from an energy-storage device at a constant volume.
Isothermal-expansion of gases. This the part of the cycle that does positive work.
Heat is transferred from the working gas to an energy storage device at a constant pressure. The heat
stored in this part of the cycle is the same as the heat absorbed in 2 -> 3 part of the cycle.
http://www.animatedengines.com/ltdstirling.html
http://www.animatedengines.com/twostroke.html
http://www.animatedengines.com/wankel.html
Brayton Cycle - Gas-Turbine Engine
Isentropic-compression of the intake air into the combustion section of the engine.
Constant-pressure combustion of fuel injected into combustion chamber.
Isentropic-expansion combustion gases through the turbine section. This the part of the cycle that does
positive work.
Constant-pressure heat exhausting into the air.
http://energy.sdsu.edu/testhome/vtAnimations/animations/chapter08/A-gasTurbines/actualBraytonCycle.html
PV DIAGRAM FOR POLYTROPIC PROCESSES
OF AN IDEAL GAS
PVn = Constant
THE STIRLING ENGINE AND THE SECOND LAW OF THERMODYNAMICS
The Stirling Engine
Carnot began working on engines in hopes of improving the efficiency of the steam engine. Although his
concept of the ideal heat engine was a rare achievement that laid the groundwork for the first and second laws
of thermodynamics, the internal combustion engine used in the cars we drive is far from ideal in its efficiency.
The Stirling Engine Cycle proposed by the Reverend Robert Stirling of the Church of Scotland in 316 is
considerably closer in its design to the Carnot engine. In a Stirling engine a piston linked to a displacement
system shuffles gas back and forth between hot and cold reservoirs. The expansion and contraction of the gas
as it is heated and cooled drives the engine. In the Stirling engine waste heat is recycled in an ingenious way
that improves efficiency. A modern working model of the early Stirling engine enables us to explore the
quantitative behavior of a Carnot-like engine.
For observing the operation of the Stirling engine you will need:
EITHER:
• 1 miniature Stirling engine
• 6 oz. of denatured alcohol
• 2 ice cubes
AND/OR
• 1 Visible Stirling Engine
• 1 ceramic coffee mug
• hot water
Activity: Stirling Engine Efficiency
a. Follow the instructions that come with the Stirling engines and operate it. Examine the engine and try to explain
the elements of a basic cycle of the engine. Where is the hot reservoir? The cold reservoir?
b. Assuming that the equation describing the efficiency of the engine is approximately the same as that for the
Carnot engine so that
Carnot = 1 
TC
TH
What do you predict will happen to the engine if an ice cube is placed in contact with the cold reservoir?
c. Place the ice cube in contact with the cold reservoir and describe what actually happens to the operation of the
engine.
The Second Law of Thermodynamics
Time does not permit you to gain direct experience with the measurement of the efficiencies of the many real
heat engines that have been devised. However, experience with hundreds of types of heat engines led scientists
in the nineteenth century to conclude the following:
It is impossible to construct a heat engine that absorbs heat energy from a high temperature reservoir and
delivers work with 100% efficiency.
Furthermore, theoretical considerations drawn from experience with heat engines have led to the
additional conclusion that
It is impossible to construct a heat engine that is more efficient than a Carnot engine operating between the same
two temperatures.
These are two of many different statements of the Second Law of Thermodynamics. There are several
other statements that require familiarity with the concept of entropy to understand. We hope you will learn
more about this concept in the future.
EFFICIENCIES OF HEAT ENGINES & HEAT PUMPS
Examples of Heat Pump & Refrigerator
First Law:
Cycle:
Compressor - Work done by the compressor goes into compressing a low-pressure gas like Freon into a highpressure superheated gas. The mechanical work is converted into creating a high-temperature, high-pressure
vapor.
Condenser - As the high-pressure superheated vapor goes through the condenser it loses heat, goes through a
phase transition to become a liquid, still at high pressure. When used as a heat pump, the heat extracted can be
used to heat a house. For a refrigerator, this heat is dumped to the environment. It is the warm air you feel on
your feet when standing in front of the refrigerator in the middle of the night. The condenser is prominent on the
backs of most refrigerators as a vertical grating of coils.
Expansion Throttling Valve - When the high-pressure liquid passes through the expansion valve it expands
rapidly into a region of lower pressure and greatly reduces both its temperature and pressure - still remaining a
liquid.
Evaporator - As the cold, low-pressure liquid goes through the evaporator it absorbs heat, and changes phase
back to a low-pressure gas. When used as a refrigerator, the absorbed heat can be used to cool down in inside of
a refrigerating compartment. When used as heat pump, this heat is drawn from some external heat reservoir
such as the air or the ground, some of which is later dumped in the heated area.
Heat Pump

Qcond is used to heat something like a
house. Then Qhouse = Qcond is heat
delivered to the house.

Qevap comes from outside source such as
the air. Then Qoutside = Qevap is the heat
extracted from the outside.
Refrigerator

Qcond is the heat dumped outside the cooling area.
For a house refrigerator, Qroom = Qcond is heat
dumped into the room where the refrigerator is
located.

Qevap is heat taken from inside the area to be cooled
- the inside of the refrigerator. Then Qevap = Qcooler is

You pay for the energy to run the
compressor Wcomp but not for Qoutside.
heat extracted.

You pay for the energy to run the compressor
Wcomp. The energy dumped into the room Wcomp +
Qcond goes into heating the air.
Working Fluid: Often the working fluid is a substance like Freon that will change phase from liquid to vapor at
the temperatures associated with environment and cooling compartment. By adjusting the pressure in the
condenser one can make Freon condense at room temperature. In a similar manner one can adjust the pressure
in the evaporator so that Freon boils at the low temperature of the cooling compartment.
EFFECTIVENESS - SECOND LAW EFFICIENCIES
Measure of Performance
Possible Choices of the Final State of the Process Used to Determine the
Maximum Output:
* Isoentropic Process (Adiabatic Reversible). The final state of the
system is usually different from that of the actual final state.
* Reversible Process with the same initial and final states as the actual
operating system.
* Reversible Process with a specifically desired final state such as the
dead state of the environment.
Heat Engines:
Example: A heat engine takes in 480 kJ of heat energy at 560 K from a large heat reservoir and exhausts 190 kJ
of heat into the environment at 290 K. Determine the second law effectiveness of this heat engine.
Although the actual thermal efficiency of this heat engine is low at 39.6%, its effectiveness is actually quite
high at 82.1%.
A Reversible Heat Engine that is Not a Carnot Engine
Two identical blocks with the same mass and heat capacity, but at different initial temperatures are used as a
heat source and a heat sink for a reversible heat engine. Block A is initially at 100 oC and block B is initially at
0 oC. The heat engine operates until the temperature of the two blocks becomes identical, the final temperature
Tf. Assume the two blocks are insulated so that there is no heat lost to the environment.
(A) Calculate the final temperature of the two blocks Tf. Compare this with the final temperature of the two
blocks if they were allowed to come to equilibrium through heat transfer alone with no work extracted.
(B) Calculate symbolically the total work accomplished by this reversible heat engine.
(C) Determine the thermal efficiency of this reversible heat engine. Compare it with the Carnot efficiency of a
reversible heat engine operating between the same two temperatures to show that the actual efficiency of
this reversible engine is less than that of the Carnot efficiency.
Sketch and Process:
A heat engine uses the energy stored in a hot block to accomplish reversible work and dumps its exhaust heat
into another colder block.
Relevant Physics:
Energy must be conserved, and since the system is reversible the total change in entropy (of the system
consisting of the two blocks and the engine) must be equal to zero
Wrev = QA - QB Ssys = 0.
(A) Find Tf by setting Ssys = 0.
When the entropy is zero
If the blocks just transfer heat until they reach a final equilibrium temperature T'f, then the conservation of
energy can be used to find T'f
The difference between the temperatures is due to the work accomplished.
(B) Find Wrev symbolically using the conservation of energy and knowing the final temperature.
(C) Find erev and compare with eCarnot
For Carnot engine operating between the same two temperatures
Thus we see that even though this engine is reversible, having the maximum possible efficiency using two
blocks as heat source and heat sink, its thermal efficiency is still less than that of a Carnot heat engine. The
property that makes the Carnot cycle produce the highest possible efficiency is that the heat source and
heat sink are infinite reservoirs whose temperatures do not change when heat is extracted or rejected.
ENTROPY CHANGE IN A PHASE TRANSITION
L = Heat of Vaporization or Fusion during the Phase Transition
Tp = Phase Transition Temperature (Melting or Boiling Point)
Entropy
(J/K)
Entropy
(Bits)
Entropy/kg
(J/kg K)
Entropy/mole
(J/mole K)
Entropy/atom
(J/atom K)
Entropy/atom
(Bits/atom)
Refrigerator Freezing Ice Problem
The cooling compartment of a refrigerator operates at a low temperature of -10.0 oC, and exhausts heat
into the air in the room at 21.0 oC. The motor of the refrigerator produces 3/4 horsepower of useful work
to operate the refrigeration cycle. If this refrigerator operates at 43.0% of its maximum coefficient of
performance:
(A) What is the maximum the maximum possible coefficient of performance of this
refrigerator?
(B) How much heat does this refrigerator exhaust into the environment each second if it
operates at 43.0% of its maximum coefficient of performance?
(C) How long would it take to cool and freeze 4.20 kg of water at 18.0 oC to ice at 0 oC when
placed in the refrigerator?
(D) What is the change in entropy of the ice, the air in the room, and the minimum change in
the entropy of the universe during the freezing process?
Sketch and Process:
More Symboliclly
Water is placed in a refrigerator with a calculable COP and is then frozen.
Relevant Physics: The first thing to recognize is that the conservation of energy for the refrigeration cycle
(or any cycle like a heat pump) can be turned into a rate equation by taking the derivative with respect to time.
For a more detailed of how a real refrigerator could be constructed see refrigerator example.
Or more simply,
For a refrigerator,
Here Ql is the heat that has to be removed from the water at 18.0 oC to freeze it.
The maximum COP is given by
Any refrigerator will violate the Second Law of Thermodynamic if
The change in entropy of the universe due to the freezing process of the water is equal to the sum of the entropy
changes of the subcomponents that make up the system.
We will assume that under the best conditions Smotor = 0 even if the system is not operating at maximum COP.
The maximum COP would only occur if SU = 0 and not just if Smotor = 0.
The entropy change of the water to ice involves two processes – cooling from 18.0 oC to 0 and freezing.
To find the change in entropy of the air we will have to attack this part from a different approach. Since we
assume the air in the room is large enough that its temperature does not change, the air acts as an isothermal
reservoir. For an isothermal process,
Using the conservation of energy we can find Qair sine it is also equal to Ql. If SU does not come out to be
positive or at least zero then it should indicate that you have made some error in your calculations since this
would violate the second law of thermodynamics.
(A) Determine is the maximum possible COP of the freezer.
Here we assume that the both the temperature of the room air outside the freezer and the air in the freezing
compartment do not change temperatures so that they both act like heat reservoirs at constant tempeatures. Then
the maximum COP can be found using the temperatures given.
We didn't covert the temperatures in bottom part of the equation in to degrees K just to remind you that T is
the same in either set of units. However, it is absolutely imperative that the top temperature has to be in degrees
K.
(B) Determine the heat exhausted into the room each second,
.
First let us calculate the actual COP since it is related to the heat removed and the work done.
Using the conservation of energy we can find the heat exhausted into the room.
At 100% efficiency this would go up to 2.66 kilojoule per second.
(C) Determine the time to freeze 4.20 kg of water at 18.0 oC placed in the freezing compartment.
The total heat that would have to be extracted from the water to freeze it can be found using the heat capacity of
water and its latent heat of fusion.
In part B we found that the rate at which heat is be extracted from the freezer was 2041.44 J/s. Thus,
(D) Find the change in entropy of the room air, the water that freezes, and the universe.
The working fluid of any cyclic process will return to the same initial state. Since entropy is a state variable, the
change in entropy of the working fluid in the refrigerator will always be zero. There will be an entropy increase
due to frictional losses in the motor driving the system. Sine there is no way we know what they will be – they
are dependent on how the system constructed – we will assume an ideal motor that has no change in entropy
change.
The total change in entropy of the universe will be equal to the entropy change of each component of the
system.
We should also add the entropy change in the power source that supplies the energy to the motor but we will
also ignore it know that we are calculating the minimum entropy change.
The entropy change of the water to ice involves two processes – cooling from 18.0 oC to 0 and freezing
The minus sign for the freezing term is necessary since the water losses heat as it freezes. You have to insert the
correct sign for the phase transitions. Unlike the first term which will be negative if Tf < Ti there is no math
equation that will give you the correct sign for phase transitions.
To find the change in entropy of the air we will have to attack this part from a different approach. Since we
assume the air in the room is large enough that its temperature does not change, the air acts as an isothermal
reservoir. For an isothermal process,
In part B we found the rate at which heat is exhausted into the air,
time it took to freeze the water, so we can find the heat dumped into the air.
Using this we can now find the entropy change of the air.
. In part C found out the
Finally the change in entropy of the universe will be the sum of all the entropy changes,

One important observation that this example shows is that if the entropy of some subsystem decreases
then there will be a greater increase in entropy in some other parts of the system so that the total change
in entropy will always be greater than zero or at the very least zero.