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Thermal Physics
Topic 10.2 Processes
The First Law of ThermodynamicsIs a statement of the Law of Conservation of
Energy in which the equivalence of work and thermal energy flow is taken into account.
It can be stated as
the heat added to a closed system equals the change in the internalenergy of the system plus the work done by the system.
That is,
Q = U + W = U + pV
or
U = Q - W
Explanation
Where `+ Q' is the thermal energy added to the system and `+ W' is the work done by the system.
If thermal energy leaves the system, then Q is negative. If work is done on the system, then W is negative.
For an isolated system, then W = Q = 0 and U = 0.
Example
If 22 J of work is done on a system and 3.4 x 102 J of heat is added, what is the change in internal energy of the system?
Solution
Using the formula, Q = U + W
We have that 340 J = U + (-22) J
So that U = 340 J + 22 J
= 362 J
That is, the change in internal energy of the system is 3.6 x 102 J.
Isobaric
A graph of pressure as a function of volume change when the pressure is kept constant is shown below
p -V diagram
p
V
p
VV1 V2
Area = work done = p (V1 - V2)
Isobaric process
Work DoneSuch a process is said to be isobaric.
Note that the work done by the gas is equal to the area under the curve
An isobaric transformation requires a volume change at constant pressure
For this to occur, the temperature needs to change to keep the pressure constant
Example
6.0 dm3 of an ideal gas is at a pressure of 202.6 kPa. It is heated so that it expands at constant pressure until its volume is 12 dm3. Find the work done by the gas.
Solution Using the formula W = pV we have that W = 202.6 kPa x(12 - 6.0) dm3
= 202.6 x 103 Pa x (12 - 6.0) x 10-3m3
= 1.216 x 103 J That is, the work done by the gas in the
expansion is 1.2 x 103 J.
Isochoric / Isovolumetric
A graph of pressure as a function of volume change when the volume is kept constant is shown below
p
V
Isochoric / Isovolumetric
Work DoneSuch a process is said to be isochoric.
When the volume is kept fixed, the curve of the transformation is said to be an isochore.Note that the work done by the gas is equal to zero as V = 0.
There is zero area under the curve on a p-V diagram.
p - V diagram
p
V
Isochoric / Isovolumetric
However
The temperature and pressure can both change and so such a transformation will be accompanied by a thermal energy change.
Example
A thermal system containing a gas is taken around the cycle as in the next slide
(Cyclic processes such as this one have important applications in heat engines that convert internal energy into useful mechanical energy).
p
V
6
2
4 10
A
B C
D
Starting at point A on the diagram, describe the cycle, and, calculate the work done by the system in the completion of one cycle.
SolutionFrom A to B, the volume is kept constant
(isovolumetric) as the pressure increases. This can be achieved by heating the gas. Since V = 0, then no work is done by the gas, W = 0.
From B to C, the gas expands (volume increases) (isobaric expansion) while the pressure is kept constant. The amount of work done by the gas is given by the area under the 6 kPa isobar.
Now, we have that W = p V
So that W = 6 kPa. (10 - 4) m3
= 36,000J
From C to D, the gas is cooled to keep the volume constant as the pressure is decreased (isovolumetric)
Again V = 0 and no work is done by the gas, W = 0
From D to A, the gas is compressed (volume decreases) (isobaric compression) while the pressure is kept constant. The amount of work done on the gas is given by the area under the 2 kPa isobar.
Again, using the fact that W = p V we have that W = 2 kPa. (4 -10) m3
= - 12,000 J
That is, the net work done by the gas is therefore 36,000 J – 12,000 J =24,000 J.
Because the cycle is traced in a clockwise direction, the net work done on the surroundings is positive. If the cycle was traced in a counter-clockwise direction, the work done would be negative
Isothermal Processes
A thermodynamic process in which the pressure and the volume are varied while the temperature is kept constant is called an isothermal process
In other words, when an ideal gas expands or is compressed at constant temperature, then the gas is said to undergo an isothermal expansion or compression.
The figure below shows three isotherms for an ideal gas atdifferent temperatures where
Tl < T2 < T3p
V
T1 T2 T3
Isothermal process
Isothermal Processes
The curve of an isothermal process represents a Boyle's Law relation
pV =constant = nRT • the moles of gas n, • the molar gas constant R, • the absolute temperature T are constant
How?
In order to keep the temperature constant during an isothermal process • the gas is assumed to be held in a thin container
with a high thermal conductivitythat is in contact with a heat reservoir - an ideal body of large mass whosetemperature remains constant when heat is exchanged with it.e.g.. a constant-temperature water bath.
And
• the expansion or compression should be done slowly so that no eddies are produced to create hot spots that would disrupt the energy equilibrium of the gas.
An Ideal GasConsider an ideal gas enclosed in a thin
conducting vessel that is in contact with a heat reservoir, and is fitted with a light, frictionless, movable piston
If an amount of heat Q is added to the system which is at point A of an isotherm, then the system will move to another point on the graph, B.
The heat taken in will cause the gas to expand isothermally and willbe equivalent to the mechanical work done by the gas
Because the temperature is constant, there is no change in internal energy of the gas
That is,
T = 0
and U = 0
Q = W
The work done by gas is equal to the Area under the curve between A and B
Isothermally
If the gas expands isothermally from A to B and then returns from B to A following exactly the same path during compression
Then the isothermal change is said to be reversible.
Adiabatic Processes
An adiabatic expansion or contraction is one in which no heat Q is allowed to flow into or out of the system
For the entire adiabatic process, Q = 0
How
To ensure that no heat enters or leaves the system during an adiabatic process it is important to:• make sure that the system is extremely
well insulated • carry out the process rapidly so that the
heat does not have the time to leave the system
p
V
T1 T2
Adiabatic compression
Adiabatic Expansion
Example
The compression stroke of an automobile engine is essentially an adiabatic compression of the air-fuel mixture
The compression occurs too rapidly for appreciable heat transfer to take place
What happensIn an adiabatic compression the work
done on the gas will lead to an increase in the internal energy resulting in an increase in temperature.
U=Q - W
but Q=O
therefore U= - W
In an adiabatic expansion the work done by the gas will lead to a decrease in the internal energy
Resulting in a decrease in temperature.
Work done
p
V
Area = work done by the gas expanding
p
VArea = work done on the gas as it is compressed
p
VArea between curves = Net work done by the gas
ExampleFor the compression stroke of an experimental diesel
engine, the air is rapidly decreased in volume by a factor of 15, the compression ratio.
The work done on the air-fuel mixture for this compression is measured to be 550 J
(a) What type of thermodynamic process is likely to have occurred?
(b) What is the change in internal energy of the air-fuel mixture?
(c) Is the temperature likely to increase or decrease?
Solution(a) Because the compression occurs
rapidly appreciable heat transfer does not take place, and the process can be considered to be adiabatic, Q = 0
(b) U = Q - W
= 0 - (-550) J
Therefore, the change in internal energy is 550J
(c) The temperature rise will be very large resulting in the spontaneous ignition of the air-fuel mix
Thermodynamic Cycle and Engines
A thermodynamic engine is a device that transforms thermal energy to mechanical energy
Cars, Steam trains, jets and rockets have engines that transform fuel energy (chemical energy) into kinetic energy of their motion
Efficiency
In all engines the conversion is accompanied by the emission of exhaust gases that waste some of the thermal energy
These engines are not very efficient as only part of the thermal energy is converted to mechanical energy
The EngineBy Vija and Bombo
Internal CombustionBy Adil and Suhayb
Schematic of a Heat Engine By Ham and Heisei
Heat PumpsBy Alex and Said
Carnot´s TheoremBy Martin and Mitul
Presenters
Schematic for a Refrigerator by Karim and Chilla
RefrigeratorsBy Aleem and Bhavik
The EngineBy Vija and Bombo
Has two crucial features:
It must work in cycles to be useful
the cyclic engine must have more than one heat reservoir
The Thermodynamic Cycle
Is the process in which the system is returned to the same state from which it started
That is, the initial and final states are the same in the cyclic process
Combustion Engines
The steam engine and the steam turbine are examples of external combustion engines
The fuel is burnt outside the engine and the thermal energy is transferred to the piston or a turbine chamber by means of steam
The Steam Engine
Water is heated in a boiler heat reservoir (high temperature) to produce steam
As the piston returns to its original position, the intake valve is closed and the exhaust valve opens
The Steam Engine cont...
The steam passes into an open intake valve where it expands causing a piston to move outwards
This allows the exhaust gas to be forced out into a condenser heat reservoir (low temperature)
The Steam Engine
Steam Turbines
Most electricity is produced using steam engines
The piston is replaced with a rotating turbine that contains blades
The rotating turbine converts mechanical energy to electrical energy via a generator
The steam is cooled in cooling towers
Internal CombustionBy Adil and Suhayb
The Intake Stroke
With the exhaust valve closed, a mixture of petrol vapour and air from the carburetor is sucked into the combustion chamber through the inlet valve as the piston moves down
The Compression Stroke
Both valves are closed and the piston moves up to squeeze the mixture of petrol vapour and air to about 1/8th its original volume
The Power Stroke
Both valves are closed the mixture is ignited by a spark from the spark plug
The mixture burns rapidly and the hot gases then expand against the piston
The Exhaust Stroke
1. The exhaust valve opens as the piston moves upwards
2. Exhaust gases are expelled
3. Then the cycle begins again
Hot reservoir at TH
Cold reservoir at TL
Engine
Schematic of a Heat Engine BBy Ham and Heisei
QL
W
QH
Fluids leave carrying energy
QL = QH - W
Fluids enter carrying energy
QH Engine does useful work, W
Efficiency
The thermal efficiency of a heat engine, , is defined as the ratio of work it does, W to the heat input QH
= W / QH
= (QH - QL) / QH
It is clear that the efficiency will be greater if QL can be made small
QuestionAn engine absorbs 230J of thermal
energy from a high temperature reservoir, does work and exhausts 140J to a cold temperature reservoir.
What is the efficiency?
How much work is done? Answer 39% and 90J
RefrigeratorsBy Aleem and Bhavik
A refrigerator is a device operating in a cycle that is designed to extract heat from its interior so as to achieve or maintain a lower temperature inside
The heat is exhausted to the surroundings normally at a higher temperature
How the Refrigerator Works
A motor driving a compressor pump provides the means by which a net amount of work can be done by the system for a cycle
Even though the cabinets are well insulated, heat from the surroundings leaks back inside
Heat PumpsBy Alex and Said1. Any device that can pump heat from a low
temperature reservoir to a high temperature reservoir is called a heat pump
2. Examples of heat pumps include refrigerator and reverse cycle air-conditioning devices used for space heating and cooling
How..
1. A volatile liquid called Freon is circulated in a closed system of pipes by the compressor pump
2. By the process of evaporative cooling, the vaporised Freon is used to remove the heat
3. The compressor maintains a high pressure difference across a throttling valve
How..Evaporation of the Freon occurs in
several loops called the evaporator pipes that are usually the coldest part of the fridge
As the liquid evaporates on the low
pressure, low temperature side, heat is added to the system
How..• In order to turn from a liquid to a gas the
Freon requires thermal energy equal to the latent heat of vaporization
• This energy is obtained from the fridge
How..• On the high pressure, high temperature
side of the throttling valve, thermal energy is removed from the system
• The vaporized Freon in the compressor pipes is compressed by the compressor pump
• It gives up its latent heat of vaporization to the air surrounding the compressor pipes
How..
• The heat fins act as a heat sink to radiate the thermal energy to the surroundings at a faster rate
• • The fins are painted black and have a
relatively large surface area for their size
A Typical Small Refrigerator
Schematic for a Refrigerator Karim and Chilla
Hot reservoir at TH
Cold reservoir at TL
QL
W
QH
Refrigerator
• The previous figure shows the energy transfers that occurs in a refrigerator cycle
• By doing work on the system, heat QL is added from the low temperature TL reservoir
• A greater amount of energy QH is exhausted to the high temperature TH
Reverse Cycle Heat Pump
room
evaporatorcondenser
TL TH
WINTER
room
condenserTH TL
evaporator
SUMMER
Winter• The evaporator heat exchanger is
outside the room • It exhausts its heat to the inside air
Summer
• The evaporator heat exchanger is inside the room
• It exhausts its heat to the outside air
In Both Cases
• Thermal energy is pumped from a low-temperature reservoir to a high-temperature reservoir
• The efficiency of refrigerators and heat pumps is defined in a different manner to heat engines
Coefficient of Performance, K
• For a refrigerator and a heat pump used for cooling
K = QL / W
• Ratio of heat absorbed from inside the refrigerator and the work supplied to the refrigerator in one cycle
Coefficient of Performance, K
• For a heat pump used for heating
• K = QH / W
• Ratio of the total positive heat exhausted to the inside and the work supplied to the heat pump in one cycle
Carnot´s Theorem By Martin and Mitul
• No engine working between two heat reservoirs can be more efficient than a reversible engine between those reservoirs
• He argued that if thermal energy does flow from a cold body to a hot body then work must be done
• Therefore no engine can be more efficient than an ideal reversible one
And...• All such engines have the same
efficiency • Therefore only a simple equation is
needed to calculate the efficiency • Consider an ideal perfectly insulated,
frictionless engine that can work backwards as well as forwards
• The p-V diagram would have the form...
The Carnot Engine
Explain
• The net work is the area enclosed by ABCDA
• Thermal energy is absorbed by the system at the high temperature TH and is expelled at the low temperature reservoir TL
and• Work is done by the system as it expands
along the top isotherm from A to B • And also along the adiabat from B to C • Work is then done on the system to compress
it along the bottom isotherm from C to D • And also along the left adiabat from D to A
Carnot’s Efficiency• Carnot was able to obtain an expression
for the efficiency in terms of temperature
• = 1 - TL / TH
• Therefore the efficiency of the Carnot cycle depends only on the absolute temperatures of the high and low temperature reservoirs
• The greater the temperature difference the greater the efficiency will be
• But engine efficiency is
• = 1 - QL / QH
• Therefore TC/TH = QC/QH
• Or QC/TC = QH/TH
• What would happen if the
temperature difference was not maintained?
Real Life
• In practice, friction and other mechanical losses and thermal losses reduce the overall efficiency
• Nearly 2/3 of the heat generated by modern power stations is released into the environment
Thermal PhysicsTopic 10.3 Second Law of Thermodynamics and Entropy
