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Basic Laws of Thermodynamics
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First Law of Thermodynamics
Conservation of Energy for Thermal
Systems
• One form of work may be converted into another,
• or, work may be converted to heat,
• or, heat may be converted to work,
• but, final energy = initial energy
The first law of thermodynamics is simply an
expression of the conservation of energy principle,
and it asserts that energy is a thermodynamic
property.
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Forms of internal energy
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THE FIRST LAW OF THERMODYNAMICS
• The internal energy of a system changes from an
initial value Ui to a final value of Uf due to heat Q and
work.
Q is positive when the system gains heat and negative
when it loses heat. W is positive when work is done
by the system and negative when work is done on the
system.
• For an isolated system there is no heat or work
transferred with the surroundings, thus, by definition
W = Q = 0 and therefore ΔU = 0.
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st
1
Law of TD
Some conventions:
For the gases perspective:
• heat added is positive, heat removed is
negative.
• Work done on the gas is positive, work done
by the gas is negative.
• Temperature increase means internal energy
change is positive.
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Positive and Negative Work
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The figure illustrates a system and its surroundings. In part a,
the system gains 1500 J of heat from its surroundings, and
2200 J of work is done by the system on the surroundings. In
part b, the system also gains 1500 J of heat, but 2200 J of
work is done on the system by the surroundings. In each case,
determine the change in the internal energy of the system.
(a)
(b)
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• For a differential change, Equation 1 becomes:
• For a cyclic process, A→B→A, when the system
returns to state A, it has the same U, thus:
• In thermodynamics, the term heat simply
means heat transfer.
• A process during which there is no heat transfer
is called an adiabatic process.
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Heat Capacity
• Before discussing isothermal or adiabatic
processes, a new term is needed to make the
calculations easier.
• Heat Capacity, C is equal to the ratio of the heat
absorbed or withdrawn from the system to the
resultant change in temperature.
q
C=
∆T
• Heat Capacity:
• The heat capacity is the amount of heat required
to raise the temperature of a mass of a system by
1°C. It is denoted by C.
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Joule Equivalent of Heat
• James Joule showed that mechanical energy
could be converted to heat and arrived at the
conclusion that heat was another form of
energy.
• He showed that 1 calorie of heat was
equivalent to 4.184 J of work.
1 cal = 4.184 J
Called: Mechanical equivalent of heat
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Energy
• Mechanical Energy: KE, PE, E
• Work is done by energy transfer.
• Heat is another form of energy.
Need to expand the conservation of energy
principle to accommodate thermal systems.
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Modes of Heat transfer
– Heat can be transferred in three different ways:
conduction, convection, and radiation.
• All modes of heat transfer require the existence
of a temperature difference, and all modes of
heat transfer are from the high-temperature
medium to a lower-temperature one.
• Work
– Work, like heat, is an energy interaction between a
system and its surroundings. As mentioned earlier,
energy can cross the boundary of a closed system in
the form of heat or work. Therefore, if the energy
crossing the boundary of a closed system is not heat,
it must be work.
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• There are two ways a process can be adiabatic:
– Either the system is well insulated so that only a
negligible amount of heat can pass through the
boundary, or
– both the system and the surroundings are at the same
temperature and therefore there is no driving force
(temperature difference) for heat transfer.
• An adiabatic process should not be confused with
an isothermal process.
– Even though there is no heat transfer during an
adiabatic process, the energy content and thus the
temperature of a system can still be changed by other
means such as work.
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Thermal Processes
quasi-static means that it occurs slowly enough
that a uniform pressure and temperature exist
throughout all regions of the system at all times.
An isobaric process is one that occurs at constant
pressure.
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For an isobaric process, a pressure-versus-volume plot is a
horizontal straight line, and the work done [W = P(V f – V i)] is
the colored rectangular area under the graph.
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(a) The substance in the chamber
is being heated isochorically
because the rigid chamber
keeps the volume constant.
(b) The pressure-volume plot for
an isochoric process is a
vertical straight line. The area
under the graph is zero,
indicating that no work is done.
isochoric process, one that occurs at constant volume.
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isothermal process, one that takes place at constant temperature. (when the
system is an ideal gas.)
There is adiabatic process, one that occurs without the transfer of heat . Since
there is no heat transfer, Q equals zero, and the first law indicates that ∆ U = Q
– W = –W. Thus, when work is done by a system adiabatically, W is positive and
the internal energy of the system decreases by exactly the amount of the work
done. When work is done on a system adiabatically, W is negative and the
internal energy increases correspondingly.
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The area under a pressurevolume graph is the work for
any kind of process.
The colored area gives the
work done by the gas for the
process from X to Y.
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Thermal Processes Using an Ideal Gas
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Example: Isobaric Expansion of Water
One gram of water is placed in the cylinder and the pressure is
maintained at 2.0 × 105 Pa. The temperature of the water is
raised by 31 C°
°. In one case, the water is in the liquid phase
and expands by the small amount of 1.0 × 10–8 m3. In another
case, the water is in the gas phase and expands by the much
greater amount of 7.1 × 10–5 m3. For the water in each case,
find (a) the work done and (b) the change in the internal
energy.
c = 4186 J/(kg·C°
°)
cP = 2020 J/(kg·C°
°).
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(a)
(b)
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ISOTHERMAL EXPANSION OR COMPRESSION
P = nRT/V
W = P ∆V = P(Vf – Vi)
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Example
Isothermal Expansion of an Ideal Gas
Two moles of the monatomic gas argon expand isothermally
at 298 K, from an initial volume of Vi = 0.025 m3 to a final
volume of Vf = 0.050 m3. Assuming that argon is an ideal gas,
find (a) the work done by the gas, (b) the change in the
internal energy of the gas, and (c) the heat supplied to the gas.
.
(a)
(b)
(c)
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ADIABATIC EXPANSION OR COMPRESSION
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[Ti = PiVi/(nR)]
[Tf = PfVf/(nR)].
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Type of Thermal
Process
Work Done
Isobaric (constant
pressure)
W = P(Vf – Vi)
Isochoric (constant
volume)
W = 0 J
Isothermal
(constant
temperature)
Adiabatic (no heat
flow)
First Law of
Thermodynamics
(∆
∆U = Q – W)
(for an ideal gas)
(for a monatomic ideal gas)
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st
1
Law of TD
• Example: 25 L of gas is enclosed in a
cylinder/piston apparatus at 2 atm of pressure
and 300 K. If 100 kg of mass is placed on the
piston causing the gas to compress to 20 L at
constant pressure. This is done by allowing
heat to flow out of the gas. What is the work
done on the gas? What is the change in
internal energy of the gas? How much heat
flowed out of the gas?
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• Po = 202,600 Pa, Vo = 0.025 m3, To = 300 K, Pf =
202,600 Pa, Vf=0.020 m3, Tf=
n = PV/RT.
W = -P∆V
∆U = 3/2 nR∆T
Q = W + ∆U
W =-P∆V = -202,600 Pa (0.020 – 0.025)m3
=1013 J energy added to the gas.
∆U =3/2 nR∆T=1.5(2.03)(8.31)(-60)=-1518 J
Q = W + ∆U = 1013 – 1518 = -505 J heat out
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Specific Heat Capacities
where the capital letter C refers to the molar specific heat
capacity in units of J/(mol·K).
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The Second Law of Thermodynamics
THE SECOND LAW OF THERMODYNAMICS: THE HEAT
FLOW STATEMENT
Heat flows spontaneously from a substance at a
higher temperature to a substance at a lower
temperature and does not flow spontaneously in the
reverse direction.
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Heat Engines
A heat engine is any device that uses heat to perform
work. It has three essential features:
1.
Heat is supplied to the engine at a relatively high
input temperature from a place called the hot reservoir.
2.
Part of the input heat is used to perform work by
the working substance of the engine, which is the
material within the engine that actually does the work
(e.g., the gasoline-air mixture in an automobile engine).
3.
The remainder of the input heat is rejected to a
place called the cold reservoir, which has a temperature
lower than the input temperature.
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These three symbols
refer to magnitudes only,
without reference to
algebraic signs.
Therefore, when these
symbols appear in an
equation, they do not
have negative values
assigned to them.
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Efficiencies are often quoted as percentages obtained by
multiplying the ratio W/QH by a factor of 100.
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Example: An Automobile Engine
An automobile engine has an efficiency of 22.0% and
produces 2510 J of work. How much heat is rejected by the
engine?
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Carnot's Principle and the Carnot Engine
A reversible process is one in which both the system and
its environment can be returned to exactly the states
they were in before the process occurred.
CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF
THE SECOND LAW OF THERMODYNAMICS
No irreversible engine operating between two reservoirs
at constant temperatures can have a greater efficiency
than a reversible engine operating between the same
temperatures. Furthermore, all reversible engines
operating between the same temperatures have the same
efficiency.
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A Carnot engine is a
reversible engine in
which all input heat QH
originates from a hot
reservoir at a single
temperature TH, and all
rejected heat QC goes
into a cold reservoir at a
single temperature TC.
The work done by the
engine is W.
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where the temperatures TC and TH must be expressed
in Kelvins .
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Example:
A Tropical Ocean as a Heat Engine
Water near the surface of a tropical ocean has a
temperature of 298.2 K (25.0 °C), whereas water 700 m
beneath the surface has a temperature of 280.2 K (7.0 °C).
It has been proposed that the warm water be used as the
hot reservoir and the cool water as the cold reservoir of a
heat engine. Find the maximum possible efficiency for such
an engine.
TH = 298.2 K and TC = 280.2 K
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Example: Limits on the Efficiency of a
Heat Engine
Consider a hypothetical engine that receives 1000 J of heat as
input from a hot reservoir and delivers 1000 J of work,
rejecting no heat to a cold reservoir whose temperature is
above 0 K. Decide whether this engine violates the first or the
second law of thermodynamics, or both.
It is the second law, not the first law, that limits
the efficiencies of heat engines to values less than
100%.
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Refrigerators, Air Conditioners, and Heat
Pumps
In the refrigeration process, work W is used to remove heat QC from the cold
reservoir and deposit heat QH into the hot reservoir.
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In a refrigerator,
the interior of
the unit is the
cold reservoir,
while the
warmer exterior
is the hot
reservoir.
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A window air
conditioner removes
heat from a room,
which is the cold
reservoir, and
deposits heat
outdoors, which is
the hot reservoir.
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In a heat pump the cold
reservoir is the wintry
outdoors, and the hot
reservoir is the inside of
the house.
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This conventional
electric heating
system is delivering
1000 J of heat to the
living room.
QH = W + QC and QC/QH = TC/TH
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Example: A Heat Pump
An ideal or Carnot heat pump is used to heat a house to a
temperature of TH = 294 K (21 °C). How much work must
be done by the pump to deliver QH = 3350 J of heat into
the house when the outdoor temperature TC is (a) 273 K
(0 °C) and (b) 252 K (–21 °C)?
(a)
(b)
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Check Your Understanding
Each drawing represents a hypothetical heat engine or a
hypothetical heat pump and shows the corresponding heats
and work. Only one is allowed in nature. Which is it?
(c)
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Entropy
In general, irreversible processes cause us to lose some, but
not necessarily all, of the ability to perform work. This partial
loss can be expressed in terms of a concept called entropy.
Reversible processes do not alter the total entropy of
the universe.
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Although the
relation ∆ S =
(Q/T)R applies to
reversible processes,
it can be used as
part of an indirect
procedure to find
the entropy change
for an irreversible
process.
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Example: The Entropy of the Universe
Increases
1200 J of heat flow
spontaneously through a
copper rod from a hot
reservoir at 650 K to a
cold reservoir at 350 K.
Determine the amount
by which this irreversible
process changes the
entropy of the universe,
assuming that no other
changes occur.
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Any irreversible process increases the entropy of the
universe.
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THE SECOND LAW OF THERMODYNAMICS STATED IN TERMS
OF ENTROPY
The total entropy of the universe does not change when a
reversible process occurs (∆ Suniverse = 0 J/K) and does
increases when an irreversible process occurs ∆
( Suniverse > 0 J/K).
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Example:
Energy Unavailable for Doing Work
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Suppose that 1200 J of heat is used as input for an engine under two
different conditions. In Figure part a the heat is supplied by a hot
reservoir whose temperature is 650 K. In part b of the drawing, the
heat flows irreversibly through a copper rod into a second reservoir
whose temperature is 350 K and then enters the engine. In either
case, a 150-K reservoir is used as the cold reservoir. For each case,
determine the maximum amount of work that can be obtained from
the 1200 J of heat.
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A block of ice is an example of an ordered system
relative to a puddle of water.
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Example: Order to Disorder
Find the change in entropy that results when a 2.3 kg block
of ice melts slowly (reversibly) at 273 K (0 °C).
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The Third Law of Thermodynamics
THE THIRD LAW OF THERMODYNAMICS
It is not possible to lower the temperature of any
system to absolute zero (T = 0 K) in a finite
number of steps.
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Concepts & Calculations Example: The
Sublimation of Zinc
The sublimation of zinc (mass per mole = 0.0654 kg/mol) occurs at
a temperature of 6.00 × 102 K, and the latent heat of sublimation
is 1.99 × 106 J/kg. The pressure remains constant during the
sublimation. Assume that the zinc vapor can be treated as a
monatomic ideal gas and that the volume of solid zinc is
negligible compared to the corresponding vapor. What is the
change in the internal energy of the zinc when 1.50 kg of zinc
sublimates?
Q = ∆U + W,
Q = mLs,
W = nRT
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Example: The Work–Energy Theorem
Each of two Carnot engines
uses the same cold
reservoir at a temperature
of 275 K for its exhaust
heat. Each engine receives
1450 J of input heat.
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The work from either of these engines is used to drive a pulley
arrangement that uses a rope to accelerate a 125-kg crate
from rest along a horizontal frictionless surface. With engine 1
the crate attains a speed of 2.00 m/s, while with engine 2 it
attains a speed of 3.00 m/s. Find the temperature of the hot
reservoir for each engine.
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