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,

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• or, heat may be converted to work,• but, final energy = initial energyThe first law of thermodynamics is simply an

expression of the conservation of energy principle,

and it asserts that energy is a thermodynamic

property.

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 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 = 0.4

1st Law of TD

Some conventions:

For the gases perspective:

• heat added is positive, heat removed is

negative.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)

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(a)

(b)

• 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 Heat Capacity, C is equal to the ratio of the heat

absorbed or withdrawn from the system to the

resultant change in temperature.

T

qC

∆=

• 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.13

• 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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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.

(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,

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isochoric process, one that occurs at constant volume.

under the graph is zero,

indicating that no work is done.

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

∆

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– 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.

The area under a pressure-

volume 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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process from X to Y.

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,

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greater amount of 7.1 ×××× 10 m . 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°°°°).

(b)

(a)

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(b)

ISOTHERMAL EXPANSION OR COMPRESSION

P = nRT/V

W = P ∆∆∆∆V = P(Vf – Vi)

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ExampleIsothermal 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.

.

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(a)

.

(b)

(c)

ADIABATIC EXPANSION OR COMPRESSION

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[Ti = PiVi/(nR)]

[Tf = PfVf/(nR)].

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Type of Thermal Process Work Done

First Law of Thermodynamics(∆∆∆∆U = Q – W)

Isobaric (constant pressure)

W = P(Vf – Vi)

Isochoric (constant volume)

W = 0 J

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volume)

Isothermal(constant temperature) (for an ideal gas)

Adiabatic (no heatflow)

(for a monatomic ideal gas)

1st 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 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 + ∆UQ = 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

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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

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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.

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

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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.

These three symbols

refer to magnitudes only,

without reference to

algebraic signs.

Therefore, when these

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Therefore, when these

symbols appear in an

equation, they do not

have negative values

assigned to them.

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

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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.

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

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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.

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

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heat engine. Find the maximum possible efficiency for such

an engine.

TH = 298.2 K and TC = 280.2 K

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.

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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%.

Refrigerators, Air Conditioners, and Heat Pumps

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In the refrigeration process, work W is used to remove heat QC from the cold

reservoir and deposit heat QH into the hot reservoir.

In a refrigerator,

the interior of

the unit is the

cold reservoir,

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cold reservoir,

while the

warmer exterior

is the hot

reservoir.

A window air

conditioner removes

heat from a room,

which is the cold

reservoir, and

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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

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reservoir is the inside of

the house.

This conventional

electric heating

system is delivering

1000 J of heat to the

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system is delivering

1000 J of heat to the

living room.

QH = W + QC and QC/QH = TC/TH

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)?

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(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?

51(c)

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.

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Reversible processes do not alter the total entropy of

the universe.

Although the

relation S =

(Q/T)R applies to

reversible processes,

it can be used as

part of an indirect

∆

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part of an indirect

procedure to find

the entropy change

for an irreversible

process.

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

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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.

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 ∆∆

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increases when an irreversible process occurs

( Suniverse > 0 J/K).

∆

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.

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

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It is not possible to lower the temperature of any

system to absolute zero (T = 0 K) in a finite

number of steps.

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

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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

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uses the same cold

reservoir at a temperature

of 275 K for its exhaust

heat. Each engine receives

1450 J of input heat.

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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