# Section 2 The First Law of Thermodynamics - Fort Thomas ...

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Objectives

Heat, Work, and Internal Energy

Thermodynamic Processes

Chapter

10

Section 1
Relationships
Between Heat and

Work

Section 1
Relationships
Between Heat and

Work

Chapter

10

Objectives

Recognize

that a system can absorb or release
energy as heat in order for work to be done on or by
the system and that work done on or by a system can
result in the transfer of energy as heat.

Compute
the amount of work done during a
thermodynamic process.

Distinguish

between isovolumetric, isothermal, and

Chapter

10

Heat, Work, and Internal Energy

Heat

and

work

are

energy transferred

to or from a
system. An object never has “heat” or “work” in it; it
has only internal energy.

A
system

is a set of particles or interacting
components considered to be a distinct physical
entity for the purpose of study.

The

environment

the combination of conditions and
influences outside a system that affect the behavior
of the system.

Section 1
Relationships
Between Heat and

Work

Chapter

10

Heat, Work, and Internal Energy,
continued

In thermodynamic systems, work is defined in terms
of

pressure
and
volume change.

Section 1
Relationships
Between Heat and

Work

This definition assumes that
P

is constant.

Chapter

10

Heat, Work, and Internal Energy,
continued

If the gas

expands,

as
shown in the figure,

V

is
positive, and

the work done
by the gas on the piston is

positive.

If the gas is

compressed,

V

is negative, and

the
work done by the gas on
the piston is

negative.

(In
other words, the piston
does work on the gas.)

Section 1
Relationships
Between Heat and

Work

Chapter

10

Heat, Work, and Internal Energy,
continued

When the gas volume remains

constant,

there is no
displacement and

no work

is done on or by the
system.

Although the pressure can change during a process,

work

is done only if the

volume

changes.

A situation in which pressure increases and volume
remains constant is comparable to one in which a
force does not displace a mass even as the force is
increased. Work is not done in either situation.

Section 1
Relationships
Between Heat and

Work

Chapter

10

Thermodynamic Processes

An

isovolumetric process

is a thermodynamic
process that takes place at constant volume so that
no work is done on or by the system.

An

isothermal process

is a thermodynamic process
that takes place at constant temperature.

An

is a thermodynamic process
during which no energy is transferred to or from the
system as heat.

Section 1
Relationships
Between Heat and

Work

Chapter

10

Energy Conservation

If

friction

is taken into account,

mechanical energy
is not conserved.

Consider the example of a roller coaster:

A steady decrease in the car’s total mechanical energy
occurs because of work being done against the friction
between the car’s axles and its bearings and between the
car’s wheels and the coaster track.

If the internal energy for the roller coaster (the system) and
the energy dissipated to the surrounding air (the
environment) are taken into account, then

the total energy
will be constant.

Section 2
The First Law of
Thermodynamics

Chapter

10

Energy Conservation

Section 2
The First Law of
Thermodynamics

Chapter

10

Energy Conservation,
continued

The principle of

energy conservation

that takes into
account a system’s internal energy as well as work
and heat is called the

first law of thermodynamics.

The first law of thermodynamics can be expressed
mathematically as follows:

U

=
Q

W

Change in system’s internal energy = energy
transferred to or from system as heat

energy
transferred to or from system as work

Section 2
The First Law of
Thermodynamics

Chapter

10

Signs of
Q

and
W

for a system

Section 2
The First Law of
Thermodynamics

Chapter

10

Sample Problem

The First Law of Thermodynamics

A total of 135 J of work is done on a gaseous
refrigerant as it undergoes compression. If the
internal energy of the gas increases by 114 J during
the process, what is the total amount of energy
transferred as heat? Has energy been added to or
removed from the refrigerant as heat?

Section 2
The First Law of
Thermodynamics

Chapter

10

Sample Problem,
continued

1. Define

Given:

W

=

135 J

U

= 114 J

Section 2
The First Law of
Thermodynamics

Tip:
Work is done
on the gas, so work
(W) has a negative
value. The internal
energy increases
during the process,
so the change in
internal energy
(

U) has a positive
value.

Diagram:

Unknown:

Q

= ?

Chapter

10

Sample Problem,
continued

2. Plan

Choose an equation or situation:

Apply the first law of thermodynamics using the values
for

U

and
W

in order to find the value for
Q
.

U

=
Q

W

Section 2
The First Law of
Thermodynamics

Rearrange the equation to isolate the unknown:

Q

=

U

+
W

Chapter

10

Sample Problem,
continued

3. Calculate

Substitute the values into the equation and solve:

Q = 114 J + (

135 J)

Q =

21 J

Section 2
The First Law of
Thermodynamics

Tip:

The sign for the value of Q is negative. This
indicates that energy is transferred as heat from
the refrigerant.

Chapter

10

Sample Problem,
continued

4. Evaluate

Although the internal energy of the refrigerant
increases under compression, more energy is
added as work than can be accounted for by the
increase in the internal energy. This energy is
removed from the gas as heat, as indicated by the
minus sign preceding the value for
Q.

Section 2
The First Law of
Thermodynamics

Chapter

10

Cyclic Processes

A

cyclic process

is a thermodynamic process in
which a system returns to the same conditions under
which it started.

Examples include

heat engines

and

refrigerators.

In a cyclic process, the final and initial values of
internal energy are the same, and the change in
internal energy is zero.

U
net
= 0 and
Q
net

=
W
ne
t

Section 2
The First Law of
Thermodynamics

Chapter

10

Cyclic Processes,
continued

A

heat engine

uses heat to do
mechanical work.

A heat engine is able to do

work

(b)

by transferring energy
from a

high
-
temperature
substance (the boiler) at

T
h

(a)

to a substance at a

lower
temperature (the air around the
engine) at

T
c

(c).

Section 2
The First Law of
Thermodynamics

The
internal
-
combustion engine

found in most
vehicles is an example of a heat engine.

Chapter

10

The Steps of a Gasoline Engine Cycle

Section 2
The First Law of
Thermodynamics

Chapter

10

The Steps of a Refrigeration Cycle

Section 2
The First Law of
Thermodynamics

Chapter

10

Thermodynamics of a Refrigerator

Section 2
The First Law of
Thermodynamics

Section 3
The Second Law of
Thermodynamics

Chapter

10

Objectives

Recognize

why the second law of thermodynamics
requires two bodies at different temperatures for work
to be done.

Calculate

the efficiency of a heat engine.

Relate

the disorder of a system to its ability to do
work or transfer energy as heat.

Chapter

10

Efficiency of Heat Engines

The

second law of thermodynamics

can be stated
as follows:

No cyclic process that converts heat entirely
into work is possible.

As seen in the last section,

W
net

=
Q
ne
t

=
Q
h

Q
c
.

According to the second law of thermodynamics,
W

can never be equal to
Q
h

in a cyclic process.

In other words, some energy must always be
transferred as heat to the system’s surroundings
(
Q
c

> 0).

Section 3
The Second Law of
Thermodynamics

Chapter

10

Efficiency of Heat Engines,
continued

A measure of how well an engine operates is given
by the engine’s

efficiency (eff ).

In general,

efficiency

is a measure of the

useful
energy

taken out of a process relative to the

total
energy

that is put into the process.

Section 3
The Second Law of
Thermodynamics

Note that efficiency is a
unitless

quantity.

Because of the second law of thermodynamics, the
efficiency of a real engine is always less than 1.

Chapter

10

Sample Problem

Heat
-
Engine Efficiency

Find the efficiency of a gasoline engine that, during
one cycle, receives 204 J of energy from combustion
and loses 153 J as heat to the exhaust.

Section 3
The Second Law of
Thermodynamics

1.

Define

Given:

Diagram:

Q
h

= 204 J

Q
c

= 153 J

Unknown

eff
= ?

Chapter

10

Sample Problem,
continued

2.

Plan

Choose an equation or situation:
The efficiency of
a heat engine is the ratio of the work done by the
engine to the energy transferred to it as heat.

Section 3
The Second Law of
Thermodynamics

Chapter

10

Sample Problem,
continued

3.

Calculate

Substitute the values into the equation and
solve:

Section 3
The Second Law of
Thermodynamics

4.

Evaluate

Only 25 percent of the energy added as heat is used
by the engine to do work. As expected, the efficiency
is less than 1.0.

Chapter

10

Entropy

In thermodynamics, a system left to itself tends to go
from a state with a very ordered set of energies to
one in which there is less order.

The measure of a system’s disorder or randomness
is called the

entropy

of the system. The greater the
entropy of a system is, the greater the system’s
disorder.

The greater probability of a disordered arrangement
indicates that an
ordered system is likely to
become disordered.

Put another way, the

entropy

of a system tends to increase.

Section 3
The Second Law of
Thermodynamics

Chapter

10

Entropy,
continued

If all gas particles moved toward the piston, all of the
internal energy could be used to do work. This
extremely well ordered system is highly improbable.

Section 3
The Second Law of
Thermodynamics

Greater disorder means there is less energy to do
work.

Chapter

10

Entropy,
continued

Because of the connection between a system’s
entropy, its ability to do work, and the direction of
energy transfer, the

second law of
thermodynamics

can also be expressed in terms of
entropy change:

The entropy of the universe increases in all
natural processes.

Entropy can decrease for parts of systems, provided
this decrease is offset by a greater increase in
entropy elsewhere in the universe.

Section 3
The Second Law of
Thermodynamics

Chapter

10

Energy Changes Produced by a Refrigerator
Freezing Water

Section 3
The Second Law of
Thermodynamics

Because of the refrigerator’s less
-
than
-
perfect efficiency, the entropy of
the outside air molecules increases more than the entropy of the
freezing water decreases.