Chapter 20 - Thermodynamics

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Oct 27, 2013 (3 years and 9 months ago)

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Chapter 20
-

Thermodynamics

A PowerPoint Presentation by

Paul E. Tippens, Professor of
Physics

Southern Polytechnic State
University

©

2007

A THERMODYNAMIC SYSTEM


A system is a closed environment in
which heat transfer can take place.
(For example, the gas, walls, and
cylinder of an automobile engine.)

Work done on
gas or work
done by gas

INTERNAL ENERGY OF
SYSTEM


The internal energy U of a system is
the total of all kinds of energy
possessed by the particles that make
up the system.

Usually the internal energy consists
of the sum of the potential and
kinetic energies of the working gas
molecules
.

TWO WAYS TO INCREASE THE
INTERNAL ENERGY,

U.


HEAT PUT
INTO

A SYSTEM
(Positive)

+

U

WORK DONE
ON

A GAS
(Positive)

WORK DONE
BY

EXPANDING GAS:

W is
positive

-

U

Decrease

TWO WAYS TO DECREASE
THE INTERNAL ENERGY,

U.


HEAT
LEAVES

A
SYSTEM



Q is
negative

Q
out

hot

W
out

hot

THERMODYNAMIC STATE

The STATE of a
thermodynamic system is
determined by four factors:


Absolute Pressure P in
Pascals


Temperature T in Kelvins


Volume V in cubic meters



Number of moles, n, of working gas

THERMODYNAMIC
PROCESS

Increase in Internal Energy,

U.

Initial State:

P
1

V
1

T
1

n
1


Final State:

P
2

V
2

T
2

n
2

Heat input

Q
in

W
out

Work by gas

The Reverse Process

Decrease in Internal Energy,

U.

Initial State:

P
1

V
1

T
1

n
1


Final State:

P
2

V
2

T
2

n
2

Work on gas

Loss of heat

Q
out

W
in

THE FIRST LAW OF
THERMODYAMICS:


The net heat put into a system is equal to
the change in internal energy of the
system minus any work done BY the
system.


U =

Q +

W

final
-

initial)


Conversely, the work done ON a system is
equal to the change in internal energy
minus any the heat gained in the process.

SIGN
CONVENTIONS
FOR FIRST LAW



Heat Q input is positive



U=

Q +

W

final
-

initial)



Heat OUT is negative


Work ON a gas is positive


Work BY a gas is negative

+Q
in

-
W
out


U

+W
in

-
Q
out


U

APPLICATION OF FIRST

LAW OF THERMODYNAMICS

Example 1:

In the figure, the
gas absorbs 400 J of heat and
at the same time does 120 J
of expansion on the piston.
What is the change in internal
energy of the system?


U =

Q +

W

Apply First Law:

Q
in

400 J

W
out

=
-
120 J

Example 1 (Cont.):
Apply First
Law





U = +280 J

Q
in

400 J

W
out

=
-
120 J


U =

Q +

W


= (+400 J) + (
-
120 J)


= +280 J


W is negative:
-
120 J (Work OUT)


U =

Q +

W


Q is positive: +400 J (Heat IN)

Example 1 (Cont.): Apply First
Law





U = +280 J

The
400 J

of input thermal
energy is used to perform
120 J

of external work,
increasing the internal
energy of the system by
280 J

Q
in

400 J

W
out

=120 J

The increase in
internal energy is:

Energy is conserved:

FOUR THERMODYNAMIC

PROCESSES:


Isochoric Process:

V = 0,

W = 0


Isobaric Process:

P = 0


Isothermal Process:

T = 0,

U = 0


Adiabatic Process:

Q = 0


Q =

U +

W


Absorbs heat Q
hot



Performs work W
out


Rejects heat Q
cold

A heat engine is any
device which through
a cyclic process:

Cold Res. T
C

Engine

Hot Res. T
H

Q
hot

W
out

Q
cold

HEAT ENGINES

THE SECOND LAW OF
THERMODYNAMICS

It is impossible to construct an
engine that, operating in a
cycle, produces no effect other
than the extraction of heat
from a reservoir and the
performance of an equivalent
amount of work.

Not only can you not win (1st law);
you can’t even break even (2nd law)!


W
out

Cold Res. T
C

Engine

Hot Res. T
H

Q
hot

Q
cold

THE SECOND LAW OF
THERMODYNAMICS

Cold Res. T
C

Engine

Hot Res. T
H

400 J

300 J

100 J



A possible engine.



An IMPOSSIBLE
engine.

Cold Res. T
C

Engine

Hot Res. T
H

400 J

400 J

EFFICIENCY OF AN
ENGINE

Cold Res. T
C

Engine

Hot Res. T
H

Q
H

W

Q
C

The efficiency of a heat engine
is the ratio of the net work
done W to the heat input Q
H
.

e = 1
-


Q
C

Q
H

e = =

W

Q
H

Q
H
-

Q
C


Q
H

EFFICIENCY EXAMPLE

Cold Res. T
C

Engine

Hot Res. T
H

800 J

W

600 J

An engine absorbs 800 J and
wastes 600 J every cycle. What
is the efficiency?

e = 1
-


600 J

800 J

e = 1
-


Q
C

Q
H

e = 25%

Question: How many joules of work is done?

EFFICIENCY OF AN IDEAL
ENGINE (Carnot Engine)

For a perfect engine, the
quantities Q of heat gained
and lost are proportional to
the absolute temperatures T.

e = 1
-


T
C

T
H

e =


T
H
-

T
C


T
H

Cold Res. T
C

Engine

Hot Res. T
H

Q
H

W

Q
C

Example 3:

A steam engine absorbs 600 J
of heat at 500 K and the exhaust
temperature is 300 K. If the actual efficiency
is only half of the ideal efficiency, how much
work is done during each cycle?

e = 1
-


T
C

T
H

e = 1
-


300 K

500 K

e = 40%

Actual e = 0.5e
i

= 20%

e =

W

Q
H

W = eQ
H

= 0.20 (600 J)

Work = 120 J

REFRIGERATORS

A refrigerator is an engine
operating in reverse:
Work is done
on

gas
extracting heat
from

cold
reservoir and depositing
heat
into

hot reservoir.

W
in
+ Q
cold
= Q
hot

W
IN
= Q
hot
-

Q
cold

Cold Res. T
C

Engine

Hot Res. T
H

Q
hot

Q
cold

W
in

THE SECOND LAW FOR
REFRIGERATORS

It is impossible to construct a
refrigerator that absorbs heat
from a cold reservoir and
deposits equal heat to a hot
reservoir with

W = 0.

If this were possible, we could
establish perpetual motion!

Cold Res. T
C

Engine

Hot Res. T
H

Q
hot

Q
cold

COEFFICIENT OF
PERFORMANCE

Cold Res. T
C

Engine

Hot Res. T
H

Q
H

W

Q
C

The COP (K) of a heat
engine is the ratio of the
HEAT Q
c

extracted to the
net WORK done W.

K =



T
H

T
H
-

T
C

For an IDEAL
refrigerator:

Q
C


W

K = =



Q
H

Q
H
-

Q
C

COP EXAMPLE

A Carnot refrigerator operates
between 500 K and 400 K. It
extracts 800 J from a cold
reservoir during each cycle.
What is C.O.P., W and Q
H

?

Cold Res. T
C

Eng
ine

Hot Res. T
H

800 J

W

Q
H

500 K

400 K

K =



400 K


500 K
-

400 K


T
C

T
H
-

T
C

=

C.O.P. (K) = 4.0

COP EXAMPLE (Cont.)

Next we will find Q
H

by
assuming same K for actual
refrigerator (Carnot).


Cold Res. T
C

Eng
ine

Hot Res. T
H

800 J

W

Q
H

500 K

400 K

K =



Q
C

Q
H
-

Q
C

Q
H

= 1000 J


800 J


Q
H

-

800 J

=

4.0

COP EXAMPLE (Cont.)

Now, can you say how much
work is done in each cycle?


Cold Res. T
C

Engine

Hot Res. T
H

800 J

W

1000 J

500 K

400 K

Work = 1000 J
-

800 J

Work = 200 J

Summary


U =

Q +

W

final
-

initial)

The First Law of Thermodynamics: The net

change in internal
energyby

a system is
equal to the sum of the
heat taken in
and
the work done on the system.


Isochoric Process:

V = 0,

W = 0


Isobaric Process:

P = 0


Isothermal Process:

T = 0,

U = 0


Adiabatic Process:

Q = 0

Summary (Cont.)


c

=


Q

n

T


U = nC
v

T

The Molar
Specific Heat
capacity, C:

Units are:Joules
per mole per
Kelvin degree

The following are true for ANY process:


U =

Q +

W

PV = nRT

Summary (Cont.)

The
Second Law of Thermo:

It is
impossible to construct an engine
that, operating in a cycle,
produces no effect other than the
extraction of heat from a reservoir
and the performance of an
equivalent amount of work.

Cold Res. T
C

Engine

Hot Res. T
H

Q
hot

Q
cold

W
out

Not only can you not win (1st law);
you can’t even break even (2nd law)!

Summary (Cont.)

The efficiency of a heat engine:

e = 1
-


Q
C

Q
H

e = 1
-


T
C

T
H

The coefficient of performance of a refrigerator
: