TO KNOW WORK , HEAT TRANSFER

1. Chapter 20 - Thermodynamics
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

2. THERMODYNAMICS
Thermodynamics is
the study of energy
relationships that
involve heat,
mechanical work,
and other aspects of
energy and heat
transfer.
Central Heating

3. Objectives: After finishing this
unit, you should be able to:
• State and apply the first and
second laws of thermodynamics.
• Demonstrate your understanding
of adiabatic, isochoric, isothermal,
and isobaric processes.
• Write and apply a relationship for determining
the ideal efficiency of a heat engine.
• Write and apply a relationship for determining
coefficient of performance for a refrigeratior.

4. 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

5. 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.

6. TWO WAYS TO INCREASE THE
INTERNAL ENERGY, U.
HEAT PUT INTO
A SYSTEM
(Positive)
+U
WORK DONE
ON A GAS
(Positive)

7. 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
Qout
hot
Wout
hot

8. 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

9. THERMODYNAMIC PROCESS
Increase in Internal Energy, U.
Initial State:
P1 V1 T1 n1
Final State:
P2 V2 T2 n2
Heat input
Qin
Wout
Work by gas

10. The Reverse Process
Decrease in Internal Energy, U.
Initial State:
P1 V1 T1 n1
Final State:
P2 V2 T2 n2
Work on gas
Loss of heat
Qout
Win

11. THE FIRST LAW OF
THERMODYAMICS:
• The net heat put into a system is equal to
the change in internal energy of the
system plus the work done BY the system.
Q = U + W final - initial)
• Conversely, the work done ON a system is
equal to the change in internal energy plus
the heat lost in the process.

12. SIGN CONVENTIONS
FOR FIRST LAW
• Heat Q input is positive
Q = U + W final - initial)
• Heat OUT is negative
• Work BY a gas is positive
• Work ON a gas is negative
+Qin
+Wout
U
-Win
-Qout
U

13. 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 work on the piston. What
is the change in internal
energy of the system?
Q = U + W
Apply First Law:
Qin
400 J
Wout =120 J

14. Example 1 (Cont.): Apply First Law
U = +280 J
Qin
400 J
Wout =120 J
U = Q - W
= (+400 J) - (+120 J)
= +280 J
W is positive: +120 J (Work OUT)
Q = U + W
U = Q - W
Q is positive: +400 J (Heat IN)

15. 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
Qin
400 J
Wout =120 J
The increase in
internal energy is:
Energy is conserved:

16. 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

17. Q = U + W so that Q = U
ISOCHORIC PROCESS:
CONSTANT VOLUME, V = 0, W = 0
0
+U -U
QIN QOUT
HEAT IN = INCREASE IN INTERNAL ENERGY
HEAT OUT = DECREASE IN INTERNAL ENERGY
No Work
Done

18. ISOCHORIC EXAMPLE:
Heat input
increases P
with const. V
400 J heat input increases
internal energy by 400 J
and zero work is done.
B
A
P2
V1= V2
P1
PA P B
TA T B
=
400 J
No Change in
volume:

19. Q = U + W But W = P V
ISOBARIC PROCESS:
CONSTANT PRESSURE, P = 0
+U -U
QIN QOUT
HEAT IN = Wout + INCREASE IN INTERNAL ENERGY
Work Out Work
In
HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY

20. ISOBARIC EXAMPLE (Constant Pressure):
Heat input
increases V
with const. P
400 J heat does 120 J of
work, increasing the
internal energy by 280 J.
400 J
B
A
P
V1 V2
VA VB
TA T B
=

21. ISOBARIC WORK
400 J
Work = Area under PV curve
W
o
r
k P V
 
B
A
P
V1 V2
VA VB
TA T B
=
PA = PB

22. ISOTHERMAL PROCESS:
CONST. TEMPERATURE, T = 0, U = 0
NET HEAT INPUT = WORK OUTPUT
Q = U + W AND Q = W
U = 0 U = 0
QOUT
Work
In
Work Out
QIN
WORK INPUT = NET HEAT OUT

23. ISOTHERMAL EXAMPLE (Constant T):
PAVA = PBVB
Slow compression at
constant temperature:
----- No change in U.
U = T = 0
B
A
PA
V2 V1
PB

24. ISOTHERMAL EXPANSION (Constant T):
400 J of energy is absorbed
by gas as 400 J of work is
done on gas.
T = U = 0
U = T = 0
B
A
PA
VA VB
PB
PAVA = PBVB
TA = TB
ln B
A
V
W nRT
V

Isothermal Work

25. Q = U + W ; W = -U or U = -W
ADIABATIC PROCESS:
NO HEAT EXCHANGE, Q = 0
Work done at EXPENSE of internal energy
INPUT Work INCREASES internal energy
Work Out Work
In
U +U
Q = 0
W = -U U = -W

26. ADIABATIC EXAMPLE:
Insulated
Walls: Q = 0
B
A
PA
V1 V2
PB
Expanding gas does
work with zero heat
loss. Work = -U

27. ADIABATIC EXPANSION:
400 J of WORK is done,
DECREASING the internal
energy by 400 J: Net heat
exchange is ZERO. Q = 0
Q = 0
B
A
PA
VA VB
PB
PAVA PBVB
TA T B
=
A A B B
P
V P
V
 


28. MOLAR HEAT CAPACITY
OPTIONAL TREATMENT
The molar heat capacity C is defined as
the heat per unit mole per Celsius degree.
Check with your instructor to
see if this more thorough
treatment of thermodynamic
processes is required.

29. SPECIFIC HEAT CAPACITY
Remember the definition of specific heat
capacity as the heat per unit mass
required to change the temperature?
For example, copper: c = 390 J/kgK
Q
c
m t



30. MOLAR SPECIFIC HEAT CAPACITY
The “mole” is a better reference for gases
than is the “kilogram.” Thus the molar
specific heat capacity is defined by:
For example, a constant volume of oxygen
requires 21.1 J to raise the temperature of
one mole by one kelvin degree.
C =
Q
n T

31. SPECIFIC HEAT CAPACITY
CONSTANT VOLUME
How much heat is required to
raise the temperature of 2 moles
of O2 from 0oC to 100oC?
Q = (2 mol)(21.1 J/mol K)(373 K - 273 K)
Q = nCv T
Q = +4220 J

32. SPECIFIC HEAT CAPACITY
CONSTANT VOLUME (Cont.)
Since the volume has not
changed, no work is done. The
entire 4220 J goes to increase
the internal energy, U.
Q = U = nCv T = 4220 J
U = nCv T
Thus, U is determined by the
change of temperature and the
specific heat at constant volume.

33. SPECIFIC HEAT CAPACITY
CONSTANT PRESSURE
We have just seen that 4220 J of
heat were needed at constant
volume. Suppose we want to also
do 1000 J of work at constant
pressure?
Q = U + W
Q = 4220 J + J
Q = 5220 J Cp > Cv
Same

34. HEAT CAPACITY (Cont.)
Cp > Cv
For constant pressure
Q = U + W
nCpT = nCvT + P V
U = nCvT
Heat to raise temperature
of an ideal gas, U, is the
same for any process.
Cp
Cv


35. REMEMBER, FOR ANY PROCESS
INVOLVING AN IDEAL GAS:
PV = nRT
U = nCv T
Q = U + W
PAVA PBVB
TA T B
=

36. Example Problem:
• AB: Heated at constant V to 400 K.
A 2-L sample of Oxygen gas has an initial temp-
erature and pressure of 200 K and 1 atm. The
gas undergoes four processes:
• BC: Heated at constant P to 800 K.
• CD: Cooled at constant V back to 1 atm.
• DA: Cooled at constant P back to 200 K.

37. PV-DIAGRAM FOR PROBLEM
B
A
PB
2 L
1 atm
200 K
400 K 800 K
How many moles
of O2 are present?
Consider point A:
PV = nRT
3
(101,300Pa)(0.002m )
0.122 mol
(8.314J/mol K)(200K)
PV
n
RT
  


38. PROCESS AB: ISOCHORIC
What is the pressure
at point B?
PA P B
TA T B
=
1 atm P B
200 K 400 K
=
P B = 2 atm
or 203 kPa
B
A
PB
2 L
1 atm
200 K
400 K 800 K

39. PROCESS AB: Q = U + W
Analyze first law
for ISOCHORIC
process AB.
W = 0
Q = U = nCv T
U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K)
B
A
PB
2 L
1 atm
200 K
400 K 800 K
Q = +514 J W = 0
U = +514 J

40. PROCESS BC: ISOBARIC
What is the volume
at point C (& D)?
VB V C
TB T C
=
2 L V C
400 K 800 K
=
B
C
PB
2 L
1 atm
200 K
400 K 800 K
D
4 L
V C = V D = 4 L

41. FINDING U FOR PROCESS BC.
Process BC is
ISOBARIC.
P = 0
U = nCv T
U = (0.122 mol)(21.1 J/mol K)(800 K - 400 K)
U = +1028 J
B
C
2 L
1 atm
200 K
400 K 800 K
4 L
2 atm

42. FINDING W FOR PROCESS BC.
Work depends
on change in V.
P = 0
Work = PV
W = (2 atm)(4 L - 2 L) = 4 atm L = 405 J
W = +405 J
B
C
2 L
1 atm
200 K
400 K 800 K
4 L
2 atm

43. FINDING Q FOR PROCESS BC.
Analyze first
law for BC.
Q = U + W
Q = +1028 J + 405 J
Q = +1433 J
Q = 1433 J W = +405 J
B
C
2 L
1 atm
200 K
400 K 800 K
4 L
2 atm
U = 1028 J

44. PROCESS CD: ISOCHORIC
What is temperature
at point D?
PC P D
TC T D
=
2 atm 1 atm
800 K TD
= T D = 400 K
B
A
PB
2 L
1 atm
200 K
400 K 800 K
C
D

45. PROCESS CD: Q = U + W
Analyze first law
for ISOCHORIC
process CD.
W = 0
Q = U = nCv T
U = (0.122 mol)(21.1 J/mol K)(400 K - 800 K)
Q = -1028 J W = 0
U = -1028 J
C
D
PB
2 L
1 atm
200 K
400 K
800 K
400 K

46. FINDING U FOR PROCESS DA.
Process DA is
ISOBARIC.
P = 0
U = nCv T
U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K)
U = -514 J
A
D
2 L
1 atm
200 K
400 K 800 K
4 L
2 atm
400 K

47. FINDING W FOR PROCESS DA.
Work depends
on change in V.
P = 0
Work = PV
W = (1 atm)(2 L - 4 L) = -2 atm L = -203 J
W = -203 J
A
D
2 L
1 atm
200 K
400 K 800 K
4 L
2 atm
400 K

48. FINDING Q FOR PROCESS DA.
Analyze first
law for DA.
Q = U + W
Q = -514 J - 203 J
Q = -717 J
Q = -717 J W = -203 J
U = -514 J
A
D
2 L
1 atm
200 K
400 K 800 K
4 L
2 atm
400 K

49. PROBLEM SUMMARY
Q = U + W
For all
processes:
Process Q U W
AB 514 J 514 J 0
BC 1433 J 1028 J 405 J
CD -1028 J -1028 J 0
DA -717 J -514 J -203 J
Totals 202 J 0 202 J

50. NET WORK FOR COMPLETE
CYCLE IS ENCLOSED AREA
B C
2 L
1 atm
4 L
2 atm
+404 J
B C
2 L
1 atm
4 L
2 atm
Neg
-202 J
Area = (1 atm)(2 L)
Net Work = 2 atm L = 202 J
2 L 4 L
B C
1 atm
2 atm

51. ADIABATIC EXAMPLE:
Q = 0
A
B
PB
VB VA
PA PAVA PBVB
TA T B
=
PAVA = PBVB
 
Example 2: A diatomic gas at 300 K and
1 atm is compressed adiabatically, decreasing
its volume by 1/12. (VA = 12VB). What is the
new pressure and temperature? ( = 1.4)

52. ADIABATIC (Cont.): FIND PB
Q = 0
PB = 32.4 atm
or 3284 kPa
1.4
12 B
B A
B
V
P P
V
 
  
 
1
.
4
(
1
a
t
m
)
(
1
2
)
B
P

PAVA = PBVB
 
A
B
PB
VB 12VB
1 atm
300 K Solve for PB:
A
B A
B
V
P P
V

 
  
 

53. ADIABATIC (Cont.): FIND TB
Q = 0
TB = 810 K
(1 atm)(12VB) (32.4 atm)(1 VB)
(300 K) T B
=
A
B
32.4 atm
VB 12VB
1 atm
300 K
Solve for TB
TB=? A A B B
A B
P V P V
T T


54. ADIABATIC (Cont.): If VA= 96 cm3
and VA= 8 cm3, FIND W
Q = 0
W = - U = - nCV T & CV= 21.1 j/mol K
A
B
32.4 atm
1 atm
300 K
810 K
Since Q = 0,
W = - U
8 cm3 96 cm3
Find n from
point A
PV = nRT
PV
RT
n =

55. ADIABATIC (Cont.): If VA= 96 cm3
and VA= 8 cm3, FIND W
A
B
32.4 atm
1 atm
300 K
810 K
8 cm3 96 cm3
PV
RT
n = =
(101,300 Pa)(8 x10-6 m3)
(8.314 J/mol K)(300 K)
n = 0.000325 mol & CV= 21.1 j/mol K
T = 810 - 300 = 510 K
W = - U = - nCV T
W = - 3.50 J

56. • Absorbs heat Qhot
• Performs work Wout
• Rejects heat Qcold
A heat engine is any
device which through
a cyclic process:
Cold Res. TC
Engine
Hot Res. TH
Qhot Wout
Qcold
HEAT ENGINES

57. 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)!
Wout
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold

58. THE SECOND LAW OF
THERMODYNAMICS
Cold Res. TC
Engine
Hot Res. TH
400 J
300 J
100 J
• A possible engine. • An IMPOSSIBLE
engine.
Cold Res. TC
Engine
Hot Res. TH
400 J
400 J

59. EFFICIENCY OF AN ENGINE
Cold Res. TC
Engine
Hot Res. TH
QH W
QC
The efficiency of a heat engine
is the ratio of the net work
done W to the heat input QH.
e = 1 -
QC
QH
e = =
W
QH
QH- QC
QH

60. EFFICIENCY EXAMPLE
Cold Res. TC
Engine
Hot Res. TH
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 -
QC
QH
e = 25%
Question: How many joules of work is done?

61. 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 -
TC
TH
e =
TH- TC
TH
Cold Res. TC
Engine
Hot Res. TH
QH W
QC

62. 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 -
TC
TH
e = 1 -
300 K
500 K
e = 40%
Actual e = 0.5ei = 20%
e =
W
QH
W = eQH = 0.20 (600 J)
Work = 120 J

63. 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.
Win + Qcold = Qhot
WIN = Qhot - Qcold
Cold Res. TC
Engine
Hot Res. TH
Qhot
Qcold
Win

64. 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. TC
Engine
Hot Res. TH
Qhot
Qcold

65. COEFFICIENT OF PERFORMANCE
Cold Res. TC
Engine
Hot Res. TH
QH W
QC
The COP (K) of a heat
engine is the ratio of the
HEAT Qc extracted to the
net WORK done W.
K =
TH
TH- TC
For an IDEAL
refrigerator:
QC
W
K = =
QH
QH- QC

66. 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 QH ?
Cold Res. TC
Eng
ine
Hot Res. TH
800 J
W
QH
500 K
400 K
K =
400 K
500 K - 400 K
TC
TH- TC
=
C.O.P. (K) = 4.0

67. COP EXAMPLE (Cont.)
Next we will find QH by
assuming same K for actual
refrigerator (Carnot).
Cold Res. TC
Eng
ine
Hot Res. TH
800 J
W
QH
500 K
400 K
K =
QC
QH- QC
QH = 1000 J
800 J
QH - 800 J
=
4.0

68. COP EXAMPLE (Cont.)
Now, can you say how much
work is done in each cycle?
Cold Res. TC
Engine
Hot Res. TH
800 J
W
1000 J
500 K
400 K
Work = 1000 J - 800 J
Work = 200 J

69. Summary
Q = U + W final - initial)
The First Law of Thermodynamics: The net
heat taken in by a system is equal to the
sum of the change in internal energy and
the work done by the system.
• Isochoric Process: V = 0, W = 0
• Isobaric Process: P = 0
• Isothermal Process: T = 0, U = 0
• Adiabatic Process: Q = 0

70. Summary (Cont.)
c =
Q
n T
U = nCv T
The Molar
Specific Heat
capacity, C:
Units are:Joules
per mole per
Kelvin degree
The following are true for ANY process:
Q = U + W
PV = nRT
A A B B
A B
P V P V
T T


71. 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. TC
Engine
Hot Res. TH
Qhot
Qcold
Wout
Not only can you not win (1st law);
you can’t even break even (2nd law)!

72. Summary (Cont.)
The efficiency of a heat engine:
e = 1 -
QC
QH
e = 1 -
TC
TH
The coefficient of performance of a refrigerator:
C C
in H C
Q Q
K
W Q Q
 

C
H C
T
K
T T



73. CONCLUSION: Chapter 20
Thermodynamics