FOUNDRY: It requires about 289 Joules of heat to melt one gram of steel. In this chapter, we will define the quantity of heat to raise the temperature and to change the phase of a substance.
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• Sadi Carnot • 1796 – 1832• French Engineer• Founder of the science
of thermodynamics• First to recognize the
relationship between work and heat
Thermodynamics is the study of processes in which energy is transferred as heat and as work.
IThe internal energy is the sum of all the energy of all the molecules in an object:· random translational kinetic energy· rotational kinetic energy· vibrational energy· intermolecular energy associated with their bonding.
ZEROTH LAW The Zeroth law states that "If bodies A and B are each separately in thermal equilibrium with body C, then A and B are in thermal equilibrium with each other."
The common property between A and B is called temperature.
Topics
4) Thermodynamic Processes
5) The Second Law of Thermodynamics
6) Heat Engines
8) Entropy
1) The First Law of Thermodynamics
7) Carnot cycle
3) Pressure - Volume Graph
2) Work Done on a Gas
The Laws of Thermodynamics (02 of 38)
First Law of Thermodynamics
SystemU
Environment
DU = Q - W
Q W
The Laws of Thermodynamics (03 of 38)
PV = NkT
Pressure(Pa)
Volume(m3)
AbsoluteTemperature
(K)
Boltzmann’sConstant
(1.38 x 10-23 J/K)
Number ofMolecules
The Ideal Gas Law
Temperature and kinetic Theory13
Thermodynamic Processes
A. Isobaric Constant Pressure
B. Iso-volumetric Constant Volume
C. Isothermal Constant Temp.
D. Adiabatic No Heat Transfer between systems
The Laws of Thermodynamics (06 of 38)
Pressure - Volume Graph
P
V
Pressure
Volume
Isotherms(lines of constant
temperature)
Area under curve represents work
T1
T4
T2
T3Internal energyis proportionalto temperature
The Laws of Thermodynamics (05 of 38)
• 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
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
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
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
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?
EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine)
maember 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
ch QQW
Work done by engine each cycle
hQW
e
J 470J 890W J 420
The efficiency of the engine
J 908J 420 % 2.47 472.0
Th= 550 K
Tc
Engine
Qh
Qc
W
= 470 J
= 890 JFor the engine
The Carnot Cycle
The Laws of Thermodynamics (26 of 38)
Th= 550 K
Tc
Engine
Qh
Qc
W = 420 J
h
c
h
cT
T
Q
Q
= 470 J
= 890 J
Temperature of the cool reservoir
h
chc Q
QTT
J 505J 470
K 550Tc K 290
The engine undergoes 22 cycles per second, its mechanical power output
tW
P Wf
s
cycles 22
cycleJ
420 kW 24.9
The Carnot Cycle
The Laws of Thermodynamics (27 of 38)
Th
Tc
Engine
Qh
Qc
W = 350 J
= 900 JA carnot engine absorbs 900 J of heat each cycle and provides 350 J of work
The heat ejected each cycle
ch QQW
hQW
e
The efficiency of the engine
J 009J 503 % 9.38 389.0
WQQ hc J 350J 900 J 550
The Carnot Cycle
The Laws of Thermodynamics (28 of 38)
Th
Tc = 283 K
Engine
Qh
Qc
W = 350 J
= 900 JA carnot engine absorbs 900 J of heat each cycle and provides 350 J of work
The engine ejects heat at 10 oCThe temperature of the hot reservoir
=550 J
h
c
h
cT
T
Q
Q
c
hch Q
QTT
J 505J 900
K 283Tc K 463 C 190 o
The Carnot Cycle
The Laws of Thermodynamics (29 of 38)
A carnot engine operates between a hot reservoir at 650 K and a cold reservoir at 300 K. If it absorbs 400 J of heat at the hot reservoir, how much work does it deliver?
Th= 650 K
Tc= 300 K
Engine
Qh
Qc
W = ?
= 400 J
hQW
e h
chT
TT
h
chh T
TTQW
K 650K 300K 650
J 400W J 215
The Carnot Cycle
The Laws of Thermodynamics (30 of 38)
Entropy is a measure of the disorder of a system. This gives us yet another statement of the second law:
Natural processes tend to move toward a state of greater disorder.
Example: If you put milk and sugar in your coffee and stir it, you wind up with coffee that is uniformly milky and sweet.
No amount of stirring will get the milk and sugar to come back out of solution.
Entropy
The Laws of Thermodynamics (33 of 38)
Another example: when a tornado hits a building, there is major damage.
You never see a tornado approach a pile of rubble and leave a building behind when it passes.
Thermal equilibrium is a similar process –
the uniform final state has more disorder than the separate temperatures in the initial state.
Entropy
The Laws of Thermodynamics (34 of 38)
Another consequence of the second law:
In any natural process, some energy becomes unavailable to do useful work.
If we look at the universe as a whole, it seems inevitable that, as more and more energy is converted to unavailable forms, the ability to do work anywhere will gradually vanish. This is called the heat death of the universe.
Entropy
The Laws of Thermodynamics (35 of 38)
WQUΔ First law of thermodynamics:
Isothermal process: temperature is constant.
Adiabatic process: no heat is exchanged.
UΔPW Work done by gas at constant pressure:
Heat engine changes heat into useful work (requires temperature difference).
HQW
e H
LQ
Q1 Efficiency of a heat engine:
Carnot efficiency:H
Lc T
T1e
Summary
The Laws of Thermodynamics (36 of 38)