Transcript Slide 1
15.1 The First Law of Thermodynamics
A system’s internal energy can be changed by doing
work or by the addition/removal of heat:
ΔU = Q - W
W is negative if work is done on the system
What is the state of the system?
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Compression of the gas
Described by P, V, T, m, U
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15.2 Thermodynamic Processes and the First Law
Isothermal: T = constant → ΔU = 0 → W = Q
Adiabatic: Q = 0 → ΔU = -W
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15.2 Thermodynamic Processes and the First Law
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If pressure is constant then
W = Fd = PAd = P ΔV
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15.2 Thermodynamic Processes and the First Law
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The total work done during a process is equal to the
area under the PV diagram
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15.4 The Second Law of Thermodynamics
Heat can flow spontaneously only from a hot object
to a cold object.
A reversible process is one that is always in
equilibrium and can return to its initial conditions
along the same path
Most natural processes are irreversible
Sets an upper limit on efficiency of heat engines
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15.5 Heat Engines
Heat engines convert U into other useful forms of
energy – mechanical, electrical, …
ΔUcycle = 0 → QH = W + QL
Automobile engines
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15.5 Heat Engines
W
QL
1
The efficiency of a heat engine is e
QH
QH
Carnot (ideal) engine
Reversible processes
Too slow for real engines
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15.6 Refrigerators, Air Conditioners and Heat Pumps
A heat engine in reverse.
QL
COP
W
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15.6 Refrigerators, Air Conditioners and Heat Pumps
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2. (a) The work done by a gas at constant pressure is found from
Eq. 15-3.
1.01105 Pa
3
3
5
W PV 1 atm
18.2
m
12.0
m
6.262
10
J
1 atm
(b) The change in internal energy is calculated from the first law
of thermodynamics
4186 J
5
6
U Q W 1400 kcal
6.262 10 J 5.2 10 J
1 kcal
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26. Find the exhaust temperature from the original Carnot
efficiency, and then recalculate the intake temperature for the
new Carnot efficiency, using the same exhaust temperature.
e1 1
TL
TL TH1 1 e 550 273 K 1 0.28 592.6 K
TH1
TL
e1 T1
TL T TH1 592.6
1 e K 550 273 K 1 0.28 592.6
o
o
L
L
T
e2 1
912 K 639 C 640 C
H1 TH2
TH2
1 e2 1 0.35
TL
TL
592.6 K
e2 1
TH2
912 K 639o C 640o
TH2
1 e2 1 0.35
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