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Thermodynamics is the study of processes in which energy
is transferred as heat and as work.
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.
FIRST LAW OF THERMODYNAMICS
ΔU = Q + W
In any thermodynamic process, the change in internal energy
(ΔU) is equal to the net heat (Q) absorbed by a system plus the
work (W) done on the system.
Q is (+) if added and (-) if removed
W (-) if done by the system and (+) if done on the system.
What does it mean for the system to do work?
Work is simply a force multiplied by the distance moved in the
direction of the force. A good example of a thermodynamic
system that can do work is the gas confined by a piston in a
cylinder, as shown in the diagram.
PRESSURE-VOLUME (PV) DIAGRAMS
Many thermodynamic processes involve energy changes that
occur to gases enclosed in cylinders. The work done by the gas
when it expands at a constant P is shown in the figure:
When a thermodynamic process involves changes in volume
and/or pressure, the work done by the system is equal to the
area under the curve of a PV diagram.
When the gas suffers changes which eventually return it to its
initial state, it is said to have undergone a cycle of operations
and the diagram is a closed loop. The net work done by the gas
in this case is represented by the enclosed area.
THERMODYNAMIC PROCESSES
There are a number of different thermodynamic processes that
can change the pressure and/or the volume and/or the
temperature of a system.
TYPES OF THERMODYNAMIC PROCESSES:
• Adiabatic
• Isovolumetric
• Isothermal
• Isobaric
ISOVOLUMETRIC PROCESS
An isovolumetric (or isochoric) process is one in which the
volume of the system remains constant.
ΔW = 0
so:
ΔQ = ΔU
All the thermal energy absorbed by a system goes to increase
its internal energy, and there is a rise in the temperature of the
system.
ISOTHERMAL PROCESS
An isothermal process is one in which the temperature of the
system remains constant.
ΔU = 0 so: -ΔQ = ΔW
All the energy absorbed by a system is converted into output
work.
ISOBARIC PROCESS
In an isobaric process the pressure remains constant and of the
heat received by the gas, some becomes internal energy as the
temperature rises, the rest is used to do work.
ΔU = Q + W
ADIABATIC PROCESS
An adiabatic process is one in which there is no exchange of
thermal energy ΔQ between a system and its surroundings.
ΔQ = 0 so: ΔW = ΔU
In an adiabatic process the work is done at the expense of
internal energy. The decrease in internal energy is usually
accompanied by a decrease in temperature.
- An example of an isobaric system is a gas, being slowly heated or
cooled, confined by a piston in a cylinder.
- An example of an isovolumetric system is a gas in a box with fixed
walls.
- An example of an isothermal system is a gas confined by a piston in a
cylinder where the gas is not heated or cooled, but the piston is slowly
moved so that the gas expands or is compressed. The temperature is
maintained at a constant value by putting the system in contact with a
constant-temperature reservoir (the thermodynamic definition of a
reservoir is something large enough that it can transfer heat into or out
of a system without changing temperature).
- An example of an adiabatic process is a gas expanding so quickly
that no heat can be transferred. The expansion does work, and the
temperature drops. This is exactly what happens with a carbon dioxide
fire extinguisher, with the gas coming out at high pressure and cooling as
it expands at atmospheric pressure.
SECOND LAW OF THERMODYNAMICS
The second law of thermodynamics can be stated in several
equivalent ways, three of which are:
1. Heat energy flows spontaneously from a hot object to a cold
object only.
2. It is impossible to construct a heat engine that is 100%
efficient. A heat engine can convert some of the input heat into
useful work, but the rest must be exhausted as waste heat.
3. The entropy of an isolated system never decreases. It can
only stay the same or increase.
First Statement of the Second Law
The first statement of the second law is a statement from
common experience. When two objects, one hot and the other
cold, come into contact, heat energy will be transferred from
the system at higher temperature to the system at lower
temperature, but not vice versa. While the first law states that
energy must be conserved, i.e., the sum of the energy lost and
gained in any process must equal zero, it does not say that heat
must flow from the hot object to the cold object. The first law
would not be violated if the hot object became hotter while the
cold object became colder. The second law states that the
direction of the heat flow must be from hot to cold.
Second Statement of the Second Law: Heat Engines
A heat engine is a device that is capable of changing thermal
energy (QH), also known as the input heat or heat of
combustion of the fuel, into useful work (W).
Heat engines cannot be made to be 100% efficient and while
part of the heat energy is converted to useful work, the
remaining heat energy will be rejected to the environment or
surroundings as waste heat (QL) like the exhaust from a car
engine for example. Therefore,
Q H = W + QL
CARNOT ENGINE
In an idealized engine, known as a Carnot engine, energy losses
due to internal friction or turbulence present in the fuel after
ignition, etc. are not considered.
The most efficient heat engine cycle is the Carnot cycle,
consisting of two isothermals and two adiabatic processes.
The Carnot cycle can be thought of as the most efficient heat
engine cycle allowed by physical laws. When the second law of
thermodynamics states that not all the supplied heat in a heat
engine can be used to do work, the Carnot efficiency sets the
limiting value on the fraction of the heat which can be so used.
Carnot determined that the
maximum efficiency (e) that
could be realized from a
heat engine depends on the
temperature of the input
heat (TH) and the exhaust
heat (TC) where TH and TC
are expressed in Kelvin.
The maximum efficiency (e) or Carnot efficiency of a heat
engine is determined by:
(QH  QL )
W
but since W = QH - QL then e 
e
QH
QH
Using the input and waste heat temperatures:
(TH  TL )
e
TH
In order for a Carnot engine to be 100% efficient, it would be
necessary for the temperature of the exhaust heat to be at
absolute zero (zero Kelvin). This is a practical as well as a
theoretical impossibility.
Refrigerators and Air Conditioners
Refrigerators and air conditioners operate by removing heat
from a low temperature (cold) reservoir and exhausting the
heat to the higher temperature (hot) reservoir. In order to
accomplish this task, work is done to cause heat to travel
opposite from its normal direction.
15.1 In a certain process, 8.00 kcal of heat is supplied to the system while
the system does 6.00 kJ of work. By how much does the internal energy of
the system change during the process?
ΔQ = 8000 cal (4.186) = 33,488 J = 33.5 kJ
ΔW = - 6 kJ
ΔU = ΔQ + ΔW
= 33.5 - 6
= 27.5 kJ
15.2 The specific heat of water is 4180 J/kg.K. By how many
joules does the internal energy of 50 g of water change if it is
heated from 21°C to 37°C?
c = 4180 J/kg.K
m = 0.05 kg
to = 21°C
tf = 37°C
ΔQ = mc ΔT
= 0.05 (4180)(37 - 21)
= 3344 J
since ΔW = 0
ΔU = ΔQ = 3344 J
15.3 Find the change in work and the change in internal energy
for a 1700 g cube of iron as it is heated from 20 °C to 300 °C at
atmospheric pressure. For iron, c = 0.11 cal/g°C and the change
in volume is 2.18x10-6 m3.
c = 0.11 cal/g°C
m = 1700 g
to = 20°C
tf = 300°C
ΔV = 2.18x10-6 m3
Patm = 1x105 Pa
ΔQ = mc ΔT = 1700 (0.11)(300 - 20)
= 52,360 cal
ΔW = P ΔV = 1x105 (2.18x10-6)
= - 0.218 J
(negative since its done BY the system)
ΔU = ΔQ - ΔW
= 52,360 cal (4.186 J/cal) - 0.218
= 2.19x105 J
15.4 In each of the following situations, find the change in internal energy
of the system
a. A system absorbs 500 cal of heat and at the same time does 400 J of
work.
Q = 500 cal
W = - 400 J
ΔU = ΔQ - ΔW
= 500 (4.186) – 400
= 1693 J = 1.69 kJ
b. A system absorbs 300 cal of heat and at the same time 420 J of work is
done on it.
Q = 300 cal
W = 420 J
ΔU = ΔQ + ΔW
= 300 (4.186) + 420
= 1675.8 J = 1.67 kJ
c. 1200 cal are removed from a gas held at constant volume.
Q = - 1200 cal
ΔU = ΔQ + ΔW
= - 1200 (4.186) + 0
= - 5023.2 J = -5.02 kJ
15.5 For each of the following adiabatic processes, find the change in
internal energy.
a. A gas does 5 J of work while expanding adiabatically.
W=-5J
Adiabatic ΔQ = 0
ΔU = ΔQ - ΔW
=0-5=-5J
b. During an adiabatic compression, 80 J of work is done on a gas.
W = 80 J
ΔU = ΔQ + ΔW
= 0 + 80 = 80 J
15.6 1 kg of steam at 100°C and 101 kPa occupies 1.68 m3.
a. What fraction of the observed heat of vaporization of water
is accounted for by the expansion of water into steam?
m = 1 kg, t = 100°C
P = 101 kPa, V = 1.68 m3
Lv = 2.256x106 J/kg
ΔW = PΔV
= 101x103 (1.68) = 169,680 J
ΔQ = mLv
= 1 (2.256x106) = 2.256x106 J
Fraction: ΔW/ ΔQ
= 169,680/2.256x106
= 0.075
= 75%
b. Determine the increase in internal energy of 1.0-kg of water
as it is vaporized at 100 °C.
m = 1 kg
t = 100°C
W = - 169,680 J
ΔU = ΔQ - ΔW
= 2.256x106 - 169,680
= 2.08x106 J
15.7 The PV diagram in the figure applies to a gas undergoing a cyclic
change in a piston-cylinder arrangement. What is the work done by the
gas:
a. In portion AB of the cycle?
Work = area under AB
W = (4 - 1.5)x106 (4x105)
=1J
b. In portion BC?
ΔV = 0 therefore W = 0 J
c. In portion CD?
ΔV (-) (contraction)
W = - (2.5x10-6)(2x105)
= -0.5 J
d. In portion DA?
W=0J
e. The net work output of the gas during the cycle
ΔW = 1 - 0.5 = 0.5 J
or area enclosed:
ΔW = (2.5x10-6)(2x105) = 0.5 J
f. The net heat flow into the gas per cycle.
ΔQ = ΔU + ΔW
ΔU = 0
ΔQ = 0 + 0.5
= 0.5 J
15.8 For the thermodynamic cycle shown in the figure, find the net work
during the cycle.
Area of the figure = Net Work
Area ≈ 22 squares
A = (0.5x105) (0.1)(22)
= 110,000 J
ΔW = 110 kJ
15.9 A steam engine operating between a boiler temperature of 220 °C and
a condenser temperature of 35 °C delivers 8.0 hp. If its efficiency is 30% of
that of a Carnot engine operating between these temperature limits,
a. How many calories are absorbed each second by the boiler?
eff = 30%
Tc = 35 + 273 = 308 K,
Th = 220 + 273 = 493 K
P = 8 hp (746 W/hp) = 5968 W
5968 J/s (1 cal/s)/4.186 J = 1425.7 cal/s
Tc
actual eff = 0.3 Carnot eff  0.3(1  )
Th
308
eff  0.3(1 
) = 0.113
493
Wout
Qin 
eff
1425.7

= 12616 cal/s = 12.6 kcal/s
0113
.
b. How many calories are exhausted to the condenser each second?
Qin = Wout + Qout
Qout/s = Qin/s - Wout/s
Wout = Qin (eff)
Qout/s = Qin/s (1 - eff)
= 12.6 (1 - 0.113)
= 11.17 kcal/s