Thermal Physics
Lecture 2
Dr.-Ing. Erwin Sitompul
President University
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Erwin Sitompul
Thermal Physics 2/1
Chapter 18
First Law of Thermodynamics
The Absorption of Heat by Solids and Liquids
 Heat Capacity
 The heat capacity C of an object is the proportionality constant
between the heat Q that the object absorbs or loses and the
resulting temperature change ΔT of the object.
Q  C T  C (T f  Ti )
Tf
Ti
C
: final temperature of the object
: initial temperature of the object
: heat capacity [cal/°C], [J/K]
 Specific Heat Capacity
 Specific heat capacity c is defined as “heat capacity per unit
mass.”
 It refers not to an object but to a unit mass of the material of
which the object is made.
Q  cmT  cm(T f  Ti )
 The specific heat of water, as the reference in defining calorie, is:
c  1cal g C  1 Btu lb F  4190 J kg  K
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Thermal Physics 2/2
Chapter 18
First Law of Thermodynamics
The Absorption of Heat by Solids and Liquids
 Molar Specific Heat
 In many instances the most convenient unit
for specifying the amount of a substance is the
mole (mol), where:
1 mol  6.02 1023 elementary units
 When quantities are expressed in moles,
specific heats must also involve moles (rather
than a mass unit.)
 They are called molar specific heats.
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Thermal Physics 2/3
Chapter 18
First Law of Thermodynamics
The Absorption of Heat by Solids and Liquids
 Heats of Transformation
 When energy is absorbed as heat by a solid or liquid, the
temperature of the sample does not necessarily rise.
 Instead, the sample may change from one phase, or state, to
another, with no change in temperature.
 The amount of energy per unit mass that must be transferred as
heat when a sample completely undergoes a phase change is
called the heat of transformation L.
 Thus, when a sample of mass m completely undergoes a phase
change, the total energy transferred is:
Q  Lm
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Thermal Physics 2/4
Chapter 18
First Law of Thermodynamics
The Absorption of Heat by Solids and Liquids
 When the phase change is from liquid to gas (the sample absorbs
heat) or from gas to liquid (the sample releases heat), then heat
of transformation is called the heat of vaporization LV.
 When the phase change is from solid to liquid (the sample absorbs
heat) or from liquid to solid (the sample releases heat), the heat
of transformation is called the heat of fusion LF.
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Erwin Sitompul
Thermal Physics 2/5
Chapter 18
First Law of Thermodynamics
The Absorption of Heat by Solids and Liquids
 Consider the energy required to convert a 1.00-g block of ice at –
30.0°C to steam at 120.0°C.
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Erwin Sitompul
Thermal Physics 2/6
Chapter 18
First Law of Thermodynamics
Problem
(a) How much heat must be absorbed by ice of mass m = 720 g at –
10°C to take it to liquid state at 15°C?
(b) If we supply the ice with a total energy of only 210 kJ (as heat),
what then are the final state and temperature of the water?
(a) Qtotal  Q1  Q2  Q3
 cice m(0  Ti )  LF m  cwater m(T f  0)
 (2220)(0.720)(0  (10))  (333 k)(0.720)  (4180)(0.720)(15  0)
 15.98 kJ  239.76 kJ  45.144 kJ
 300 kJ
(b) 210 kJ is not enough to melt all the ice.
The remaining heat for melting is
210 kJ – 15.98 kJ = 194.02 kJ
= 80.92% of 239.76 kJ
Thus, 80.92% of ice (582.62 g) will to water
while 19.08% of ice (137.38 g) will stay as ice.
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Thermal Physics 2/7
Chapter 18
First Law of Thermodynamics
Problem
A copper slug whose mass mc is 75 g is heated in a laboratory oven
to a temperature T of 312°C. The slug is then dropped into a glass
beaker containing a mass mw =220 g of water.
The heat capacity Cb of the beaker is 45 cal/K. The initial
temperature Ti of the water and the beaker is 12°C. Assuming that
the slug, the beaker and water are an isolated system and the water
does not vaporize, find the final temperature Tf of the system at
thermal equilibrium.
cc  386 J kg  K
Cb  45cal K  188.55J K
Qc  Qw   Qb
cw  4180 J kg  K
cc mc (Tc  T )  cw mw (T  Tw )  Cb (T  Tb )
(386)(0.075)(312  T )  (4180)(0.22)(T  12)  (188.55)(T  12)
(9032.4  28.95T )  (919.6T  11035.2)  (188.55T  2262.6)
1137.1T  22330.2
T  19.64C
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Thermal Physics 2/8
Chapter 18
First Law of Thermodynamics
A Closer Look at Heat and Work
 Consider a gas –as a system– confined to a
cylinder with a movable piston.
 The system starts from an initial state i, with
pressure pi, volume Vi, and temperature Ti.
 We want to change the system to a final state f,
described by pressure pf, volume Vf, and
temperature Tf.
 The procedure of bringing the system from its
initial state to its final state is called a
thermodynamic process, where:
 Heat Q can be added to the gas (positive
heat) or withdrawn from it (negative heat), by
regulating the temperature of the adjustable
thermal reservoir.
 Work W can be done by the gas by raising the
piston (positive work) or lowering it (negative
work).
 We assume that all such change occur slowly  the system is
always in (approximate) thermal equilibrium  the temperature
of every part of the gas is always the same.
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Chapter 18
First Law of Thermodynamics
A Closer Look at Heat and Work
 Suppose a few lead shot is removed, allowing the
gas to push the piston and remaining shot upward
→
through a differential →
displacement ds with an
upward force F.
 The differential work dW done by the gas during
the displacement is:
dW  F  d s  ( pA)(ds)
 ( p)( Ads )
 ( p )(dV )
 If the gas changes its volume from Vi to Vf, the
total work done by the gas is:
W   dW 
Vf
 pdV
Vi
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Erwin Sitompul
Thermal Physics 2/10
Chapter 18
First Law of Thermodynamics
A Closer Look at Heat and Work
 A system can be taken from a given initial state
to a given final state by an infinite number of
processes.
 Heat may or may not be involved.
 In general, the work W and the heat Q will have
different values from different processes.
 Heat and work are path-dependent quantities.
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Erwin Sitompul
Thermal Physics 2/11
Chapter 18
First Law of Thermodynamics
A Closer Look at Heat and Work
ia : Q > 0
ib : Q < 0
af : Q < 0
b
Positive work
More positive work
bf : Q > 0
Less positive work
Wighf > Wicdf
Amount of work
depends on path
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Negative work
Net positive work
Erwin Sitompul
Thermal Physics 2/12
Chapter 18
First Law of Thermodynamics
Checkpoint
The p-V diagram on the right shows six curved paths (connected by
vertical paths) that can be followed by a gas. Which two of the
curved paths should be part of a closed cycle if the net work done by
the gas during the cycle is to be at its maximum positive value?
 c and e, which will result the
maximum area enclosed by a
clockwise cycle
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Thermal Physics 2/13
Chapter 18
First Law of Thermodynamics
The First Law of Thermodynamics
 Previously we have seen, that when a system changes from a
given initial state to a given final state, both the work W and the
heat Q depend on the nature (path) of the process.
 The quantity (Q–W) is however the same for all process. It
depends only on the initial and final stated.
 The quantity (Q–W) must represent a change in some intrinsic
property of the system. We call it the internal energy Eint and we
write:
Eint  Eint, f  Eint,i  Q  W
First Law of Thermodynamics
 If the thermodynamic system undergoes only a differential change,
we can write the first law as:
dEint  dQ  dW
First Law of Thermodynamics
 The kinetic energy or the potential energy of the system as a
whole is assumed to be unchanged, ΔK = ΔU = 0.
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Chapter 18
First Law of Thermodynamics
Checkpoints
The figure on the right shows four paths on a
p-V diagram along which a gas can be taken
from state i to state f. Rank the path
according to:
(a) the change ΔEint in the internal energy of
the gas,
(b) the work W done by the gas, and
(c) the magnitude of the energy transferred
as heat Q between the gas and its
environment,
greatest first.
(a) All tie
(b) 4, 3, 2, 1. (area under the curve)
(c) 4, 3, 2, 1. (Q = ΔEin + W)
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Thermal Physics 2/15
Chapter 18
First Law of Thermodynamics
Some Special Cases of the First Law of Thermodynamics
 Adiabatic processes. An adiabatic process is one that occurs so
rapidly or occurs in a system that is so well insulated that no
transfer of energy as heat occurs between the system and its
environment.
Eint  W
 Constant-volume processes. If the volume of a system is held
constant, that system can do no work, since W = pΔV.
Eint  Q
 Cyclical processes. There are processes in which, after certain
interchanges of heat and work, the system is restored to its initial
state. In this case, no intrinsic property of the system can change.
Q W
 Free expansions. These are adiabatic processes in which no
transfer of heat occurs between the system and its environment
and no work is done on or by the system.
Eint  0
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Chapter 18
First Law of Thermodynamics
Some Special Cases of the First Law of Thermodynamics
● Free expansion
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● Adiabatic process
Thermal Physics 2/17
Chapter 18
First Law of Thermodynamics
Checkpoint
For one complete cycle as shown in the p-V diagram here, are
(a) ΔEint for the gas and
(b) the net energy transferred as heat Q
positive, negative or zero?
(a) ΔEint = 0
(closed cycle)
(b) Q < 0
(Q = W, while W < 0 for ccw cycle)
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Chapter 18
First Law of Thermodynamics
Problem
Let 1.00 kg of liquid water at 100 °C be converted
to steam at 100°C by boiling at standard
atmospheric pressure (= 1.00 atm or 1.01×105
Pa) in the arrangement as shown below. The
volume of that water changes from an initial value
of 1.00×10–3 m3 as a liquid to 1.671 m3 as steam.
(a) How much work is done by the system during
this process?
W  pV
 (1.01105 )(1.671 1.00 103 )
 168771 J  168.77 kJ
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Thermal Physics 2/19
Chapter 18
First Law of Thermodynamics
Problem
Let 1.00 kg of liquid water at 100 °C be converted
to steam at 100°C by boiling at standard
atmospheric pressure (= 1.00 atm or 1.01×105
Pa) in the arrangement as shown below. The
volume of that water changes from an initial value
of 1.00×10–3 m3 as a liquid to 1.671 m3 as steam.
(b) How much energy is transferred as heat during
the process?
Q  LV m  (2256 k)(1.00)  2256 kJ
(c) What is the change in the system’s internal energy during the
process?
Eint  Q  W
 2256 k  168.77 k
 2087.23 kJ  23087 MJ
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Thermal Physics 2/20
Chapter 18
First Law of Thermodynamics
Class Group Assignments
1. A 0.050-kg ingot of metal is heated to 200.0°C and then dropped into a beaker
containing 0.400 kg of water initially at 20.0°C. The final equilibrium temperature
of the mixed system is 22.4°C. The specific heat of the metal is (in J/kg·K):
(a) 217.8
(d) 681.0
(b) 451.9
(e) 1368.1
(c) 562.2
2. A series of thermodynamic processes is shown in the next
p-V diagram. Heat is added to the system, during process
ab as much as 150 J and during process bd 600 J. The
internal energy changes (ΔEint) of process ab and abd are,
respectively,:
(a) 90 J and 510 J
(d) 150 J and 360 J
(b) 310 J and 360 J
(e) 310 J and 240 J
(c) 150 J and 510 J
3. A system undergoes an adiabatic process in which its internal energy increases by
20 J. Which of the following statement is true?
(a) 20 J of work was done on the system
(b) 20 J of work was done by the system
(c) The system received 20 J of energy as heat
(d) The system lost 20 J of energy as heat
(e) None of the above are true
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Chapter 18
First Law of Thermodynamics
Homework 2
1. (18-41) (a) Two 50-g ice cubes are dropped into 200 g of water in a thermally
insulated container. If the water is initially at 25°C, and the ice comes directly from
a freezer at –15°C, what is the final temperature at thermal equilibrium? (b) What
is the final temperature if only one ice cube is used?
2. (18-49) When a system is taken from state i to state f
along path iaf, Q = 50 cal and W = 20 cal. Along path
ibf, Q = 36 cal.
(a) What is W along path ibf?
(b) If W = –13 cal for the return path fi, what is Q for
this path?
(c) If Eint,i = 10 cal, what is Eint,f?
(d) If Eint,i = 10 cal and Eint,b = 22 cal, what is Q for
path ib and bf?
 Deadline: 23 April 2015.
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