The First Law of Thermodynamics
Modern Physics
August 31, 2016
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Energy Conservation
— 
— 
In this section, we will discuss the concepts of heat,
internal energy, and work.
In PHY 140, we had talked about conservation of
energy (Chapters 7 and 8 of Serway) and had said
that for a system
◦  ΔKE + ΔPE + ΔEint =
W + Q + TMW + TMT + TET + TER
◦  Change in kinetic energy + potential E. + internal E. =
work + heat + mechanical waves, e.g., sound + matter +
electrical + radiation
— 
We’ll explore the relationship between these terms,
in particular these two:
◦  Internal Energy (Eint): All the energy of a system associated
with its microscopic components, e.g., atoms/molecules
◦  Heat (Q): This is the process of transferring energy across
the boundary of a system due to a temperature difference
between the system and its surroundings
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Work
—  Units
of work: Force * distance = N-m =
Joule
—  Heat: Originally, was defined by means of
◦  1 calorie = energy required to raise the
temperature of 1 g of water from 14.5 to
15.5°C
◦  Now, we use the Joule [1 cal = 4.186 J]
–  Note that when we eat food or do exercise, we talk
about consuming/burning off Calories. Actually, this
usage of Calorie (food Calorie) = 1,000 calories (c)
—  We
did not understand this fully initially as
same
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Example
—  If
you lift 50 kg (~110 pounds) through
2.0 m (~6.6 feet), then you’re performing
work
◦  (a) How much work? (In J, cal, and kcal/Cal)
◦  (b) To burn off 2,000 (food) Calories, how
many times would you have to lift this weight?
◦  (c) How long would part (b) take? (Estimate)
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Specific Heat
— 
When we add / subtract heat to a system, usually only
its temperature changes
◦  Except when you have a phase transition: solid->liquid,
liquid->gas, etc.
◦  Also, assuming KE and PE remain the same
— 
Smaller the value it
means it “heats up”
faster, and can
transfer heat away
from a hot object
efficiently (quickly!)— 
c
Water(liq)
Iron
Copper
Silver
Gold
Mercury(liq)
— 
4186
448
387
234
129
140
The amount of energy needed to raise the
temperature of a system by, say, 1°C, depends on what
it is made out of.
◦  For 1 kg of water, we need 4,186 J(oules)
◦  For 1 kg of copper, we need 387 J
We define Q = C0 * ΔT, where C0 is the heat capacity
◦  Change in Eint of system, if no work et al. done
OR, Q = m * c * ΔT, where m is mass and c is called
the SPECIFIC HEAT (not light speed..)
◦  That is, c = Q / ( m * ΔT ) -> Units of J / ( kg * °C)
◦  Think: if very high, what are the implications? And if is low?
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Calorimetry
— 
How do we measure the SPECIFIC HEAT of
something? Consider this:
Qcold
Qhot
mwater
Twater ,cwater
Boundary, which isolates the system
This object has mX, TX, cX (unknown).
We immerse an object of mass mX and temperature
TX and specific heat cX in a bath of water with mW,
TW, and cW
—  If TX > TW, then heat will be transferred from the
object to the water, and after a while they will come
to equilibrium, i.e., the same temperature Tf
—  So, -Qhot = Qcold -> the negative sign just means that
Qhot is *leaving from* mX
— 
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continued
Energy transfer for water = mW * cW * ( Tf – TW )
[>0]
—  For hot object = mX * cX * ( Tf – TX ) [ < 0 ]
— 
— 
Equating the two
◦  mW * cW * ( Tf – TW ) = -mX * cX * ( Tf – TX ) =
mX * cX * ( TX – Tf )
— 
Re-arranging to get cX, we have
◦  cX = mW * cW * ( Tf – TW ) / [ mX * ( TX – Tf ) ]
(Here, we have ignored the container. If you include
the mass/temperature/specific heat of the container,
then the equations will become slightly modified)
—  You will be getting a homework problem like this
— 
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Example
—  We
put a metal block of mass = 0.05 kg
and at a temperature of 200°C into a bath
of water with mass = 0.400 kg and initial
temperature of 20°C. The equilibrium
temperature is 22.4°C.
◦  (a) What is c of the metal?
◦  (b) From our table, it appears it is made of…?
◦  (c) Why didn’t we get the “correct” answer?
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Latent Heat
In previous discussions, temperature increased
when we transferred heat into a system
—  But, when we have a phase transition, i.e., the
state of a system changes from solid -> liquid, or
liquid->gas, the heat goes into changing the state
without changing the temperature.
—  Imagine that you have 1 kg of ice at -5°C. Now,
transfer heat into the ice.
— 
◦  The temperature of the ice goes from -5°C to 0°C
◦  Ice -> Water @ 0°C (PHASE TRANSITION, which is
governed by a quantity called LATENT HEAT)
◦  Temperature of water goes from 0°C -> final value
— 
— 
Q = L * mi or L = Q / mi (mass of ice @ 0°C)
Similarly, we have latent heat when water @
100°C turns into steam/water vapor @ 100°C
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—  Note
that 1 kg of steam at 100°C will
burn you more severely than 1 kg of
water at 100°C. Why?
Examples
—  We
have 25.0 liters of LN2 [density = 0.81
kg/L] @ its boiling point. We put a 25 W
heating element in this container and
leave it there for 4 hours. How much of
the LN2 boils off? (Note 1 Watt = 1 J/sec)
MelBngPt. Latentheat BoilingPt.
Water(Liq) 0°C
Copper
1,083°C
Silver
960.8°C
LiquidNitrogen(LN2)
offusion
[Jperkg]
3.33E+05100°C
1.34E+051,187°C
8.82E+042,193°C
-196°
Latentheat
of
vaporizaBon
[Jperkg]
2.26E+06
5.06E+06
2.33E+06
2.00E+05
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Work and Heat and the First Law
—  Some
definitions
◦  State variables define the state of a system, e.g.,
Pressure, Temperature,Volume, Eint (also KE/PE).
–  A state of a system can be specified only when it is in
thermal equilibrium internally, i.e., all parts of the system
are at the same temperature/pressure
◦  Transfer variables describe the transfer of energy
in / out of the system => these variables cause a
change in the state of the system
–  So far, we have discussed heat as a transfer variable
— someHow
gas
about work? What would happen if we
were to push down on the piston, compress
the gas inside it?
piston
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Work as it Relates to a Gas
We push the piston “quasi-statically,” that is, the
system is assumed to be in internal and thermal
area = A
equilibrium at all times
—  How much work is done on the gas? At any
gas
instant |F| = P * A (force due to gas on piston, and
vice versa)
P, V —  dW = F * dr (vector dot product symbol * here)
—  dW = -F ^j * dy ^j = -P * A * dy = - P * dV
—  If dV is negative then the gas is being compressed
=> work done on gas, i.e., dW > 0
—  If the gas were to expand, dV > 0 and dW < 0
—  If the volume Vremains the same, then dW = 0
—  W = ∫dW = -∫P dV. In general, P will depend on V.
— 
dy
->
->
f
Vi
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Examples on Board
—  Area
under the curve is the total work
performed/done.
—  Work done depends on initial / final
points, but it also depends on the path
◦  At constant volume, no work is done
◦  When we increase the pressure at constant
volume, we can do so by heating up the gas
—  This
is the connection between work and
heat -- both may be used for changing the
state of a system
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Part II: 1st Law of Thermo
= Q + W (this is a special case of the
energy conservation law, where only Q and
W are permitted in the equation)
—  ΔEint
—  In
an isolated system, no work is done on it
and no heat is transferred => ΔEint = 0 or
Eint_initial = Eint_final
—  Consider a non-isolated system that starts
and ends up at same state (cyclic process)
◦  ΔEint = 0. Remember that Eint is a state variable
◦  Q = -W for a cyclical process. Law!
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Example on Board
—  (a)
Find the net energy transferred to the
system by heat as we go from A->B->C->A,
i.e., a cyclic process
◦  Units: 1 atm = 1.013e5 N / m ^ 2 = 1.013e5 Pa
= 101.3 kPa (Remember: e or E is short-hand or
calculator/computer notation for “x 10 ^”)
◦  As we go around the closed curve, ΔEint = 0
(we start from point A and end up at point A)
◦  Let’s analyze each leg separately now
—  (b)
Exercise for you: what happens if we go
the other way, i.e., A->C->B->A
◦  Would the answer change? Why or why not??
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Applications and Terminology
ADIABATIC process: One in which no heat
enters or leaves the system, that is Q = 0 =>
ΔEint =W (from first law)
—  ISOTHERMAL process: One which occurs at
constant temperature
— 
◦  For an ideal gas, ΔT = 0 => ΔEint = 0 or Q = W
◦  Heat can enter a system and leave as work, keeping
the temperature constant
— 
ISOBARIC process: At constant pressure. Both Q
and W are, in general NOT equal to zeroes
◦  See A<-> C from the previous example
— 
ISO-VOLUMETRIC process: At constant volume
=> W = 0 and ΔEint = Q.
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Energy Transfer Mechanisms
— 
3 main --
◦  Conduction -> Two systems in contact with each
other and one is hotter than the other. Heats flows
from hot->cold.
–  Usually due to atomic vibrations
◦  Convection -> Energy transfer by movement of hot
materials
–  Take a pot of water and put it on an electric stove
–  Stove transfers energy to bottom of pot by conduction.
–  Bottom of pot heats up a thin layer of water right next to it
(again mainly by conduction)
–  This warm water has lower density (than cool water), so it
rises and the denser, colder water at the top of the water
sinks and gets heated
–  This process is convection
◦  Radiation
— 
Can mix
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Radiation
— 
Why do you get warmed by the sun?
— 
ALL objects give off electromagnetic radiation
◦  After all, there is vacuum between the earth and sun!
◦  The frequency/wavelength of this radiation depends on
the temperature of the object
◦  The temperature of the outer surface of the sun is
~6,000 K, and the peak of its EM radiation is ~yellow
— 
The rate at which a body radiates energy
Power=Energy/Time=σ * A * ε * T4 -> Stefan’s law.
◦  Sigma: constant = 5.67e-8 W / ( m^2 * K^4 )
◦  A: Area
◦  Epsilon: emissivity (=absorptivity) of object (0 to 1)
◦  T: temperature in Kelvin
— 
When ε = 1 -> Known as BLACKBODY
◦  We’ll return to this later. A BIG DEAL!
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Final Example
—  Assume
that a 100-Watt light bulb has a
tungsten filament, and radiates off 2 W of
power (other 98 W is carried away by
conduction and convection). If the area of a
filament is 0.250 mm^2, and the emissivity is
0.950, what is its temperature? (This is a
simple “plug & chug” problem.)
◦  Note that the melting point of Tungsten is 3,683
K (fun fact – not needed to solve problem)
—  Supposedly, it
took Thomas Edison a while
to “perfect” the light bulb, because he could
not find an element that would be stable for
a long period of time! (Did not invent)
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Homework
—  5
problems due next Wednesday
—  Copies available here
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