Download Thermodynamics: C l i t H t alorimetry, Heat

Thermodynamics:
C l i t
Calorimetry,
H t
Heat
• Physics 1
Internal Energy
Internal energy (also called thermal energy) is the energy an
object
j or substance is due to the kinetic and potential
p
energies
g
associated with the random motions of all the particles that make
it up. The kinetic energy is, of course, due to the motion of the
particles.
i l
To
T understand
d
d the
h potential
i l energy, imagine
i
i a solid
lid in
i
which all of its molecules are bound to its neighbors by springs.
As the molecules vibrate
vibrate, the springs are compressed and
stretched. (Liquids and gases are not locked in a lattice structure
like this.))
The hotter something is, the faster its
molecules are moving or vibrating, and the
hi h its
higher
i temperature. Temperature
T
i
is
proportional to the average kinetic energy
of the atoms or molecules that make up a
substance.
Internal Energy vs. Heat
The term heat refers is the energy
gy that is transferred from one
body or location due to a difference in temperature. This is similar
to the idea of work, which is the energy that is transferred from
one body
b d to another
h due
d to forces
f
that
h act between
b
them.
h
H
Heat iis
internal energy when it is transferred between bodies.
Technically, a hot potato does not possess heat; rather it possesses
a good deal of internal energy on account of the motion of its
molecules If that potato is dropped in a bowl of cold water,
molecules.
water we
can talk about heat: There is a heat flow (energy transfer) from the
hot potato to the cold water; the potato
potato’ss internal energy is
decreased, while the water’s is increased by the same amount.
Isolated, Closed and Open Systems 1
• A system is the portion of the physical world being studied.
• The system plus surroundings comprise a universe.
• The boundaryy between a system
y
and its surroundings
g is the
system wall.
• If heat cannot p
pass through
g the system
y
wall,, it is termed an
adiabatic wall, and the system is said to be thermally isolated
or thermally insulated.
• If heat can pass through the wall, it is termed a diathermal
wall.
• Two systems connected by a diathermal wall are said to be in
thermal contact.
Isolated,, Closed and Open
p Systems
y
2
• An isolated system cannot exchange mass or energy with its
surroundings.
di
• The wall of an isolated system
y
must be adiabatic.
• A closed system
y
can exchange
g energy,
gy, but not mass,, with
its surroundings.
• The energy exchange may be mechanical (associated with a
volume change) or thermal (associated with heat transfer
through a diathermal wall).
• An open
p system
y
can exchange
g both mass and energy
gy with
its surroundings.
Isolated, Closed and Open Systems 3
Isolated
System
Neither energy nor mass
can be exchanged.
Closed
System
Open
System
Energy, but not mass can
be exchanged.
Both energy and mass can be
exchanged.
Units ffor Heat
Like any type of energy, the SI unit for heat is the Joule.
Another common unit is the calorie, which is approximately
the amount of heat energy needed to raise one gram one degree
Celsius. 1000 calories are in a Calorie, which is used to
measure the energy in foods (that the human body can make
use of)
of). The British thermal unit (BTU) is approximately the
energy needed to raise one pound of water one degree
Fahrenheit.
a e e t.
1 cal = 4.186 J
1 BTU = 1055 J = 252 cal
Internal vs. “External” Energy
Suppose a 1 kkg bl
S
block
k off iice is
i sliding
lidi att 7 m/s.
/
This is the speed of the center of mass of the
block, not the speed of each individual water
molecule. To calculate the total kinetic energy
of the water molecules of the block directly, we would have to know
the speed of each molecule as it vibrates, all 33.4 trillion trillion of
them! (In practice we would just measure the temperature & mass of
the ice.)
ice ) The internal energy of the ice does not depend on the motion
of the whole body relative to Earth. What matters is the motion of the
molecules in the reference frame of the block. O
Otherwise,, it would be
impossible for a cold object to move quickly or a hot one to move
slowly.
Note: If friction is present, it could do work on the ice
and convert some of the “uniform” kinetic energy of the block into
“random” kinetic energy of its molecules (internal energy).
Regardless the total energy of the block is the kinetic energy of the
Regardless,
center of mass + the internal energy: Ktotal = Kcm + Eint
vcm = 7 m/s
/
Temperature Scales
F h h i water ffreezes at 32 °F;
Fahrenheit:
°F bboils
il at 212 °F
Celsius: water freezes at 0 °C; boils at 100 °C
Kelvin: water freezes at 273.15 K; boils at 373.15 K
A change
h
off 100 °C corresponds
d to a change
h
off 180 °F.
°F Thi
This means
5 C° = 9 F° or 1 C° = 1.8 F° Note that the degree symbol is on the
opposite side of the letter,
letter indicating that we
we’re
re talking about
temperature differences. In other words, five steps on the Celsius
q
to nine steps
p on the Fahrenheit scale, but 5 °C is
scale is equivalent
certainly not equal to 9 °F. Since these scales are linear, and they’re
offset by 32 °F, we get the conversion formula: F = 1.8C + 32
One step on the Kelvin scale is the same as one step on the Celsius
scale. These scales are off by 273.15 K, so: K = C + 273.15
Room temperature is around 293 kelvins, which is 20 °C, or 68 °F.
Absolute Zero & the Kelvin Scale
The Kelvin
Th
K l i scale
l is
i setup
t so that
th t its
it zero point
i t is
i the
th coldest
ld t possible
ibl
temperature--absolute zero, at which point a substance would have
zero internal energy. This is -273.15
273.15 °C,
C, or -459.69
459.69 °F.
F. Absolute zero
can never be reached, but there is no limit to how close we can get to
it. Scientists have cooled substances to within 10-5 kelvins of absolute
zero. How do we know how cold absolute zero is, if nothing has ever
been at that temperature? The answer is by graphing Pressure vs.
Temperature for a variety of gases and extrapolating.
extrapolating
P
A gas exerts no
ppressure when at
absolute zero.
T (°C)
-273.15 °C
0 °C
Thermal Equilibrium
Two bodies
T
b di are said
id to
t be
b att thermal
th
l equilibrium
ilib i
if they
th are att the
th
same temperature. This means there is no net exchange of thermal
energy between the two bodies. The top pair of objects are in
contact, but since they are at different temps, they are not in thermal
equilibrium, and energy is flowing from the hot side to the cold side.
heat
hot
26 °C
cold
26 °C
No net heat flow
The
and,
are in
Th two
t purple
l objects
bj t are att the
th same temp
t
d therefore
th f
i
thermal equilibrium. There is no net flow of heat energy here.
Heat Transfer Processes
Heat energy can be transferred from one body to another in three
different ways.
• Conduction: Energy is transferred when two objects are in direct
contact Molecules of the hotter object bump into molecules of the
contact.
colder object and cause them to speed up, warming the colder object.
• Convection: Energy is transferred from one body to a cooler one
via currents in a fluid (a gas or liquid).
• Radiation:
R di ti
All objects,
bj t att any temperature,
t
t
radiate
di t electromagnetic
l t
ti
radiation (light of visible and invisible wavelengths). Unlike
conduction & convection,
convection no medium (matter of any type) is
necessary for heat transfer through radiation. Objects absorb
radiation as well. At thermal equilibrium it will absorb as much as it
radiates.
Thermal Conductivity, k
Heatt transfer
H
t
f via
i conduction
d ti was described
d
ib d a few
f slides
lid back.
b k Thermal
Th
l
conductivity, k, refers how easily heat can move through a material.
Metals have high thermal conductivity, meaning heat passes through
them readily. Wood is a fairly good insulated of heat, and styrofoam is
even better. These materials have low thermal conductivities. k is
very low for air as well. (Attic insulation and styrofoam cups trap air,
making them good insulators.) Heat from a boiler passes through all
sides of its metal enclosure.
enclosure The rate at which heat is transferred is
given by:
H = k A ( T2 - T1 )
L
T2
T1
heat
eat
A = area of side wall
L = thickness
thi k
off wall
ll
k = thermal conductivity of the metal
T2 - T1 = temperature difference
H is simply power, and its SI unit is the Watt.
SI Units ffor Thermal Conductivityy
k A (T - T )
H=
2
1
L
k must have units that cancel out all the units on the right,
leaving only the units for H. The units are:
W
m·K
m
K
or equivalently,
W
m·°C
m
C
Since one kelvin is as big a change in temp as one degree
Celsius, these units are equivalent.
Note: k for thermal conductivity is not the same as the k
in Hooke’s Law in which it represents the spring constant!
Thermopane Windows
heat
heat
In a house we often want to prevent heat from getting
in or getting out. Windows can be problematic.
Thermopane windows have two or more panes of
glass with air or some other gas between the panes.
Which type of window
window, a double pane or a thick
single pane, is better for minimizing heat transfer, if
the total thickness is the same?
answer:
There is more glass in the single pane window to
block the heat, but the air in between the panes of the
do ble pane window
double
indo has thermal conductivity
cond cti it that is
about 35 times lower than that of the glass itself. So
much more heat would be transferred through the
single pane.
Triple pane vs. Double pane
If they are of the same total thickness and pane
thickness,, which is better at minimizingg heat transfer,,
a double or triple pane window?
heat
answer:
heat
The double pane window has more air between the
outer panes
panes, so its thermal conductivity is lower.
lower
However, air is a mobile medium, and convection
currents can shuttle warm air from the warm side to
the cold side. On the warm side the air rises, moves
across the the cold side, and sinks, moving in a loop
andd carrying
i its
i energy from
f
the
h warm side
id to the
h
cold side. The middle pane in the triple pane window
reduces the energy transfers due to convection and is
the better window (but probably more expensive).
R Value
The R value of a material is its “thermal resistance” and refers to
how good an insulator is. Here’s how it’s defined:
L
R =
k
As in previous equations:
L = the thickness of the material
k = thermal conductivity of the material
Note that the R value is inversely proportional to thermal
conductivity, meaning good heat conductors have a low R value
and are poor insulators. Also, the R value is directly proportional
to the thickness of the material, meaning the thicker it is, the better
it insulates.
i l t
Thus,
Th more insulation
i l ti in
i the
th attic
tti can save energy.
Wind & Heat Loss
A breeze can cool us off in the summer, and wind can make us feel
colder in the winter. Why is this?
answer:
When we sweat the perspiration absorbs body heat, and when it
evaporates, it takes this heat with it. This is called evaporative
cooling. A steaming cup of hot chocolate cools in the same way.
p us warm in the winter is because it traps
p air
The reason a coat keeps
that is heated by our bodies. (Wearing layers is like having a triple
pane window.) A thin layer of stagnant air also surrounds the outside
walls
ll off buildings
b ildi
andd helps
h l insulate
i l them.
h
Wind
i d tends
d to blow
bl this
hi
warm air away, along with its heat.. The windier it is, the colder it feels
to us,
us and the greater the heat loss from a building
Trees around you home can save energy in two ways: blocking wind
in the winter;
inter; and shielding your
o r home from excess
e cess solar radiation in
the summer.
Laws of Thermodynamics
(examples upcoming)
• Zeroth Law: If object A is in thermal equilibrium with object B,
and if object B is in thermal equilibrium with object C,
C then objects
A and C are also in equilibrium. This is sort of a “transitive
pproperty
p y of heat.”
• First Law: Energy is always conserved. It can change forms:
kinetic potential,
kinetic,
potential internal etc.,
etc but the total energy is a constant.
constant
Another way to say it is that the change in thermal energy of a
system is equal to the sum of the work done on it and the amount
of heat energy transferred to it.
• Second Law: Duringg any
y natural pprocess the total amount of
entropy in the universe always increases. Entropy can be defined
informally as a measure of the randomness or disorder in a system.
Heat flows naturally from a hot to cooler surroundings as a
consequence of the second law.
Change in Entropy Equation
Because most systems are many up off so many particles,
i l calculating
l l i
entropy via probabilities would be very difficult. Fortunately, we are
normally concerned only with changes in entropy
entropy. If we have a
system in which energy is not changing forms, the change in entropy
is defined as:
∆Q
∆S =
T
∆S = change in entropy
∆Q = change in internal energy (heat flow)
T = absolute temperature
The 2nd Law of Thermodynamics says that during any process:
∆Suniverse = ∆Ssystem + ∆Ssurroundings ≥ 0
Change in Entropy Example
A glass
l rodd is
i heated
h t d andd then
th blown
bl
by
b a
glassblower. When it is at 185°C it is brought
outside to cool. 3200 J of heat are transferred
from the glass to the air, which is at 18°C.
Find the change in entropy of the universe:
∆Suniverse = ∆Ssystem + ∆Ssurroundings
= ∆S
Sglass + ∆S
Sair
∆Qglass
∆Q
Qair
+
= T
Tair
glass
-3200
3200 J 3200 J
=
+
458 K 291 K
= -7 J/K + 11 J/K = +4 J/K
Change in Entropy Example
(cont.)
A th
As
the glass
l cooled
l d we assumedd that
th t the
th air
i temp
t
didn’t
did ’t go up
appreciably due after the heat transfer, which would have complicated the problem. Important points:
• The temps were converted to kelvins.
• The
Th glass
l lost
l as muchh thermal
h
l energy as air
i gained,
i d as the
h 1stt Law
L
requires.
• ∆Q
Qglass is negative since the glass lost thermal energy so ∆S
Sglass is
also negative.
• ∆Qair is positive since the air gained thermal energy so ∆Sair is
also positive.
• Even though the ∆Q’s are the same size, the ∆S’s aren’t, since the
temps are different.
• The positive ∆S is greater than the negative ∆S, as the 2nd Law
requires.
Second Law Consequences
• Heat
H t will
ill nott flow
fl from
f
a cold
ld body
b d to
t a hot
h t body.
b d
• “Reverse diffusion” is a no-no (such as smoke from a fire
isolating itself in a small space).
• An object or fluid of uniform temperature (no matter how hot)
cannot do useful work. (There must be temperature difference
so that there will be a heat flow, which can be used to do work.)
• The various forms of energy tend to degrade over time to
thermal energy. This represents useful, low probability forms of
energy converting into an unusable, high probability form.
• Without input
p of energy,
gy, bodies tend to reach thermal
equilibrium. (We can maintain temperature differences via
refrigerators or heating units, but this requires energy.)
continued on next slide
Second Law Consequences
(cont.)
• Any time we do something that decreases the entropy of a
system the energy we expend in doing it increases the entropy
system,
of the surroundings even more.
• A perpetual motion machine is impossible to make.
make A
perpetual motion machine is a device that would absorb thermal
energy from a hot body and do as much work as the energy it
absorbed.
• During any process the entropy of the universe cannot
decrease. Expending energy to decrease the entropy of a system
will lead to an increase in entropy for the surrounding by a
greater amount.
Thermal Equilibrium and the Zeroth Law
• If warm
a a
and
d coo
cool objects a
are
ep
placed
aced in tthermal
e a co
contact,
tact,
energy, known as heat, flows from the warm to the cold object
until thermal equilibrium is established.
• Zeroth Law of Thermodynamics
Two systems
systems, separately in thermal equilibrium with a third
system, are in thermal equilibrium with each other.
• The property which the three systems have in common is
known as temperature θ.
• Thus the zeroth law may be expressed as follows:
if θ1 = θ2 and θ1 = θ3, then θ2 = θ3.
25
Thermodynamic Variables
• Thermodynamic variables are the observable macroscopic
variables of a system
system, such as P
P, V and T
T.
• If the are used to describe an equilibrium state of the system,
they are known as state variables
variables.
• Extensive variables depend on the size of the system; e.g.
mass, volume,
l
entropy,
t
magnetic
ti moment.
t
• Intensive variables do not depend on size; e.g. pressure,
temperature magnetic field
temperature,
field.
• An extensive variable may be changed to an intensive
variable, known as a specific
f value, by dividing it by a
suitable extensive variable, such as mass, no.of kmoles, or no.
of molecules.
molecules
• Example: the specific heat is normally (heat capacity)/(mass).
Equilibrium States
• An equilibrium
q
state is one in which the p
properties
p
of the
system do not change with time.
• In many cases, an equilibrium state has intensive variables
which
hi h are uniform
if
th
throughout
h t th
the system.
t
• A non-equilibrium state may contain intensive variables which
vary in
i space and/or
d/ titime.
• An equation of state is a functional relationship between the
state
t t variables;
i bl
e.g. if P
P,V
V and
d T are th
the state
t t variables,
i bl
th
then
the equation of state has the form f(P, V, T) =0.
• In
I 3-dimensional
3 di
i
lP
P-V-T
V T space,
an equilibrium state is represented by a point,
and the equation of state is represented by a surface
surface.
27
Processes 1
• A process refers to the change of a system from one
equilibrium state to another.
• The initial and final states of a process are its end-points.
end points
• A quasistatic process is one that takes place so slowly that
th system
the
t
may be
b considered
id d as passing
i th
through
ha
succession of equilibrium states.
• A quasistatic process may be represented by a path (or line)
on the equation-of-state surface.
• If it is non-quasistatic, only the end-points can be shown.
• A reversible process is one the direction can be reversed by
an infinitessimal change of variable.
• A reversible process is a quasistatic process in which no
dissipative forces, such as friction, are present.
• A reversible change must be quasistatic, but a quasistatic
process need not be reversible; e.g. if there is hysteresis.
28
Processes 2
• An isobaric process is one in which the pressure is constant.
• An isochoric process is one in which the volume is constant
constant.
• An isothermal process is one in which the temperature is
constant.
• An adiabatic process is one in which no heat enters or leaves
y
i.e. Q = 0.
the system;
• An isentropic process is one in which the entropy is constant.
• It is a reversible adiabatic process.
• If a system is left to itself after undergoing a non-quasistatic
process, it will reach equilibrium after a time t much longer than
the longest relaxation time τ involved; i.e. t » τ.
• Metastable equilibrium occurs when one particular relaxation
time τ0 is much longer than the time ∆t for which the system is
observed; i.e. τ0» ∆t .
29
Three Types of Process
Isothermal process
System
Adiabatic process
P
Adiabat
Isotherm
Heat bath or reservoir
Adiabatic free expansion
V
P
●
1
End points
●
2
V
30
Boyle’s Law and the Ideal Gas Scale
h
P
P = PA + gρh
P = a(TC + 273.15)
• Boyle’s Law
• At sufficiently
ffi i tl low
l
pressure,
the product PV was found to
be constant for gases held at
a given temperature θ; i.e.
PV = f(θ) for P → 0
0.
• Fixed points (prior to 1954)
• The
Th iice and
d steam
t
points
i t
were defined to be 0oC and
100oC exactly
exactly.
• The ideal gas (or kelvin)
scale
l was d
defined
fi d as
TK = TC + 273.15.
31
The Ideal Gas Law
• Fixed point (1954)
• The triple point of water is Ttr = 273.16 K, Ptr = 6.0 x 10–3 atm.
• Ideal gas law
PV = nRT or Pv = RT,
•
where n is the no. of kmoles, v is the volume per kmole, T is the absolute
temperature in K
K, and the gas constant R = 8.314
8 314 x 103 J/(K.kmol).
J/(K kmol)
• For a constant quantity of gas, P1V1/T1 = P2V2/T2.
P
P
V
T increasing
V increasing
V
T
P increasing
32
T
Ideal g
gas: Macroscopic
p description
p
• Consider a gas in a container of volume V, at pressure P,
and at temperature T
• Equation of state
– Links these quantities
q
– Generally very complicated: but not for ideal gas
l Equation of state for an ideal gas
™ Collection of atoms/molecules moving randomly
™ No long-range forces
™ Their size (volume) is negligible
PV = nRT
RT
R iis called
ll d the
h universal
i
l gas constant
In SI units,
units R =8.315
=8 315 J / mol·K
mol K
n = m/M : number of moles
Boltzmann’s constant
l
Number of moles: n = m/M
m=mass
M=mass of one mole
™ One mole contains NA=6.022 X 1023 particles :
Avogadro’ss number = number of carbon atoms in 12 g of carbon
Avogadro
• In terms of the total number of particles N
PV = nRT = ((N/NA ) RT
PV = N kB T
kB = R/NA = 1.38 X 10-23 J/K
kB is called the Boltzmann’s constant
• P,
P V,
V and
d T are the
h thermodynamics
h
d
i variables
i bl
Th Ideal
The
Id l G
Gas L
Law
pV
V = nRT
RT
Whatt iis th
Wh
the volume
l
off 1 moll off gas att STP ?
T = 0 °C = 273 K
5
p = 1 atm = 1.01
1 01 x 10 Pa
V RT
=
n
P
8.31 J / (mol ⋅ K ) 273 K
=
1.01 x 10 5 Pa
= 0.0224 m 3 = 22.4 l
Specific Heat
Specific heat is defined as the amount of thermal energy needed
p
Its
to raise a unit mass of substance a unit of temperature.
symbol is C.
For example, one way to express the specific heat of water is
one calorie per gram per degree Celsius: C = 1 cal / (g·ºC), or
4.186 J / (g·ºC). This means it would take 20 cal of thermal
energy to raise 4 grams of water 5 ºC.
Water has a veryy high
g specific
p
heat,, so it takes more energy
gy to
heat up water than it would to heat up most other substances (of
the same mass) by the same amount. Oceans and lake act like
“h sinks”
“heat
i k ” storing
i thermal
h
l energy absorbed
b b d in
i the
h summer andd
slowing releasing it during the winter. Large bodies of water
thereby help to make local climates less extreme in temperature
from season to season.
Specific Heat Equation
Q = mC ∆T
Q = thermal energy
m = mass
C = specific heat
∆T = change in temp
Ex: The specific
p
heat of silicon is 703 J / ((kg·
g ºC).
) How much
energy is needed to raise a 7 kg chunk of silicon 10 ºC?
answer:
Q = 7 kg · 703 J ·10 ºC = 49 210 J
kg· ºC
Note that the units do indeed work out to be energy units.
Calorimetry
Schmedrick
S
h d i k takes
k another
h hhorseshoe
h out off the
h fire
fi when
h it’s
i ’ at
275 ºC, drops in his bucket of water, and this time covers the bucket.
The bucket and cover are made of an insulating material.
material The bucket
contains 2.5 L of water originally at 25 ºC. The 1.9 kg shoe is made of
iron, which has a specific
p
heat of 448 J / ((kg·ºC).
g ) Let’s find the temp
p
of the horseshoe and water once equilibrium is reached.
Let’s assume that the container allows no
h to escape. Then
heat
h the
h 1stt Law implies
i li
that all heat the shoe loses is gained by the
water Since one milliliter of water has a
water.
mass of one gram, the bucket contains
2.5 kgg of water. At thermal equilibrium
q
the water and shoe are at the same temp.
The total thermal energy in the bucket
d
does
not change,
h
b it
but
i is
i redistributed.
di ib d
continued on next slide
Calorimetry
(cont.)
Let T = the equilibrium temperature.
Q lost byy iron = Q ggained byy water
miron Ciron ∆Tiron = mwater Cwater ∆Twater
(1.9 kg)(448 J / kg·ºC)(275 ºC - T) = (2.5 kg)(4186 J / kg·ºC)(T - 25 ºC)
Note
N
t hhow the
th ∆T terms
t
are written
itt so that
th t
each side is positive. We’ve got a simple
linear equation with T on both sides.
sides
Solving it gives us T = 43.8 ºC. This is
the equilibrium temp--the final temp for
both the shoe and water. If T had come
out over 100 ºC, the answer would have
b
been
invalid,
i lid since
i
the
th specific
ifi heat
h t for
f
steam is different than that of water.
Latent Heat
The word “latent” comes from a Latin word that means “to lie
hidden.” When a substance changes phases (liquid ↔ solid or
gas ↔ liquid) energy is transferred without a change in
temperature. This “hidden energy” is called latent heat. For
example,
l to
t turn
t
water
t ice
i into
i t liquid
li id water,
t energy mustt be
b
added to bring the water to its melting point, 0 ºC. This is not
enough however
enough,
however, since water can exist at 0 ºC
C in either the liquid
or solid state. Additional energy is required to change 0 ºC ice
into 0 ºC
C water.
water The energy increases the internal energy of the
water but does not raise its temp. When frozen, water molecules
are in a crystalline
y
structure,, and energy
gy is needed to break this
structure. The energy needed is called the latent heat of fusion.
Additional energy is also needed to change water at 100 ºC to
steam at 100 ºC, and this is called the latent heat of vaporization.
Latent Heat Formula
Q = m Lf or
Q = m Lv
Q = thermal energy
m = mass
L = heat of fusion or vaporization
L is the energy per unit mass needed to change the state
of a substance from solid to liquid or from liquid to gas.
Ex: Lf (the latent heat of fusion) for gold is 6440 J/kg.
J/kg
Gold melts at 1063 ºC. 5 grams of solid gold at this
temp will not become liquid until additional heat is
added. The amount of heat needed is:
(6440 J/kg)
J/k ) (0.005
(0 005 kkg)) = 32 JJ. Th
The li
liquid
id gold
ld will
ill still
till be
b
at 1063 ºC.
Latent Heat / Specific Heat Example
Superman vaporizes
S
i
a 1800 kkg iice monster
t with
ith
his heat ray vision. The ice monster was at
-20
20 ºC
C. After being vaporized he is steam at
135 ºC. How much energy did Superman expend?
Substance
Specific Heat (in J / kg
kg·ºC)
C)
ice
2090
liquid water
4186
steam
1970
For water: Lf = 3.33 ·105 J / kg; Lv = 2.26 ·106 J / kg
Q = (1800 kg)(2090 J / kg·ºC)(20 ºC) heating ice to melting pt.
+ (1800 kg)(3.33 ·10
105 J / kg) ice to water, const. temp of 0 ºC
C
+(1800 kg)(4186 J / kg·ºC)(100 ºC) heating water to boiling pt.
+ (1800 kg)(2.26 ·106 J / kg) water to steam, const. temp of 100 ºC
+ (1800 kg)(1970 J / kg·ºC)(35 ºC) heating steam to 135 ºC
= 5.62 ·109 J total energy expended by Superman
Latent Heat & Entropy
A 10 g ice cube at 0 ºC
C falls to the ground and melts.
melts The temp
outside is 10 ºC. Calculate the change in entropy of the universe
due to the meltingg of the ice only.
y answer:
For the cubie: Q = m Lf = (0.01 kg)(3.33 ·105 J / kg) = + 3330 J.
Thi is
This
i the
th energy absorbed
b b d by
b the
th ice
i from
f
the
th surroundings.
di
∆Sice = ∆Qice /Tice = +3330 J / 273 K = +12.198 J/K.
For the
h surroundings:
di
Q = -3330 J, since
i
the
h surroundings
di
lost
l as
much thermal energy as the cubie gained. The temperature of the
backyard does not decrease significantly
significantly, though
though, with such a small
energy loss. ∆Ssurr = ∆Qsurr /Tsurr = -3330 J / 283 K = -11.767 J/K.
For the universe:
ni erse: ∆Suniv = ∆Ssurr + ∆Sice = 12.198
12 198 J/K - 11.767
11 767 J/K
= +0.431 J/K. Thus, the 2nd Law is satisfied.
Thermal Expansion
As a material heats upp its atoms/molecules move or vibrate more
vigorously, and the average separation between them increases. This
results in small increases in lengths and volumes. Buildings, railroad
t k bridges,
tracks,
b id
andd highways
hi h
contain
t i thermal
th
l expansion
i joints
j i t to
t
prevent cracking and warping due to expansion. The amount of expansion depends
p
on the original
g
length/volume,
g
, the type
yp of material,, and the
change in temp. L is length, V is volume, T is temp, α is the coefficient of linear expansion, and β is the coef. of volume expansion.
Wh a solid
When
lid off a single
i l material
t i l expands,
d it does
d
so proportionally
ti ll in
i
all directions. Since volume has 3 dimensions and length is only 1,
β = 3 α.
L th expansion:
Length
i
∆L
= α ∆T
L
Volume expansion:
cold solid
hot solid
∆V
= β ∆T
V
Thermal expansion
• In most liquids or solids, when temperature rises
– molecules have more kinetic energy
• they are moving faster, on the average
– consequently, things tend to expand (works for a gas)
• amount of expansion ∆L depends on…
– change in temperature ∆T
– original length L0
L0
∆L
– coefficient of thermal expansion
• L0 + ∆L = L0 + α L0 ∆T
• ∆L = α L0 ∆T (linear expansion)
V
• ∆V = β V0 ∆T ((volume expansion)
p
)
V + ∆V
• α ≅ 3β
Thermostats
Bimetallic strips are used in thermostats, at least in some older
ones. When the temperature changes, the strip bends, making or
b ki an electrical
breaking
l t i l circuit,
i it which
hi h causes the
th furnace
f
to
t turn
t
on or shut off. In this model when the strip bends it tilts a bulb
of mercury
mercury, which then bridges two wires and allows current to
flow.
Thermal Expansion & The Concorde
The Concorde
Th
C
d is
i a supersonic
i jet
j t made
d off a heat
h t tolerant
t l
t aluminum
l i
alloy. Its nose tilts down on takeoff and landing so the pilot can see
the runway
runway. In flight the nose comes up to reduce drag,
drag but at a
speed of around 1,350 mph, friction with the air causes significant
heatingg of the plane,
p
,
enough to make the
Concorde grow in
length by 7 inches!
(To maintain this
speed
d ffor one hhour, the
th
Concorde must burn
over 6,700
6 700 gallons of
fuel.)
Thermal Expansion and Teeth
Thermal Expansion and Teeth
Coefficients of linear expansion:
p
Enamel: 11.4 x 10-6 °C
Dentin: 8.3 x 10-6 °C
If you quickly switch from eating/drinking something hot to
somethingg cold,, the brittle enamel will contract more than the
dentin, and develop small cracks called crazes.
Crazing:
Thermal expansion matching is important in choosing
materials for fillings
fillings.
Exercise
Thermal Expansion
• On a cold winter day with T = -10
10 °C
C,
the Eiffel Tower is 300 m tall.
• How tall is it on a hot summer day when
T = 40 °C?
• The Eiffel Tower is made from steel
with a thermal expansion coefficient
α =11 x 10-6/°C
R ll ∆L/L = α ∆T
Recall:
(A) 300.001 m
(B) 300.16
300 16 m
(C) 316 m
(D) 420 m