Internal energy, Temperature,
and Heating
Physics 116
Tues. 2/28, Thurs. 3/2
Macro and Microscopic descriptions
• So far we’ve dealt only with macroscopic gas properties, Pressure,
Volume, and Temperature. How do we connect these to microscopic
properties of the gas? In particular, how can we think about Pressure
and Temperature at the atomic scale?
•A real gas consists of a vast number of molecules, each moving
randomly and undergoing millions of collisions every second.
• Despite the apparent chaos, averages, such as the average number
of molecules in the speed range 600 to 700 m/s, have precise,
predictable values.
• The “micro/macro” connection is built on the idea that the
macroscopic properties of a system, such as temperature or
pressure, are related to the average behavior of the atoms and
molecules.
Creating a mathematical model: How do the macroscopic quantities of pressure
and temperature depend on the kinetic energies of the particles?
1) The pressure arises from the particles hitting the
wall and transferring momentum.
2) Characterize gas by its density
3) How many particles hit a wall in a time Dt?
4) What is the pressure exerted? Recall that force is
the rate of change of the momentum:
F = total momentum change /Dt = Dmv/Dt
And Pressure = Force/Area, so
P = Dmv/(Dt*A)
5) Now relax assumptions that all particles have
same speed and move in the + direction:
vx2  ½ vaverage(x)2 (two directions)
Now relax assumptions that all particles move in
the x-direction:
vaverage(x)2  ⅓ vrms2 (three dimensions)
Where vrms2 = vaverage(x)2 + vaverage(y)2 + vaverage(z)2
Temperature measures average Kinetic
Energy – how fast the particles move!
1) Momentum change for each hit = 2mvx
2) particle density = N/V
3) #hits in a time interval Dt:
#hits = (N/V)(vx*Dt)(Area of wall)
• Total momentum change in time Dt:
Dmv = (2mvx) [(N/V)(vx*Dt)(Area of wall)]
4) Pressure (P) = Force / (Area of wall)
= Dmv / Dt*(Area of wall)
P =(2mvx) (N/V)(vx*Dt)(Area of wall)/Dt*(Area of
wall)
P
= (2mvx) *(N/V)(vx)
P
= (2mvx2) *(N/V)
5) Pressure
= (⅓mvrms2) *(N/V)
P
= (2/3)(½mvrms2) *(N/V)
P
= (2/3)(KEave) *(N/V)
From the ideal gas law: PV=NkBT or P = kBT*(N/V)
• T=(2/3kB)(KEave)
How does this relate to Pressure? Recall: P= NkT/V = nRT/V
Macro-micro connection
• Assumptions for ideal gas:
– # of molecules N is large
– They obey Newton’s laws
– Short-range interactions with
elastic collisions
– Elastic collisions with walls (an
impulse…pressure)
P  Nk BT / V
T=
2
(KE)avg/particle
3kB
2N
P
( KE ) avg/particle
3V
l What we call pressure P is a direct
measure of the number density of
2
B
molecules, and how fast they are
rms
avg
moving (vrms)
l What we call temperature T is a
https://phet.colorado.edu/en/simulation/legac
direct measure of the average
y/gas-properties
translational kinetic energy
v
 (v )
3k T

m
The distribution of molecular speeds in a sample of N2 gas
vrms  (v ) avg
2
3k BT

m
Side note: What is “root mean
square (rms)” and why do we use it?
Why not just use “average”?
A simple (not complete) reason:
Velocities are vectors with sign and
direction, some can be negative.
Calculate the average of the
following particle velocities:
[-2, 5, -8, 9, -4] = 0
But we know the particles have
velocity!
RMS process squares each number then takes
the average, and then square roots that whole
thing, so we get information about magnitudes.
Distribution of velocities for various temperatures of N2 gas –
distribution broadens and the peak shifts
Flux
Flux is how much of something flows through a
particular 2D slice of space in some amount of time.
Examples: water flux out of a faucet, wind (or rain) flux
through your window, flux of rain on your umbrella.
• http://www.shutterstock.com/video/clip13665980-stock-footage-animated-rain-ofcigarettes-against-transparent-background-ink-low-angle-shot-alphachannel.html?src=rel/13677485:2/3p
Catching cigarettes
Marco and Donald are at a celebration sponsored by Phillip
Morris, where cigarettes are just falling from the sky. They
are discussing the quickest way to fill up their identical empty
containers:
Marco: You need to hold the opening of the box
horizontally so it fills fastest.
Donald: It doesn’t matter if you tilt your box, the
cigarettes fall at the same rate either way.
They hold their boxes out, the opening to Marco’s box is
horizontal, and Donald’s is tilted at q degrees to the
horizontal. Whose box fills faster?
A. Donald
B. Marco
C. They fill at the same rate.
Catching cigarettes
Marco and Donald are at a celebration sponsored by Phillip
Morris, where cigarettes are just falling from the sky. They
are discussing the quickest way to fill up their identical empty
containers:
Marco: You need to hold the opening of the box
horizontally so it fills fastest.
Donald: It doesn’t matter if you tilt your box, the
cigarettes fall at the same rate either way.
They hold their boxes out, the opening to Marco’s box is
horizontal, and Donald’s is tilted at q degrees to the
horizontal. Whose box fills faster?
A. Donald
B. Marco
C. They fill at the same rate.
iClicker!
Catching cigarettes
Marco and Donald are at a celebration sponsored by Phillip
Morris, where cigarettes are just falling from the sky. They
are discussing the quickest way to fill up their identical empty
containers:
Marco: You need to hold the opening of the box
horizontally so it fills fastest.
Donald: It doesn’t matter if you tilt your box, the
cigarettes fall at the same rate either way.
They hold their boxes out, the opening to Marco’s box is
horizontal, and Donald’s is tilted at q degrees to the
horizontal. Whose box fills faster?
A. Donald
B. Marco
C. They fill at the same rate.
Let’s break this problem down step by step
Limiting cases
What would be a “limiting case” argument that
supports the outcome of this experiment?
How many cigarettes go through each
unit of area of the opening in each unit
of time?
• It depends on the rate at which the cigarettes
fall – higher rate will result in more cigarettes
• It depends on the size of the opening – a
bigger opening will fill faster
• It depends on the orientation – horizontal will
fill fastest and vertical won’t fill at all – how to
quantify?
How many cigarettes go through each
unit of area of the opening in each unit
of time?
• It depends on the rate at which the cigarettes
fall – higher rate will result in more cigarettes
• It depends on the size of the opening – a
bigger opening will fill faster
• It depends on the orientation – horizontal will
fill fastest and vertical won’t fill at all – how to
quantify?
How many cigarettes go through each
unit of area of the opening in each unit
of time?
• It depends on the rate at which the cigarettes
fall – higher rate will result in more cigarettes
• It depends on the size of the opening – a
bigger opening will fill faster
• It depends on the orientation – horizontal will
fill fastest and vertical won’t fill at all – how to
quantify?
Flux: How many cigarettes go through
each unit of area of the opening in
each unit of time?
Number
that falls in
each time
interval
X
Area of
plane it
falls
through
X
Cosine of
“tilting”
angle
The flux characterizes the flow by combining the
rate and the crowdedness
Flux: How many cigarettes go through
each unit of area of the opening in
each unit of time?
Number
that falls in
each time
interval
X
Area of
plane it
falls
through
X
Cosine of
“tilting”
angle
(selects the
fraction of
Area that
faces the
falling stuff)
The flux characterizes the flow by combining the
rate and the crowdedness
Flux: How many cigarettes go through
each unit of area of the opening in
each unit of time?
Number
that falls in
each time
interval
X
Area of
plane it
falls
through
X
Cosine of
“tilting”
angle
(selects the
fraction of
Area that
faces the
falling stuff)
The flux characterizes the flow by combining the
rate and the crowdedness
How might you characterize rainfall
onto an umbrella using the idea of
flux?
• http://www.shutterstock.com/video/clip9063377-stock-footage-animated-very-heavyrainfall-on-transparent-background-alphaembedded-with-hd-pngfile.html?src=rel/9063365:2
Consider Pressure to be a “momentum change” flux as the
particles hit a wall
• Total momentum change in a time interval =(2mvx (N/V)(vx*Dt)(wall
Area), so
• Pressure = total momentum change per unit area per unit of time
• Pressure = (2mvx) *(N/V)(vx)
= (2mvx2) *(N/V)
• Pressure = (⅓mvrms2) *(N/V) (considering two directions and 3
dimensions)
= (2/3)(½mvrms2) *(N/V)
= (2/3)(KEave) *(N/V)
From the ideal gas law: PV=NkT
• T=(2/3kB) (KEave)
(kB=1.38 × 10-23 J/K) Boltzmann constant
Internal energy and the 1st law of
thermodynamics
3
• The average kinetic energy of each
( KE ) avg/particle  k BT
particle is:
2
• Summing up the energy of the particles
is the same as multiplying the average
energy per particle times the total
3
number of particles (N). This total
U internal  Nk BT
energy is called the internal energy of
2
the system:
• Energy can be added to/removed from
the system either by doing work on/by
the system or heating/cooling the
system. This is the first law of
DU internal  Won _ system  Q
thermodynamics:
Last semester: ΔKEtotal = W. If Q=0 above, all of the positive work done on the gas increases
the gas temperature, and so increases the total internal energy, ΔUinternal
Changing the system energy
DUint =
The algebraic sign (+) or (-) of W and Q indicate if energy is gained or lost by
the system, respectively.
Heating and Temperature
Q: What happens to matter when it is heated or
cooled?
“heating” and “cooling” are processes which add or
take away energy from the system, so they increase
or decrease the kinetic energy of the particles in an
ideal gas at a constant volume. What about other
matter, like solids and liquids?
There is electrical potential energy involved
between particles, and so the response to heating is
more complicated. Back to the macroscopic world…
In real matter, it depends…
Heating is a way of adding energy to a system,
so it can:
• cause the particles to become more energetic.
• cause them to change their average positions
relative to each other, which will change the
potential energy of the system and result in a
phase change.
Phases (states) of Matter
Cooling: Remove energy from system
Heating: Add energy to system
Phase Transitions
Heating Matter
• When matter is heated, its temperature changes OR its
phase changes, but they don’t happen simultaneously.
• For example: a solid is heated, its temperature
changes until it reaches the melting temperature for
that substance. Continued heating melts the substance
and the temperature stays constant at the melting
temperature until the entire solid is melted. Once the
matter is in a liquid form entirely then continued
heating results in an increase in temperature, until the
liquid reaches the vaporization temperature, etc.
Heating Matter
• Matter’s resistance to temperature change is characterized
by an index, called the specific heat capacity. The symbol
lower case “c” is commonly used and its value gives the
energy required for every 1 kg of mass to increase in
temperature by 1 K. (Units: J / Kkg )
Q = M c ΔT
• Side note – there’s another quantity you developed in
recitation which describes how much heat it takes to
change the temperature of a substance by 1 Kelvin, the
units were J/K. This is known as the heat capacity
(sometimes written as upper-case C).
• Heat capacity differs from specific heat capacity because it
changes based on how much of the substance you have – it
will take more heat to heat up a larger quantity of
substance, whereas specific heat capacity is a material
property and doesn’t depend on how much you have.
The specific heat capacity of aluminum is about twice
that of iron. Consider two blocks of equal mass, one
made of aluminum and the other one made of iron,
initially in thermal equilibrium.
Both blocks are heated at the same constant rate until
reaching a temperature of 500 K. Which of the
following statements is true?
a) The iron takes less time than the aluminum to
reach 500 K
b) The aluminum takes less time than the iron to
reach 500 K
c) The two blocks take the same amount of time to
reach 500 K
• Matter’s resistance to temperature change is characterized
by an index, called the specific heat . The symbol lower
case “c” is commonly used and its value gives the energy
required for every 1 kg of mass to increase in temperature
by 1 K. (Units: J / Kkg )
cAl = 0.897 J/Kg or 897 J/Kkg
Q = M c ΔT
cFe = 0.412 J/Kg or 412 J/Kkg
• When matter changes phase (e.g., between solid and liquid
or between liquid and gas) that phase change has energy
associated with it. The phase change index, or latent heat
of transformation is the energy required to complete the
phase change for every kg of the substance. The symbol
upper case “L” is commonly used and its units are (J / kg)
Q = ±ML
LAl = 398 J/g or 398,000 J/kg
LFe = 272 J/g or 272,000 J/kg
• When matter changes phase (e.g., between solid and liquid
or between liquid and gas) that phase change has energy
associated with it. The phase change index, or latent heat
of transformation is the energy required to complete the
phase change for every kg of the substance. The symbol
upper case “L” is commonly used and its units are (J / kg)
Q = ±ML
LAl = 398 J/g or 398,000 J/kg
https://www.youtube.com/watch?v=If5h3CMzdrw
LFe = 272 J/g or 272,000 J/kg
Since LAl is so large (it
takes a lot of energy to
melt Al) it’s cheaper to
ship Al as a liquid and
keep it hot than ship it as
a room temp solid.
iClicker!
40 g of water at 100 °C and 60 g of water at 0 °C
are mixed together. When the mixture reaches
thermal equilibrium, the final temperature will
be:
a) greater than 50 °C .
b) equal to 50 °C .
c) less than 50 °C .
Can reason what the
answer must be without
considering equations?
Let’s work through the physics together AFTER
you’ve thought through the problem.
iClicker!
50 g of steam at 100 °C and 50 g of liquid water
at 80 °C are mixed together in an insulated
container. When the combination reaches
thermal equilibrium, the final temperature will
be:
a) greater than 90 °C .
b) equal to 90 °C .
c) less than 90 °C .
iClicker!
60 g of ice at 0 °C and 40 g of liquid water at 20 °C
are mixed together in an insulated container.
When the combination reaches thermal
equilibrium, the final temperature will be:
a) greater than 10 °C .
b) equal to 10 °C .
c) less than 10 °C .
Does the water melt the ice?
How much energy does it take to melt the ice?
How much energy can the liquid water supply as it cools to 0°C?
iClicker!
60 g of ice at 0 °C and 40 g of liquid water at 20°C are mixed
together in an insulated container. When the combination reaches
thermal equilibrium, the final temperature will be…
Does the water melt the ice? Let’s see:
How much energy does it take to melt the ice?
It takes Q = mLf energy to melt a mass “m” of ice, in this case. So we
Q=(0.06 kg)(334000 J/kg) = 20,040 J to melt the ice entirely.
have
How much energy can the liquid water supply as it cools from 20°C to 0°C?
If the 40g of liquid water starts at 20C, the most energy it can give to the ice is if
it cools to 0C as well, so Qwater,max = mwcw(273K – 293K), where we’ve converted
the final temp of 0C to 273K, and the initial temp of 20C to 293K.
Qwater,max = (0.04 kg)(4180 J/kgK)(-20 K) = 3344J
Since the warm water only supplies 3344 J max, it will only melt a bit of the ice, not
nearly all of it. The amount of ice melted can be calculated using Q=mL where Q is
3344J, then solving for m gives m=Q/L = 3344 J / (334000 J/kg) = 0.01 kg, or 10g. The
warm water melts just 10g of the ice, so in the final state we end up having a nice glass
of ice-water with 50g of ice and 50g of water.