# PPTX

```Clicker Question
Room Frequency BA
On which side of the Coors Event Center arena will the PHYS2010 final
exam take place?
A)
B)
C)
D)
E)
North
South
East
West
I don’t know!
Flatirons are west of campus!
1
Announcements
• CAPA assignment #15 is due on tonight at 10 pm.
• By Sunday final Clicker scores will be posted on
CULearn
• By Monday all Recitation/Lab scores will be posted.
• If you find a discrepancy, send me email Monday
2
Final Exam News
• Final Exam is Tuesday morning, Dec 13, 10:30am – 1pm
• Exam will be held in Coors Event Center, on the West
(mountain) side of the arena. Calc-based physics is on
the east side; don’t sit down in the wrong side!
• We are in sections 20, 21, 22, with individual seat
assignments. Your seat assignment are posted on CU
Learn.
• Bring student ID. We will be checking that people are in
their assigned seat.
• Practice Exams and Formula sheet are posted on
CULearn; you can ignore Prob. 37 on the practice exam
3
Solving Calorimetry Problems
• Problem solution is based on energy conservation: heat lost by
hot object must equal heat gained by cold object
• Heat (transfer without phase change) Q = mcΔT
• Latent Heat (Phase change) Q = mL
• In problems with phase changes, you have to figure out the
final phase or phases
• Final state can be all one phase or a mixture of several phases
QMatter cooling off  QMatter heating up
4
Room Frequency BA
Clicker Question
An ice cube is placed in a cup containing some liquid water. The water
and ice exchange energy with each other but not with the outside
world. After the water and ice come to the same temperature, is it
possible the ice could freeze the water rather than the water melt
some of the ice?
ice
liquid water
A) Yes, the ice could freeze the water.
B) No, the water will always melt some of the ice.
If the ice is cold enough initially (well below zero), then the
heat lost by the liquid to cool it and freeze it could be more
than the heat gained by the initial cold ice!
5
Calorimetry Example with Ice - 1
1 kg of ice at temperature -200&deg;C is placed is a container of 100 mL
of water at 20&deg;C. What is the final temperature and phase of the
system?
Preliminary analysis: Will the final state be all ice, all water or
a mixture of ice and water?
Energy loss to cool water to 0&deg;C: Qwater=mwatercwater(0-20)
with cwater = 4186 J/(kg&deg;C) and mwater = 0.1 kg, Qwater = 8372 J
Energy gain to heat ice to 0&deg;C: Qice=micecice(200-0)
with cice = 2100 J/(kg&deg;C) and mice = 1 kg, Qice = 420,000 J
Ice will heat up, water will cool to 0&deg;C and start to freeze;
how much of the water will freeze? Look at total water freeze.
Qwater to ice = mwaterLice-water = 0.1 kg*333,000 = 33,000 J
Original ice would gain 8372+33,000 J = 41372 J &lt; 420,000 J
Not enough to bring ice up to 0&deg;C =&gt; All ice in the final state
6
Calorimetry Example with Ice - 2
1 kg of ice at temperature -200&deg;C is placed is a container of 100 mL
of water at 20&deg;C.
Final analysis: Final state will be all ice. What is Tf?
Qice=micecice(Tf - -200)
cice = 2100 J/(kg&deg;C)
Qwater=mwatercwater(0-20) - mwaterLwater-ice + mwatercice(Tf - 0)
|Qice| =micecice(Tf + 200 ) = |Qwater| =mwatercwater(20) + mwaterLwater-ice
+ mwatercice(0 – Tf )
Solve for Tf and You’ll get T f  mwater cwater 20  mwater Lwaterice  micecice 200
cice (mice  mwater )
Using cwater = 4186 J/(kg&deg;C), Lwater-ice = 333,000 J/kg
Tf = -164&deg;C
Is the final T between -200 and 0? Yes!
7
Calorimetry Example with Less Ice - 1
0.05 kg of ice at temperature -200&deg;C is placed is a container of 100
mL of water at 20&deg;C. What is the final temperature/phase of the
system?
Preliminary analysis: Will the final state be all ice, all water or
a mixture of ice and water?
Energy loss to cool water to 0&deg;C: Qwater=mwatercwater(0-20)
with cwater = 4186 J/(kg&deg;C) and mwater = 0.1 kg, Qwater = 8372 J
Energy gain to heat ice to 0&deg;C: Qice=micecice(0 - -200)
with cice = 2100 J/(kg&deg;C) and mice = 0.05 kg, Qice = 21,000 J
Ice will heat up, water will cool to 0&deg;C and start to freeze;
how much of the water will freeze? Look at total water freeze.
Qwater to ice = mwaterLice-water = 0.1 kg*333,000 = 33,000 J
Heat Loss to freeze all water is 8372+33,000 J = 41372 J &gt; 21,000 J
Not enough to freeze all water =&gt; Ice water mixture in the final state8
Calorimetry Example with Less Ice - 2
Final analysis: Final state will be ice-water mixture (at 0 &deg;C).
How much ice and water will there be?
Qice=micecice(0 - -200)
cice = 2100 J/(kg&deg;C)
Qwater=mwatercwater(0-20) - mfreezeLwater-ice + mwatercice(0 - 0)
|Qice| =micecice(200 ) = |Qwater| =mwatercwater(20) + mfreezeLwater-ice
Solve for mfreeze :
m freeze 
micecice 200  mwater cwater 20
Lwaterice
Using cwater = 4186 J/(kg&deg;C), Lwater-ice = 333,000 J/kg
mfreeze = 0.038 kg Must be less than original 0.1 kg of water
mtotal-ice = 0.088 kg, mliquid water = 0.062 kg (62 mL)
9
Gases and Absolute Temperature
• In 1600’s Robert Boyle find gas Pressure P times its Volume V
is a constant at constant temperature: PV = constant
• In late 1700’s Jacques Charles find that at constant pressure,
the Volume of the gas changes linearly with temperature T(&deg;C)
giving V = constant*(T+267)
• Inventing a new temperature scale Tabs = T(&deg;C) +267, we get
Charles’ law V = constant * Tabs
• In early 1800’s, Joseph Gay-Lussac finds Pressure P is
proportional to T(K): P = constant*Tabs
• In 1834 Clapeyron put all this together to get PV = nRTabs
• R is the same for many gases! n is “amount” of gas
• Avogadro says n is the number of molecules
• 1 mole = 6.02 x 1023 molecules
• More refined Tabs is Kelvin scale T(K) = T(&deg;C) + 273.15
10
Gas Constants
• In SI units, R = 8.314 J/(mole*K)
• If you want to see this law in terms of individual gas molecules,
you divide R by Avogadro’s number to get k, the Boltzmann
constant
• In SI units, k = 1.38 x 10-23 J/K
• Now Ideal Gas Law is PV = NkT, where N is the number of gas
molecules
• In many problems, N is a constant, so we have PV/T = constant
11
Clicker Question
Room Frequency BA
A balloon full of gas of volume 1 m3 at pressure 1 atm at temperature
20&deg;C. If I heat the balloon to 40&deg;C and double the pressure to 2 atm,
what is the approximate volume of the balloon?
A)
B)
C)
D)
E)
0.5 m3
1.0 m3
2.0 m3
4.0 m3
None of the above
PV/T = constant = (1 atm)(1 m3)/293 K = (2 atm)(V)/313 K
V =(313/293)/2 m3 = 0.534 m3
12
Mechanics and Ideal Gas Law
• Consider a hollow cube of side length L filled with N atoms
• Each atom has mass m and average velocity v
• If the atoms bounces off a wall of the cube it exerts a force on
the wall, and the wall exerts a force on the atom
F
L
v
13
Atom Force on Wall
• The average force is found from the change in the atom momentum
FaveΔt = mΔv ≈2mv, so Fave = 2mv /Δt (single atom)
• For N atoms in cube, about 1/6 are heading towards a given wall
• Then on average for Δt = L/v ,N/6 hit the wall each giving an average
force Fave .
• Average total force FT on a wall is then (N/6)*(2mv)/(L/v) = Nmv2/3L
• Pressure P = FT/L2 = Nmv2/3L3
• P = Nmv2/3V or
• PV = N(2/3)(mv2/2) = N(2/3)(KEatom) F
Δv ≈ 2v
v
14
Kinetic Theory!
• PV = NkT from experiment
• PV = N*number*KE from theory of molecules
• kT is average kinetic energy of molecules!!!
• Using proper exact averages you find
1
2
mv2  23 kT
• In a real gas, there is a distribution of velocities
• square root of average of v2 is called vrms and we now get
formula
vrms  v 
2
3kT
.
m
At same T, molecule speed is faster for lighter molecules
15
Clicker Question
Room Frequency BA
Two boxes, one filled with helium (m = 4u) and one filled with nitrogen
(m =30 u) are brought to the same temperature T. The box holding
nitrogen is twice as large as that holding helium.
How are the average kinetic energies of the individual molecules of the
two gases related?
A) KEHe &lt; KEN
B) KEHe = KEN
C) KEHe &gt; KEN
D) not enough information
3kT/2 = average KE so same T, same average KE
16
Clicker Question
Room Frequency BA
Two boxes, one filled with helium (m = 4u) and one filled with nitrogen
(m =30 u) are brought to the same temperature T. The box holding
nitrogen is twice as large as that holding helium.
How are the average molecular speeds of the two gases related?
A) vHe &lt; vN
B) vHe = vN
C) vHe &gt; vN
D) not enough information
Nitrogen is heavier, hence vN is smaller
17
On behalf of Profs. Nagle and Uzdensky
Thank you for being a great class!
Stay calm and rested and do well on the
final exam!!!
18
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