Clicker Question
Room Frequency BA
When people breathe helium and then speak, their voices sound
much higher in pitch! Given that a person’s throat and vocal chords,
which produce the sounds, do not change, how does the speed of
sound in helium compare to that in regular air?
A) sound in helium is faster
B) sound in helium is slower
C) sound has the same speed in air and helium; the change in pitch is
from a different effect.
v = fλ : since λ is determined by the throat size, and pitch
goes up, f is also larger so v is larger (by about a factor of 3!)
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Announcements
• CAPA assignment #15 is due on Friday Dec 9th at 10
pm.
• This week: Review in Section, Make-up Labs arranged
in other sections. Last chance to make up labs! 5 labs
must be completed in order to pass this course.
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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 will be 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
• Prof. Nagle will hold a review session on Thursday evening,
Dec 8th, 7-9pm, in this room (Duane G1B30).
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Temperature and Thermal Energy
• Thermal Energy is in the KE and PE of the individual
atoms in matter from their individual random motion!
• Temperature is a measure of the thermal energy per
atom, not the total thermal energy of a piece of matter
• Conversion formulas:
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T (oF)  T (oC)  32
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T (K)  T (oC)  273.15
• We measure temperature by letting thermal energy
“flow” between the thermometer and the system whose
temperature we are trying to measure.
• When the flow stops, we say the thermometer and
system have reached thermal equilibrium. At this point
the average thermal energy of the thermometer atoms is
equal to the average thermal energy of system atoms.
• This or any transfer of thermal energy is called heat.
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How is Heat Transferred?
Heat is transferred mainly through 3 different mechanisms:
• Conduction: Heat transfer direct inter-atom interaction
without bulk movement of atoms.
• Convection: Heat transfer by bulk movement of atoms, either
forced (for example, stirring) or natural (boiling, magma)
• Radiation: Heat transfer by electromagnetic radiation (light,
especially infrared part of the spectrum)
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Units of Heat
• Heat is measured by a change in temperature, so the standard
unit took a common material (water) and related the heat to
the change in temperature of a certain amount of water:
• 1 calorie (cal) is the amount of heat required to heat 1 gram of
water 1 degree Celsius.
• 1 kcal heats 1 kg of water by 1 °C
• Warning: US “food energy” calories are actually kcal ( = 1 Cal)
• British Thermal Unit (Btu) is 1 lb of water up 1 °F
•The calorie is not the SI unit of energy! Using transfer of energy by
stirring (mechanical action), one measures that 1 cal = 4.186 J
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Heat: Thermal Energy Transfer
• Heat is the thermal equivalent of mechanical work. It is a
transfer of energy which changes the energy of the system.
• The amount of heat transferred to some mass m is
proportional to its temperature change ΔT.
• For the same heat, different materials will have different
changes in temperature ΔT: the structure of the matter
determines how much thermal energy an atom can absorb
• We define a quantity called the specific heat or heat capacity c
which quantifies how much heat corresponds to a change in
temperature of a given amount ΔT for unit mass.
• Specific heats depend on material and the phase of the
material: gas, liquid, solid, and even solid structure (crystal)
• Heat Q = mcΔT
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Clicker Question
Room Frequency BA
A kg of aluminum and a kg of iron at the same initial temperature
receive the same heat Q. The final temperature of the iron is higher
than that of the aluminum. What is the relation of their specific heats,
ciron and caluminum?
A) ciron > caluminum
B) ciron = caluminum
C) ciron < caluminum
Q = mcΔT: since Q was the same for each and ΔT was larger
for iron, its heat capacity must be smaller
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Specific Heats
• Generally specific heats of gases and liquids are much greater
than those of solids
• Specific heats can also depend on temperature, not just phase
• Watch out for units! cal/(g °C) = kcal/(kg °C) = 4.186 kJ/(kg °C)
• Liquid water has one of the largest heat capacities!
• Humans (= water): Their temperature doesn’t change fast
• Scalding fluids burn badly; they hold lots of thermal energy!
• Combining materials at different temperatures and finding
final temperature is called calorimetry. Used to measure heat
released in chemical reactions.
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Heating Power Example
We want to change the temperature of the air in a room of
volume 5 m x 4 m x 5 m with a heater which can deliver 1500 W.
The initial temperature of the air is 15°C and we want the final
temperature to be 20°C. How long will it take?
First: What is the heat Q for this ΔT?
Q  mair cair T
mair  airVroom
Using ρair =1.29 kg/m3 and cair = 1050 J/(kg°C) we get Q = 6.8 x 105 J
Q = (Heating Power)*(Δ time) so Δ time = Q/(Heating Power)
Time to heat is 6.8 x 105 J/1500 W = 450 s = 7.5 min
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Calorimetry Problems
• A calorimeter is an insulated, sealed chamber that does not let
heat or matter flow in or out
• Often in simple problems, they do not mention using a
calorimeter => so you must assume matter or heat flow out of
system can be neglected!!!
•Thermal energy by simple conduction always flows from hot to
cold! This is independent of the conservation of energy…
• Problem solution is based on energy conservation: heat lost by
hot object must equal heat gained by cold object
QMatter cooling off  QMatter heating up
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Calorimetry Example
1 kg of gold at temperature 100°C is placed in a container of 1
liter of water at 25°C. What is the final temperature of the gold
and water?
Q gold=mgoldcgold(Tf-100)
Qwater=mwatercwater(Tf-25)
cgold = 130 J/(kg°C)
cwater = 4186 J/(kg°C)
Using ρwater =1000 kg/m3 and 1 liter = 1000 cm3 = 0.001 m3 we have
mwater = 1kg
|Q gold| =mgoldcgold(100 – Tf) = |Qwater| =mwatercwater(Tf-25)
Solve for Tf ! You’ll get
Tf = 27.26°C
mgold cgold 100  mwater cwater 25
Tf 
mgold cgold  mwater cwater
Is the final T between 25 and 100? Yes!
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Latent Heats
• Sometimes heat transfer does not change the temperature!
• Usually the energy is causing a change in phase of the
material, for example solid to liquid, or liquid to gas, but
sometimes internal solid structure is changing.
• The phase changes occur at some particular temperature, for
example water changes from solid to liquid at 0°C and from
liquid to gas at 100°C
• Since the temperature doesn’t change, these are called latent
heats and correspond to the amount of heat to change a certain
amount of material of mass m between one phase and the
other: Q = mL
• Units of L are just kcal/kg ( = 4186 J/kg)
• In calorimetry problems, you must account for energy to
change phase of something if you reach its temperature of a
phase change
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Clicker Question
Room Frequency BA
Data:
cice = 0.50 cal/(g oC) ,
cwater = 1.00 cal/(g oC),
csteam = 0.48 cal/(g oC),
L(ice-water)= 80 cal/g ,
L(water-steam)=539 cal/g
Which requires the most energy?
A) Raise the temperature of 1 gram of liquid water 50 oC. Q = 50 cal
B) Melt 1 gram of ice at T=0 oC. Q = 80 cal
C) Raise the temperature of 1 gram of ice from 73K to 273K . Q = 100 cal
D) Vaporize 0.5 gram of water at T=100 oC. Q = 269.5 cal
Q = mcΔT for no phase change, Q = mL for phase change.
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Melting Ice Example
How much energy is required to change 1 kg of ice, which is
initially at -200°C, into water at 25°C?
We will have three different steps:
1) Heat the ice from -200°C to 0°C
2) melt the ice at 0°C
3) then heat water from 0°C to 25°C
cice = 2100 J/(kg°C), cwater = 4186 J/(kg°C), and Lice-water = 333 kJ/kg
QTotal = (1)(2100)(200) +(1)(333,000)+(1)(4186)(25) = 858 kJ
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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!
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