Transcript Gas Laws

Temperature vs. Heat vs. Internal Energy
1.
2.
3.
4.
Work in groups in 2-4
Collect a whiteboard and pens
Make a Venn Diagram showing
how the terms Temperature,
Heat and Internal Energy are
similar and different
Use your notes and phones for
resources!
Thermal Physics
AP Physics B
Thermal Physics

B. Temperature and heat 2%

Mechanical equivalent of heat



Heat transfer and thermal expansion




Zeroth Law of Thermodynamics
Latent Heat
Conduction
Convection
Radiation
C. Kinetic theory and thermodynamics 7%

Ideal gases



Kinetic model
Ideal gas law
Laws of thermodynamics


First law (including processes on pV diagrams)
Second law (including heat engines)
Temperature and Heat
Temperature: physical property of matter that
quantitatively expresses the common notions of
hot and cold.

The temperature varies with the microscopic
speed of the fundamental particles that it
contains (or their kinetic energy).
All particles have internal movement... ALWAYS
1.
2.
Temperature measures
the average Kinetic
Energy of the particles
within a substance
NOT the total energy of
the substance, which
depends on it’s mass
Basis for Temp Scales



Fahrenheit: Oldest scale, Freezing point is 32oF, Boiling
point 212oF ~ makes no sense…
Celsius: Water Freezes/Melts at 0oC, and boils at 100oC
Kelvin: 0oK is the coolest theoretical temperature
possible, no negative Kelvins. Same increments at
Celsius Scale. Basically an updated version of the
Celsius scale
Temperature

Scales:
T (  F )  9 T (  C )  32 
5
T (  C )  5 [T (  F )  32  ]
9
T ( K )  T (  C )  273 . 15
Absolute Zero – 0 Kelvin
Q: What would have to happen in order to reach absolute
zero?
Atoms
and subatomic particles would have to stop moving…
impossible.
2003 - MIT scientists cooled sodium gas to the lowest
temperature ever recorded -- only half-a-billionth of a degree
above absolute zero.
Zeroth Law of Thermodynamics
If Tx = Ty and Ty = Tz,
then Tx = Tz
Well
Why
… Duh!
Zeroth Law????
Made after Laws 1,2,
and 3
Heat

Heat: Transmission of
energy from one body to
another due to
temperature difference
(hot to cold) – unit is
Calorie or Joule

We will talk about Heat
Transfer when we
discuss the Mechanical
Equivalent of Heat
Internal Energy

Compare TOTAL INTERNAL ENERGY of
gas to liquid/solid

Study liquids/solids now…ideal gases later
Heat

Take Home Message(s)

Heat is a process


Matter contains internal energy NOT HEAT
Heat is the transfer or conversion of energy

Compare to Work and Mechanical Energy…
In-Class Work + Homework

Collect the worksheet from the front of the room
Please watch the Video – 2 for homework! + MC Q’s: 6, 24,
30, 32, 35, 39, 58

ALSO: Watch Specific Heat Video if you haven’t already done so!

What’s next?

How? – Next Class! Quiz! + …



Conduction: molecular collisions
Convection: motion of fluid
Radiation: no medium necessary (EM waves)
AP Phys 12 – Class Starter
Quiz: Temperature and Heat Transfer!
1.
2.
3.
Please clear everything off your desk except a pencil
and calculator
Collect Data Sheets and a scrap piece of paper from
the front of the room
Wait for the quiz!
When finished collect a new Quiz Log from the front!
Whiteboarding!



Please work in groups of 2
Collect a Whiteboard and Pens
Let’s review a few concepts!
Heat, Thermal Energy and Internal Energy:
Words of caution!
“To describe the energy that a high temperature object has, it is not a
correct use of the word heat to say that the object "possesses heat" - it
is better to say that it possesses internal energy as a result of its
molecular motion. The word heat is better reserved to describe the
process of transfer of energy from a high temperature object to a lower
temperature one. You can take an object at low internal energy and
raise it to higher internal energy by heating it. But you can also increase
its internal energy by doing work on it, and since the internal energy of
a high temperature object resides in random motion of the molecules,
you can't tell which mechanism was used to give it that energy.”
Heat Transfer Due to ΔT

All materials are not created equally in terms
of heat transfer


Would you rather touch your tongue to a 0 C
metal pole or wooden pole?
Why?
Thermal Conductivity, k

k is a measure of an object’s ability to
conduct heat (transfer)


Higher k means faster rate of transfer
Materials of high thermal conductivity are widely used in
heat sink applications and materials of low thermal
conductivity are used as thermal insulation



H – Rate of Heat Transfer (J/s or kcal/s)
A – Area
L – thickness
H 
kA  T
L
Question: Find H of the glass below in J/s
Table 14-4
Thermal
Conductivities
Question: Find H of the glass below in J/s
Ans: 7.9 x 102 J/s
Question
What is a better insulator, an object with a
larger k, or smaller k?
Question
What is a better insulator, an object with a
larger k, or smaller k?

Ans: Smaller k!
Question
If air has such a low thermal conductivity (0.22),
why do we need to wear clothes? (Other than
for decency reasons…)
Question
If air has such a low thermal conductivity (0.22), why
do we need to wear clothes? (Other than for decency
reasons…)
ANS: Air is always moving, we use clothes to trap air
close to our bodies. The thicker the clothes the more
air we trap.
Specific Heat: Heat Transfer with
Changes in Temperature
Q = mc∆T

Heat transfer depends on amount of material (m),
temperature difference (∆T), and material property (c)


c = specific heat – amount of heat/mass required to raise
temperature by 1 degree
Note: Specific Heat was covered in Physics 11
Water vs. Copper

cwater = 1 cal/g oC
= 4186 J / kg oC
ccopper = 0.093 cal/g oC = 389 J / kg oC

1 gram of copper at 0 oC and 1 gram of water at 0 oC



If heated by 1oC….
Which has a greater internal energy?


Same kinetic energy (temperature is the same)
Water has more internal energy due to higher specific heat
(requires more heat to increase kinetic energy of particles)
Specific Heat: Heat Transfer with
Changes in Temp
ΔQ = 0 (closed system)
Q heat IN = Q heat OUT
Same

Q transfers, different objects (ignore -/+!)
Big mass vs. small mass: small mass will get hotter
Metal (lower c) vs. Wood (higher c): metal gets hotter

∆T= Q/mc

Hot Object Added to Cool Liquid
-Qhot = Qcold
mhotchot(Thot-Tfinal) = mcoldccold(Tfinal-Tcold)
Question
500. grams of 20.0o C water is added to 700. g
of 85o C water. What is temperature of the
mixture?
ANSWER
500. grams of 20.0o C water is added to 700. g of
85o C water. What is temperature of the mixture?
Qhot = Qcold
mhot chot (Thot - T final ) = mcold ccold (T final - Tcold )
1.00cal
1.00cal
o
o
(700.g)(
)(85
T
)
=
(500.g)(
)(T
20
)
final
final
o
o
gC
gC
1.4(85 o - T final ) = (T final - 20 o )
139 = 2.4T final
o
T final = 58 o
Question
We wish to determine the specific heat of a new alloy. A
0.150 kg sample of the alloy is heated too 540o C. It is then
quickly placed in 400. g of water at 10.0o , which is
contained in a 200. g aluminum calorimeter cup. (Assume
that the insulating jacket insulates well, so the temperature
does not change significantly). The final temperature of the
mixture is 30.5o C. Calculate the specific heat of the alloy.
cw = 4186 J/kg oC ; ccal = 900.0 J/kg oC
-mscsΔT = mwcwΔT + mcalccalΔT
Remember, Heat LOST = Heat GAINED (this changes your ΔT)
ANSWER – 497 J/kg
oC
We wish to determine the specific heat of a new alloy. A 0.150 kg sample of
the alloy is heated too 540o C. It is then quickly placed in 400. g of water at
10.0o , which is contained in a 200. g aluminum calorimeter cup. (Assume
that the insulating jacket insulates well, so the temperature does not change
significantly). The final temperature of the mixture is 30.5o C. Calculate the
specific heat of the alloy.
cw = 4186 J/kg oC ; ccal = 900.0 J/kg oC
-Qhot = Qcold
-ms cs DT = mw cw DT + mcal ccal DT
( 0.150 kg ) cs (540 oC
– 30.5 o C) = (0.40 kg)(
Heat lost
4186J
900.0J
o
o
)(30.5
C
–
10.0
C)
+
(0.20kg)(
)(30.5 oC – 10.0 oC)
o
o
kg C
kg C
Heat gained
Remember, Heat LOST = Heat GAINED (this changes your ΔT)
Phases and Phase Changes
Question:
Why doesn’t the temp increase at 0o C and 100o C?
Energy Input goes to enabling the Phase Change
Heat Transfer w/out Changes in T
(Phase Changes)


If No ∆T: Phase change MUST be occurring
Input of energy is used to break intermolecular
bonds

Q=mxL

Heat transfer depends on how much (mass), and
material property (L)

L = latent heat of fusion (melting/solidifying) or
vaporization (vaporizing/condensing)
 Lf water = 3.33x105 J/kg
 Lv water =2.26x106 J/kg
Total Heat

Go from -50 C to 120 C…
1.
Qtotal = Q-50-0 + Q0 + Q0-100 + Q100 + Q100-120
2.
Qtotal = mc(50) + mLf +mc(100) + mLv + mc(20)
3.
Qtotal = mc(170) + mLf + mLv
Question
How much energy does a refrigerator have to remove
from 1.5 kg of water at 20.0o C to make ice at -12o C.
The Heat of Fusion is 3.33 x105 J/Kg, the specific heat
of water is 4180 J/(kgoC), the specific heat of ice is
2100 J/(kgoC).
ANSWER
How much energy (in kJ) does a refrigerator have to
remove from 1.5 kg of water at 20.0o C to make ice at 12o C. The Heat of Fusion is 3.33 x105 J/Kg, the
specific heat of water is 4180 J/(kgoC), the specific
heat of ice is 2100 J/(kgoC).
Q = -Q20-0 + Q0 + -Q0-( -12)
Q = mcDt + mL + mcDt
J
J
o
o
5 J
Q = - (1.5kg)(4180 o )(0 C - 20 C) + (1.5kg)(3.33 x10 ) - (1.5kg)(2100 o )(-12 o C - 0 o C)
kg C
kg
kg C
Q = 660kJ
Joule’s Experiment - GENIUS
Video – Mechanical Equivalent of
Heat
For the Rest of Class!


Complete the Heat Transfer Worksheet
Heat Transfer – AP Questions: MC: 10, 26,
28, 42, 43, 53, 56, 65, 81
Now what…
1.
2.
3.
Next Video: Ideal Gases
MC Question’s: 2, 5, 7, 9, 21 -23, 27, 31, 38, 44,
45, 47, 48, 51, 52, 61-64, 73, 76, 78
Next Class: Quiz on Ideal Gases + Gas Law
Simulation (Please bring a computer!)