Transcript Notes - Intro to Thermodynamics Presentation
Temperature vs. Heat vs. Internal Energy
1.
2.
3.
4.
Work in groups in 2-3 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!
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
.
Thermal Physics
AP Physics B
Thermal Physics
Temperature and Heat Mechanical equivalent of heat Zeroth Law of Thermodynamics Latent Heat Heat transfer and thermal expansion Conduction Convection Radiation Kinetic Theory and Thermodynamics 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
contains (or their kinetic energy). that it
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 32 o F, Boiling point 212 o F ~
makes no sense…
Celsius: Water Freezes/Melts at 0 o C, and boils at 100 o C Kelvin: 0 o K 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 5
T
(
C
)
32
T
(
C
)
5 9
[
T
(
F
)
32
]
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 T x = T y and T y then T x = T z = T z ,
Well … Duh!
Why Zeroth Law???? Made
after
and 3 Laws 1,2,
Heat
Heat:
Transmission
of energy from one body to another due to temperature difference (hot to cold) – unit is Calorie or Joule
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
If you haven’t already
… please watch the Video – 2 for homework! + MC Q’s: 6, 24, 30, 32, 35, 39, 58 ALSO: Watch Specific Heat Video if you need a review from Phys 11!
Heat Transfer Due to
Δ
T
All materials are not created equally in terms of heat transfer Would you rather touch your tongue to a 0 o C metal pole or wooden pole?
Why? (Splinters?)
Thermal Conductivity, k
k is a measure of an object ’ s ability to conduct heat (transfer) – RATE QUANTITY 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 of surface in contact L – Thickness
H
=
Q
D
t
=
kA
D
L T
Question: Find H of the glass below in J/s
Ans: 7.9 x 10 2 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! Some questions are not difficult
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
TOTAL 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 (
K or C!
)
Note: Specific Heat was covered in Physics 11 as well as AP Chem
Water vs. Copper
c water = 1 cal/g o C = 4186 J / kg o C c copper = 0.093 cal/g o C = 389 J / kg o C 1 gram of copper at 0 o C and 1 gram of water at 0 o C If heated by 1 o C….
Specific Heat: Heat Transfer with Changes in Temp
∆T= Q/mc
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
Hot Object Added to Cool Liquid
ΔQ = 0 (closed system) Q heat IN = Q heat OUT Q Lost = Q Gained In this case….
-Q hot = Q cold -m hot c hot (T final - T hot ) = m cold c cold (T final -T cold )
Question
500. grams of 20.0
o C water is added to 700. g of 85 o C water. What is temperature of the mixture?
ANSWER
500. grams of 20.0
o 85 o C water is added to 700. g of C water. What is temperature of the mixture? -
Q hot
=
Q cold
-
m hot c hot
(
T final
-
T hot
) =
m cold c cold
(
T final
-
T cold
) (700.
g
)( 1.00
cal g C o
)(
T final
85
o
) = (500.
g
)( 1.00
cal g C o
)(
T final
20
o
) 1.4(
T final
85
o
) = (
T final
20
o
) 139
o
= 2.4
T final T final
= 58
o
Question –
Connect to AP Chem Lab!
We wish to determine the specific heat of a new alloy. A 0.150 kg sample of the alloy is heated too 540 o C. It is then quickly placed in 400. g of water at 10.0
o , 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 30.5 o C
.
Calculate the specific heat of the alloy
.
c w = 4186 J/kg o C ; c cal = 900.0 J/kg -m s c s ΔT = m w c w ΔT + m cal c cal ΔT o C Remember, Heat LOST = Heat GAINED (this changes your ΔT)
ANSWER – 497 J/kg
o C
We wish to determine the specific heat of a new alloy. A 0.150 kg sample of the alloy is heated too 540 o C. It is then quickly placed in 400. g of water at 10.0
o , 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 30.5
o C. Calculate the specific heat of the alloy.
c w
Q hot
=
Q cold m s c s
D
T
=
m w c w
D
T
+
m cal c cal
D
T
= 4186 J/kg o C ; c cal = 900.0 J/kg o C ( 0.150
kg
)
c s
(540
o C
– 30.5
o C
) = (0.40
kg
)( 4186
kg o C J
)(30.5
o C
– 10.0
o C
) + (0.20
kg
)( 900.0
kg o C J
)(30.5
o C
– 10.0
o C
) Heat lost Heat gained Remember, Heat LOST = Heat GAINED (this changes your ΔT)
Phases and Phase Changes
Question: Why doesn’t the temp increase at 0 o C and 100 o 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
inter
molecular bonds Q = m x L Heat transfer depends on how much (mass), and material property (L) L =
latent heat
of
fusion
(melting/solidifying) or
vaporization
(vaporizing/condensing) L f water = 3.33x10
5 J/kg L v water =2.26x10
6 J/kg
Total Heat
Go from 50 C to 120 C… 1.
Q total = Q -50-0 + Q 0 + Q 0-100 + Q 100 + Q 100-120 2.
Q total = mc(50) + mL f +mc(100) + mL v + mc(20) Note: water and ice have DIFFERENT specific heats!
Question
How much energy does a refrigerator have to remove from 1.5 kg of water at 20.0
o C to make ice at -12 o C. The Heat of Fusion is 3.33 x10 5 J/Kg, the specific heat of water is 4180 J/(kg o C), the specific heat of ice is 2100 J/(kg o C).
ANSWER
How much energy (in kJ) does a refrigerator have to remove from 1.5 kg of water at 20.0
o C to make ice at 12 o C. The Heat of Fusion is 3.33 x10 5 J/Kg, the specific heat of water is 4180 J/(kg o C), the specific heat of ice is 2100 J/(kg o C).
Q
=
Q
=
Q
20 0
mc
D
t
-
Q
0 +
Q
0 - 12 +
mL
+
mc
D
t Q
= (1.5
kg
)(4180
kg J o C
)(0
o C
20
o C
) (1.5
kg
)(3.33
x
10 5
Q
= 660
kJ J kg
) + (1.5
kg
)(2100
kg J o C
)(0
o C
12
o C
)
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.
Next Video: Ideal Gases 2.
MC Question’s: 2,5,7,9,21-23,27,31,38,44,45, 47,48,51,52,61-64,73,76,78 3.
Next Class: Quiz on Ideal Gases + Gas Law Simulation (
Please bring a computer!
)