AP2 Ch13-15 OVERVIEW PPTX Format
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Transcript AP2 Ch13-15 OVERVIEW PPTX Format
Heat & Thermo
Important Shtuff
© 2015 D Taylor
Heat &
Thermodynamics
Or, “Let’s try to remember all that
stuff we learned last year, K!”
Mini-LAB
100 Revs.
MEASURE T.
-Record on WhiteBd
-Discuss
Basic LAWS
• 0th Law
–Two systems individually in thermal
equilibrium with a 3rd system are in
equilibrium with each other.
–Sorta like
• If A=B & B=C, then A=C
13-2: Thermal Expansion
L Lo T
13-2: Thermal Expansion
L Lo T
A 2Ao T
V Vo T
13-4 Gas Laws
• Ideal Gas Laws
–Boyle’s Law (1700)
• P1V1=P2V2
–Charles’ Law
• V1/T1=V2/T2
–Guy-Lusacs Law
• P1/T1=P2/T2
Put them all
together and…
P1V1 P2V2
T1
T2
Which means?
PV nRT NkT
P1V1 n # mol
CONSTANT
T1 N # atoms
J
R 8.31
mol K
23 J
k 1.38 x10
K
13-4: Kinetic Theory
113-4: Kinetic2 Theory
PV Nm v
3
1 2
KE mv
2
3PV 3 Nkt 3
KE
kT
2N
2N
2
Basic Stuff
• Kinetic Theory
–Look on pg 452+ for pix & examples
1 2
3
KE mv rms kT
2
2
13-4: Derivation of U
1 2
U N mv
2
3
3 3
U N kt Nkt nRT
2
2 2
AP2 Ch14
Heat & Heat Transfer
EZ PZ
14.2 Temperature Change and Heat
Capacity
Q T
Heat transfer : Fig 14-4
Qm
Qc
Calorimetry
• Since heat, Q, is a form of E…
• QL = QG
• QTotal = 0 = mcT
• So,
m1c1T1 m2 c2 T2
Calorimetry
• Q L = QG
• Phase Change
• So,
Q mL
14.4: Heat Transfer Methods
Q kAT
t
d
Ch15:
Thermodynamics
Or, “Let’s try to remember all that
stuff we learned last year, K!”
Figure 15.3
•
WRONG!
The first law of thermodynamics is the conservation-of-energy principle stated for a system where heat and work are the
methods of transferring energy for a system in thermal equilibrium. 𝑄 represents the net heat transfer—it is the sum of
all heat transfers into and out of the system. 𝑄 is positive for net heat transfer into the system. 𝑊 is the total work done
on and by the system. 𝑊 is positive when more work is done by the system than on it. The change in the internal energy
of the system, Δ𝑈 , is related to heat and work by the first law of thermodynamics, Δ𝑈 = 𝑄 − 𝑊 .
•
1st
Law
Basic LAWS
–Internal energy of a system changes
from an initial value Ui to a final value
Uf due to heat Q and work W:
• U = Uf – Ui = Q + W
• Q>0 when system gains heat & <0 when
it loses heat.
• W>0 when work is done ON system & <0
when gas does work.
QIN > 0
QOUT < 0
WON > 0
WBY < 0
Using the 1st LAW
• 2500J of heat is added to a system, and 1800J
of work is done on the system. Find change in
internal energy of the system?
U Q W
U 2500 J 1800
U 4300 J
Using the 1st LAW
• Work done on a system in one cycle?
U Q W
0 Q W
W Q (QH QC )
–T of 3 mol is reduced from 540K to
350K by adding 5500J of heat to it.
• Find U & W done by or on gas.
3
U nRT
2
3
(3mol ) 8.31 J
(350 540 K )
mol K
2
7100 J
–T of 3 mol is reduced from 540K to
350K by adding 5500J of heat to it.
• Find U & W done by or on gas.
U W Q
W U Q
7100 J 5500 J
by or on gas?
12,600 J
Using the 1st LAW
• Work done on or by a gas?
WBY Fd PAd
WBY PV
Basic LAWS
• 2nd Law
–Heat flows from a substance at
higher temp to a substance at
lower temp and does NOT flow in
the reverse direction. Ever.
Carnot (IDEAL) Engine
Carnot (IDEAL) Engine
eCarnot
TH TL
TL
1
TH
TH
Basic LAWS
• 3rd Law
–Absolute Zero is impossible to
reach in a finite number of steps.