3b Thermodynamics Specific heat capacity, c Latent heat capacity, L Change of phase, evaporation...
-
Upload
priscilla-dawson -
Category
Documents
-
view
247 -
download
11
Transcript of 3b Thermodynamics Specific heat capacity, c Latent heat capacity, L Change of phase, evaporation...
3b Thermodynamics
•Specific heat capacity, c
•Latent heat capacity, L
•Change of phase, evaporation
•First law of thermodynamics,
byin WQU
Real Life Application of specific heat capacity, c
Thinking QuestionCheese is still hot a long time, after being taken out of the oven. Why?
Heat capacity, Cnote: C = m.c
m = Mass of the substance, c = Specific heat capacity
The heat capacity, C, of a body is defined as
the amount of heat Q required to raise its
temperature by 1 Kelvin without going
through a change in state.
The S.I. unit is J K–1
QC
Why is C worth learning?
• Demo video on Heat capacity, C
• Water Balloon Heat Capacity.MPG
Nice explanation
• http://www.asknlearn.com/contentpackaging/14297/RS14/Container.html
Specific Heat Capacity
The specific heat capacity, c, of a body is defined as the amount of heat required to raise the temperature of 1 kg of the body by 1 Kelvin.
The S.I. unit is J kg–1 K–1
m
Qc
Substance J/ kg oC
Aluminum 900Bismuth 123
Copper 386
Brass 380
Gold 126
Lead 128Silver 233
Tungsten 134
Zinc 387
Mercury 140
Alcohol (ethyl) 2400
Water 4186Ice (-10 0C) 2050
Specific Heat Capacity, c
of some substances at 25 oC and atmospheric pressure
What does larger specific heat capacity mean?
• Specific Heat with Rods and Wax.MPG• Hint:
• For the same mass, m, • cAl > c steel> c lead means• QAl > Q steel> Q lead for the same ∆θ
mcQ
To solve problems mcQ
if
0gainQ
Example 1• How much energy is required when a
piece of copper of mass 0.275 kg is heated from 14.0 C to 100.0 C? (Specific heat capacity of copper = 380 J kg–1 K–1)
J 31099.8
14100380275.0
,
mcQcopperbygainedHeat
m = 0.275 kgc = 380 J/(kg.K)
θf - θi = (100 – 14)Q
Example 2
• An electric heater of 5 kW is used to heat up a piece of copper of mass 0.5 kg from 10 C to 90 C. Calculate the time taken. (ccopper = 380 J kg–1 K–1)
m = 0.5 kgc = 380 J/(kg.K)
θf - θi = (90 – 10)
P= 5x103 W
heater
• Q = m c (θf - θi )
• (5x103)(t)= (0.5)(380)(90-10)
• t = 3.0 s
m = 0.5 kgc = 380 J/(kg.K)
θf - θi = (90 – 10)
P= Q/t = 5x103 W
heater
Example 3• An ethnic restaurant serves coffee in copper mugs. A waiter
fills a cup having a mass of 0.10 kg, initially at 20 °C, with 0.20 kg of coffee initially at 70 °C. What is the final temperature after the coffee and the cup has attained thermal equilibrium? Assume that there is no heat loss to the surroundings.
• (ccoffee = 4200 J kg–1 K–1; ccopper = 380 J kg–1 K–1)
H2Om = 0.2 kg
c = 4200 J/(kg.K)θf - θi = (θ – 70)
Q= ?cupm = 0.1 kg
c = 380 J/(kg.K)θf - θi = (θ – 20)
• ∑Q = 0
• mcup.ccup (θ – 20) + mH20.cH20 (θ – 70) = 0
• (0.1)(380)(θ – 20) + (0.2)(4200)(θ – 70)=0
• Solve for θ
• θ = 67.8 0C = 68 0C
H2Om = 0.2 kg
c = 4200 J/(kg.K)θf - θi = (θ – 70)
Q= ?cupm = 0.1 kg
c = 380 J/(kg.K)θf - θi = (θ – 20)
Flash Simulation• heat_metal.swf
• Thinking and reflection• The random fluctuation of the thermometer
instrument, relate back to chapter 1 called uncertainty or random error
• the assumption no heat loss to surrounding is shown in simulation but not easily explain with clarity with words
• The concept of thermal equilibrium is vividly show in simulation when calculations alone fail to show the process of thermal equilibrium
Electrical method to determine c
AV
Heater & material under
test
Connect to power supply
Remember PWS6 electrical setup for heater power?Term 1 practical YJC.
• Example 4• Hand write
Phase Changes
Q= mcsolid∆θ
Q= mcliquid∆θ
Q= mcgas∆θ
Change of phase-melting
• Latent heat of fusion, Lf
• Heat is supplied, Q
• Temperature remains constant. T=constant
Q= mLf
Change of phase-boiling• Latent heat of vaporisation, Lv
• Heat is supplied, Q• Energy (Work done by system) is needed to
expand against atmosphere Wby
• Temperature remains constant, T =constantQ= mLv
Change of phaseSpecific Latent heat of vaporisation >
Specific latent heat of fusion. Lv > Lf
Energy is needed to work against atmosphere, Wby
• Difference in PE between molecules in liquid to gaseous > in solid to liquid
Q= mLv
Q= mLf
Select 2
• http://www.asknlearn.com/contentpackaging/14297/RS14/Container.html
cooling effect during evaporation
Molecules with higher KE and are near to surface, are able to overcome intermolecular force and escape from liquid surface.
Molecules with lower KE are left behind in the liquid.
As a result, the average molecular KE decreases and hence the temperature drops.
Evaporation
• Occurs at all temperature (including boiling)
• Average KE of liquid decreases
• Cooling process
http://www.colorado.edu/physics/2000/applets/bec.html
Video on Real Life Application of Evaporation & Condensation
• Drinking Bird.MPG
• Clever synthesis of many Physics ideas
• You can read up about it by “Goggling” Dippy Bird
• http://science.howstuffworks.com/question608.htm
Now let’s test yourselvesQn 1For a given liquid at atmospheric pressure, which process can occur at any temperature?
A) BoilingB) EvaporationC) MeltingD) Solidification
Now let’s test yourselvesQn 2Latent heat of vaporisation is the energy required to
A) Separate the molecules of the liquidB) Force back the atmosphere to make space for the
vapourC) Increase the average molecular speed in the liquid
phase to that in the gas phaseD) Separate the molecules and to force back the
atmosphereE) Separate the molecules and to increase their
average molecular speed to that in the gas phase.
Latent heat, L
The S.I. unit for l is J kg–1.
mLQ
Specific latent heat of fusion, L f
Specific latent heat of vaporisation, L v
Q = m L f Q = m L v
No change in temperature
• Example 5
Total heat gained = Total heat lost / supplied
Or
∑ (heat) = 0
summary of Specific heat capacity, Specific latent heat
Specific heat capacity, c : Q=mc ∆θ
Specific latent heat of fusion, Lf : Q= m Lf
Specific latent heat of vaporisation, Lv: Q= m Lv
In a closed system,
3 most important equations:
Phase Change of water
Electrical method to determine L
AV
Heater & material under test
Connect to power supply
• Example 6
• Example 7
• Suggested to hand write, for ease of following the chain of thought
Example 6An electric kettle contains 1.5 kg of water at 100 C and is powered by a 2.0 kW electric element. If the thermostat of the kettle fails to operate, calculate the time taken for the kettle to boil dry.(lv of water = 2000 kJ kg–1)
t = 1500 s
heat gained by water (boil) = heat supplied by kettle
Example 7 (a) better to use h/t
Therefore, lv = 1.98 x 106 J Kg-1
1st set up:
Mass = 0.1 kgPower = 240 WTime taken = 15 min
2nd set up:
Mass = 0.15 kgPower = 350 WTime taken = 15 min
a) Calculate the specific latent heat of vaporisation.
Heat supplied = heat gained by liquid + heat loss
P t = m lv + h h = P t – m lv
Since time taken is the same, h will also be the same.
(240)(15 x 60) – (0.1) lv = (350)(15 x 60) – (0.15) lv
Example 7 (b)
= 8.3 %
b) Find out what percentage of the heat supplied was lost in 1st attempt.
Percentage of heat loss =
Heat loss, h = (240)(15 x 60) – (0.1)(1.98 x 106)
Using h = P t – m lv for 1st set up:
= 1.80 x 10 4 J
=
First law of thermodynamics
• Conservation of energy
• To change the internal energy of system
1. Forces to do work
2. By means of heat transfer
3. Combination of both
byin WQU
First law of Thermodynamics
The increase in internal energy of a closed system is the sum of the heat supplied to the system subtract the work done by the system.
byWQU Increase in internal energy of system
Heat supplied to system
Work done by system
(+)(+) (+)
Volume will
increase.
Temperature will
increase.
First law of Thermodynamics
The increase in internal energy of a closed system is the sum of the heat supplied to the system subtract the work done by the system.
byWQU Increase in internal energy of system
Work done by system
Heat supplied to system
↑ to
In a closed system, the mass is fixed. Therefore no mass can be added into
or removed from the system.
Recap: Internal energy
Internal energy, U of a system is the total sum of the kinetic energy, Ek and the potential energy, Ep of the molecules in the system.
Kinetic energy, Ek is associated with the temperature.
Potential energy, Ep depends on the separation of molecules.
moleculespk )EE(U
Mechanical Equivalent of Heat.MPG
• Video showing Q is a form of energy.
Nice explanation
• http://www.asknlearn.com/contentpackaging/14297/RS15/Container.html
byin WQU
onin WQU
onby WW
Remember one form
First law of Thermodynamics
byWQU Sign convention:
Descriptors Meaning Signs
Heat enters system
Heat supplied to system
+ Q
Heat leaves system
Heat given out by system
– Q
onby WW
First law of Thermodynamics
Descriptors Meaning Signs
Volume of system decreases
Work done on system
- Wby
Volume of system increases
Work done by system
+ Wby
Sign convention:byWQU
onby WW
First law of Thermodynamics
Sign convention:
Descriptors Meaning Signs
Temperature increases
Increase in internal energy
+ ΔU
Temperature decreases
Decrease in internal energy
– ΔU
byWQU
onby WW
Sign conventions
conventions Opposite convention
∆U Increase in internal energy
Decrease in internal energy
Q Heat supplied to the system
Heat loss by the system
Wby Work done by the system
Work done on the system
byin WQU
onby WW
Internal energy of ideal gas
• Recall internal energy • does not possess internal potential energy• PE=0 by definition for gas• is the sum of the molecular KE of particles• KE temperature
nRT2
3 U gas ideal NkT
2
3 U gas ideal
Internal energy of ideal gas
nRT2
3 U gas ideal NkT
2
3 U gas ideal
Average translational KE per molecule = kT2
3
Total KE for N molecules in the gas = NkT2
3
nRT2
3
Work done by system
V
A
B
p
Work done by gas in compression is a negative number
p
V
A
B
Work done by gas in expansion is a positive number
Work done by system (gas)
V
p
Wby = area under the p-v graph =
dvp
Wby= p ∆V = p (Vf – Vi) = negative number
Work done-cycle
p
V
By gas
On gas
A B
CD
Process
A B : by gas
B C : zero
C D : on gas
D A : zero
Overall : work done by gas
WD in a cycle = area of enclosed regionWD ABCDA = WD by gas = area of enclosed regionWD ADCBA = - WD by gas = - area of enclosed region
Video on Fire Syringe
• After watching the video, think about how this video is related to the table
Process What it means
Iso-baric Constant pressure
Iso-choric /Iso-volumetric Constant volume
Iso-thermal Constant temperature
AdiabaticNo heat enters or leaves the system
Notice isothermal lines.
• http://www.lon-capa.org/~mmp/applist/pvt/pvt.htm
• http://mysite.verizon.net/pmrenault/thermo.html
T=200K
T=150K
• http://mysite.verizon.net/pmrenault/thermo.html
P=constant
• http://mysite.verizon.net/pmrenault/thermo.html
V=constant
Thermodynamic processesThere are 4 special processes to consider:
Process What it means
Iso-baric Constant pressure
Iso-choric /Iso-volumetric Constant volume
Iso-thermal Constant temperature
AdiabaticNo heat enters or leaves the system
p
V
T2T1
Isothermal processes
Thermodynamic ProcessesThere are 4 special thermodynamic processes:
Process What it means Implications
Iso-baric Constant pressure P = constant
Iso-choric/ Iso-volumetric
Constant volume V = constant
∆V = 0 m3
W = P ∆V = 0 J
Iso-thermal Constant temperature
T = constant
∆U = 0 J
Adiabatic No heat enters or leaves system
Q = 0 J
∆U = W
Thermodynamic ProcessesThere are 4 special thermodynamic processes:
Process What it means Implications
Iso-baric Constant pressure P = constant
Iso-choric/ Iso-volumetric
Constant volume V = constant
∆V = 0 m3
W = P ∆V = 0 J
Iso-thermal Constant temperature
T = constant
∆U = 0 J
Adiabatic No heat enters or leaves system
Q = 0 J
∆U = W
Thermodynamic ProcessesThere are 4 special thermodynamic processes:
Process What it means Implications
Iso-baric Constant pressure P = constant
Iso-choric/
Iso-volumetric
Constant volume V = constant
∆V = 0 m3
Wby = P ∆V = 0 J
Iso-thermal Constant temperature
T = constant
∆U = 0 J
Adiabatic No heat enters or leaves system
Q = 0 J
∆U = -Wby
Thermodynamic ProcessesThere are 4 special thermodynamic processes:
Process What it means Implications
Iso-baric Constant pressure P = constant
Iso-choric/
Iso-volumetric
Constant volume V = constant
∆V = 0 m3
W = P ∆V = 0 J
Iso-thermal Constant temperature
T = constant
∆U = 0 J
Adiabatic No heat enters or leaves system
Q = 0 J
∆U = -Wby
Cool video on Q=0
• Adiabatic Expansion.MPG
pV diagramIndicator diagram or pV diagram is simply a Pressure vs Volume graph
p / Pa
V / m3
V = volume of containerp = Pressure exerted on the walls by gas molecules
pV diagramIndicator diagram or pV diagram is simply a Pressure vs Volume graph
p / Pa
V / m3
∆U = Q - Wby
Heat flow in
U= (3/2).nRT
So
ΔU = (3/2)nRΔT pV diagram
Wby = P ∆V= area under pV diagram
Iso-volumetric processConstant volume
p / Pa
V / m3
V remains unchanged when P changes.
∆U = Q - Wby
pV diagram
Area = 0 J
Wby = 0 J
∆U = Q
60
Iso-baric processConstant pressure
p / Pa
V / m3
P remains unchanged when V changes.
∆U = Q - Wby
pV diagram
100
∆V = 50
Area = 100 x 50 = 5000 J
Wby = + 5000 J or – 5000 J?
→ V increases→ Work done BY gas
→ + Wby
Wby = + 5000 J
Iso-baric processConstant pressure
p / Pa
V / m3
P remains unchanged when V changes.
∆U = Q -Wby
pV diagram
100
∆V = 50
Area = 100 x 50 = 5000 J
Wby = + 5000 J or – 5000 J?
→ V decreases→ Work done ON gas
→ Wby negative number
Wby = - 5000 J
Iso-thermal processConstant Temperature
p / Pa
V / m3
∆U = Q - Wby
pV diagram
Area = W
No need to calculate W!
T = 80 oc
T remains unchanged along the curve.
However, calculation of W in Iso-thermal process is not in A level syllabus!
Iso-thermal processConstant Temperature
p / Pa
V / m3
T remains unchanged along the curve.
∆U = Q - Wby
pV diagram
T remains constant
0= Q – Wby
∆U = 0 J
T = 80 oc
T = 90 oc
U also remains constant
[ Wby = negative number if process is decrease in volume
Wby = postive number if process is increase in volume]
50
60
70 Higher T?
Finding Wby from pV diagram
Important points:
1. Find the area under graph = magnitude of W.
p / Pa
V / m3
2. If volume decreases (compression), work is done ON gas → Wby is negative number
3. If volume increases (expansion), work is done BY gas → Wby is positive number
4. Work done ON gas = – Work done BY gas
• Example 8
• Example 9
• Example 10
• Example 11
• Example 12
• Example 13
• Example 14
C200
500
Example 8
Find the work done by the gas in going from the situation represented by point A, through point B, to point C.
300 500 800 V/cm3
p/KPa
ABArea = (200+500)(200/2) + (500)(300)
= 2.20 x 105 J
Magnitude of Wby:
Wrong answer! Can you spot the mistake?
Draw this!
C200
500
Example 8
Find the work done on the gas in going from the situation represented by point A, through point B, to point C.
Wby gas = + 220 J
= 220 J
300 500 800 V/cm3
p/KPa
AB
Since volume decreases (compression) → Wby is a positive number
Area = (2 x 105 + 5 x 105)(200 x 10-6)/2 + (5 x 105)(300 x 10-6)
Magnitude of W:
Example 12 – (a)A thermodynamic system is taken from initial state A to state B and back again to A via state C.
V/cm3
p/KPa
A B
C
Initial state
31
20
40
∆U = Q - Wby
A → B
B → C
C → A – 2
+0.04
0
-0.06
a) Fill up the table with ‘+’, ‘–’ or ‘0’.
+
+
–
+
+
pV diagramIso-thermal curve1st law of thermodynamics
Example 12 – (b)A thermodynamic system is taken from initial state A to state B and back again to A via state C.
V/cm3
p/KPa
A B
C
Initial state
31
20
40
2
b) Calculate net work done by the system for one complete cycle.
WDA → B = + (20 x 103)(2 x 10-6) = 0.04 J
WDB → C = 0 J
WDC → A = – [(20 + 40) x 103)](2 x 10-6)/2 = – 0.06 JTherefore net work done by system = 0.04 + 0 + (– 0.06)
= – 0.02 J
(area enclosed by loop)
Example 13An ideal gas undergoes a cycle of changes A → B → C → D as shown in the figure below. Complete the table.
V/10-3 m3
p/KPa
AB
DC
Initial state
0.30.1
100
200
∆U = Q - Wby
A → B
B → C
C → D
D → A
– 50
25
140
+ 20
0
– 40
0
1. Draw table2. Known
values3. Find W from
pV diagram
Example 13An ideal gas undergoes a cycle of changes A → B → C → D as shown in the figure below. Complete the table.
V/10-3 m3
p/KPa
AB
DC
Initial state
0.30.1
100
200
∆U = Q - Wby
A → B
B → C
C → D
D → A
– 50 – 70
25
140
– 20
+ 25
100
0
+40
0– 75– 75
4. Find ∆U 3. Find W from pV diagram
5. Finally, find QQ may be found
using mc∆θ
Summary
For cycle, total ∆U = 0 J
Example 13: Won explanationAn ideal gas undergoes a cycle of changes A → B → C → D as shown in the figure below. Complete the table.
V/10-3 m3
p/KPa
AB
DC
Initial state
0.30.1
100
200
∆U = Q - Wby
A → B
B → C
C → D
D → A
– 50 – 70
25
140
– 20
+ 25
100
0
+40
0– 75– 75
Summary
+ 20
0
– 40
0
Won
onby WW onin WQU
Example 14 – (a)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc
V/m3
p/ x 105 Pa
A
BC
V2V1
1.01
a) Illustrates these changes on a pV diagram.
A: the gas is heated at constant pressure to 127 oc
B: compressed isothermally to its original volume
C: the gas is cooled at constant volume to its original temperature27 oc
127 oc
P2
Initial state?
Example 14 – (b)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc
V/m3
p/ x 105 Pa
A
BC
V2V1
1.01
b) How much work does the gas do in pushing the piston during stage A?
P2
Find V1:
331
15
10371
2732731818
110011
m.V
))(.(V).(
nRTpVUse
V1 = 1.37 x 10-3 m3
Example 14 – (b)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc
b) How much work does the gas do in pushing the piston during stage A?
Find V2:
332
23
2
2
1
1
10831
27312727327
10371
m.V
V.
T
V
T
V
V/m3
p/ x 105 Pa
A
BC
V2V1
1.01
P2
V1 = 1.37 x 10-3 m3
V2 = 1.83 x 10-3 m3
Example 14 – (b)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc
b) How much work does the gas do in pushing the piston during stage A?
Therefore work done by gas
= + (area under graph)
V/m3
p/ x 105 Pa
A
BC
V2V1
1.01
P2
V1 = 1.37 x 10-3 m3
= (1.01 x 105)[(1.83 – 1.37) x 10-3]
= 46.5 J
V2 = 1.83 x 10-3 m3
Example 14 – (c)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc
c) What is the change in the internal energy in stage A?
V/m3
p/ x 105 Pa
A
BC
V2V1
1.01
P2
= + 69.3 J
))(.(
TnRU
2712731818
1
2
32
3
Assuming negligible potential energy,
∆U = ∆KE = (3/2) NK ∆T = (3/2) nR ∆T
27 oc127 oc
Example 14 – (d)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc
d) What is the heat input to the cylinder in stage A?
V/m3
p/ x 105 Pa
A
BC
V2V1
1.01
P2
∆U = Q – Wby
69.3 = Q – 46.5
115.8 J = Q
Note: Wby gas = + 46.5 J → Won gas = – 46.5 J
Example 14 – (e)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc
e) What is the change in internal energy of the gas in stage B?
V/m3
p/ x 105 Pa
A
BC
V2V1
1.01
P2
Since ∆T = 0,
therefore ∆U = 0 J,
since ∆U = (3/2) nR ∆T
Is there any change in temperature in stage B?
Example 14 – (f)Mass of gas = 1 g Initial pressure = 1.01 x 105 PaMolar mass = 18 Initial temperature = 27 oc
f) How much heat must be extracted from the gas in stage C?
V/m3
p/ x 105 Pa
A
BC
V2V1
1.01
P2
Therefore ∆U for stage C = – 69.3 J
For a cycle, total ∆U = 0 J
∆U = + 69.3 J
∆U = 0 J
∆U = ? J
∆U = Q - Wby
– 69.3 = Q - 0– 69.3 J =Q = Qin
Q to be released = +69.3 J