Transcript Lecture25-12

Lecture 25
Thermal Processes
Course Evaluations
Now through December 7: you have the
opportunity to submit a course
evaluation
You may drop your lowest problem set score
IF AND ONLY IF
you submit a course evaluation
Water and Ice
You put 1 kg of ice at 0°C
together with 1 kg of water at
50°C. What is the final
temperature?
LF = 80 cal/g
cwater = 1 cal/g °C
a) 0°C
b) between 0°C and 50°C
c) 50°C
d) greater than 50°C
Water and Ice
You put 1 kg of ice at 0°C
together with 1 kg of water at
50°C. What is the final
temperature?
a) 0°C
b) between 0°C and 50°C
c) 50°C
d) greater than 50°C
LF = 80 cal/g
cwater = 1 cal/g °C
How much heat is needed to melt the ice?
Q = mLf = (1000 g)  (80 cal/g) = 80,000 cal
How much heat can the water deliver by cooling from 50°C to 0°C?
Q = cwater m x T = (1 cal/g °C)  (1000 g)  (50°C) = 50,000 cal
Thus, there is not enough heat available to melt all the ice!!
Follow-up: How much more water at 50°C would you need?
Ice Cold Root Beer
You have neglected to chill root
beer for your son’s 5th-birthday
party. You submerge the cans
in a bath of ice and water as you
start dinner. How can you hurry
the cooling process?
a) Add more ice to the icewater
b) add salt to the icewater
c) hold the icewater in an
evacuated chamber (vacuum)
d) Jump in the car and drive to a
nearby convenience store
Ice Cold Root Beer
You have neglected to chill root
beer for your son’s 5th-birthday
party. You submerge the cans
in a bath of ice and water as you
start dinner. How can you hurry
the cooling process?
a) Add more ice to the icewater
b) add salt to the icewater
c) hold the icewater in an
evacuated chamber (vacuum)
d) Jump in the car and drive to a
nearby convenience store
Not a), because ice water at 1 atm is zero degrees, no matter the
proportion of water and ice
Not c), because ice is less dense than water so you will raise the melting
point when you reduce the pressure. This will allow the water to get a
little warmer than 0o
Not d), because you’ll forget your wallet and it will end up taking more
time
b) because salt interferes with the formation of ice. This barrier to the
solid phase lowers the fusion temperature, and so reduces the
temperature of the ice water
The larger ΔT, the more heat transfers
per unit time. Thus, the colder the ice
bath, the faster the root beer will chill,
and the warmer the bath, the slower
the root beer will chill
1
ΔP
2
ΔT
When two states exist in the same
system (like, ice and water), the
system MUST be on the equilibrium
curve (in the case, the fusion curve).
Fusion curve for water
As pressure goes lower, the ice/water
mixture will ride the fusion curve from
point 1 to point 2.
This implies that temperature goes up.
The Zeroth Law of Thermodynamics
If object A is in thermal equilibrium with object B,
and object C is also in thermal equilibrium with
object B, then objects A and C will be in thermal
equilibrium if brought into thermal contact.
Object B can then be a thermometer,
providing a scale to compare objects:
Temperature
That is, temperature is the only factor that
determines whether two objects in thermal
contact are in thermal equilibrium or not.
Kinetic Energy and Temperature
Comparing pressure in the kinetic theory (monatomic
ideal gas) with the ideal gas law allows us to relate
average kinetic energy and temperature
Internal Energy
The internal energy of an ideal monatomic gas is
the sum of the kinetic energies of all its molecules.
In the case where each molecule consists of a single
atom, this is all linear kinetic energy of atoms:
Conservation of Energy
If a system does work on the external world, and no
heat is added, its internal energy decreases.
Internal energy changes with heat input
If heat is added to a system, this is an increase in
internal energy.
Assuming constant volume (so W = 0):
The First Law of Thermodynamics
Combining these gives the first law of thermodynamics.
The change in a system’s internal energy is related to the
heat Q and the work W by conservation of energy:
It is vital to keep track of the signs of Q and W.
The First Law of Thermodynamics
Jogger is warm: heat transfer to the environment
She is doing work on the environment (force*distance)
Internal energy is decreasing
The First Law of Thermodynamics
State function of a system depend only on the state of the
system (temperature, pressure, etc), not on how a system
arrived in that state.
The internal energy of a system depends only on its
temperature. It is a state function.
The work done and heat added are specific to a process.
There is no “work” or “heat” in a system... those are just
terms to describe the change in internal energy.
Boiling water:
When 1 g of water boils at 100o C under 1 Atm. The volume of the steam at
100o C is 1671 cm3. Find the work done in the expansion and calculate the
change in internal energy of the system
Lv = 22.6 x 105 J/kg
Boiling water:
When 1 g of water boils at 100o C under 1 Atm. The volume of the steam at
100o C is 1671 cm3. Find the work done in the expansion and calculate the
change in internal energy of the system
Lv = 22.6 x 105 J/kg
Q = 0.001 kg x 22.6 x 105 J/kg = 2260 J
W = (101 x 103 N/m2) x (1671 cm3 -1 cm3)x(10-6 m3 /cm3) = 169 J
ΔU = Q - W = 2091 J
Reversible Thermal Processes
We will assume that all processes we discuss are
“quasi-static” – they are slow enough that the
system is always “in equilibrium” (fluid volumes
have the same temperature throughout, etc.)
We also assume they are reversible (frictionless pistons, etc.):
For a process to be reversible, it must be possible to return both the
system and its surroundings to the same states they were in before the
process began.
We will discuss 4 idealized processes with Ideal Gases:
•Constant Pressure
•Constant Volume
•Constant Temperature
•Q= 0 (adiabatic)
Constant pressure
Isobaric process
Work done by an expanding gas, constant pressure:
Work is area
under the PV
graph
Examples: piston against atmosphere, or
vertical piston with constant weight on top
so changing volume implies
changing temperature
imagining any general process as approximated by
a number of constant pressure processes:
Work is area under
the PV graph
Constant Volume
Isovolumetric process
If the volume stays constant, nothing moves and
no work is done.
Change in internal energy is
related only to the net heat input
so changing pressure implies
changing temperature
Constant Temperature
Isothermal processes
If the temperature is constant, the
pressure varies inversely with the
volume.
Constant Temperature
A system connected to a large heat reservoir is
usually thought to be held at constant temperature.
Volume can change, pressure can change, but the
temperature remains that of the reservoir.
T = constant
if W < 0 (work done on the system)
than Q<0 (heat flows out of the system)
W=Q
if W > 0 (work done by the system)
than Q>0 (heat flows out into the system)
Work in an Constant-Temperature Process
The work done is the area under the curve:
For you calculus junkies:
Adiabatic Process
An adiabatic process is one in which no heat
flows into or out of the system.
One way to ensure that a process is adiabatic is
to insulate the system.
Q=0
The adiabatic P-V curve is
similar to the isothermal
one, but is steeper.
Rapid Adiabatic Process
Another way to ensure
that a process is
effectively adiabatic is
to have the volume
change occur very
quickly.
In this case, heat has no
time to flow in or out of
the system.
Thermal Processes
The different types of ideal thermal processes
Specific Heat for an Ideal Gas at Constant Volume
Specific heats for ideal gases must be quoted
either at constant pressure or at constant
volume. For a constant-volume process,
First Law of Thermodynamics
for an ideal gas (from the
kinetic theory)
Specific Heat for an Ideal Gas at Constant Pressure
At constant pressure, (some work is done)
Some of the heat energy goes into the mechanical work, so
more heat input is required to produce the same ΔT
First Law of Thermodynamics
for an ideal gas (from the
kinetic theory)
Specific Heats for an Ideal Gas
Both CV and CP can be
calculated for a
monatomic ideal gas
Although this calculation was
using the first law of
done for an ideal, monatomic gas,
thermodynamics.
the difference Cp - Cv works well
for real gases.
Specific Heats and Adiabats In Ideal Gas
The P-V curve for an adiabat is
given by
for
monotonic
gases
Work of a Thermal Cycle
In the closed thermodynamic
cycle shown in the P-V diagram,
the work done by the gas is:
a) positive
b) zero
c) negative
P
V
Work of a Thermal Cycle
In the closed thermodynamic
cycle shown in the P-V diagram,
the work done by the gas is:
a) positive
b) zero
c) negative
The gas expands at a higher pressure
and compresses at a lower pressure.
P
In general, clockwise = positive work;
counterclockwise = negative work.
V
a) 4 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much work is done by the gas in this
process, in terms of the initial pressure and
volume?
b) 7 P1V1
c) 8 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 4 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much work is done by the gas in this
process, in terms of the initial pressure and
volume?
Area under the curve:
(4 V1)(P1) + 1/2 (4V1)(2P1)
= 8 V 1P 1
b) 7 P1V1
c) 8 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much internal energy is gained by the
gas in this process, in terms of the initial
pressure and volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much internal energy is gained by the
gas in this process, in terms of the initial
pressure and volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
Ideal monatomic gas: U = 3/2 nRT
Ideal gas law: PV = nRT
U = 3/2 PV
P2V2 = 15 P1V1
P2 = 3P1
Δ(PV) = 14 P1V1
P1
ΔU = 21 P1V1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much heat is gained by the gas in this
process, in terms of the initial pressure and
volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
P2 = 3P1
P1
V1
V2 =5V1
a) 7 P1V1
One mole of an ideal monatomic gas undergoes
the reversible expansion shown in the figure,
where V2 = 5 V1 and P2 = 3 P1.
How much heat is gained by the gas in this
process, in terms of the initial pressure and
volume?
b) 8 P1V1
c) 15 P1V1
d) 21 P1V1
e) 29 P1V1
First Law of Thermodynamics
W = 8 P 1 V1
P2 = 3P1
P1
V1
V2 =5V1
The Second Law of Thermodynamics
We observe that heat always flows spontaneously from
a warmer object to a cooler one, although the opposite
would not violate the conservation of energy.
This direction of heat flow is one of the ways of
expressing the second law of thermodynamics:
When objects of different temperatures are brought
into thermal contact, the spontaneous flow of heat
that results is always from the high temperature
object to the low temperature object. Spontaneous
heat flow never proceeds in the reverse direction.
Heat Engines
A heat engine is a device that converts heat into work. A classic
example is the steam engine. Fuel heats the water; the vapor
expands and does work against the piston; the vapor condenses
back into water again and the cycle repeats.
All heat engines have:
a working substance
a high-temperature reservoir
a low-temperature reservoir
a cyclical engine
PHYS2010 practice for Chap14-18, Fall 2009
Provided in lecture
notes on: 12/1
12) An ideal monatomic gas undergoes a reverrsible expansion to 1.5 times its original
volume. In which of these processes does the gas perform the least amount of work?
A) at constant temperature
B) at constant pressure
C) if the pressure decreases in proportion to the volume (i.e. PV=constant)
D) adiabatically
E) if the pressure increases in proportion to the volume (i.e., P/V=constant)
suggested time: 1-2 minutes
PHYS2010 practice for Chap14-18, Fall 2009
Provided in lecture
notes on: 12/1
13) An expandable container holds 1.50 mole of He gas with an initial pressure of 650
kPa and an initial volume of 2.10 L. The gas expands isothermally to a final pressure
of 350 kPa. How much heat is gained by the gas in this process?
A) 1370 J
B) 685 J
C) 792 J
D) 1280 J
E) 1700 J
suggested time: 4 minutes
PHYS2010 practice for Chap14-18, Fall 2009
Provided in lecture
notes on: 12/1
14) 1.50 moles of an ideal monatomic gas are initially at a temperature of 317 K. If the
gas gains 2670 J of heat and performs 770 J of work, what is its final temperature?
A) 419 K
B) 756 K
C) 687 K
D) 359 K
E) 526 K
suggested time: 3-4 minutes