E =W + Q

Physics 207 – Lecture 24
Lecture 24
Goals:
• Chapter 17
Employ heat (Q) and energy transfer in materials
Recognize adiabatic processes (i.e., Q=0)
• Chapter 18
Follow the connection between temperature, thermal
energy, and the average translational kinetic energy
molecules
Understand the molecular basis for pressure and the idealgas law.
To predict the molar specific heats of gases and solids.
• Assignment
HW11, Due Wednesday 9:00 AM
For Thursday, Read through all of Chapter 18
Physics 207: Lecture 24, Pg 1
1st Law of Thermodynamics
Eth =W + Q
(if ∆K & ∆U =0 )
W & Q with respect to the system
Work W and heat Q depend on process by which the
system is changed (path dependent).
The change of energy in the system,
Eth depends
only on the total energy exchanged W+Q, not on the
process.
Physics 207: Lecture 24, Pg 2
Page 1
Physics 207 – Lecture 24
1st Law: Work & Heat
Work done on system (an ideal
gas….notice minus sign, V is in
reference to the system)
Won = −
final
∫ p dV = −(area under curve )
initial
W on system < 0 Moving left to right
[where (Vf > Vi)]
If
ideal gas, pV = nRT
W by system
> 0 Moving left to right
Physics 207: Lecture 24, Pg 3
1st Law: Work & Heat
Work:
Depends on the path taken in the pV-
diagram
(It is not just the destination but the path…)
Won system > 0 Moving right to left
Physics 207: Lecture 24, Pg 4
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Physics 207 – Lecture 24
1st Law: Work (“Area” under the curve)
Work depends on the path taken in the pV-diagram :
3
W23= -pf (Vf -Vi)
W12= -pi (Vf -Vi)
2
3
1
1
2
(a) W a = W 1 to 2 + W 2 to 3 (1st p then V constant, T changes)
W a (on) = - pi (Vf - Vi) + 0 > 0
(b) W b = W 1 to 2 + W 2 to 3 (1st p then V constant, T changes)
W b (on) = 0 - pf (Vf - Vi) > W a > 0
(c) Need explicit form of p versus V but W c (on) > 0
Physics 207: Lecture 24, Pg 5
Paths on the pV diagram (with an idea gas)
W = - p ∆V
????
W=0
????
(1) Isobaric
(2) Isothermal
(3) Isochoric
(4) Adiabatic
1
p
3
T1
2
4
T2
Ideal
gas
T4 T3
V
Physics 207: Lecture 24, Pg 6
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Physics 207 – Lecture 24
Isothermal processes (in an ideal gas)
Work done when PV = nRT = constant P = nRT / V
W =−
final
∫ p dV
= − (area under curve )
initial
Vf
Vf
Vi
Vi
W = − ∫ nRT dV / V = − nRT ∫ dV / V
W = − nRT ln ( Vf /Vi )
T1
3
p
T2
T4 T3
For this we need access
to thermal energy
V
Physics 207: Lecture 24, Pg 7
Adiabatic Processes (in an ideal gas)
An adiabatic process is process in which there is no
thermal energy transfer to or from a system (Q = 0)
A reversible adiabatic
process involves a
“worked” expansion in
which we can return all of
the energy transferred.
p
In this case
Cp
γ =C
PVγ = const.
V
All real processes are not.
4
2
T1
3
1
T2
T4 T3
V
We need to know Cp & CV
Physics 207: Lecture 24, Pg 8
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Physics 207 – Lecture 24
Work and Ideal Gas Processes (on system)
Isothermal
W = − nRT ln ( Vf /Vi )
Isobaric
W = − p ( Vf - Vi )
Isochoric
W =0
γ
FYI: Adiabatic (and reversible) PV = const.
W = − ∫VV PdV = − ∫VV
2
1
2
1
const dV
Vγ V
(V2−γ − V1−γ )
= const
γ
Physics 207: Lecture 24, Pg 9
Two process are shown that take an ideal gas from state 1 to
state 3. (“by” means “by the system on the world”)
Compare the work done by process A to the work done by
process B.
A. WA > WB
B. WA < WB
C. WA = WB = 0
D. WA = WB but neither is zero
p3
p2
p1
ON
A 1 3 W12 = 0 (isochoric)
B 1 2 W12 = -½ (p1+p2)(V2-V1) < 0
B 2 3 W23 = -½ (p2+p3)(V1-V2) > 0
B 1 3
= ½ (p3 - p1)(V2-V1) > 0
Page 5
BY
-W12 > 0
-W23 < 0
< 0
Physics 207: Lecture 24, Pg 10
Physics 207 – Lecture 24
Two process are shown that take an ideal gas from state 1 to
state 3.
Compare the work done by process A to the work done by
process B.
A. WA > WB
B. WA < WB
C. WA = WB = 0
D. WA = WB but neither is zero
A13
B12
B23
B 1 3
ON
W12 = 0 (isochoric)
W12 = -½ (p1+p2)(V2-V1) < 0
W23 = -½ (p2+p3)(V1-V2) > 0
= ½ (p3 - p1)(V2-V1) > 0
BY
-W12 > 0
-W23 < 0
< 0
Physics 207: Lecture 24, Pg 11
Both Q and W can change T
We know how work changes the mechanical energy of
a solid “system”
Here our system is an ideal gas….only the
temperature can change
For real materials we can change the temperature or
the state
We must quantify the response of a system to thermal
energy transfer (Q)
Physics 207: Lecture 24, Pg 12
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Physics 207 – Lecture 24
What about Q?
Specific
Heat
Latent Heat
What are the relationships between ∆ETh and T.
Latent Heat
Physics 207: Lecture 24, Pg 13
Heat and Latent Heat
Latent heat of transformation L is the energy required for 1 kg of
substance to undergo a phase change. (J / kg)
Q = ±ML
Specific heat c of a substance is the energy required to raise the
temperature of 1 kg by 1 K. (Units: J / K kg )
Q=Mc T
Molar specific heat C of a gas at constant volume is the energy
required to raise the temperature of 1 mol by 1 K.
Q = n CV T
If a phase transition involved then the heat transferred is
Q = ±ML+M c T
Physics 207: Lecture 24, Pg 14
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Physics 207 – Lecture 24
Q : Latent heat and specific heat
The molar specific heat of gasses depends on the
process path
CV= molar specific heat at constant volume
Cp= molar specific heat at constant pressure
Cp= CV+R (R is the universal gas constant)
Physics 207: Lecture 24, Pg 15
Q : Latent heat and specific heat
The molar specific heat of gasses depends on the
process path
CV= molar specific heat at constant volume
Cp= molar specific heat at constant pressure
Cp= CV+R (R is the universal gas constant)
γ
Cp
=
CV
Page 8
Physics 207: Lecture 24, Pg 16
Physics 207 – Lecture 24
Latent Heat
Most people were at least once burned by hot water
or steam.
An equal amount (by mass) of boiling water and
steam contact your skin.
Which is more dangerous, the water or the steam?
Physics 207: Lecture 24, Pg 17
Mechanical equivalent of heat
Heating
liquid water:
Q = amount of heat that must be supplied to
raise the temperature by an amount ∆ T .
[Q] = Joules or calories. 1 cal = 4.186 J
1 kcal = 1 cal = 4186 J
calorie: energy to raise 1 g of water from
14.5 to 15.5 °C
(James Prescott Joule found the mechanical
equivalent of heat.)
Sign convention:
+Q : heat gained
- Q : heat lost
Physics 207: Lecture 24, Pg 18
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Physics 207 – Lecture 24
Exercise
The specific heat (Q = M c T) of aluminum is about twice that
of iron. Consider two blocks of equal mass, one made of
aluminum and the other one made of iron, initially in thermal
equilibrium.
Heat is added to each block at the same constant rate until it
reaches a temperature of 500 K. Which of the following
statements is true?
(a) The iron takes less time than the aluminum to reach 500 K
(b) The aluminum takes less time than the iron to reach 500 K
(c) The two blocks take the same amount of time to reach 500 K
Physics 207: Lecture 24, Pg 19
Exercise
The specific heat (Q = M c T) of aluminum is about twice that
of iron. Consider two blocks of equal mass, one made of
aluminum and the other one made of iron, initially in thermal
equilibrium.
Heat is added to each block at the same constant rate until it
reaches a temperature of 500 K. Which of the following
statements is true?
(a) The iron takes less time than the aluminum to reach 500 K
(b) The aluminum takes less time than the iron to reach 500 K
(c) The two blocks take the same amount of time to reach 500 K
Physics 207: Lecture 24, Pg 20
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Physics 207 – Lecture 24
Heat and Ideal Gas Processes (on system)
Isothermal Expansion/Contraction
∆ETh = 0 = W + Q
Q = −W = nRT ln ( Vf /Vi )
Isobaric
Q = nC p ∆T = n (CV + R ) ∆T
Isochoric
∆ETh = 0 + Q
Q = nCV ∆T
Adiabatic
Q=0
∆ETh = W + 0
Physics 207: Lecture 24, Pg 21
Combinations of Isothermal & Adiabatic Processes
All engines employ a thermodynamic cycle
W = ± (area under each pV curve)
Wcycle = area shaded in turquoise
Watch sign of the work!
Physics 207: Lecture 24, Pg 22
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Physics 207 – Lecture 24
Exercise Latent Heat
Most people were at least once burned by hot water or steam.
Assume that water and steam, initially at 100°C, are cooled down
to skin temperature, 37°C, when they come in contact w ith your
skin. Assume that the steam condenses extremely fast, and that
the specific heat c = 4190 J/ kg K is constant for both liquid water
and steam.
Under these conditions, which of the following statements is true?
(a) Steam burns the skin worse than hot water because the thermal
conductivity of steam is much higher than that of liquid water.
(b) Steam burns the skin worse than hot water because the latent
heat of vaporization is released as well.
(c) Hot water burns the skin worse than steam because the thermal
conductivity of hot water is much higher than that of steam.
(d) Hot water and steam both burn skin about equally badly.
Physics 207: Lecture 24, Pg 23
Exercise Latent Heat
Most people were at least once burned by hot water or steam.
Assume that water and steam, initially at 100°C, are cooled down to
skin temperature, 37°C, when they come in contact with your skin.
Assume that the steam condenses extremely fast, and that the
specific heat c = 4190 J/ kg K is constant for both liquid water and
steam.
Under these conditions, which of the following statements is true?
(b) Steam burns the skin worse than hot water because the latent
heat of vaporization is released as well.
How much heat H1 is transferred to the skin by 25.0 g of steam?
The latent heat of vaporization for steam is L = 2256 kJ/kg.
H1 = 0.025 kg x 2256 kJ/kg = 63.1 kJ
How much heat H2 is transferred to the skin by 25.0 g of water?
H2 = 0.025 kg x 63 K x 4190 J/ kg K = 6.7 kJ
Physics 207: Lecture 24, Pg 24
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Physics 207 – Lecture 24
Energy transfer mechanisms
Thermal conduction (or conduction)
Convection
Thermal Radiation
Physics 207: Lecture 24, Pg 25
Energy transfer mechanisms
Thermal conduction (or conduction):
Energy transferred by direct contact.
e.g.: energy enters the water through
the bottom of the pan by thermal
conduction.
Important: home insulation, etc.
Rate of energy transfer ( J / s or W )
Through a slab of area A and
thickness ∆x, with opposite faces at
different temperatures, Tc and Th
Q / ∆t = k A (Th - Tc ) / ∆x
k :Thermal conductivity (J / s m °C)
Physics 207: Lecture 24, Pg 26
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Physics 207 – Lecture 24
Thermal Conductivities
J/s m °C
J/s m °C
J/s m °C
Aluminum
238
Air
0.0234
Asbestos
0.25
Copper
397
Helium
0.138
Concrete
1.3
Gold
314
Hydrogen
0.172
Glass
0.84
Iron
79.5
Nitrogen
0.0234
Ice
1.6
Lead
34.7
Oxygen
0.0238
Water
0.60
Silver
427
Rubber
0.2
Wood
0.10
Physics 207: Lecture 24, Pg 27
Home Exercise
Thermal Conduction
Two identically shaped bars (one blue
and one green) are placed between two
different thermal reservoirs . The thermal
conductivity coefficient k is twice as large
for the blue as the green.
You measure the temperature at the joint
between the green and blue bars.
100 C
Tjoint
300 C
Which of the following is true?
(A) Ttop > Tbottom
(B) Ttop= Tbottom
(C) Ttop< Tbottom
(D) need to know k
Physics 207: Lecture 24, Pg 28
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Physics 207 – Lecture 24
Home Exercise Thermal Conduction
Two identically shaped bars (one blue
and one green) are placed between
two different thermal reservoirs . The
thermal conductivity coefficient k is
twice as large for the blue as the
green.
100 C
Tjoint
300 C
Top: Pgreen = Pblue = Q / ∆t = 2 k A (Thigh - Tj ) / ∆x= k A (Tj - Tlow ) / ∆x
2 (Thigh - Tj ) = (Tj - Tlow ) 3 Tj(top)
By analogy for the bottom:
= 2 Thigh – Tlow
3 Tj(bottom) = 2 Tlow – Thigh
3 (Tj(top) - Tj(bottom) = 3 Thigh – 3 Tlow > 0
(A) Ttop > Tbottom
Physics 207: Lecture 24, Pg 29
Exercise Thermal Conduction
Two thermal conductors are butted
together and in contact with two thermal
reservoirs held at the temperatures
100 C
shown.
Which of the temperature vs. position
plots below is most physical?
(C)
Temperature
Temperature
Temperature
(B)
(A)
Position
300 C
Position
Position
Physics 207: Lecture 24, Pg 30
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Physics 207 – Lecture 24
Exercise Thermal Conduction
Two thermal conductors are butted
together and in contact with two thermal
reservoirs held at the temperatures
shown.
100 C
Which of the temperature vs. position
plots below is most physical?
(C)
Temperature
Temperature
Temperature
(B)
(A)
Position
300 C
Position
Position
Physics 207: Lecture 24, Pg 31
Energy transfer mechanisms
Convection:
Energy is transferred by flow of substance
1. Heating a room (air convection)
2. Warming of North Altantic by warm waters
from the equatorial regions
Natural convection: from differences in density
Forced convection: from pump of fan
Radiation:
Energy is transferred by photons
e.g.: infrared lamps
Stefan’s Law
P = σ A e T4 (power radiated)
σ = 5.7×10-8 W/m2 K4 , T is in Kelvin, and A is the surface area
e is a constant called the emissivity
Physics 207: Lecture 24, Pg 32
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Physics 207 – Lecture 24
Minimizing Energy Transfer
The Thermos bottle, also called a
Dewar flask is designed to minimize
energy transfer by conduction,
convection, and radiation. The
standard flask is a double-walled
Pyrex glass with silvered walls and
the space between the walls is
evacuated.
Vacuum
Silvered
surfaces
Hot or
cold
liquid
Physics 207: Lecture 24, Pg 33
Anti-global warming or the nuclear winter scenario
Assume P/A = P = 1340 W/m2 from the sun is incident on
a thick dust cloud above the Earth and this energy is
absorbed, equilibrated and then reradiated towards space
where the Earth’s surface is in thermal equilibrium with
cloud. Let e (the emissivity) be unity for all wavelengths of
light.
What is the Earth’s temperature?
P
= σ A T4= σ (4π r2) T4 = P π r2 T = [P / (4 x σ )]¼
σ=
5.7×10-8 W/m2 K4
T = 277 K (A little on the chilly side.)
Physics 207: Lecture 24, Pg 34
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Physics 207 – Lecture 24
Lecture 24
• Assignment
HW11, Due Wednesday (9:00 AM)
Tuesday review
Reading assignment through all of Chapter 18
Physics 207: Lecture 24, Pg 35
Page 18