Physics 207: Lecture 2 Notes

Report
Physics 207, Lecture 21, Nov. 12
Goals:
• Chapter 16
•
 Use the ideal-gas law.
 Use pV diagrams for ideal-gas processes.
Chapter 17
 Employ energy conservation in terms of 1st law of TD
 Begin understanding the concept of heat.
 Demonstrate how heat is related to temperature change
 Apply heat and energy transfer processes in real situations
 Recognize adiabatic processes.
• Assignment
 HW9, Due Wednesday, Nov. 19th
 HW10, Due Sunday, Wednesday, Read through 18.3
Physics 207: Lecture 22, Pg 1
Fluids: A tricky problem
 A beaker contains a layer of oil (green) with density ρ2 floating
on H2O (blue), which has density ρ3. A cube wood of density ρ1
and side length L is lowered, so as not to disturb the layers of
liquid, until it floats peacefully between the layers, as shown in
the figure.
 What is the distance d between the top of the wood cube (after
it has come to rest) and the interface between oil and water?
 Hint: The magnitude of the buoyant force
(directed upward) must exactly equal the
magnitude of the gravitational force
(directed downward). The buoyant force
depends on d. The total buoyant force
has two contributions, one from each of
the two different fluids. Split this force
into its two pieces and add the two
buoyant forces to find the total force
Physics 207: Lecture 22, Pg 2
Example problem: Air bubble rising
 A diver produces an air bubble underwater, where the absolute
pressure is p1 = 3.5 atm. The bubble rises to the surface, where
the pressure is p2 = 1 atm. The water temperatures at the
bottom and the surface are, respectively, T1 = 4°C, T2 = 23°C
 What is the ratio of the volume of the bubble as it reaches the
surface,V2, to its volume at the bottom, V1? (Ans.V2/V1 = 3.74)
 Is it safe for the diver to ascend while holding his breath?
No! Air in the lungs would expand, and the lung could rupture.
This is addition to “the bends”, or decompression sickness, which is due to the
pressure dependent solubility of gas. At depth and at higher pressure N2 is
more soluble in blood. As divers ascend, N2 dissolved in their blood stream
becomes gaseous again and forms N2 bubbles in blood vessels, which in
turn can obstruct blood flow, and therefore provoke pain and in some cases
even strokes or deaths. Fortunately, this only happens when diving deeper
than 30 m (100 feet). The diver in this question only went down 25 meters.
How do we know that?
Physics 207: Lecture 22, Pg 3
PV diagrams: Important processes
 Isochoric process:
V = const (aka isovolumetric)
 Isobaric process:
p = const
pV
 Isothermal process: T = const
 constant
T
1
p1V1  p2V2
1
Volume
Pressure
p1 p2

T1 T2
Isobaric
Isothermal
Pressure
Pressure
Isochoric
2
V1 V2

T1 T2
1
2
2
Volume
Volume
Physics 207: Lecture 22, Pg 4
1st Law of Thermodynamics
ΔEth =W + Q
W & Q with respect to the system
 Thermal energy Eth : Microscopic energy of moving




molecules and stretched molecular bonds. ΔEth depends on
the initial and final states but is independent of the process.
Work W : Energy transferred to the system by forces in a
mechanical interaction.
Heat Q : Energy transferred to the system via atomic-level
collisions when there is a temperature difference.
Work W and heat Q depend on the process by which the
system is changed.
The change of energy in the system, ΔEth depends only on
the total energy exchanged W+Q, not on the process.
Physics 207: Lecture 22, Pg 5
1st Law: Work & Heat
 Work done on system (an
ideal gas)
W 
final
 p dV  (area under curve )
initial
 Won system < 0 Moving left to right
[where (Vf > Vi)]
 If ideal gas, PV = nRT, and given Pi
& Vi fixes Ti
 Wby system > 0 Moving left to right
Physics 207: Lecture 22, Pg 6
 Work:
1st Law: Work & Heat
 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 22, Pg 7
1st Law: Work (Area under the curve)
 Work depends on the path taken in the PV-diagram :
3
3
2
1
2
1
(a) Wa = W1 to 2 + W2 to 3 (here either P or V constant)
 Wa (on) = - Pi (Vf - Vi) + 0 > 0
(b) Wb = W1 to 2 + W2 to 3 (here either P or V constant)
 Wb (on) = 0 - Pf (Vf - Vi) > Wa > 0
(c) Need explicit form of P versus V but Wc (on) > 0
Physics 207: Lecture 22, Pg 8
Combinations of Isothermal & Adiabatic Processes

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.
 In this case

PVg = const.
 All real processes are not.
Example: Opening a valve
between two chambers, one
with a gas and one with a
vacuum.
 Isothermal PV= const.=nRT
Physics 207: Lecture 22, Pg 9
Isothermal processes
 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 n(Vf /Vi )
Physics 207: Lecture 22, Pg 10
First Law of Thermodynamics
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 22, Pg 11
Work, Heat & Themodynamics
Something in common:
a thermodynamic cycle
with work and heat
Physics 207: Lecture 22, Pg 12
Q : Latent heat and specific 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 / °C kg )
Q = M c ΔT
 Molar specific heat C of a substance is the energy required to
raise the temperature of 1 mol by 1 K.
Q = n C ΔT
If a phase transition involved then the heat transferred is Q = ±ML+M c ΔT
 The molar specific heat of gasses depends on the process
 CV= molar specific heat at constant volume
 Cp= molar specific heat at constant pressure
 Cp= CV+R (R is the universal gas constant)
g
Cp

CV
Physics 207: Lecture 22, Pg 13
Mechanical equivalent of heat

Heat: Q = C  T (internal energy transferred)
 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
 Energy to raise 1 g of water from 14.5 to 15.5 °C
(James Prescott Joule found the mechanical
equivalent of heat.)
C ≡ Heat capacity (in J/ K)
Q=cmT
 c: specific heat (heat capacity per
units of mass)
 amount of heat to raise T of 1 kg
by 1 °C
 [c] = J/(kg °C)
Sign convention:
+Q : heat gained
- Q : heat lost
Physics 207: Lecture 22, Pg 14
Exercise
 The specific heat 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 22, Pg 15
Exercise
 When the two materials have reached thermal equilibrium, the
block of aluminum is cut in half and equal quantities of heat are
added to the iron block and to each portion of the aluminum
block. Which of the following statements is true?
(a) The three blocks are no longer in thermal equilibrium; the
iron block is warmer.
(b) The three blocks are no longer in thermal equilibrium; both
the aluminum blocks are warmer.
(c ) The blocks remain in thermal equilibrium.
Physics 207: Lecture 22, Pg 16
Latent Heat
 Latent heat: amount of internal energy needed to add or to
remove from a substance to change the state of that substance.
 Phase change: T remains constant but internal energy
changes
 Heat does not result in change in T ( latent = “hidden”)
 e.g. : solid  liquid or liquid  gas
Lfusion (J / kg)
33.5 x 104
Lvapor. (J / kg)
22.6 x 105
Physics 207: Lecture 22, Pg 18
Latent Heats of Fusion and Vaporization
Question: Can you identify the heat capacity?
T (oC)
120
100
80
60
40
Water + Steam
20
Steam
0
-20
-40
Water
+
Water
Ice
62.7
396
815
3080
Energy added (J) (per gm)
Physics 207: Lecture 22, Pg 19
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?
(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 22, Pg 20
Energy transfer mechanisms
 Thermal conduction (or conduction)
 Convection
 Thermal Radiation
For a material of cross-section area A and length L,
spanning a temperature difference ΔT = TH – TC, the rate
of heat transfer is
Q / t = k A T / x
where k is the thermal conductivity, which characterizes
whether the material is a good conductor of heat or a poor
conductor.
Physics 207: Lecture 22, Pg 21
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 22, Pg 22
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 22, Pg 23
Exercise 2
Thermal Conduction
 Two identically shaped bars (one blue
100 C
Tjoint
300 C
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.
Which of the following is true?
(A) Ttop > Tbottom
(B) Ttop= Tbottom
(C) Ttop< Tbottom
(D) need to know k
Physics 207: Lecture 22, Pg 24
Exercise Thermal Conduction
 Two thermal conductors (possibly
inhomogeneous) are butted together and
in contact with two thermal reservoirs
100 C
held at the temperatures 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 22, Pg 25
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 = s A e T4 (power radiated)
 s = 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 22, Pg 26
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 22, Pg 27
Anti-global warming or the nuclear winter scenario
 Assume P/A = 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 = s A T4= s (4p r2) T4 = I p r2  T = [I / (4 x s )]¼
 s = 5.710-8 W/m2 K4
 T = 277 K (A little on the chilly side.)
Physics 207: Lecture 22, Pg 28
Physics 207, Lecture 20, Nov. 10
• Assignment
 HW9, Due Wednesday, Nov. 19th
 Wednesday: Read through Chapter 18.3
Physics 207: Lecture 22, Pg 29

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