Physics 207: Lecture 22, Pg 1 Physics 207, Lecture 21, Nov. 12 Goals: Chapter 16 Chapter 16 Use the...

Physics 207: Lecture 22, Pg 1

Physics 207, Lecture 21, Nov. 12Goals:Goals:

• Chapter 16Chapter 16 Use the ideal-gas law. Use pV diagrams for ideal-gas processes.

• Chapter 17Chapter 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.

• AssignmentAssignment HW9, Due Wednesday, Nov. 19th HW10, Due Sunday, Wednesday, Read through 18.3


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


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?


PV diagrams: Important processes

Isochoric process: V = const (aka isovolumetric) Isobaric process: p = const Isothermal process: T = const constant

TpV

Volume

Pre

ssur

e

2

2

1

1 T

p

T

p

1

2Isochoric

Volume

Pre

ssur

e

2

2

1

1 T

V

T

V

1 2

Isobaric

Volume

Pre

ssur

e

2211 VpVp 1

2

Isothermal


1st Law of Thermodynamics

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.

ΔEth =W + Q

W & Q with respect to the system


1st Law: Work & Heat

Work done on system (an ideal gas)

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

final

initial

)curveunder area( dVpW


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


1st Law: Work (Area under the curve) Work depends on the path taken in the PV-diagram :

(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

2 1

3

1

23


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

PV = 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


Isothermal processes

Work done when PV = nRT = constant P = nRT / V

final

initial

)curveunder area( dVpW

f

i

f

i

V

V

V

V

/ nRT/ nRT VdVVdVW

)/VV( nRT ifnW


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!


Work, Heat & Themodynamics

Something in common: a thermodynamic cycle with work and heat


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 ΔTIf 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)

VC

Cp


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.

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)

1 Cal = 4.186 J

1 kcal = 1 Cal = 4186 J

Q = c m 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


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


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.


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


Latent Heats of Fusion and Vaporization

Energy added (J) (per gm)

T (oC)

120

100

80

60

40

20

0

-20

-40Water

Water+

Ice

Water + Steam Steam

62.7 396 815 3080

Question: Can you identify the heat capacity?


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.

Exercise Latent Heat


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

where k is the thermal conductivity, which characterizes whether the material is a good conductor of heat or a poor conductor.

Q / t = k A T / x


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)


Thermal Conductivities

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

J/s m °C J/s m °C J/s m °C


(B) Ttop= Tbottom

(A) Ttop > Tbottom (C) Ttop< Tbottom

100 CTjoint

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.

Which of the following is true?

Exercise 2Thermal Conduction

300 C

(D) need to know k


100 C

Two thermal conductors (possibly inhomogeneous) are butted together and in contact with two thermal reservoirs held at the temperatures shown.

Which of the temperature vs. position plots below is most physical?

Exercise Thermal Conduction

300 C

Position

Te

mp

era

ture

Position

Te

mp

era

ture

Position

Te

mp

era

ture

(A) (B) (C)


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

= 5.710-8 W/m2 K4 , T is in Kelvin, and A is the surface area e is a constant called the emissivity

P = A e T4 (power radiated)


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.

VacuumVacuum

SilveredSilveredsurfacessurfaces

Hot orHot orcoldcoldliquidliquid


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 = A T4= (4 r2) T4 = I r2 T = [I / (4 x )]¼ = 5.710-8 W/m2 K4 T = 277 K (A little on the chilly side.)


Physics 207, Lecture 20, Nov. 10

• AssignmentAssignment HW9, Due Wednesday, Nov. 19th Wednesday: Read through Chapter 18.3

Date post:	18-Jan-2016
Category:	Documents
Upload:	edmund-jennings
View:	213 times
Download:	0 times

Physics 207: Lecture 22, Pg 1 Physics 207, Lecture 21, Nov. 12 Goals: Chapter 16 Chapter 16 Use the...

Documents