+ All Categories
Home > Documents > Lecture 24 Temperature and Heat Phases and Phase Changes

Lecture 24 Temperature and Heat Phases and Phase Changes

Date post: 05-Jan-2016
Category:
Upload: sharla
View: 25 times
Download: 0 times
Share this document with a friend
Description:
Lecture 24 Temperature and Heat Phases and Phase Changes. Heat Transfer Mechanisms. Thermal equilibrium is reached by means of thermal contact, which in turn can occur through three different mechanisms. - PowerPoint PPT Presentation
63
Lecture 24 Temperature and Heat Phases and Phase Changes
Transcript
Page 1: Lecture 24 Temperature and Heat Phases and Phase Changes

Lecture 24Temperature and Heat

Phases and Phase Changes

Page 2: Lecture 24 Temperature and Heat Phases and Phase Changes

Thermal equilibrium is reached by means of thermal contact, which in turn can occur through three different mechanisms

Heat Transfer Mechanisms

conduction : it occurs when objects at different temperature are in physical contact (e.g. when holding a hot potato). Faster moving molecules in the hotter object transfer some of their energy to the colder one

convection : this occurs mainly in fluids. In a pot of water on a stove, the liquid at the bottom is heated by conduction. The hot water has lower density and rises to the top, cold water from the top falls to the bottom and gets heated, etc.

radiation : any object at non-zero temperature emits radiation (in the form of electromagnetic waves). The effect is more noticeable when standing next to a red-hot coal fire, or in the sun rays

Page 3: Lecture 24 Temperature and Heat Phases and Phase Changes

ConductionConduction is the flow of heat directly through a

physical material

The amount of heat Q that flows through a rod:• increases proportionally to the cross-sectional area A• increases proportionally to ΔT from one end to the other• increases steadily with time• decreases inversely with the length of the rod

The constant k is called the thermal conductivity of the material

Page 4: Lecture 24 Temperature and Heat Phases and Phase Changes

Some Typical Thermal Conductivities

Substances with high thermal conductivities are good conductors of heat; those with low thermal conductivities are good insulators.

Page 5: Lecture 24 Temperature and Heat Phases and Phase Changes

Two metal rods—one lead, the other copper—are connected in series, as shown. Note that each rod is 0.525 m in length and has a square cross section 1.50 cm on a side. The temperature at the lead end of the rods is 2.00°C; the temperature at the copper end is 106°C. (a) The average temperature of the two ends is 54.0°C. Is the temperature in the middle, at the lead-copper interface, greater than, less than, or equal to 54.0°C? Explain. (b) find the temperature at the lead-copper interface.

kPb = 34.3 W / (kg-m)kCu = 395 W / (kg-m)

Page 6: Lecture 24 Temperature and Heat Phases and Phase Changes

Two metal rods—one lead, the other copper—are connected in series, as shown. Note that each rod is 0.525 m in length and has a square cross section 1.50 cm on a side. The temperature at the lead end of the rods is 2.00°C; the temperature at the copper end is 106°C. (a) The average temperature of the two ends is 54.0°C. Is the temperature in the middle, at the lead-copper interface, greater than, less than, or equal to 54.0°C? Explain. (b) find the temperature at the lead-copper interface.

Assumptions: •The end points are infinite heat reservoirs... so their temperature doesn’t change for this exercise •The temperature is constant in time at every point. This is not true at moment of thermal connection. We are solving the “steady state” condition, when the temperature at each point doesn’t change.

kPb = 34.3 W / (kg-m)kCu = 395 W / (kg-m)

Page 7: Lecture 24 Temperature and Heat Phases and Phase Changes

Two metal rods—one lead, the other copper—are connected in series, as shown. Note that each rod is 0.525 m in length and has a square cross section 1.50 cm on a side. The temperature at the lead end of the rods is 2.00°C; the temperature at the copper end is 106°C. (a) The average temperature of the two ends is 54.0°C. Is the temperature in the middle, at the lead-copper interface, greater than, less than, or equal to 54.0°C? Explain. (b) find the temperature at the lead-copper interface.

- The heat (per unit time) through the lead must equal that through the copper

- The lead has a smaller thermal conductivity than the copper

The lead requires a larger temperature difference across it than the copper, to get the same heat flow. So TJ > 54o C

kPb = 34.3 W / (kg-m)kCu = 395 W / (kg-m)

(a)

Page 8: Lecture 24 Temperature and Heat Phases and Phase Changes

Two metal rods—one lead, the other copper—are connected in series, as shown. Note that each rod is 0.525 m in length and has a square cross section 1.50 cm on a side. The temperature at the lead end of the rods is 2.00°C; the temperature at the copper end is 106°C. (a) The average temperature of the two ends is 54.0°C. Is the temperature in the middle, at the lead-copper interface, greater than, less than, or equal to 54.0°C? Explain. (b) find the temperature at the lead-copper interface.

kPb = 34.3 W / (kg-m)kCu = 395 W / (kg-m)

(b)

Page 9: Lecture 24 Temperature and Heat Phases and Phase Changes

ConvectionConvection is the flow of fluid due to a difference in temperatures, such as warm air rising. The fluid “carries” the heat with it as it moves.

Page 10: Lecture 24 Temperature and Heat Phases and Phase Changes

Objects that are hot enough will glow – first red, then yellow, white, and blue.

RadiationAll objects give off energy in the form of radiation, as electromagnetic waves (light) – infrared, visible light, ultraviolet – which, unlike conduction and convection, can transport heat through a vacuum.

Page 11: Lecture 24 Temperature and Heat Phases and Phase Changes

RadiationThe amount of energy radiated by an object due to its temperature is proportional to its surface area and also to the fourth (!) power of its temperature.

It also depends on the emissivity, which is a number between 0 and 1 that indicates how effective a radiator the object is; a perfect radiator would have an emissivity of 1.

Here, e is the emissivity, and σ is the Stefan-Boltzmann constant:

Page 12: Lecture 24 Temperature and Heat Phases and Phase Changes
Page 13: Lecture 24 Temperature and Heat Phases and Phase Changes

The surface of the Sun has a temperature of 5500 oC. (a) Treating the Sun as a perfect blackbody, with an emissivity of 1.0, find the power that it radiates into space. The radius of the sun is 7.0x108 m, and the temperature of space can be taken to be 3.0 K (b) the solar constant is the number of watts of sunlight power falling on a square meter of the Earth’s upper atmosphere. Use your result from part (a) to calculate the solar constant, given that the distance from the Sun to the Earth is 1.5x1011 m.

Page 14: Lecture 24 Temperature and Heat Phases and Phase Changes

The surface of the Sun has a temperature of 5500 oC. (a) Treating the Sun as a perfect blackbody, with an emissivity of 1.0, find the power that it radiates into space. The radius of the sun is 7.0x108 m, and the temperature of space can be taken to be 3.0 K (b) the solar constant is the number of watts of sunlight power falling on a square meter of the Earth’s upper atmosphere. Use your result from part (a) to calculate the solar constant, given that the distance from the Sun to the Earth is 1.5x1011 m.

emissivity (a)

(b)

Page 15: Lecture 24 Temperature and Heat Phases and Phase Changes

Heat Conduction Heat Conduction

Given your experience of

what feels colder when you

walk on it, which of the

surfaces would have the

highest thermal

conductivity?

a) a rugb) a steel surfacec) a concrete floord) has nothing to do with

thermal conductivity

Page 16: Lecture 24 Temperature and Heat Phases and Phase Changes

Heat Conduction Heat Conduction

Given your experience of

what feels colder when you

walk on it, which of the

surfaces would have the

highest thermal

conductivity?

a) a rugb) a steel surfacec) a concrete floord) has nothing to do with

thermal conductivity

The heat flow rate is k A (T1 − T2)/l. All things being

equal, bigger k leads to bigger heat loss.From the book: Steel = 40, Concrete = 0.84,Human tissue = 0.2, Wool = 0.04, in units of J/(s.m.C°).

Page 17: Lecture 24 Temperature and Heat Phases and Phase Changes

Phases and Phase Changes

Page 18: Lecture 24 Temperature and Heat Phases and Phase Changes

Ideal Gases

Gases are the easiest state of matter to describe, as all ideal gases exhibit similar behavior.

An ideal gas is one that is thin enough, and far away enough from condensing, that the interactions between molecules can be ignored.

Page 19: Lecture 24 Temperature and Heat Phases and Phase Changes

Soda Bottle Soda Bottle

a) it expands and may burst

b) it does not change

c) it contracts and the sides collapse

inward

d) it is too dark in the fridge to tell

A plastic soda bottle is empty and sits out in the sun, heating the air inside. Now you put the cap on tightly and put the bottle in the fridge. What happens to the bottle as it cools?

Page 20: Lecture 24 Temperature and Heat Phases and Phase Changes

The air inside the bottle is warmair inside the bottle is warm, due to heating by the

sun. When the bottle is in the fridge, the air coolsthe air cools. As

the temperature drops, the pressure in the bottle also pressure in the bottle also

dropsdrops. Eventually, the pressure inside is sufficiently

lower than the pressure outside (atmosphere) to begin

to collapse the bottle.

Soda Bottle Soda Bottle

a) it expands and may burst

b) it does not change

c) it contracts and the sides collapse

inward

d) it is too dark in the fridge to tell

A plastic soda bottle is empty and sits out in the sun, heating the air inside. Now you put the cap on tightly and put the bottle in the fridge. What happens to the bottle as it cools?

Page 21: Lecture 24 Temperature and Heat Phases and Phase Changes

If the volume of an ideal gas is held constant, we find that the pressure increases with temperature:

Constant Volume

Page 22: Lecture 24 Temperature and Heat Phases and Phase Changes

Constant Temperature

If the volume and temperature are kept constant, but more gas is added (such as in inflating a tire or basketball), the pressure will increase:

If the temperature is constant and the volume decreases, the pressure increases:

(fixed volume V) (fixed number N)

Page 23: Lecture 24 Temperature and Heat Phases and Phase Changes

where k is called the Boltzmann constant:

Equation of State for an Ideal Gas

Combining these observations:

Page 24: Lecture 24 Temperature and Heat Phases and Phase Changes

Instead of counting molecules, we can count moles.

Equation of State for an Ideal Gas

for n moles of gas:

Page 25: Lecture 24 Temperature and Heat Phases and Phase Changes

Charles’s law says that the volume of a gas increases with temperature if the pressure is constant.

Properties of Ideal Gases: Constant Pressure

Page 26: Lecture 24 Temperature and Heat Phases and Phase Changes

Boyle’s law says that the pressure varies inversely with volume. These curves of constant temperature are called isotherms.

Properties of Ideal Gases: Constant Temperature

Page 27: Lecture 24 Temperature and Heat Phases and Phase Changes

Kinetic Theory

Assumptions:

• N identical molecules of mass m are inside a container of volume V; each acts as a point particle.

• Molecules always obey Newton’s laws and are moving randomly.

• Collisions with other molecules and with the walls are elastic.

The kinetic theory relates microscopic quantities (position, velocity) to macroscopic ones (pressure, temperature).

Page 28: Lecture 24 Temperature and Heat Phases and Phase Changes

Containing an Ideal Gas

What is the impulse to turn a molecule around?

What is the period to time between such impulses?

I = 2(mvx)

t = 2L / vx

What is the average force on the surface from one molecule?

F = I/t = 2(mvx) / (2L/vx) = 2 mvx2 / L

Page 29: Lecture 24 Temperature and Heat Phases and Phase Changes

Pressure and an Ideal Gas

Pressure is the result of collisions between the gas molecules and the walls of the container.

It depends on the mass and speed of the molecules, and on the container size:

Page 30: Lecture 24 Temperature and Heat Phases and Phase Changes

Distribution of Molecular Speeds

Not all molecules in a gas will have the same speed; their speeds are represented by the Maxwell distribution, and depend on the temperature and mass of the molecules.

Page 31: Lecture 24 Temperature and Heat Phases and Phase Changes

Pressure and Kinetic EnergyWe replace the speed in the previous expression for pressure with the average speed-squared:

Therefore, the pressure in a gas is proportional to the average kinetic energy of

its molecules.

Including the other two directions, and all N particles:

Page 32: Lecture 24 Temperature and Heat Phases and Phase Changes

Kinetic Energy and TemperatureCompare to ideal gas law:

average kinetic energy is related to temperature

The square root of is called

the root mean square (rms) speed.

Page 33: Lecture 24 Temperature and Heat Phases and Phase Changes

r.m.s. Speed

The rms speed is slightly greater than the most probable speed and the average speed.

Page 34: Lecture 24 Temperature and Heat Phases and Phase Changes

Internal Energy

In the case where each molecule consists of a single atom, this is all linear kinetic energy of atoms:

The internal energy of an ideal monotonic gas is the sum of the kinetic energies of all its

molecules.

Page 35: Lecture 24 Temperature and Heat Phases and Phase Changes

Distribution of Molecular SpeedSome molecules will have speeds exceeding the planetary escape velocity!

Lighter molecules will have higher speeds (at the same temperature) and so will leave the planet more quickly.

This is why less massive planets have thin, or no, atmosphere...

and why earth has little H2 in the atmosphere, but Jupiter has plenty

Page 36: Lecture 24 Temperature and Heat Phases and Phase Changes

Solids

Page 37: Lecture 24 Temperature and Heat Phases and Phase Changes

Solids and Elastic Deformation

Solids have definite shapes (unlike fluids), but they can be deformed. Pulling on opposite ends of a rod can cause it to stretch:

Page 38: Lecture 24 Temperature and Heat Phases and Phase Changes

Stretching / Compression of a SolidThe amount of stretching will depend on the force; Y is Young’s modulus and is a property of the material:

The stretch is proportional to the force, and also to the original length

The same formula works for stretching or compression (but sometimes with a different Young’s modulus)

Page 39: Lecture 24 Temperature and Heat Phases and Phase Changes

Shear ForcesAnother type of deformation is called a shear deformation, where opposite sides of the object are pulled laterally in opposite directions.

The “lean” is proportional to the force, and also to the original height

Page 40: Lecture 24 Temperature and Heat Phases and Phase Changes

Shear Modulus

S is the shear modulus.

Page 41: Lecture 24 Temperature and Heat Phases and Phase Changes

Uniform Compression

Here, the proportionality constant, B, is called the bulk modulus.

Under uniform pressure, an object will shrink in volume

Page 42: Lecture 24 Temperature and Heat Phases and Phase Changes

Stress and Strain

The applied force per unit area is called the stress, and the resulting deformation is the strain. They are proportional to each other until the stress becomes too large; permanent deformation will then occur.

Page 43: Lecture 24 Temperature and Heat Phases and Phase Changes

Phase Changes

Page 44: Lecture 24 Temperature and Heat Phases and Phase Changes

EvaporationMolecules in a liquid can sometimes

escape the binding forces and become vapor (gas)

Page 45: Lecture 24 Temperature and Heat Phases and Phase Changes

Phase Equilibrium

If a liquid is put into a sealed container so that there is a vacuum above it, some of the molecules in the liquid will vaporize. Once a sufficient number have done so, some will begin to condense back into the liquid. Equilibrium is reached when the numbers in each phase remain constant.

Page 46: Lecture 24 Temperature and Heat Phases and Phase Changes

Vapor PressureThe pressure of the gas when it is in equilibrium with the liquid is called the equilibrium vapor pressure, and will depend on the temperature.

A liquid boils at the temperature at which its vapor pressure equals the external pressure.

Page 47: Lecture 24 Temperature and Heat Phases and Phase Changes

a) Charlottesville

b) Denver (the “mile high” city)

c) the same in both places

d) I’ve never cooked in Denver, so I really don’t know

e) you can boil potatoes?

Boiling PotatoesBoiling Potatoes

Will boiled potatoes cook faster in Charlottesville or in Denver?

Page 48: Lecture 24 Temperature and Heat Phases and Phase Changes

a) Charlottesville

b) Denver (the “mile high” city)

c) the same in both places

d) I’ve never cooked in Denver, so I really don’t know

e) you can boil potatoes?

Boiling PotatoesBoiling Potatoes

Will boiled potatoes cook faster in Charlottesville or in Denver?

The lower air pressure in Denver means that the water will boil at a lower temperature... and your potatoes will take longer to cook.

Page 49: Lecture 24 Temperature and Heat Phases and Phase Changes

Phase Diagram

The vapor pressure curve is only a part of the phase diagram.

When the liquid reaches the critical point, there is no longer a distinction between

liquid and gas; there is only a “fluid” phase.

There are similar curves describing the pressure/temperature of transition from solid to liquid, and solid to gas

Page 50: Lecture 24 Temperature and Heat Phases and Phase Changes

Fusion CurveThe fusion curve is the boundary between the solid and liquid phases; along that curve they exist in equilibrium with each other.

One of these two fusion curves has a shape that is typical for most materials, but the other has shape specific to water.

Which is which?

(a) Curve 1 is the fusion curve for water

(b) Curve 2 is the fusion curve for water

(c) Trick question: there is no fusion curve for water!

Curve 1

Curve 2

Page 51: Lecture 24 Temperature and Heat Phases and Phase Changes

Fusion CurveThe fusion curve is the boundary between the solid and liquid phases; along that curve they exist in equilibrium with each other.

One of these two fusion curves has a shape that is typical for most materials, but the other has shape specific to water.

Which is which?

(a) Curve 1 is the fusion curve for water

(b) Curve 2 is the fusion curve for water

(c) Trick question: there is no fusion curve for water!

Curve 1

Curve 2

Page 52: Lecture 24 Temperature and Heat Phases and Phase Changes

Fusion curve for water

Ice melts under pressure!This is how an ice skate works

Page 53: Lecture 24 Temperature and Heat Phases and Phase Changes

Phase EquilibriumThe sublimation curve marks the boundary between the solid and gas phases.

The triple point is where all three phases are in equilibrium.

Page 54: Lecture 24 Temperature and Heat Phases and Phase Changes

Heat and Phase ChangeWhen two phases coexist, the temperature remains the same even if a small amount of heat is added. Instead of raising the temperature, the heat goes into changing the phase of the material – melting ice, for example.

Page 55: Lecture 24 Temperature and Heat Phases and Phase Changes

Latent HeatThe heat required to convert from one phase to another is called the latent heat.

The latent heat, L, is the heat that must be added to or removed from one kilogram of a substance to convert it from one phase to another. During the conversion process, the temperature of the system remains constant.

Page 56: Lecture 24 Temperature and Heat Phases and Phase Changes

Latent HeatThe latent heat of fusion is the heat needed to go from solid to liquid;

the latent heat of vaporization from liquid to gas.

Page 57: Lecture 24 Temperature and Heat Phases and Phase Changes

You’re in Hot Water!You’re in Hot Water!

Which will cause more severe

burns to your skin: 100°C water or

100°C steam?

a) water

b) steam

c) both the same

d) it depends...

Page 58: Lecture 24 Temperature and Heat Phases and Phase Changes

Although the water is indeed hot, it releases only 1 cal/1 cal/gg of

heat as it cools. The steam, however, first has to undergo a

phase changephase change into water and that process releases 540 cal/g540 cal/g,

which is a very large amount of heat. That immense release of

heat is what makes steam burns so dangerous.

You’re in Hot Water!You’re in Hot Water!

Which will cause more severe

burns to your skin: 100°C water or

100°C steam?

a) water

b) steam

c) both the same

d) it depends...

Page 59: Lecture 24 Temperature and Heat Phases and Phase Changes

a) Yes

b) No

c) Wait, I’m confused. Am I still in Denver?

Boiling PotatoesBoiling Potatoes

Will potatoes cook faster if the water is boiling faster?

Page 60: Lecture 24 Temperature and Heat Phases and Phase Changes

Boiling PotatoesBoiling Potatoes

Will potatoes cook faster if the water is boiling faster?

The water boils at 100°C and remains at that temperature until allof the water has been changed into steam. Only then will the steam increase in temperature. Because the water stays at the same temperature, regardless of how fast it is boiling, thepotatoes will not cook any faster.

Follow-upFollow-up: : How can you cook the potatoes faster?

a) Yes

b) No

c) Wait, I’m confused. Am I still in Denver?

Page 61: Lecture 24 Temperature and Heat Phases and Phase Changes

Phase Changes and Energy Conservation

Solving problems involving phase changes is similar to solving problems involving heat transfer, except that the latent heat must be included as well.

Page 62: Lecture 24 Temperature and Heat Phases and Phase Changes

Water and Ice 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/gcwater = 1 cal/g °C

a) 0°C

b) between 0°C and 50°C

c) 50°C

d) greater than 50°C

Page 63: Lecture 24 Temperature and Heat Phases and Phase Changes

How much heat is needed to melt the ice?

QQ = = mLmLff = (1000 = (1000 gg) ) (80 cal/ (80 cal/gg) = 80,000 cal) = 80,000 cal

How much heat can the water deliver by cooling from 50°°C to 0°°C?

QQ = = ccwaterwater m xm xTT = (1 cal/ = (1 cal/gg °°C) C) (1000 (1000 gg) ) (50 (50°°C) = 50,000 calC) = 50,000 cal

Thus, there is not enough heat available to melt all the ice!!

Water and Ice 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/gcwater = 1 cal/g °C

a) 0°C

b) between 0°C and 50°C

c) 50°C

d) greater than 50°C

Follow-upFollow-up: : How much more water at 50°C would you need?


Recommended