Lecture 25 Phases & Phase Changes Thermal Processes

Lecture 25 Phases & Phase Changes 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

Solids

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:

Stretching / Compression of a Solid The 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)

Shear Forces Another 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

Shear Modulus S is the shear modulus.

Uniform Compression Under uniform pressure, an object will shrink in volume Here, the proportionality constant, B, is called the bulk modulus.

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.

Phase Changes

Evaporation Molecules in a liquid can sometimes escape the binding forces and become vapor (gas)

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.

Vapor Pressure The 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.

Boiling Potatoes Will boiled potatoes cook faster in Charlottesville or in Denver? 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 Potatoes Will boiled potatoes cook faster in Charlottesville or in Denver? 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? The lower air pressure in Denver means that the water will boil at a lower temperature... and your potatoes will take longer to cook.

Phase Diagram The vapor pressure curve is only a part of the phase diagram. There are similar curves describing the pressure/temperature of transition from solid to liquid, and solid to gas When the liquid reaches the critical point, there is no longer a distinction between liquid and gas; there is only a “fluid” phase.

Fusion Curve Curve 1 Curve 2 The 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!

Fusion curve for water Ice melts under pressure! This is how an ice skate works

Phase Equilibrium The sublimation curve marks the boundary between the solid and gas phases. The triple point is where all three phases are in equilibrium.

Heat and Phase Change When 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.

Latent Heat The 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.

Latent Heat The latent heat of fusion is the heat needed to go from solid to liquid; the latent heat of vaporization from liquid to gas.

You’re in Hot Water! a)water b)steam c)both the same d) it depends... Which will cause more severe burns to your skin: 100°C water or 100°C steam?

You’re in Hot Water! a)water b)steam c)both the same d) it depends... Which will cause more severe burns to your skin: 100°C water or 100°C steam? Although the water is indeed hot, it releases only1 cal/gof heat as it cools. The steam, however, first has to undergo aphase changeinto water and that process releases540 cal/g, which is a very large amount of heat. That immense release of heat is what makes steam burns so dangerous.

Boiling Potatoes Will potatoes cook faster if the water is boiling faster? • Yes • No • c) Wait, I’m confused. Am I still in Denver?

Boiling Potatoes Will potatoes cook faster if the water is boiling faster? • Yes • No • c) Wait, I’m confused. Am I still in Denver? 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-up: How can you cook the potatoes faster?

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.

Water and Ice a)0°C b)between0°C and 50°C c)50°C d) greater than 50°C 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

Water and Ice a)0°C b)between0°C and 50°C c)50°C d) greater than 50°C 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 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 = cwaterm xT = (1 cal/g°C) × (1000 g) × (50°C) = 50,000 calThus, 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

1 ΔP 2 ΔT Fusion curve for 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 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). 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.

Thermal Processes

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)

so changing volume implies changing temperature Work is area under the PV graph Constant pressure Isobaric process Work done by an expanding gas, constant pressure: Examples: piston against atmosphere, or vertical piston with constant weight on top

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.

T = constant W = Q 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. if W < 0 (work done on the system) than Q<0 (heat flows out of the system) if W > 0 (work done by the system) than Q>0 (heat flows out into the system)

Lecture 25 Phases & Phase Changes Thermal Processes