Lecture 25 Thermal Processes

Lecture 25 Thermal Processes

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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.

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)

Work in an Constant-Temperature Process The work done is the area under the curve: For you calculus junkies:

Adiabatic Process An adiabatic process is one in which no heat flows into or out of the system. One way to ensure that a process is adiabatic is to insulate the system. The adiabatic P-V curve is similar to the isothermal one, but is steeper. Q = 0

Rapid Adiabatic Process Another way to ensure that a process is effectively adiabatic is to have the volume change occur very quickly. In this case, heat has no time to flow in or out of the system.

Thermal Processes The different types of ideal thermal processes

Specific Heat for an Ideal Gas at Constant Volume Specific heats for ideal gases must be quoted either at constant pressure or at constant volume. For a constant-volume process,

for an ideal gas (from the kinetic theory) First Law of Thermodynamics

Specific Heat for an Ideal Gas at Constant Pressure At constant pressure, (some work is done) Some of the heat energy goes into the mechanical work, so more heat input is required to produce the same ΔT

for an ideal gas (from the kinetic theory) First Law of Thermodynamics

Specific Heats for an Ideal Gas Both CV and CP can be calculated for a monatomic ideal gas using the first law of thermodynamics. Although this calculation was done for an ideal, monatomic gas, the difference Cp - Cv works well for real gases.

Specific Heats and Adiabats In Ideal Gas The P-V curve for an adiabat is given by for monotonic gases

P V Work of a Thermal Cycle a)positive b)zero c)negative In the closed thermodynamic cycle shown in the P-V diagram, the work done by the gas is:

P Work of a Thermal Cycle V a)positive b)zero c)negative In the closed thermodynamic cycle shown in the P-V diagram, the work done by the gas is: The gas expands at a higher pressure and compresses at a lower pressure. In general, clockwise = positive work; counterclockwise = negative work.

P2 = 3P1 P1 V1 V2 =5V1 a)4 P1V1 b)7 P1V1 c) 8 P1V1 d)21 P1V1 e) 29 P1V1 One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1. How much work is done by the gas in this process, in terms of the initial pressure and volume?

P2 = 3P1 P1 V1 V2 =5V1 a)4 P1V1 b)7 P1V1 c) 8 P1V1 d)21 P1V1 e) 29 P1V1 One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1. How much work is done by the gas in this process, in terms of the initial pressure and volume? Area under the curve: (4 V1)(P1) + 1/2 (4V1)(2P1) = 8 V1P1

P2 = 3P1 P1 V1 V2 =5V1 a)7 P1V1 b)8 P1V1 c) 15 P1V1 d)21 P1V1 e)29 P1V1 One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1. How much internal energy is gained by the gas in this process, in terms of the initial pressure and volume?

P2 = 3P1 P1 V1 V2 =5V1 a)7 P1V1 b)8 P1V1 c) 15 P1V1 d)21 P1V1 e)29 P1V1 One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1. How much internal energy is gained by the gas in this process, in terms of the initial pressure and volume? Ideal monatomic gas: U = 3/2 nRT Ideal gas law: PV = nRT U = 3/2 PV P2V2 = 15 P1V1 Δ(PV) = 14 P1V1 ΔU = 21 P1V1

P2 = 3P1 P1 V1 V2 =5V1 a)7 P1V1 b)8 P1V1 c) 15 P1V1 d)21 P1V1 e)29 P1V1 One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1. How much heat is gained by the gas in this process, in terms of the initial pressure and volume?

P2 = 3P1 P1 V1 V2 =5V1 a)7 P1V1 b)8 P1V1 c) 15 P1V1 d)21 P1V1 e)29 P1V1 One mole of an ideal monatomic gas undergoes the reversible expansion shown in the figure, where V2 = 5 V1 and P2 = 3 P1. How much heat is gained by the gas in this process, in terms of the initial pressure and volume? First Law of Thermodynamics W = 8 P1V1

The Second Law of Thermodynamics We observe that heat always flows spontaneously from a warmer object to a cooler one, although the opposite would not violate the conservation of energy. This direction of heat flow is one of the ways of expressing the second law of thermodynamics: When objects of different temperatures are brought into thermal contact, the spontaneous flow of heat that results is always from the high temperature object to the low temperature object. Spontaneous heat flow never proceeds in the reverse direction.

Heat Engines A heat engine is a device that converts heat into work. A classic example is the steam engine. Fuel heats the water; the vapor expands and does work against the piston; the vapor condenses back into water again and the cycle repeats. All heat engines have: a working substance a high-temperature reservoir a low-temperature reservoir a cyclical engine

Provided in lecture notes on: 12/1 PHYS2010 practice for Chap14-18, Fall 2009 12) An ideal monatomic gas undergoes a reverrsible expansion to 1.5 times its original volume. In which of these processes does the gas perform the least amount of work? A) at constant temperature B) at constant pressure C) if the pressure decreases in proportion to the volume (i.e. PV=constant)‏ D) adiabatically E) if the pressure increases in proportion to the volume (i.e., P/V=constant)‏ suggested time: 1-2 minutes

Provided in lecture notes on: 12/1 PHYS2010 practice for Chap14-18, Fall 2009 13) An expandable container holds 1.50 mole of He gas with an initial pressure of 650 kPa and an initial volume of 2.10 L. The gas expands isothermally to a final pressure of 350 kPa. How much heat is gained by the gas in this process? A) 1370 J B) 685 J C) 792 J D) 1280 J E) 1700 J suggested time: 4 minutes

Provided in lecture notes on: 12/1 PHYS2010 practice for Chap14-18, Fall 2009 14) 1.50 moles of an ideal monatomic gas are initially at a temperature of 317 K. If the gas gains 2670 J of heat and performs 770 J of work, what is its final temperature? A) 419 K B) 756 K C) 687 K D) 359 K E) 526 K suggested time: 3-4 minutes

Lecture 25 Thermal Processes