2Thermodynamic Systems and Their Surroundings Thermodynamics is the branch of physics that is builtupon the fundamental laws that heat and work obey.The collection of objects on which attention is beingfocused is called the system, while everything elsein the environment is called the surroundings.Walls that permit heat flow are called diathermal walls,while walls that do not permit heat flow are calledadiabatic walls.To understand thermodynamics, it is necessary todescribe the state of a system.
3The Zeroth Law of Thermodynamics Two systems are said to be in thermalequilibrium if there is no heat flowbetween then when they are broughtinto contact.Temperature is the indicator of thermalequilibrium in the sense that there is nonet flow of heat between two systemsin thermal contact that have the sametemperature.
4The Zeroth Law of Thermodynamics Two systems individually in thermal equilibriumwith a third system are in thermal equilibriumwith each other.
5The First Law of Thermodynamics Suppose that a system gains heat Q and that is the only effect occurring.Consistent with the law of conservation of energy, the internal energyof the system changes:Heat is positive when the system gains heat and negative when the systemloses heat.
6The First Law of Thermodynamics If a system does work W on its surroundings and there is no heat flow,conservation of energy indicates that the internal energy of the systemwill decrease:Work is positive when it is done by the system and negative when it is doneon the system.
7The First Law of Thermodynamics The internal energy of a system changes due to heat and work:Heat is positive when the system gains heat and negative when the systemloses heat.Work is positive when it is done by the system and negative when it is doneon the system.
8The First Law of Thermodynamics Example 1 Positive and Negative WorkIn part a of figure, the system gains 1500J of heatand 2200J of work is done by the system on itssurroundings.In part b, the system also gains 1500J of heat, but2200J of work is done on the system.In each case, determine the change in internal energyof the system.
10The First Law of Thermodynamics Example 2 An Ideal GasThe temperature of three moles of a monatomic ideal gas is reducedfrom 540K to 350K as 5500J of heat flows into the gas.Find (a) the change in internal energy and (b) the work done by thegas.
12A quasi-static process is one that occurs slowly enough that a uniform Thermal ProcessesA quasi-static process is one that occurs slowly enough that a uniformtemperature and pressure exist throughout all regions of the system at alltimes.isobaric: constant pressureisochoric: constant volumeisothermal: constant temperatureadiabatic: no transfer of heat
13An isobaric process is one that occurs at constant pressure. Thermal ProcessesAn isobaric process is one that occurs atconstant pressure.Isobaric process:
14Example 3 Isobaric Expansion of Water Thermal ProcessesExample 3 Isobaric Expansion of WaterOne gram of water is placed in the cylinder andthe pressure is maintained at 2.0x105Pa. Thetemperature of the water is raised by 31oC. Thewater is in the liquid phase and expands by thesmall amount of 1.0x10-8m3.Find the work done and the change in internalenergy.
18Example 4 Work and the Area Under a Pressure-Volume Graph Thermal ProcessesExample 4 Work and the Area Under aPressure-Volume GraphDetermine the work for the process inwhich the pressure, volume, and temp-erature of a gas are changed along thestraight line in the figure.The area under a pressure-volume graph isthe work for any kind of process.
19Since the volume increases, the work is positive. Thermal ProcessesSince the volume increases, the workis positive.Estimate that there are 8.9 coloredsquares in the drawing.
20Thermal Processes Using and Ideal Gas ISOTHERMAL EXPANSION OR COMPRESSIONIsothermalexpansion orcompression ofan ideal gas
21Thermal Processes Using and Ideal Gas Example 5 Isothermal Expansion of an Ideal GasTwo moles of the monatomic gas argon expand isothermally at 298Kfrom and initial volume of 0.025m3 to a final volume of 0.050m3. Assumingthat argon is an ideal gas, find (a) the work done by the gas, (b) thechange in internal energy of the gas, and (c) the heat supplied to thegas.
23Thermal Processes Using and Ideal Gas ADIABATIC EXPANSION OR COMPRESSIONAdiabaticexpansion orcompression ofa monatomicideal gasAdiabaticexpansion orcompression ofa monatomicideal gas
24Specific Heat Capacities To relate heat and temperature change in solids and liquids, weused:specific heatcapacityThe amount of a gas is conveniently expressed in moles, so we write thefollowing analogous expression:molar specificheat capacity
25Specific Heat Capacities For gases it is necessary to distinguish between the molar specific heatcapacities which apply to the conditions of constant pressure and constantvolume:first law ofthermodynamicsconstant pressurefor a monatomicideal gas
26Specific Heat Capacities first law ofthermodynamicsconstant pressurefor a monatomicideal gasmonatomicideal gasany ideal gas
27The Second Law of Thermodynamics The second law is a statement about the natural tendency of heat toflow from hot to cold, whereas the first law deals with energy conservationand focuses on both heat and work.THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW STATEMENTHeat flows spontaneously from a substance at a higher temperature to a substanceat a lower temperature and does not flow spontaneously in the reverse direction.
28A heat engine is any device that uses heat to Heat EnginesA heat engine is any device that uses heat toperform work. It has three essential features.Heat is supplied to the engine at a relativelyhigh temperature from a place called the hotreservoir.Part of the input heat is used to performwork by the working substance of the engine.The remainder of the input heat is rejectedto a place called the cold reservoir.
29The efficiency of a heat engine is defined as Heat EnginesThe efficiency of a heat engine is defined asthe ratio of the work done to the input heat:If there are no other losses, then
30Example 6 An Automobile Engine Heat EnginesExample 6 An Automobile EngineAn automobile engine has an efficiency of 22.0% and produces2510 J of work. How much heat is rejected by the engine?
31Carnot’s Principle and the Carnot Engine A reversible process is one in which both the system and theenvironment can be returned to exactly the states they were inbefore the process occurred.CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECONDLAW OF THERMODYNAMICSNo irreversible engine operating between two reservoirs at constant temperaturescan have a greater efficiency than a reversible engine operating between the sametemperatures. Furthermore, all reversible engines operating between the sametemperatures have the same efficiency.
32Carnot’s Principle and the Carnot Engine The Carnot engine is usefule as an idealizedmodel.All of the heat input originates from a singletemperature, and all the rejected heat goesinto a cold reservoir at a single temperature.Since the efficiency can only depend onthe reservoir temperatures, the ratio ofheats can only depend on those temperatures.
33Carnot’s Principle and the Carnot Engine Example 7 A Tropical Ocean as a Heat EngineWater near the surface of a tropical ocean has a temperature of K, whereasthe water 700 meters beneath the surface has a temperature of K. It hasbeen proposed that the warm water be used as the hot reservoir and the cool wateras the cold reservoir of a heat engine. Find the maximum possible efficiency forsuch and engine.
35Carnot’s Principle and the Carnot Engine Conceptual Example 8 Natural Limits on the Efficiency of a Heat EngineConsider a hypothetical engine that receives 1000 J of heat as input from ahot reservoir and delivers 1000J of work, rejecting no heat to a cold reservoirwhose temperature is above 0 K. Decide whether this engine violates the firstor second law of thermodynamics.
36Refrigerators, Air Conditioners, and Heat Pumps Refrigerators, air conditioners, and heat pumps are devices that makeheat flow from cold to hot. This is called the refrigeration process.
38Refrigerators, Air Conditioners, and Heat Pumps Conceptual Example 9 You Can’t Beat the Second Law of ThermodynamicsIs it possible to cool your kitchen by leaving the refrigerator door open or tocool your room by putting a window air conditioner on the floor by the bed?
39Refrigerators, Air Conditioners, and Heat Pumps Refrigerator orair conditioner
40Refrigerators, Air Conditioners, and Heat Pumps The heat pump uses work to makeheat from the wintry outdoors flowinto the house.
41Refrigerators, Air Conditioners, and Heat Pumps Example 10 A Heat PumpAn ideal, or Carnot, heat pump is used to heat a house at 294 K. How muchwork must the pump do to deliver 3350 J of heat into the house on a day whenthe outdoor temperature is 273 K?
43a concept called entropy. In general, irreversible processes cause us to lose some, but not necessarilyall, of the ability to do work. This partial loss can be expressed in terms ofa concept called entropy.Carnotengineentropychangereversible
44Consider the entropy change of a Carnot engine. The entropy of the Entropy, like internal energy, is a function of the state of the system.Consider the entropy change of a Carnot engine. The entropy of thehot reservoir decreases and the entropy of the cold reservoir increases.Reversible processes do not alter the entropy of the universe.
45What happens to the entropy change of the universe in an irreversible process is more complex.
46Example 11 The Entropy of the Universe Increases The figure shows 1200 J of heat spontaneously flowing througha copper rod from a hot reservoir at 650 K to a coldreservoir at 350 K. Determine the amount by whichthis process changes the entropy of theuniverse.
48Any irreversible process increases the entropy of the universe. THE SECOND LAW OF THERMODYNAMICS STATEDIN TERMS OF ENTROPYThe total entropy of the universe does not change when areversible process occurs and increases when an irreversibleprocess occurs.
49Example 12 Energy Unavailable for Doing Work EntropyExample 12 Energy Unavailable for Doing WorkSuppose that 1200 J of heat is used as input for an engineunder two different conditions (as shown on the right).Determine the maximum amount of work that can be obtainedfor each case.
50The maximum amount of work will be achieved when the EntropyThe maximum amount of work will be achieved when theengine is a Carnot Engine, where(a)(b)The irreversible process of heat through the copperrod causes some energy to become unavailable.