ecture 26 Second law of thermodynamics. Heat engines and refrigerators.
The Second aw of Thermodynamics Introduction The absence of the process illustrated above indicates that conservation of energy is not the whole story. If it were, movies run backwards would look perfectly normal to us!
The Second aw of Thermodynamics Introduction The second law of thermodynamics is a statement about which processes occur and which do not. There are many ways to state the second law; here is one: Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object.
Heat engine = device with a working substance (eg. gas) that operates in a thermodynamic cycle. In each cycle, the net result is that the system absorbs heat (Q > 0) and does work (W > 0). Examples: - Car engine: burns fuel, heats air inside piston. Piston expands, does mechanical work to move car - Animal: burns food to be able to move Steam engine Car engine (internal compusion engine)
Hot and cold reservoirs Stages of the cycle Absorb heat from hot reservoir (QH) Perform mechanical work (W ) Dump excess heat into cold reservoir (Q < 0) Reservoir = large body whose temperature does not change when it absorbs or releases heat.
Energy flow of Heat engines Working substance in engine completes a cycle, so ΔU = 0: (Q H + Q ) W = 0 W = QH + Q = QH Q This relation follows naturally from the diagram (QH splits ). Draw it every time!
Energy flow diagrams in heat engines
imitations We are not saying that you can absorb 10 J of heat from a hot source (a burning fuel) and produce 10 J of mechanical work... You can absorb 10 J of heat from a hot source (a burning fuel) and produce 7 J of mechanical work and release 3 J into a cold source (cooling system). so at the end you absorbed 10 J but used (= converted to work) only 7 J. (We ll see later that it is impossible to make QH = W, or Q = 0)
Efficiency what you use Efficiency what you pay for W For a heat engine: e QH e = 0 e 1 QH Q QH = 1 Q QH Example: A heat engine does 30 J of work and exhausts 70 J by heat transfer. What is the efficiency of the engine? W = 30 J Q = 70 J Q = 70 J Q H = W Q = 100 J e= W QH = 0.3 (or 30 %)
ACT: Two engines Two engines 1 and 2 with efficiencies e1 and e2 work in series as shown. et e be the efficiency of the combination. Which of the following is true? TH A. e > e1 + e2 Q1 B. e = e1 + e2 C. e < e1 + e2 W1 e1 Q2 e2 W2 Q3 TC
Carnot engine QH QH Q W e= = QH QH = TH T Q TH Another way to write it: e= QH Q QH =1 = 1 T TH Q QH Carnot engine is ideal (reversible)
Refrigerators Absorb heat from cold reservoir (Q > 0) Work done on engine (W < 0) Dump heat into hot reservoir (QH < 0) Energy balance: W = QH + Q W = Q H Q (We want as much Q while paying for the smallest possible W.) Coefficient of performance (refrigerator) Carnot Refrigerator COP = Q = Q W Q H Q COP = T T H T 0 < COP < 0 < COP <
Heat pumps A very efficient way to warm a house: bring heat from the colder outside. Inside of house TH This time we are interested in QH : Heat pump Coefficient of performance (heat pump) Q Q COP = = W Q Q H H 1 < COP < Outside of house TC Same energy diagram as refrigerator H Carnot pump COP = TH T H T 1 < COP <
ACT: eaving the fridge open If you leave the door of your fridge open, you will get a heart-stopping electricity bill, but you will also: A. Freeze the kitchen B. Warm up the kitchen
Another way of stating the Second law of thermodynamics It is impossible for any system to undergo a process in which it absorbs heat form a reservoir at a single temperature and convert the heat completely into mechanical work, with the system ending in the same state as it began ( engine or Kelvin-Plank statement) i.e., It is impossible to build a 100%-efficient heat engine ( e = 1) Or It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body ( refrigerator or Clausius statement) i.e., It is impossible to build a workless refrigerator (COP )
Clausius Kelvin-Plank Workless refrigerator 100%-efficient engine Kelvin-Plank Clausius
Kelvin-Plank Clausius Workless refrigerator 100%-efficient engine Clausius Kelvin-Plank
New state variable: entropy TH TC Q positive large QC QH 0 QH QC TC TH (QC > 0), (QH < 0) : same magnitude Q since TC < TH negative small QC QH 0 TC TH This is positive (> 0) for all types of interacting systems. Define new state variable entropy (S ): S Q T The change of entropy gives the direction of the process (direction of energy dispersion): entropy must increase S SC SH 0
ACT: Glass of cold water A glass of cold water is placed in a hot room. Consider a small heat transfer Q. Which of the following statements is a true? A. SW > 0, SR > 0 B. SW < 0, SR > 0 C. SW > 0, SR < 0 Q TW TR
Terminology Q TC TH Glass of cold water (system) in hot room (environment) Interacting components naturally exchange energy: real, spontaneous processes. Often there are two interacting parts labeled as system and environment Δ Sinteracting parts = ( Δ Ssystem + Δ Senvirenment ) 0 Second law of thermodynamics in terms of entropy
Reversibility The processes we have analyzed are irreversible: spontaneous dispersion of energy (in one direction) the entropy of the universe increases If a system and its environment are almost at the same temperature: a small change in any temperature can reverse the direction of the heat flow reversible process the entropy of the universe (not of each part) remains constant Tsys Q Tenv=Tsys+ Tsys Q Tenv=Tsys
Change of entropy in an adiabatic process That s the easiest case: since Q = 0, ΔS = 0 We can now write the list of thermodynamics processes: Isobaric: Constant p Isochoric: Constant V Isothermal: Constant T Adiabatic: Constant S ( S = 0 )
Overall, ΔSall = ΔSgas + ΔS'environment > 0 this cycle is an irreversible process Any process involving exchange of heat between bodies at different temperature is irreversible. Only ideal isothermal and adiabatic processes can be reversible. And this is why the Carnot cycle is reversible Exercise: Prove that ΔSall = 0 for a Carnot cycle.
In-class example: Second law Which of the following is NOT an accurate statement of the second law of thermodynamics? A. The change of entropy in an object can never be less than zero. B. A perfect heat engine (e = 1) cannot exist. C. A perfect refrigerator (COP = ) cannot exist. D. ΔStotal (isolated system) > 0 E. All of the above are accurate statements of the second law of thermodynamics.
In-class example: Second law Which of the following is NOT an accurate statement of the second law of thermodynamics? A. The change of entropy in an object can never be less than zero. B. A perfect heat engine (e = 1) cannot exist. C. A perfect refrigerator (COP = ) cannot exist. D. ΔStotal (isolated system) > 0 E. All of the above are accurate statements of the second law of thermodynamics. Entropy of a part can decrease. Entropy of the total (universe) cannot decrease.