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Lecture 15 Heat Engines Review & Examples

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Lecture 15 Heat Engines Review & Examples Lecture 15 Heat Engines Review & Examples p b Hot reservoir at Th pb Q Heat h c leak p Heat W a pump a adiabats Qc d Cold reservoir at Tc V1 V2 V Lecture 15, p 1 Irreversible Processes Entropy-increasing processes are irreversible, because the reverse processes would reduce entropy. Examples: • Free-expansion (actually, any particle flow between regions of different density) • Heat flow between two systems with different temperatures. Consider the four processes of interest here: Isothermal: Heat flow but no T difference. Reversible Adiabatic: Q = 0. No heat flow at all. Reversible Isochoric & Isobaric : Heat flow between different T’s. Irreversible (Assuming that there are only two reservoirs.) p p p isotherm adiabat isochor p isobar V V V V reversible reversible irreversible irreversible Lecture 15, p 2 ACT 1 1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment? a) ∆Senv < 0 b) ∆Senv = 0 c) ∆Senv > 0 2) Consider instead the ‘free’ expansion ( i.e. , not quasi-static ) of a gas. What happens to the total entropy during this process? a) ∆Stot < 0 b) ∆Stot = 0 c) ∆Stot > 0 gas vacuum Remove the barrier Lecture 15, p 3 Solution 1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment? a) ∆Senv < 0 b) ∆Senv = 0 c) ∆Senv > 0 Energy (heat) leaves the environment, so its entropy decreases. In fact, since the environment and gas have the same T, the two entropy changes cancel: ∆Stot = 0. This is a reversible process. 2) Consider instead the ‘free’ expansion ( i.e. , not quasi-static ) of a gas. What happens to the total entropy during this process? a) ∆Stot < 0 b) ∆Stot = 0 c) ∆Stot > 0 gas vacuum Remove the barrier Lecture 15, p 4 Solution 1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment? a) ∆Senv < 0 b) ∆Senv = 0 c) ∆Senv > 0 Energy (heat) leaves the environment, so its entropy decreases. In fact, since the environment and gas have the same T, the two entropy changes cancel: ∆Stot = 0. This is a reversible process. 2) Consider instead the ‘free’ expansion ( i.e. , not quasi-static ) of a gas. What happens to the total entropy during this process? a) ∆Stot < 0 b) ∆Stot = 0 c) ∆Stot > 0 There is no work or heat flow, so Ugas is constant. ⇒ T is constant. However, because the volume increases, so does the number of available states, and therefore Sgas increases. Nothing is happening to the environment. Therefore ∆Stot > 0. This is not a reversible process. Lecture 15, p 5 Review Entropy in Macroscopic Systems Traditional thermodynamic entropy: S = k ln Ω = kσ We want to calculate S from macrostate information (p, V, T, U, N, etc .) Start with the definition of temperature in terms of entropy: 1∂σ 1 ∂ S ≡, or ≡ kT∂ UVN, T ∂ U VN , The entropy changes when T changes : (We’re keeping V and N fixed.) dU CdTT2 CdT dS==V⇒ ∆= S V T T∫ T T1 T2 dT T2 If C V is constant: =CV = C V ln ∫ T T T1 1 Lecture 15, p 6 ACT 2 Two blocks, each with heat capacity* T1 = 250K C = 1 J/K are initially at different T1 T2 T2 = 350K temperatures, T1 = 250 K , T2 = 350 K . They are then placed into contact, and eventually reach a final temperature of Tf Tf Tf = 300 K 300 K. (Why?) What can you say about the total change in entropy ∆S ? Two masses each tot with heat capacity C = 1J/K* a) ∆Stot < 0 b) ∆Stot = 0 c) ∆Stot > 0 *For a solid, C = C V = C p, to good approximation, since ∆∆∆V ≈≈≈ 0. Lecture 15, p 7 Solution Two blocks, each with heat capacity* T1 = 250K C = 1 J/K are initially at different T1 T2 T2 = 350K temperatures, T1 = 250 K , T2 = 350 K . They are then placed into contact, and eventually reach a final temperature of Tf Tf Tf = 300 K 300 K. (Why?) What can you say about the total change in entropy ∆S ? Two masses each tot with heat capacity C = 1J/K* a) ∆Stot < 0 b) ∆Stot = 0 c) ∆Stot > 0 This is an irreversible process, so there must be a net increase in entropy. Let’s calculate ∆S: Tf T f ∆Stot = Cln + C ln T1 T 2 300 300 The positive term is slightly =Cln + C ln 250 350 bigger than the negative term. 300 2 =Cln = (0.028) C = 0.028 J/K 250× 350 *For a solid, C = C V = C p, to good approximation, because ∆V ≈ 0. Lecture 15, p 8 Example Process To analyze heat engines, we need to be able to calculate ∆U, ∆T, W, Q, etc . for the processes that they use. How much heat is absorbed by 3 moles of helium when it expands from Vi = 10 liters to Vf = 20 liters and the temperature is kept at a constant 350 K ? What are the initial and final pressures? Lecture 15, p 9 Solution To analyze heat engines, we need to be able to calculate ∆U, ∆T, W, Q, etc . for the processes that they use. How much heat is absorbed by 3 moles of helium when it expands from Vi = 10 liters to Vf = 20 liters and the temperature is kept at a constant 350 K ? What are the initial and final pressures? st Q = Wby The 1 law. For an ideal gas, ∆T = 0 → ∆U = 0. Positive Q means heat flows into the gas. Wby = nRT ln(V f/V i) = 6048 J An expanding gas does work. 5 pi = nRT/V i = 8.72 ×10 Pa Use the ideal gas law, pV = nRT 5 pf = p i/2 = 4.36 ×10 Pa Where is the heat coming from? In order to keep the gas at a constant temperature, it must be put in contact with a large object ( a “heat reservoir”) having that temperature. The reservoir supplies heat to the gas (or absorbs heat, if necessary) in order to keep the gas temperature constant. Very often, we will not show the reservoir in the diagram. However, whenever we talk about a system being kept at a specific temperature, a reservoir is implied. Lecture 15, p 10 Example Process (2) Suppose a mole of a diatomic gas, such as O 2, is compressed adiabatically so the final volume is half the initial volume . The starting state is Vi = 1 liter , Ti = 300 K . What are the final temperature and pressure? Lecture 15, p 11 Solution Suppose a mole of a diatomic gas, such as O 2, is compressed adiabatically so the final volume is half the initial volume . The starting state is Vi = 1 liter , Ti = 300 K . What are the final temperature and pressure? γ γ pVii= pV ff Equation relating p and V for adiabatic process in α-ideal gas 7 / 2 γ = = 1.4 γ is the ratio of C /C for our diatomic gas ( =( α+1)/ α). 5 / 2 p V γ Vi pf= p i Solve for p f. Vf γ nRT V = i i We need to express it in terms of things we know. Vi V f = 6.57x 106 Pa p V T =f f = 395 K Use the ideal gas law to calculate the final temperature. f nR α α TVii= TV ff Alternative: Use the equation relating T and V for an adiabatic 1 process to get the final temperature. α = 5/2 α Vi Tf= T i = 395 K Vf Lecture 15, p 12 Helpful Hints in Dealing with Engines and Fridges Sketch the process (see figures below). Define Qh and Q c and Wby (or W on ) as positive and show directions of flow. Determine which Q is given. Write the First Law of Thermodynamics (FLT). We considered three configurations of Carnot cycles: T T T h h h Q = Q Qh Qh Qh leak h Wby Won Won Q Q Q Tc c Tc c Tc c Qleak = Q C Engine: Refrigerator: Heat pump: We pay for Qh, We pay for W on , We pay for W on , we want Wby . we want Q c. we want Qh. Wby = Qh -Qc = εQh Qc = Qh -Won = ΚWon Qh = Q c + W on = ΚWon Carnot: ε = 1 - Tc/T h Carnot: Κ = Tc/(T h -Tc) Carnot: Κ = Th/(T h -Tc) This has large ε These both have large Κ when T -T is small. when T -T is large. h c h c Lecture 15, p 13 ACT 3: Entropy Change in Heat Pump Consider a Carnot heat pump. 1) What is the sign of the entropy change of the Hot reservoir at Th hot reservoir during one cycle? Qh a) ∆Sh < 0 b) ∆Sh = 0 c) ∆Sh > 0 Engine W Qc 2) What is the sign of the entropy change of the Cold reservoir at T cold reservoir ? c a) ∆Sc < 0 b) ∆Sc = 0 c) ∆Sc > 0 3) Compare the magnitudes of the two changes. a) | ∆Sc| < | ∆Sh| b) | ∆Sc| = | ∆Sh| c) | ∆Sc| > | ∆Sh| Lecture 15, p 14 Solution Consider a Carnot heat pump. 1) What is the sign of the entropy change of the Hot reservoir at Th hot reservoir during one cycle? Qh a) ∆Sh < 0 b) ∆Sh = 0 c) ∆Sh > 0 Engine Energy (heat) is entering the hot reservoir, so the W number of available microstates is increasing. Qc 2) What is the sign of the entropy change of the Cold reservoir at T cold reservoir ? c a) ∆Sc < 0 b) ∆Sc = 0 c) ∆Sc > 0 3) Compare the magnitudes of the two changes.
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