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Lecture 1: Energy • Reading: Zumdahl 9.1 • Outline – Energy: Kinetic and Potential – System vs. Surroundings – Heat, Work, and Energy • Problems: – Z9.10, Z9.15, Z9.18, Z9.19,Z9.20, Z9.21 – Working these problems are the key to knowing the material and testing whether you understand the ideas. 1 Energy: Kinetic vs. Potential • Potential Energy (PE) – Energy due to position or composition. – Equals (mgh) in example. m • Kinetic Energy (KE) h v – Energy due to motion. – Equals mv2/2 in example. • Total Energy is Always Conserved 2 Energy: E = KE + PE • Energy is the sum of kinetic energy and potential energy. 1 mv 2 KE = – KE (Kinetic Energy) 2 – PE (Potential Energy) PE = mgh • Energy is readily converted between these two forms. • If the system of interest is isolated (no exchange with surroundings), then total energy is constant, or conserved. 3 Problem: Z9.15 Ball A (2kg) swings down (falls) a height of 10 m, and hits B (4kg), transfers all energy to ball B which climbs a 3 m hill. What is the velocity of B at the top (of the 3 m hill)? How do we solve the: 1) What is the energy of A when it hits B? Hint: Potential energy is converted to kinetic energy. ΔE = 0 ⇒ KE = − PE = − m gz = 2 ⋅ 9.8 ⋅ 10 J ∼ 200 J A A 2) What is the initial Kinetic energy of B? Hint: All Energy is transferred to KE. 3) How much Potential Energy is gained in climbing 3m? PE = mB gzB = 4 ⋅ 9.8 ⋅ 3J ∼ 120 J 4) What is the total energy of B? Hint: Energy is conserved. 5) What is the KE of B at the top of the hill? E = KE + PE 200 = KE + 120 J 4 Ex. for Conserved Energy: Mass on a Spring Units of Energy in SI system is the Joule, J. 1J ≡ kg − m 2 sec 2 • Initial PE = 1/2 kx2 • At x = 0: 0 E = KE + PE – PE = 0 – KE = 1/2mv2 = mv + kx 1 2 • Example: 2 – Init. PE = 10 J (at a displacement) – M = 10 kg – Vmax = [2(PE)/M]1/2 = 1.4m/s (at zero displacement) Energy is Conserved: E = KE (is0) + PE ( x 1 2 2 = Init ) = 10 J 2 E = KE ( @ x = 0 ) + 0(isPE ) = 12 mvmax 2 vmax 2 2 ⎛ m 5⎞ 2 = ⋅ 10m = ⋅ 10 = 2 ⎜ ⎟ 10 sec m ⎝ ⎠ 2 First Law of Thermodynamics First Law: Energy of the Universe is Constant E = q + w (remember this) q = heat. Transferred between two bodies of differing temperature. Note: q ≠ Temp! w = work (transferred). Force acting over a distance (F x d) (do not do work if something does not move) Energy (of the system) can be found by the state of the system; q and w depend to the transfer and cannot be deduced from the state of the system. The change in PE is the work done. Force is mg and distance is h or z. 6 Applying the First Law • Need to differentiate between the system and surroundings. Surroundings System q transfer • System: That part of the universe you are interested in (i.e., you define it). • Surroundings: The rest of the universe. w transfer The change (DELTA) in energy of the system ΔE ≡ ΔEsystem 7 Conservation of Energy • Total energy is conserved. P = 1atm • Energy gained by the system must be lost by the surroundings. Initial • Energy exchange can be in the form of q, w, or both. • How much heat was withdrawn to cool the system? Great Question, we will answer it now. P = 1atm Final 8 Heat Exchange: Exothermic Ene rgy Water @ 80° C E initial Water @ 20° C q E final E final < E initial • Exothermic Reaction. Chemical process in which system evolves resulting in heat transfer to the surroundings. • q < 0 (heat is lost) • Heat goes out of system, so energy of system drops. 9 Another Example of Exothermic The loss in potential energy within the system results in a release of heat into the environment (q<0), no work (w=0) Thermite Reaction! ΔE = ΔKE + ΔPE = q + w = q < 0 ΔKE = 0 (isothermal) ΔE = q < 0 10 Heat Exchange: Endothermic Ene rgy Water @ 80° C E final Water @ 20° C q • Endothermic Reaction: Chemical process in which system evolves resulting in heat transfer to the system. E initial E final > E initial • q > 0 (heat is gained) 11 Another Example of Endothermic NO is unstable w.r.t. air. So why do we get any NO from our cars? This question presages what is to come: There is more to chemical equilibrium that just whether the reaction is endo- or exo- thermic!!! 12 Energy and Sign Convention Ene rgy E initial Eout E final ΔE < 0 Ene rgy E final E initial Ein • If system loses energy: Efinal < Einitial Efinal-Einitial = ΔE < 0. • If system gains energy: Efinal > Einitial Efinal-Einitial = ΔE > 0. ΔE > 0 13 Heat and Work Sign Convention ΔE = q + w < 0 If system gives up heat q < 0 (q is negative) If system gets heat q > 0 (q is positive) If system does work w < 0 (w is negative) If work is done on system w > 0 (w is positive) ΔE = q + w > 0 Which of these two cases best describes the thermite reaction? 14 Problem Z9.20 • A mixture of gas and air (in a cylinder, 50 cc) The explosion releases 1kJ of energy. What is the final volume of the new product (water and CO2) expanding against one Atmosphere (or 10^5 Pascals). Assume all of the energy is converted to work to expand the cylinder. • How to solve: – – – – Know the energy released, and that goes into work only (so q=0) Know that work and volume are related: w = − PdV First law: ΔE = q + w = 0 − PΔV Energy is released. This is important, it tells you the system gives up this much energy, so the change in energy is negative. ΔE = −1 ⋅ 103 J = − PΔV = −1 ⋅ 105 ⋅ ΔV ΔV = +10−2 m 3 = 10 = 104 cc ΔV = VF − VI = 104 cm 3 VF = VI + 104 = 50 + 10,000 ~ 10,000cm 3 15 Example: Piston Expansion • Figure 9.4, expansion against a constant external pressure • No heat exchange: q = 0 (adiabatic) • The amount of work the System does is negative: w<0 Because the system is doing the work, expanding the gas. Pext < P ΔV = Vf −Vi = Ahf − Ahi = AΔh 16 Example (cont.) • How much work does the system do? • Pext = force/area= mass*g/area • w = force * distance = (m*g)*Δh = -Pext * A * Δh = -Pext * ΔV • w = - Pext ΔV (note sign) • When will it stop expanding? Z9.10 Why the negative sign on work? 17 Example 9.1, and Problem Z9.19 • A hot-air balloon is inflated from 4 * 106 l to 4.5 x 106 l by the addition of 1.3 x 108 J of heat. If the balloon expands against an external pressure of 1 atm, what is ΔE for this process? – (FYI) A sphere 20 meters in diameter has a volume about 4 * 106 l . (just over 20 yards; c.f. football field). • Solve: – First, define the system: the balloon and the air inside the balloon. – Key: Some of the heat energy (q) will go into work (w), and the rest goes into internal energy of the gas in the balloon. 18 Example 9.1 (cont.) ΔE = q + w = (1.3 x 108 J) + (-PΔV) w= (-1 atm (Vfinal − Vinit)) = (-0.5 x 106 l.atm) 1 − Atm = 1.01 ⋅ 105 ⋅ 10−3 Pa ⋅ m 3 ∼ 102 J • Conversion: 101.3 J per l.atm w=(-0.5 x 106 l.atm) x (101.3 J/l.atm) = -5.1 x 107 J The SI unit of pressure is the Pascal. 1 J is a Pa*m3 1 bar is also an SI unit and is exactly 1 x 105 Pa, and 1.013 bars is 1 Atmosphere. 19 Example 9.1 (cont.) ΔE = (13 x 107 J) + (-5 x 107 J) = 8 x 107 J (Ans.) In English: the system gained more energy through heat than it lost doing work; which had better be the case or it would not expand on heating. Therefore, the overall energy of the system has increased. 20