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Chapter 12 the Laws of Thermodynamics Animals Have

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Chapter 12 the Laws of Thermodynamics Animals Have Chapter 12 The Laws of Thermodynamics Animals have been doing work using biologic form of energy since their inception. What has it take so long to get machines to do work and how do they do it? Work and the Ideal gas. Work done at constant pressure P See figure 12.1 page Assume the pressure is constant in the cylinder and it piston is depressed Δy W=Fd=FΔy =PAΔy=PΔV work at constant pressure. It takes work to compress a gas so work is done on the gas when the volume is decreased. Eg. The gas has more energy If the gas expands it volume increases and we get work out of the gas (the piston can do work like move a car). Eg The gas has less energy. Example 12.1 page 386 All you want to know about PV diagrams. See Quick Quiz 12.1 page 387 for times when pressure is not constant. Dr. Popp’s statement of the First Law the Thermodynamics ΔU=± Q ±W In words the change of internal energy in any reversible process (ΔU=3/2nRΔT) Is equal to the heat add or removed and the work done by the gas or on the gas € € Q is plus if we add heat and minus if heat is removed W is plus if we do work on the gas (compress it ) and minus if the gas does work (expands) Example 12.3 page 289 Specific heat at constant volume Analogy with solids Q=mcΔT, ΔU=3/2nRΔT=nCvΔT where Cv =3/2R and is the specific heat a constant volume. Difference between Monatomic, Diatomic, and Polyatomic gases Cv depend on the number a way a gas molecule can store energy 3 ways for monatomic, 5 ways for diatomic and variable ways for polyatomic gases ½ R for each degree of freedom Specific heat at constant pressure ΔU= ± Q ±W => Q=ΔU+W =ΔU+PΔV=3/2nRΔT+nR ΔT = 5/3nRΔT = CpΔT Type of thermal processes 1. Isobaric constant pressure € € 2. Isothermal constant temperature 3. Isovolumetric constant volume 4. Adiabatic not heat transferred See Quick Quiz 12.2 page 398 See Table 12.2 Page 399 See Example 12.9 page 397 Getting work out of an ideal gas (Engines) Schematics diagram of an engine Hot Reservoir at TH QH Engine Work QC Cold Reservoir at TC For any cycle ΔU = 0 => W=QH-QC Efficiency is defined as W Q − Q Q e = = H C =1− C QH QH QH € Otto Cycle Car’s Internal Combustion Engine Top Dead Center Otto Cycle 35% Efficiency The four-stroke engine was first patented by Eugenio Barsanti and Felice Matteucci in 1854, followed by a first prototype in 1860. However, the German engineer Nicolaus Otto was the first to develop a functioning four-stroke engine, which is why the four-stroke principle today is commonly known as the Otto cycle and four-stroke engines using spark plugs often are called Otto engines. The Otto Cycle consists of adiabatic compression, heat addition at constant volume, adiabatic expansion and rejection of heat at constant volume. Internal combustion engine power primarily originates from the expansion of gases in the power stroke. Compressing the fuel and air into a very small space increases the efficiency of the power stroke, but increasing the cylinder compression ratio also increases the heating of the fuel as the mixture is compressed (following Charles's law). A highly flammable fuel with a low self-ignition temperature can combust before the cylinder reaches top-dead-center, potentially forcing the piston backwards against rotation. Alternately, a fuel which self- ignites at top-dead-center but before the cylinder has started downwards can damage the piston and cylinder due to the extreme thermal energy concentrated into a very small space with no relief. This damage is often referred to as engine knocking and can lead to permanent engine damage if it occurs frequently. The octane rating is a measure of the fuel's resistance to self-ignition, by increasing the temperature at which it will self-ignite. A fuel with a greater octane rating allows for a much higher compression ratio without the risk of damage due to self-ignition.. Cooling engines Hot Reservoir at TH QH Engine Work QC Cold Reservoir at TC Coefficient of performance COP = QC/W a good refrigerators or air conditioner has a COP of 5 or 6. See example 12.12 page 403. Carnot Engine Carnot engine is the most efficient engine and Q T e =1− C =1− C QH TH See figure 12.16 page 407 € Some reflection on Entropy Entropy has two definition 1. S= ΔQ/T 2. S is a measure of the disorder in a system Entropy always increases implying system more toward more disorder Metabolic Rate ΔU/Δt=-Q/Δt -W/Δt The internal energy of the body (temperature) = heat lost plus work done. To offset this we take food to increase the internal energy. Metabolic Rate can be measure by the rate of burning of oxygen. ΔU/Δt = 4.8 ΔV(oxygen)/Δt See example 12.17 page 415 Chapter 13 Vibration and Waves Revisit Hooke’s Law F=-kx F=ma => ma=-kx and a=-kx/m Look at the Periodic Motion of a spring position, velocity, acceleration using Figure 13.4 page 429 Spring Energy = E=½kx2 Archer Problem: An archer pulls back a simple bow .4 meters applying a maximum force of 52 lb and shoot a 10 gram arrow. Assume the bow is approximated by Hooke’s Law 1. What is the spring constant of the bow? 2. What is the maximum horizontal distance the arrow can travel? Answer: 1. 52lb. =230 N From Hooke’s Law k =F/x =230/.4 =575 N/m. 2. Use energy to find initial velocity ½ mv2=½kx2 => v2=k/m x2 = 575(.4) 2 /.010 =>v=96m/sec Range = v2 /g=940m Define parameter and units 1. Amplitude A as maximum(meters) 2. Period T as the time it take to make 1 trip (sec) 3. Frequency f=1/T (1/sec) Find v and any point x using energy. 1 1 1 KA2 = Kx 2 + mv 2 2 2 2 k v = ± A2 − x 2 m ( ) Does this check out with what we know? Example 13.4 page 431 € The connection between Simple Harmonic Motion and Uniform Circular Motion θ x = ACos(θ) F = ma = −kx k a = − ACos(θ) m y(opposite) Sin(θ) = A y = A2 − x 2 −v k A2 − x 2 Sin(ωt) = = m A A k k v = −A Sin(ωt) where v = A m 0 m v = −v0Sin(ωt) 2πA = v0T m T = 2π k 1 1 k f = = T 2π m k ω = m € Applying Physics 13.1 Bungee Jumping Motion of a Pendulum Ft = −mgsin(θ ) x sin(θ ) ≈ L ma = −mgLx L θ T Compare to ma = -kx k = mgL X mgsin(θ) k g ω = ⇒ω = m L mgcos(θ) L T = 2π g mg 1 g f = 2π L Compound Pendulum €τ = Iα mgsin( )L τ = − θ −mgsin(θ )L = Iα L For small θ , sin(θ ) =θ −mgL CM α = θ I mgL ω = I For sepecial case of pendulum 2 I = mL g ω = L Three Types of Damping 1. Over damped 2. Under damped € 3. Critically damped Types of Waves and Interference 1. Transverse 2. Longitudinal 3. Shock 4. Circulars Fundamental Law of all Waves v = fλ € T Speed of a wave on a string v = µ € Chapter 14 Sound Sound waves are longitudinal pressure waves 1. Audio 20-20,00 HZ Hertz are another name for cycles per sec 2. Ultrasound 20,000 >Hz silent dog whistles, medical and therapeutic 3. Infrasound <20 Hz earthquakes 4. Shock waves 2-4 time speed of sound Piezoelectric Effect. Certain crystals vibrate with the frequency of electric voltage applied to them and produce pressure waves of the same frequency. These crystals also give a voltage change when they vibrate in response to sound waves. They can be used a ultrasound generator and receivers. Sonograms use ultrasound that is reflected differently from areas of different densities and so can give a picture similar to an x-ray but ultrasound is non-ionizing radiation. What is the purpose of the gel used on the tip of the ultrasound probe? Speed of Sound Waves T 1. In air v = 331 273 2. Liquids ~1,500 m/s 3. Solids 5,000 m/s Energy€ in Sound Waves Intensity is Power/Area watts/sq meter Sound Intensity W/m2 ΔPressure Pa Displacement m dB Low audible 10-12 3x10-5 10-11 0 Pain 1 29 10-5 120 Decibel 0 is lowest the ear can hear 120 is painful I log(ab) = loga + logb β =10log I 0 log(a/b) = loga − logb I =10−12 0 log(ab ) = bloga log1= 0 100 b log10 =1101 a = b ⇒ 10 =10 2 log100 = 2 10 log x = a ⇒ 10log x =10a 3 log1000 = 2 10 10log x = x =10a log1010 =10 −12 log 10 = −12 € € What is the Intensity level of painful sound? 120 =10log(I /10−12 ) 12 = log(I /10−12 ) −12 1012 =10log(I /10 ) = I /10−12 I =1.0w /m2 See Example 14.2 Page 465 € Doppler Effect Think of the shore and the frequency of the wave hitting the shore. Wave per sec. Now run into the water the frequency of the wave increases as you run out of the water the frequency decreases. Moving towards a source frequency increases moving away it decreases.
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