Physics 9 Wednesday, February 29, 2012

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Christiana Bishop
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1 Physics 9 Wednesday, February 29, 2012 learningcatalytics.com class session ID: Today: heat pumps, engines, etc. Aim to cover everything you need to know to do HW #8. Friday: start electricity (lots of fun demos) I ll put Friday video online, in case you desperately need to get a head start on your drive down to Daytona Beach (If you won t be here: please have a safe & fun spring break.) I will be in 3W2 as usual at 7pm this Thursday, at least for the first half hour or so. Zoey will do a study session 4pm-7pm on Mar 11 (next Sunday), but not on Mar 4 (i.e. not this Sunday). Q#1 right away: why is earth warmer than 18 C?

2 Entropy The logarithm of the number of different microstates that contribute to a given macrostate is called the entropy of that macrostate: S = k B ln(ω) for the coin toss, the macrostate corresponds to knowing N and m, and Ω corresponds to N! m! (N m)! Interesting nerd fact: as N, ln(n!) N (ln(n) 1) N ln(n) So the statistical entropy for m coins heads-up is ( ) N! ln(ω) = ln N ln N m ln m (N m) ln(n m) m! (N m)!

3 ( ) N! ln(ω) = ln N ln N m ln m (N m) ln(n m) m! (N m)!

4 Entropy Suppose I start out with a jar of 1000 coins that are carefully arranged to be entirely heads-up, and I shake the jar for a long time. What will I find?

5 Entropy Suppose I start out with a jar of 1000 coins that are carefully arranged to be entirely heads-up, and I shake the jar for a long time. What will I find? Shaking will tend toward more probable (higher entropy) states. After shaking, I will find that the entropy has increased to the largest possible value (plus or minus very small fluctuations).

6 Entropy Suppose I start out with a jar of 1000 coins that are carefully arranged to be entirely heads-up, and I shake the jar for a long time. What will I find? Shaking will tend toward more probable (higher entropy) states. After shaking, I will find that the entropy has increased to the largest possible value (plus or minus very small fluctuations). Once you reach equilibrium, i.e. the m heads 500 state, you will never spontaneously go back to the m heads 0 state. It s just too improbable. Once you shake for long enough to get close to 500, you ll stay pretty close to 500 (roughly within 500 ± 500 or so).

7 Entropy Suppose I start out with a jar of 1000 coins that are carefully arranged to be entirely heads-up, and I shake the jar for a long time. What will I find? Shaking will tend toward more probable (higher entropy) states. After shaking, I will find that the entropy has increased to the largest possible value (plus or minus very small fluctuations). Once you reach equilibrium, i.e. the m heads 500 state, you will never spontaneously go back to the m heads 0 state. It s just too improbable. Once you shake for long enough to get close to 500, you ll stay pretty close to 500 (roughly within 500 ± 500 or so). Similarly, the second law of thermodynamics states that the combined entropy of system + environment will never decrease with time. (It can stay the same or can increase.)

8 Entropy Increases in entropy are associated with irreversible processes, like the transformation of mechanical energy into heat when I dropped the lead buckshot last week. For system in equilibrium with thermal reservoir at temperature T, or in other words, S = Q T Q = T S

9 Entropy and latent heat In HW7, you needed to use the fact that the latent heat involved in melting ice is L = 333 kj/kg To melt 1 kg of ice at 0 C, you need to add heat Q = ml = 333 kj What is the difference in entropy between 1 kg of ice and 1 kg of liquid water at 0 C?

10 Entropy and latent heat In HW7, you needed to use the fact that the latent heat involved in melting ice is L = 333 kj/kg To melt 1 kg of ice at 0 C, you need to add heat Q = ml = 333 kj What is the difference in entropy between 1 kg of ice and 1 kg of liquid water at 0 C? S = Q 333 kj = = 1.22 kj/k T 273 K Liquid water is a higher entropy ( more disordered ) state than solid ice, because the liquid state has a much larger number of equivalent ways to arrange the same water molecules.

11 The image shows an ideal gas cycle consisting of two isotherms ( T = 0) and two isochores ( V = 0). A complete cycle starts at state 4, then goes to states 1, 2, 3, and back to 4. On which parts of the cycle is positive work done ON the gas (i.e. W in > 0 )?

13 On which parts of the cycle is positive work done BY the gas (i.e. W out > 0)? Over one complete cycle, is W out larger than, smaller than, or equal to W in? On which parts of the cycle is the internal energy of the gas constant? Once I have determined W out W in for one complete cycle of the engine (which is known as a steady device ), how can I go about finding Q in Q out for one complete cycle?

15 Efficiency of heat engine The whole point of a heat engine is to convert thermal energy into mechanical energy, i.e. to turn some net input of Q into some net output of work: W = W out W in So the efficiency is what you get what it costs η = W out W in Q in (You don t bother to subtract Q out in the denominator, because you have to pay for the high-temperature heat Q in, but you can t do anything useful with the low-temperature heat Q out.)

16 Efficiency of heat engine η = W out W in Q in Conservation of energy (over one complete cycle): Q in + W in = Q out + W out Q in Q out = W out W in so we can rewrite η = Q in Q out Q in In the special, ideal case of a reversible heat engine (i.e. an engine for which S env = 0, so the movie is allowed to run backwards), η ideal = T in T out T in (Basically this works if the S cancels out for Q = T S, which is the case if Q in /T in = Q out /T out.)

17 Efficiency of heat engine η = Q in Q out Q in In the special, ideal case of a reversible heat engine (i.e. an engine for which S env = 0, so the movie is allowed to run backwards), we have S env = Q out T out Q in T in = 0 Q in = Q out T in T out Plugging in and canceling, we get η ideal = Q out (T in /T out ) Q out Q out (T in /T out ) = T in T out T in

18 Coefficient of performance of heat pump COP heating = Q out W T out T out T in (equality in the ideal (reversible, Carnot, S env = 0) case) COP cooling = Q in W T in T out T in where W means W in W out over the whole cycle.

19 HW8 problem 3 What is the efficiency η = Wout W in Q in, if Q out = 43.5 kj per cycle?

Physics 9 Friday, November 2, 2018

Physics 9 Friday, November 2, 2018 Turn in HW07. Pick up handout for HW08, due next Friday. For Monday, read Eric Mazur s chapter 22 (Electric Interactions) PDF on Canvas. I have a large number of supplemental