MidTerm. Phys224 Spring 2008 Dr. P. Hanlet

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1 MidTerm Name: Show your work!!! If I can read it, I will give you partial credit!!! Correct answers without work will NOT get full credit. Concept 5 points) 1. In terms of the First Law of Thermodynamics and in your own words, what is an adiabatic process? The First Law of Thermodynamics states: de int = dq dw An adiabatic process is one which happens so rapidly that there is no time for heat to be transferred. This means that Q = 0, such that: de int = dw 2. In your own words, state the equipartition theorem. The equipartion theorem states that every degree of freedom of a molecule has, on average, associated with it an energy of 1 2kT ; where a degree of freedom is an independent way that a molecule can move. 3. In your own words, what is the difference between heat capacity and specific heat? Heat capacity is the material dependent proportionality constant which relates the heat transferred to/from the material to the temperature of the material. In other words, it is the capacity of the material to maintain a certain temperature, when heat can flow to/from the material. For example, a cup of hot coffee will stay warmer longer than a plate of rice at the same initial temperature since the coffee has a higher heat capacity. Specific heat is heat capacity normalized to either mass or number of moles. 4. In your own words, what is entropy? Entropy has a number of descriptions. It is: a measure of the unavailability of a system s energy to do work a measure of the randomness of molecules in a system the quantity which determines the direction of an irreversible process the quantity that takes a system from the initial state to the final state: f dq S = T i 1

2 5. Based on the equation: S = S f S i = nr ln V f V i + nc V ln T f T i why is S a state function? This equation shows that the entropy depends only on the initial and final states of V and T namely V i, V f, T i, and T f ), and not on the path taken to get from the initial to the final states. Hence S is a state function. 6. Whether for a spherical mirror or a spherical refracting surface, if a ray enters the interface parallel to the symmetry axis, what point will the exiting reflected or refracted, real or virtual) ray go through? Any ray entering a reflecting or refracting surface parallel to the axis of symmetry of the surface will exit through a focus; or the extensions of the rays will exit through a focus if the image is virtual. 7. If a light ray goes from medium with index of refraction n 1 to medium with index of refraction n 2, with n 2 > n 1, will the refracted ray be bent towards or away from the normal at the point of incidence? If the ray goes from medium with n 1 to medium with n 2 and n 2 >n 1, then the refracted ray will be bent towards the normal. 8. The following solid objects are made of the same material and are maintained at a temperature of 300K in an environment whose temperature is 350K: a cube of edge length r, a sphere of radius r, and a hemisphere of radius r. Rank the objects according to the net rate at which thermal radiation is exchanged with the environment. Answer with greatest first, and a brief description as to why. The power net radiated by an object is given by: P rad = σεat env T body ) 4 where σ is a constant, ε depends on the material, and T is the difference in the temperatures of the environment and the radiating bodys. Since all of these are the same, then the difference in power raditated between the objects is in their surface areas: a) Cube: A = 6r 2 b) Sphere: A = 4πr 2 c) Hemisphere: A = 1 2 4πr2 + πr 2 = 3πr 2 Since these all have the factors r 2, then the coefficients will distinguish the ranking, and: A sphere > A hemisphere > A cube such that since power is the rate of energy transfer: P sphere > P hemisphere > P cube 2

3 9. Does the termperature of an ideal gas increase, decrease, or stay the same during a) an isothermal expansion Isothermal means constant temperature, therefore the temperature stays the same. b) an isobaric expansion Isobaric means constant pressure. Solving the ideal gas law for temperature: T = pv nr so if V expands, then T increases. c) an adiabatic expansion Adiabatic means that no heat is transferred in or out of the system. Hence, by the first law of thermodynamics, if Q = 0, then E int = W = p dv, which means that the internal energy of the gas decreases; i.e. the temperature of the gas decreases. d) increase in pressure at constant volume? Constant volume is Isochoric. From before, the ideal gas law solved for temperature shows that if p expands, then T increases. 10. One mole of gas A, with molecular diameter 2d 0 and average molecular speed v 0, is placed inside a certain container. One mole of gas B, with molecular diameter d 0 and average molecular speed 2v 0 is placed inside a different, but identical container. Which gas has the greater average collision rate with its container, and why? Note that there are two competing effects. B molecules have twice the speed, so they would have 2 the collision rate. The other effect is how the rate is affected by the mean free path. The mean free path of a molecule of gas is given by: λ = 2πd 2 N/V With all other parameters namely N and V ) being equal, then λ d 2, such that larger diameter d, implies smaller mean free path λ. Here, since gas A has diameter d = 2d 0 and gas B has diameter d = d 0, then gas A has the shorter mean free path by a factor of 4, and hence will have a higher collision rate over the molecules of gas B by 4 for this effect. Finally, gas A has a higher rate by 4 and gas B has a higher rate by 2. So, gas A has the higher collision rate. ) 1 3

4 11. A box contains one mole of a gas. Consider two configurations: a) each half of the box contains half of the molecules, and b) each third of the box contains one third of the molecules. Which configuration has more microstates, and why? In order to solve this, one needs to use the statistical mechanics form of the entropy equation: S = k ln W = k ln N! n!n n)! where here N = 1 mol of gas. For the two configurations: a) 2 halves: b) 3 thirds: S 3 = k ln S 2 = k ln N! ln N 2! N 2! = k N! ln N2! ln N2 )! N! ln N 3! N 3! N 3! = k N! ln N3! ln N3! ln N3 )! Using Stirlings formula, which is valid for large N here N is VERY large): ln N! = N ln N N, then: a) 2 halves: S 2 = k ln N! ln N2! ln N2 )! N = k N ln N N) 2 ln N 2 N ) N 2 2 ln N 2 N 2 = k N ln N N N 2 ln N 2 + N 2 N 2 ln N ) 2 + N 2 = k N ln N N 2 ln N 2 N 2 ln N ) 2 = k N ln N N2 ln N ln 2) N2 ) ln N ln 2) = k N ln N N ln N + N ln 2) = k N ln 2 )) 4

5 b) 3 thirds: S 3 = k ln N! ln N3! ln N3 )! N = k N ln N N) 3 ln N 3 N ) N 3 3 ln N 3 N ) N 3 3 ln N 3 N 3 = k N ln N N N 3 ln N 3 + N 3 N 3 ln N 3 + N 3 N 3 ln N ) 3 + N 3 = k N ln N N 3 ln N 3 N 3 ln N 3 N 3 ln N ) 3 = k N ln N N3 ln N ln 3) N3 ln N ln 3) N3 ) ln N ln 3) = k N ln N N ln N + N ln 3) = k N ln 3 )) In summary, S 3 = k N ln 3 > S 2 = k N ln 2, so the entropy, and therefore, number of microstates is greater for the triply divided box than the double divided box. Note that one could simply have calculated W, and not S. 12. Does the entropy per cycle increase, decrease, or remain the same for a a) Carnot engine A Carnot engine is a reversible engine, therefore the entropy remains the same. b) real engine A real engine has losses due to friction, etc. will increase. therefore the entropy c) perfect engine? A perfect engine will actually have entropy decrease; just goes to show how real a perfect engine is. 5

6 Problem 10 points) 13. Ignoring radiation, what is the rate of energy loss in watts through a 1.5m 2.5m 5mm glass window thermal conductivity k = 1.0 W/m/K) thick if the outside temperature is 35 F and the inside temperature is 70 F? Since there is no flow of fluid and no radiation, the rate of heat flow via conduction is: P cond = Q t = ka T H T C L where: A is the cross-sectional area, here A = 1.5m 2.5m = 3.75m 2 and: L = 5mm = m. Furthermore, the temperature for all thermodynamics problems must be in kelvin. However, since the scale for Celsius and Kelvin is the same, and the temperature difference removes the dependence on offset, then one can use the conversion to Celsius. Explicitly: T = T H T C ) ) 5 5 = 9 T F H 32 ) T F C 32 ) = 5 9 T F H T 5 F C = 5 9 T F H T FC ) = ) = K = 58.3 K So: Finally: P cond = 1.0 W m K 3.75 m 2 P cond = kw 58.3 K m 6

7 moles of ideal Oxygen gas at 27 C and pressure P a occupies what volume? If it then expands to double this volume at the same pressure, what is the final temperature of the gas? C V O 2 ) = 20.8J/mol/K) In what follows, one uses the Ideal Gas Law: pv = nrt For the first part, solve the Ideal Gas Law for V : V = nrt = 150 mol 8.31 J/ mol/ K 300 K = 3.7 J m2 p N/m 2 N = 3.7 Nm m 2 N such that: V = 3.7 m 3 For the second part, solve the Ideal Gas Law for T : ) T = pv nr = N/ m m mol 8.31 N m/ mol/k such that: T = K 15. A 3.5mol sample of ideal gas O 2 expands reversibly and isothermally at 420K until its volume is doubled. What is the increase in entropy of the gas? Using the following equation for entropy, and noting that isothermal means that T i = T f, and the volume doubling means that V f = 2V i, then substituting these results into the equation, one has: S S = S f S i = nr ln V f V i + nc V ln T f = nr ln 2 V i V i 0 = nr ln 2 + ln 1 1 T i + nc V ln T 1 i = 3.5 mol J 8.31 mol K ln 2 Hence, the increase in entropy of the gas is: S = 20.2 J/K T i 7

8 16. A Carnot engine whose low-temperature reservoir is at 17 C has an efficiency of 40%. By how much should the temperature of the high-temperature reservoir be increased to increase the efficiency to 50%? Let T h be the temperature of the hot reservoir, and T l the temperature of the cold reservoir. Then, for a Carnot engine, using the efficiency of the engine and solving for T h : ε C T h ε C = T h T l T h = T h T l ε C T h T h = T l T h ε C 1) = T l T h = T l ε C 1 = T l 1 ε C Converting T l to kelvin which one should almost automatically do), then T l = 273K + 17K = 290K For 40% efficiency: T h = 290K = 290K 0.6 = 483.3K For 50% efficiency: T h = 290K = 290K 0.5 = 580.0K Finally, the hot reservoir should be increased by T h = 580.0K 483.3K = 96.7K 17. Given the following cartoon of a lens and an image, draw 3 rays to determine where the object must be on the left side of the lens. Is the image real of virtual? IO c f f c 1 Converging convex) Lens Object: inside focus Image: virtual, same side, same orientation 8

9 18. 10pts) You are putting up an external wall measuring 12f t by 24f t and consisting of a layer of white pine of thickness L p = 3 in and a layer of rock wool. You want to put enough rock wool so that the 4 thermal conduction rate, P cond 120W during a winter night when the external temperature is 96.8 F below the indoor temperature. What thickness, L w, of rock wool is needed? k p = 0.11 W/m/K and k w = W/m/K) Heat going through walls is done by conduction; therefore the conduction equation for multiple layers is required. Here, there are two layers: Solving for L w : Converting: such that: P cond = AT H T C ) = A T L p k p 12 ft = 12 ft 24 ft = 24 ft 3 4 in = 3 4 in and from problem 13): i L i k i L p k p + L w = A T k w P cond L w = A T L p k w P cond k p A T L w = k w P cond + L w k w L p k p 12 in 2.54 cm 1 ft 1 in 1 m 100 cm = 3.66 m 12 in 2.54 cm 1 ft 1 in 1 m 100 cm = 7.32 m 2.54 cm 1 in 1 m 100 cm = 0.02 m A = 3.66 m 7.32 m = m 2 T = 5 9 T F H T FC ) = F = C = K ) Then: A T L w = k w Finally: P cond L p k p ) = W m K L w = 0.51 m m K 120 W ) 0.02 m m K 0.11 W 9

10 19. Suppose that Young s experiment is performed with light of wavelength, λ = 400 nm. The slits are 1.2 mm apart and the viewing screen is 3.5 m from the slits. How far apart are the bright fringes? In order that light have maxium constructive inteference from a two-slit system, the phase shift from the light between the two slits must have integral wave lengths; i.e.: d sin θ = mλ, where m is an integer, d is the distance between the slits, and λ is the wavelength of the light. d θ m θ m+1 y D Assuming that D >> d, then each ray can be considered parallel and the angle θ to be the same for each ray here we use the angle half way between). Furthermore, for large D, one can assume that θ is small, such that the small angle approximation is valid: sin θ θ. Then, from the figure, adjacent bright fringes are at: d sin θ m dθ m = mλ = θ m = m λ d d sin θ m+1 dθ m+1 = m + 1)λ = θ m+1 = m + 1) λ d such that the angular separation between adjacent fringes is: θ = θ m+1 θ m = m + 1) λ ) m λ ) = λ d d d The first m=0) fringe occurs at θ 0 =0. Again, for large D: m=0 θ m=1 D y tan θ = Finally: sin θ cos θ θ 1 = y D = y = D θ = Dλ d y = 1.17mm = 3.5m m m 10

11 20. 15pts) I ll walk you through this one. During an extended wilderness hike, you have a terrific craving for ice. Unfortunately, the air temperature drops to only 6.0 C each night, which is too warm to freeze water. However, because a clear, moonless night sky acts like a black body radiator at a temperature T s = 23.0 C, perhaps you can make ice by letting a shallow layer of water radiate energy to such a sky. To start, you place a layer of thermal insulation straw or foam rubber) on the ground then place a broad bottomed container with area A = 9.0cm 2 on top of this insulation. Into this container, you pour m = 4.5g of water with a thin, d = 5.0mm layer of water. If the initial water temperature is 6.0 C, and the emissivity is ɛ = 0.9, how long would it take to freeze the water via radiation? Can this occur in a single night? For water: L F = 333 kj/kg, c = 4190 J/kg/K) a) First determine how much heat needs to be removed to change liquid water into ice. First, one must remove the heat and change the liquid to ice: i. 6 C 0 C: Q i = m c T = kg) ii. liquid to ice: 4190 Q ii = m L f = kg) J kg K ) 6 K) = J 333 kj ) = 1, J kg So, the total heat that must be removed is: Q = Q i + Q ii = 1, J b) Next, note that power is the rate of change of energy: P = E/t and that heat is a form of energy. Using this fact, the results of part a), and the equation for thermal radiation: P = σɛat 4, you can determine the time to radiate enough energy to freeze the water. Note that you need the net rate of thermal radiation. The net rate of thermal radiation means that: P = σ ε A T 4 T 4) env where T env = 23 C = 250 K and T = 6 C = 279 K Solving for t: t = E P = Q σ ε A T 4 env T 4 ) 1, J = W/ m 2 / K 4 )0.9) m 2 ) ) K 4 = 16, J = 16, s J/s or t = 4.53 hr so, the water can be frozen overnight. 11

18.13 Review & Summary

5/2/10 10:04 PM Print this page 18.13 Review & Summary Temperature; Thermometers Temperature is an SI base quantity related to our sense of hot and cold. It is measured with a thermometer, which contains