PES 2130 Exam 1/page 1. PES Physics 3 Exam 1. Name: SOLUTIONS Score: / 100

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1 PES 2130 Exam 1/page 1 PES Physics 3 Exam 1 Name: SOLUTIONS Score: / 100 Instructions Time allowed or this is exam is 1 hours 15 minutes 10 written problems For written problems: Write all answers in the spaces provided. An extra sheet is included at the back i you need more space or your calculations. Answers to questions involving calculations should be evaluated to appropriate signiicant igures and given in decimal orm. Despite an incorrect inal result on written problems, credit may be obtained or method and working, provided these are clearly and legibly set out. Only required equipment may be placed on the desk. No talking permitted. No questions. I you are unsure o what is being asked, take your best guess and state your reasoning. Cell phones must be turned o. Equipment This Question/Answer booklet Equation sheet (need to be turned in with Question/Answer booklet) Pens, pencil, eraser, ruler, water bottle Graphing or scientiic calculator (cell phones may not be used as a calculator)

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3 PES 2130 Exam 1/page 3 Question 1 (10 points) a) () A solid steel object has a hole in it. Which o these illustrations (#1 or #2 or neither) more correctly shows how the size o the object and the hole change as the temperature increases? : Illustration #1 b) (8 points) A hole in a steel plate (coeicient o linear expansion = 12.0 x 10-6 K -1 ) has a diameter o 20.0 cm at 0.00 o C. What is the radius o the hole when the steel plate is heated red hot, 700. o C? New diameter: L = cm Radius = diameter/2 R = cm = 10.1 cm 0.5 points 0.5 points -0.5 units -0.5 sig igs -0.5 math mistake - 1 did not calculate radius

4 PES 2130 Exam 1/page 4 Question 2 (10 points) 10.0g o ice at 0.00 o C is mixed with 100 g water at 10.0 o C in a thermos. The Speciic heat o water is c w = 4187 J kg -1 K -1 and or ice c i = 2108 J kg -1 K -1. The latent heat o usion to melt ice is 333 kj/kg. a) () What is the composition o this mixture at thermal equilibrium? Choose rom ice, water, or ice-water. The inal composition is all water. () b) (8 points) What is the inal temperature o the composite? The ice will irst melt and then the water rom the ice will get warmer as the initial water will get colder. Q 0 Q Q 0 m L m c T m c T 0 m L m c ( T T ) m c ( T T ) 0 m L m c T m c T m c T m c T mi cwtii mwcwtiw mi L T m c m c T ii 0 i i i w i w w w w i i w ii w w iw i i w i w ii w w w w iw i w w w mwcwtiw mi L T m c m c T i w w w 3 (0.1)(4187)(10) (0.01)( ) T K K (0.01)(4187) (0.1)(4187) C + + rearrange Final answer sig igs units wrong numerical result else correct

5 PES 2130 Exam 1/page 5 Question 3 (10 points) 2.0 L o an ideal gas is at 10.0 o C. It is heated at constant pressure to 40.0 o C. a) (3 points) Draw the corresponding p-v diagram and label the graph appropriately. shape s correct labels b) (7 points) Calculate the inal volume o the gas. pressure is constant hence: pv = nrt () V V i nr T p T V i T Vi () T i V (2) L 2.2L (3 points) sig igs units wrong numerical result else correct -1 temperature not converted into Kelvin 40 V (2) L 8.0L not in Kelvin: 10

6 PES 2130 Exam 1/page 6 Question 4 (10 points) a) (4 points) State the "Equipartition Theorem" or 1 mole o gas. Every "degree o reedom" () that a microscopic object has is associated with energy o 1 RT per mole ( ) 2 - i answered 1 k T per molecule 2 B b) (6 points) You have 1.00 mol o an ideal monatomic gas and 1.00 mol o an ideal diatomic gas whose molecules can rotate. Initially both gases are at room temperature. I the same amount o heat lows into each gas, which gas will undergo the greatest increase in temperature and why? Use equations to explain. Answer: the monatomic gas () detailed explanation (4 points) (- i answer was: KE tr depends on temperature only) 3 1 mole o monatomic gas: KEtr RT av mole o diatomic gas: KEtr RT av 2 Assume W = 0: 1st law: E Q W int E Q KE int tr av 1 mole o monatomic gas: 3 3 Q RT RTm T 2 2 2Q T T m 3R 1 mole o diatomic gas: 5 5 Q RT RTm T 2 2 2Q T T d 5R Hence T m > T d

7 PES 2130 Exam 1/page 7 Question 5 (10 points) A straight-line process is done on a monatomic ideal gas, as shown. The temperature o the gas at point A is 500 K and the temperature at point B is 300 K. a) () For the straight-line process shown, is work done ON the gas or BY the gas? For ull points you need to explain your reasoning without doing a calculation. answer: Work is done BY the gas. b/c gas expands b) (6 points) What is the magnitude o work done? Work = are underneath p-v diagram Work = area o bottom rectangle + area o top triangle W = (2.0 m 3 )(1.0 Pa)+ (0.5) (2.0 m 3 )(4.0 Pa) W = 6.0 J sig igs units wrong numerical result c) () Is heat added to the gas or removed rom the gas? For ull points you need to explain your reasoning. answer: Heat is added Explanation: Work is done by the system and thereore energy must be added into the system in orm o heat. Calculation (was not required) Q= E int + W = 3/2 nr T +W 3 p V Q T2 T1 W J 6J 3J 6J 3J 2 T

PES 2130 Exam 1/page 8 Question 6 (10 points) The igure below shows a low diagram or a heat engine. T L =300K a) (6 points) Does the heat engine violate the irst law o thermodynamics?

8 PES 2130 Exam 1/page 8 Question 6 (10 points) The igure below shows a low diagram or a heat engine. T L =300K a) (6 points) Does the heat engine violate the irst law o thermodynamics? You must show a calculation to explain your answer. irst law: Eint Q W or cycle: 0 Q W Q = Q H + Q L = 30J - 20J = 10J Q=W=10J First law OK s b) (4 points) What is the eiciency (in %) o this engine? Wout heat engine eiciency: QH Wout 10J Q 30J 3 H The eiciency is 33% answer not in % - 1 used T H and T L instead o Q H and Q L in part a)

9 PES 2130 Exam 1/page 9 Question 7 (10 points) Two 250g potatoes at 295K are thrown into a 2.00 L pot o water at 375K. The inal equilibrium temperature o the potato-water system is 363 K. Assume that this system is isolated rom the surroundings. a) () Will the change in entropy o this system increase or decrease? For ull points you need to explain your reasoning without doing a calculation. (s) The entropy o the system will increase (s) - The second law o thermodynamics can be stated in terms o entropy: No process is possible in which the total entropy o an isolated system decreases. - Because heat does not low spontaneously rom cold to hot. b) (8 points) Calculate the change in entropy o this system using speciic heat o potato, c p = 3430 J kg -1 K -1, and speciic heat o water, c w = 4187 J kg -1 K -1. You will also need dx ln( x) x. S S S S total p w i T T Stotal mpc p ln mwcw ln T ip Tiw Stotal (0.5)(3430)ln (2)(4187) ln S J / K total total dq T dq mcdt dt T S mc mcln i T Ti S J / K 83.4 J / K - 1 sig igs - 1 units - 1 wrong numerical result else correct

10 PES 2130 Exam 1/page 10 Question 8 (10 points) a) () How did we deine temperature o a substance? Temperature is a measure o the average (translational) kinetic energy o the molecules, characterized by the Maxwell-Boltzmann distribution. give ull points i "characterized by the Maxwell-Boltzmann distribution" part is missing. b) (5 points) Calculate the translational root-mean-square (rms) velocity o water vapor molecules at room temperature, i.e. at T = 293 K. The molar mass o a water vapor molecule is 18.0 g/mol. v rms 3RT M M = molar mass (needs to be in kg/mol) M = 18.0 x 10-3 kg/mol 3RT 3(8.31)(293) vrms m / s 637 m / s 3 M wrong answer i M not converted into kg/mol 3RT 3(8.31)(293) vrms m / s 20.1 m / s M sig igs - 1 units c) (3 points) I a water molecule at room temperature is traveling on average at 587 m/s, briely explain why a puddle o water will eventually evaporate at room temperature. Temperature is a measure o the average kinetic energy o particles, characterized by a Maxwell-Boltzmann distribution. At room temperature, a large percentage o its particles have a temperature close to 25 C, but some have a temperature arther away. When some o those particles that happen to have a temperature o 100 C are at the surace o the water, they have enough energy to break away and evaporate. This cools the rest o the puddle down, but the puddle is constantly in interaction with other substances around it (ground, air) which put that heat back into the puddle. So, the puddle maintains a constant temperature (roughly) and shrinks smaller and smaller as particles near the surace get enough kinetic energy to evaporate.

11 PES 2130 Exam 1/page 11 Question 9 (10 points) The igure below shows several isotherms (lines o equal temperature) o an ideal gas on a p-v diagram. Starting at point a clearly indicate (draw path with label) in the igure the ollowing thermodynamics processes: a) () constant pressure volume expansion b) () constant volume change in pressure c) () constant temperature volume expansion d) () adiabatic volume expansion a) b) c) d) e) () In which process o these our processes is the work done by the gas zero? Is it process a), b), c) or d) and why? Answer: process b) area below p-v diagram is zero: W = 0

12 PES 2130 Exam 1/page 12 Question 10 (10 points) (Do this problem last.) An ideal gas is expanded adiabatically rom state 1 (p 1, V 1, T 1 ) to state 2 (p 2, V 2, T 2 ). Staring with an ininitesimal change in the system pdv nr dt (Equation 1) proo by showing a detailed derivation that 1/ 1/ T V T V, (Equation 2) where α is a constant. [Hint: Which state variable (p,v, or T) in Equation 1 does not appear in Equation 2? dx You will need ln( x) x.] no pressure p in Eq 2, hence use ideal gas law to sub in p: pv = nrt p = nrt/v nrt sub into Eq. 1: nr dt dv V dt dv rearrange and simpliy: T V integrate both sides rom state 1 to state 2: T2 V2 dt dv T V T1 V1 solve evaluate integrals, take exponential and rearrange: T2 V2 ln ln T V T V V T V V / T2 V1 T 1 V 2 T T T V V V 1/ TV 1/ 1/ / END OF EXAM

13 PES 2130 Exam 1/page 13 EXTRA SPACE BELOW FOR CALCULATIONS

14 PES 2130 Exam 1/page 14

UNIVERSITY OF SOUTHAMPTON

UNIVERSITY OF SOUTHAMPTON PHYS1013W1 SEMESTER 2 EXAMINATION 2014-2015 ENERGY AND MATTER Duration: 120 MINS (2 hours) This paper contains 8 questions. Answers to Section A and Section B must be in separate