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Semester I Final Exam Review

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1 Semester I Final Exam Review
Physics Semester I Final Exam Review

2 Write the following information in your notebook
It is to be turned in the day of final exams as a major grade We are to cover the Conservation of Energy, Conservation of Momentum, and Thermal Energy in this block of instruction. This is most of your review for the final exam

3 Science Physics Unit 01: Laboratory Management
1. A student carefully opens the gas valve and attempts to light a Bunsen burner using a striker. After several attempts, the student is unable to light the burner. The student believes the striker is broken. Which is the best course of action for the student? A. Disconnect the Bunsen burner from the gas valve. B. Open the gas valve further and continue trying to light the burner. C. Attempt to light the burner using another lit Bunsen burner. D. Close the gas valve and look for another striker.

4 2. You are using a concentrated solution of copper (II) sulfate to generate a current for electroplating. When finished, you need to dispose of the solution. Which of the following is the best method of disposal? A. Pour the solution down the drain and run the water for five minutes. B. Pour the solution into a sealed bottle and place in the storeroom. C. Discard the solution in accordance with the school's waste disposal plan. D. Take the solution outside and pour it out on a grassy area.

5 3. In the drawing, approximately how many
volts does the voltmeter indicate? A V B. 0.7 V C. 6.7 V D. 67 V

6 4. Which of the following is the BEST sequence for the Laboratory Report format?
A. Title, Purpose/Problem, Introduction, Materials, Procedures, Results/Analysis, Discussion, Conclusion B. Title, Introduction, Purpose/Problem, Procedures, Materials, Results/Analysis, Discussion, Conclusion C. Title, Purpose/Problem, Introduction, Materials, Procedures, Results/Analysis, Conclusion, Discussion D. Title, Purpose/Problem, Discussion, Introduction, Materials, Procedures, Results/Analysis, Conclusion

7 5. A 100 kg gymnast comes to a stop after
Science Physics Unit 07: Conservation of Energy and Energy Transformations 5. A 100 kg gymnast comes to a stop after tumbling. Her feet do 5,000 J of work to stop her. Which of the following was the girl's velocity when she began to stop? A. 5 m/sec B. 10 m/sec C. 50 m/sec D. 100 m/sec KE = 1/2 mv2 V = (2KE/m)1/2 = (10000/100) 1/2

8 6. Which of the following does not contribute to
the potential energy of an object? A. height of the object above the Earth's surface B. gravitational constant or g of the Earth C. velocity of the object D. mass of the object GPE = mgh

9 7. A 1000 kg truck moving at 10 m/s runs into a concrete wall
7. A 1000 kg truck moving at 10 m/s runs into a concrete wall. It takes 1.0 sec for the truck to completely stop. What is the magnitude of force exerted on the truck during the collision? Given m, vi, vf, and t W = F * d = KE F = W/d F = (KE)/d KE = 1/2 mv2 = (.5)(1000 kg)(10 m/s)2 = 50,000 J d = vt = (10 m/s)(1.0 s) = 10 m F = (KE)/d = (50,000 J)/(10 m) = 5,000 N

10 8. A football player pushes a 300 N coach riding a blocking sled up a 6.0 m long ramp using a force of 150 N. The coach was raised a vertical distance of 3.0 m. The actual work done by the worker was — A. 300 N B. 600 N C. 900 N D N W = Fd Win = Wout

11 9. Which of the following is the best example of
kinetic energy being transformed into potential energy? A. Winding a grandfather clock B. Bending a paper clip C. Coasting down hill on a bicycle D. Starting an automobile engine

12 10. Which of the following is not associated with
mechanical energy? A. A ball rolling down a ramp B. A diver on a spring board waiting to dive C. A loaded spring D. A heat lamp

13 11. A student designs a machine that he claims
has at least a 95% efficiency. His machine based on a simple lever requires 100 N of force applied through a distance of 3 m to raise a 50 N object 5.8 m. Can you support his claim? Show your calculations eff = Wout/Win x 100 = [(50N)(5.8m)]/[(100N)(3m)] x 100 = 96.7 % YES

14 12. If a power lifter raises a 1000 N weight a
distance of 2.0 meters in 0.5 seconds, what is his power output? P = work/time W = Fd P = Fd/t = (1000N)(2m)/0.5s = 4000Watts

15 13. An ice skater glides across a skating rink, slowly coming to a stop. The total work done by friction to stop this skater is known. Which of the following information is needed to calculate the initial velocity of the skater? A. The mass of the skater B. The acceleration of the skater C. The force that starts the skater sliding D. The amount of time it takes the skater to stop W = KE = ½ mv v = (2KE/m)1/2

16 14. For the lever apparatus shown, the weight raises 2 cm for each 20 cm that the end of the bar raises. If the weight is 100 N, how far would the end move to do 40 J of work on the ball? A. 1 m B. 2 m C. 3 m D. 4 m W = Fd d =W/F = 40J/ 100N = .4m Class II lever mechanical advantage = dout/din = 20cm/2cm .4m x 20cm/2cm = 4m dout din

17 15. The speed for an object to escape the Earth's gravity is called the escape velocity. This speed is calculated from energy considerations. An object must have enough kinetic energy to overcome the gravitational potential energy. For a 20 kg satellite in a circular orbit around the Earth at a distance of 7x105 m from the center, we find the orbital speed to be 6.75x103 m/sec. From that location calculate the gravitational PE, the kinetic energy and the escape velocity. m1 = 20kg m2 = 6x1024 d = 7x105 MEarth = 6x1024 kg G = 6.67x10-11 GPE = -G(m1)(m2)/d = -[(6.67x10-11) (6x1024)(20)] /(7x105)= 1.14x1010 J Orbital KE = (1/2mv2) = (.5)(20)(6.75x103)2 =4.56x108 J v = [2(GPE)/m] 1/2 = [2(1.14x1010)/20] 1/2 =3.38x104 m/sec

18 Science Physics Unit 08: Momentum and Energy in Collisions
Momentum is defined as p = mv where p is the symbol for momentum, m is the symbol for mass, and v is the symbol for velocity. Momentum is a vector quantity and points in the direction of the direction of the velocity. Momentum has units of kg m/sec. A large mass moving at a high velocity has a large momentum. Impulse is defined as J = Fav Δt where J is the impulse, Fav is the average force over an event (collision) over a time period Δt. The time interval is normally short but that is not a real restriction. Impulse uses the same units as momentum and, in normal usage, an impulse is given to an object producing a change in momentum. Thus J = Δp for an object involved in a collision type event.

19 Conservation of Momentum is a law regarding the total momentum of a system when external forces can be neglected. The most common use of this law is in describing features of collisions. The simplest of these events is a head on collision between two objects. In that situation the total momentum of the colliding objects before the collision is equal to the total momentum after the collision. Totally Elastic Collision is a collision where kinetic energy is conserved. This is an ideal situation because some kinetic energy is almost always lost in a collision. Totally Inelastic Collision is a collision where the two (or more) colliding objects stick together after the collision. Coefficient of Restitution (e) is a term sometimes used to indicate the relative elasticity of collisions. It is defined as the ratio of the departing relative velocity of colliding object to the approaching relative velocity. The values range from 1 for a perfectly elastic collision to 0 for a perfectly inelastic collision. Kinetic energy will be lost in this type collision.

20 16. A 2 kg cart rolls across the floor and runs into the wall and sticks.. Before the collision the cart has a velocity of 4 m/sec in the x direction. The collision took place in 0.1 sec. The bold quantities below are vectors and require a direction as part of the answer. 1. What was the cart’s momentum before the collision? m = 2kg, v = 4m/s, t = .1s P=mv=2kg x 4m/s = 8 kg m/sec in the x direction 2. What was the cart’s KE before the collision? KE= 1/2mv2 = ½ (2kg x 42) = 16J 3. What was the cart’s momentum after the collision? 4. What was the change in the cart’s momentum? 8 kg m/sec in the negative x direction

21 17. What was the impulse delivered to the car by the wall
17. What was the impulse delivered to the car by the wall? ? m = 2kg, v = 4m/s, t = .1s J = Ft = mat = m x v/t x t = 2kg x 4m/s/.1s x .1s = kg m/sec in the negative x direction 18. What was the average force exerted on the car by the wall? F = J/t = 8kgm/s / .1s = 80 N in negative x direction 19. What was the average force exerted on the wall by the car? 80 N in the positive x direction 20. Describe the collision as partially elastic, totally elastic, totally inelastic. totally inelastic

22 KEi = ½ m1v1i2 = ½ (1)(1) = .5 J KEf = ½ (m1 + m2)Vf2 =
21. A 1 kg mass moving at 1 m/sec has a totally inelastic collision with a 0.7 kg mass. What is the speed of the resulting combined mass after the collision? What is the ratio of final kinetic energy to the initial total kinetic energy for the collision? KEi = ½ m1v1i2 = ½ (1)(1) = .5 J KEf = ½ (m1 + m2)Vf2 = ½ (1.7)(.59)2 = .3J Ratio = .3/.5 m1 = 1 kg v1i = 1 m/sec m2 = 0.7 kg m1v1i = (m1 + m2) vf Vf = (m1v1i) / (m1 + m2) = 1 / 1.7 = .59m/s Vf = m1V1i/m1 + m2 = .59m/s

23 22. An excellent golfer can provide an impulse of nearly 2 kg m/sec impulse to a kg ball. What momentum and speed result from such a hit? J = 2 kg m/sec m = kg The impulse is equal to the momentum thus p=2 kg m/sec p=mv v=p/m =2/0.045 =44.4m/s P = mv = .045kg x 44.4m/s = 2 kg m/s

24 m = 1000 kg vi = 20 m/sec Δt = 10 sec FΔt = Δp and F = Δp/ Δt Δp = mv
23. A car with a mass of 1000 kg moves at 20 m/sec. What braking force is needed to bring the car to a halt in 10 sec? m = 1000 kg vi = 20 m/sec Δt = 10 sec FΔt = Δp and F = Δp/ Δt Δp = mv F = mv/ Δt F = (1000kg)(20m/s)/10s = 2000N

25 24. Air bags help to minimize injuries by providing a large area for the force of impact to be exerted. In terms of the impulse momentum theory, what is another advantage of an air bag? A. Decreases the momentum of the passenger B. Increases the amount of time the force is applied C. Decreases the net force applied by the air bag D. Increases the impulsive force applied to the passenger

26 25. A billiard ball moving at 0
25. A billiard ball moving at 0.5 m/s strikes another identical billiard ball, which is at rest, in an elastic head-on collision on a level table. Which of the following statements best describes the billiard balls after the collision? A. The moving ball rebounds directly back at 0.5 m/s and the other ball remains at rest. B. The balls both move off together in the same direction at 0.25 m/s. C. The moving ball continues to move at a 0.2 m/s in its original direction and the other ball moves at 0.4 m/s in the same direction. D. The moving ball stops and the other ball moves off at 0.5 m/s in the original direction of the moving ball.

27 26. A 7 kg bowling ball traveling at 2
26. A 7 kg bowling ball traveling at 2.0 m/s collides with a stationary 0.5-kg beach ball in an elastic collision. The bowling ball leaves the collision with a velocity of 1.5 m/s traveling in the same direction as the beach ball. Calculate the speed of the beach ball. Momentum before the collision equals momentum after the collision. mb = momentum of bowling me = momentum of beach ball pi = pf mbvbi + mevei = mbvbf + mevef (7kg)(2 m/s) + (0.5 kg)(0 m/s) = (7kg)(1.5 m/s) + (0.5 kg)(vef) vef = (3.5)/(0.5) = 7 m/s

28 27. Which of the following has the least momentum?
A. .01 kg mass moving at 200 m/s B. 1 kg mass moving at 35 m/s C. 100 kg mass moving at 3 m/s D kg mass moving at 0.1 m/s P = mv

29 28. A hammer strikes a nail with a 10 N force for
0.01 seconds. Calculate the impulse of the hammer. J = Ft = (10 N)(0.01 s) = 0.1 kg m/s

30 29. A force of of 500 N is exerted on a baseball
by the bat for .001 s. What is the change in momentum of the baseball? A. 0.5 kg m/s B. 5 kg m/s C. 50 kg m/s D. 400 kg m/s J = Ft

31 vA = velocity of student one vB = velocity of student two
30. Two students on roller skates stand face to face and push each other away. One student(A) has a mass of 90 kg and the other (B) has a mass of 60 kg.What must the resulting speed of student (A) be relative to student (B) in order for momentum to be conserved? (Assume that the momentum before the push is zero.) In otherwords, looking for a ratio of student B to student A vA = velocity of student one vB = velocity of student two mA = mass of student one = 90kg mB = mass of student two = 60kg mAvA = mBvB or vA/vB = mB/mA = 60/90 = .667

32 31. A student using an air track with two carts records the result of an inelastic collision. One cart has a mass of 0.2 kg and the other a mass of 0.4 kg. They moved off together at 0.2 m/s after the collision. The 0.4 kg cart was initially at rest. m1 = .2kg, m2 = .4kg, Vf = .2m/s -What was the final momentum of the carts? The final momentum Pf was (m1 + m2)(Vf) = ( )kg (0.2 m/s) = 0.12 kg m/s. -What was the initial velocity of the 0.2 kg cart? m1v1 + m2v2 = (m1 + m2)(Vf) so (.2kg) v1 = (0.12kg m/s) v1 = (0.12kg m/s) / .2kg = 0.6 m/s. -If the collision lasted 0.02 s, what force did each cart exert on the other? The force between the carts is obtained by dividing the change in momentum by the time.(0.12 kg m/s)/ .02 s = 6 N.

33 32. A baseball player hits a 2. 5-kg ball with a force of 20 N
32. A baseball player hits a 2.5-kg ball with a force of 20 N. The duration of the force was .05 s. m = 2.5kg, F = 20N, t = .05s -Calculate the impulse delivered to the ball. J =Ft = (20N)(.05s) = 1 kg m/s -What was the change in momentum of the ball? 1 kg m/s -What was the velocity of the ball as it left the bat? P =J = mv, v = J/m = (1 kg m/s)/ (2.5 kg) = .4 m/s

34 33. -What was its initial potential energy?
In an auditorium with a 8.5-m ceiling, a large pendulum was constructed using a 7.0-kg bowling ball suspended on an 8.0-m steel wire. The bowling ball was pulled to the side of its lowest hanging point. It was raised a distance 2.0 m vertically above its lowest point and released. m = 7kg, h = 2m 33. -What was its initial potential energy? The initial potential energy is mgh = (7.0 kg)(9.8 m/s2)(2.0 m) = 137 J. 34. After dropping 1.0 m vertically, what is the bowling ball's velocity? The ball's velocity after falling 1.0 m is calculated from 1/2 mv2 = 137 J. This gives 4.4 m/s. 35. What is the bowling ball's maximum possible velocity? KE = PE = 1/2 mv2 , v = (2PE/m)1/2 = (2(137)/7) 1/2 = 6.3 m/s. 36. At what height above its lowest point is the bowling ball's velocity one-half of the maximum velocity? The velocity would be 3.15 m/s at 1.5 m above the lowest point. Substitute into mgh = 1/2 mv h = 1/2 mv2 / mg = ½ v2/g h = ½(3.15m/s)2/9.8m/s/s = .50m 2.0m - .5m = 1.5m above its lowest point

35 A 5,000-kg railroad car moving at 2 m/s collides and connects to another identical car initially at rest. 37. What is the momentum of the moving train car before the collision? p = mv = (5000 kg)(2 m/s) = 10,000 J 38. What is the final velocity of the two connected train cars? P = mv mf = m1 + m2 vf = p/(m1 + m2) = 10,000 J / 10,000 kg = 1 m/s

36 At scout camp, a 1. 00-kg arrow is shot at 50m/s
At scout camp, a 1.00-kg arrow is shot at 50m/s. It hits and is embedded into a 2.00-kg block of wood that slides to a stop. 39. Calculate the kinetic energy of the arrow. KE= 1/2 mv2 = .5(1.00 kg)(50 m/s)2 =1250 J 40. Calculate the momentum of the arrow before it hits the block of wood. p= mv = (1.00 kg)(50 m/s) = 50 J 41. Calculate the initial velocity of the arrow and wood after the inelastic collision. v = p/(ma + mw) = 50/3kg = 16.7 m/s 42. Calculate the energy of the block and arrow after the collision. KE= 1/2 mv2 = .5(3.00 )(16.7 m/s)2 = J

37 Science Physics Unit 09: Thermodynamics
Equations and Constants 1. KEave = (½ m v2)ave = 3/2 kT: k = 1.38 x 10-23J/K 2. 0C = ( 0F -32) (5/9): 0F = (0C)9/ : 0C = K Q = m c ΔT 4. Work = PΔV 5. 1 cal = J = 3.97 x 10-3 BTU 6. Cwater = 1 cal/gram oC Latent Heat Fusion = 80 cal/g0C Latent Heat Vaporization = 540 cal/g0C

38 1. Temperature: how hot or cold
 Temperature is related to average KE of molecules: KEave = (1/2 mv2)average = 3/2 kT : k = 1.38 x J/K: T is Kelvin temperature. Concept –higher temperature (macroscopic) is faster molecules (microscopic) Three popular temperature scales:  Fahrenheit scale  Celsius scale –0 freezing ice, 100 boiling water, value is Kelvin  Kelvin scale –0 K is absolute coldest. Absolute temperature scale

39 2. Thermal Energy: energy of temperature  U –Internal energy of a system (like an engine) is the total of KE and PE of molecules –formally more than the thermal energy but often the same. Official term in engines and gasses.  Q –Heat –transfer of energy due to a temperature difference – officially only called heat in transit Heat "flows" from high temperature to low temperature until thermal equilibrium (same temperature) is reached.  Conduction –requires matter but mass does not move, similar to electricity, electrons carry energy  Convection –requires matter- mass moves carrying energy  Radiation –does not require matter –electromagnetic waves carry energy –black absorbs and emits better. Mirror –white reflect. Hot emits more energy

40 Heat units are calorie, Calorie, Joule, BTU (British Thermal Unit)
Heat units are calorie, Calorie, Joule, BTU (British Thermal Unit).  1 calorie = heat to raise temperature of 1 gram of water by 1o Celsius  1 Calorie = 1000 cal (diet calorie)  1 cal = J = 3.97 x 10-3 BTU Heat as a form of energy was established by James Joule in an experiment where paddles converted mechanical motion into heating water.

41 3. Specific Heat Capacity: how efficient at absorbing, storing, or giving up internal energy  C –defined by the equation Q = m c ΔT: cal/gram oC is unit –or other appropriate units m = mass, c = specific heat, T = temperature  Materials with high heat capacity require lots of energy to raise the temperature much  Water has a relatively high heat capacity of 1 cal/gramoC  Is one of the primary variables in heat transfer and calorimetry problems

42 4. Expansion: most substances expand as temperature rises: solids least, gasses most  Used in thermometers, thermostats  ΔL = αL0 ΔT –change in length proportional to length L0 and ΔT: α is constant depending upon type of material. Bi-metallic strip had metals of different α : linear coefficient of expansion  ΔV = 3αV0 ΔT –change in volume is proportional to original volume V0 and ΔT: 3α is volume expansion coefficient – for gasses and liquids β substitutes for 3α in the equation  Water's behavior is unusual since it has maximum density (minimum volume) at 40C. Thus as it gets near freezing temperature it floats and freezes –from the top down, most things freeze from the bottom up. Water also expands upon freezing –forming hollow ringed structures

43 5. Phase Change: Energy is required to change a solid to a liquid and from a liquid to a vapor  Phases are solid, liquid, and vapor: for water it takes 80 calories per gram to melt or freeze water and 540 calories to vaporize water. Since this energy is hidden (doesn’t change the temperature) it is called latent heat of –fusion, vaporization  The heat of fusion and vaporization are basically the energy necessary to break the solid structure bonds for fusion and the bonds between molecules for vaporization. This is analogous to breaking through the surface tension energy for evaporation. In a related thought, boiling involves forming bubbles in the depth of the liquid. This temperature increases with increased air pressure. Water boils at lower temperature in the mountains

44 43. Kinetic energy is a microscopic concept. The
equivalent macroscopic concept is — A. mass B. entropy C. temperature D. heat

45 44. The Second Law of Thermodynamics has a number of equivalent definitions, all of which give an indication of which way nature flows (in a time fashion). Which of these statements is not generally derivable from the concept that a particular engine is the most efficient engine possible and that it is reversible? A. Heat will not flow spontaneously from a cold reservoir to hot reservoir. B. An engine or combination of engines alone cannot make heat flow from a cold reservoir to a hot reservoir. C. Heat will flow spontaneously from a cold reservoir to hot reservoir. D. It is not possible to convert 100% of heat into work.

46 Practice: Thermochemistry
Convert 500C to Kelvin temperature. add 273 and get 323 Thus 500C = 323 Kelvin Helium gas atoms have a mass of 6.8 x kg. What is the speed of a He atom in a gas at 00C? (1/2 mv2)average = 3/2 kT : .5 (6.8 x 10-27)(v2) = 1.5(1.38 x 10-23)(273) vave = 1.3 km/sec: a heavier atom would be slower, and at a higher temperature faster. 47. Sand has a specific heat of 0.5cal/gram0C. How many calories does it take to raise the temperature of 100 g of sand from 600C to 700 C? Q = m c ΔT Q = 100(.5)(10) = 500 calories.

47 48. How much would this amount of heat energy (example 3) raise the temperature of 100 grams of water? Q = m c ΔT : ΔT = Q / (m c) = 500/100 = 5 0C 49. How much energy does is required to take 10 grams of ice at 00C and produce 10 grams of steam at 1000C. For 1 gram it takes 80 calories to melt, 100 calories to heat the water to 1000C and 540 calories to change it into steam. Thus, for 1 gram it takes 720 calories. For 10 grams it would take 7,200 calories or 7.2 Calories (with a capital C).

48 50. Heat flows from a hotter source to a cooler region by which methods? Mark all that apply.
a. convection b. radiation c. conduction d. expiration e. reduction

49 51. Water has a higher specific heat capacity than sand
51. Water has a higher specific heat capacity than sand. Which will cool down faster in the evening and warm up faster in the morning Sun? Sand heats and cools faster.  How does this influence the direction of the winds on the beach in the day and night? Hot air rises, so in the morning as the sand temperature increases the beach air rises pulling in air from the ocean. So the wind blows in from the ocean in the day and out at night

50 52. Water boils at a lower temperature in the mountains than at sea level. The primary reason is that a. The temperature in the mountains is cooler b. The air pressure is less in the mountains c. The air is cleaner with less pollution d. The water is purer in the mountains than from the faucet

51 53. Compare the temperature of a glass of ice water (stirred completely) nearly full of ice compared to a glass of ice water (stirred completely) half full of ice. a. The nearly full of ice glass is colder b. The half full of ice glass is colder c. They are at the same temperature Explain: The ice water mixture will be at the melting-freezing temperature of water.

52 54. The temperature of helium gas is directly related to the kinetic energy of the gas. If a quantity of helium in a closed container at 100 0C has the kinetic energy of the helium atoms doubled, what is the resulting temperature of the gas? a C b. 746 Kelvin = 2( ) c C d. 540 Kelvin

53 55. Calculate the energy to convert 50g of ice at 00C to water at 500C.
M = 50 grams ΔT = 500C Q = (Latent Heat of fusion )( mass) + m c ΔT = 80(50) + 50 (1) (50) = 6500 calories = 6.5 kcal

54 56. A bag of Peanut M&M’s has 220 Calories
56. A bag of Peanut M&M’s has 220 Calories. If this was used to heat the body of a 50 kg girl (assume all water), how many degrees Celsius would it raise her body temperature? (These are kilocalories.) Q = 220 Calories (big calories) Q = m c ΔT thus ΔT = Q/mc = 220,000cal /(50,000 g)(1) ΔT = 4.4 0C

55 m2 = 75 g T2 = 40 0C m1T1 + m2T2 = (m1 + m2 ) Tfinal
57. What is the final temperature when 100g of water at 250C is mixed with 75g of water at 400C ? Since no latent heats, we can consider this a straight conservation of energy. The number of calories add to a total. Q1 + Q2 = Qtotal Q = m c T and c is 1 for water m1 = 100 g T1 = 25 0C m2 = 75 g T2 = 40 0C m1T1 + m2T2 = (m1 + m2 ) Tfinal (100) (20) + (75)( 40) = 175 (Tfinal) Tfinal = C

56 58. Steel expands 1 part in 100,000 for each 10C increase in temperature (α = 1 x 10-5 /0C). If the 1.5-km main span of a steel suspension bridge has no expansion joints, how much longer will it be for a temperature increase of 200C? α = 1 x 10-5 /0C ΔL = αL0 ΔT ΔT = 200C ΔL = αL0 ΔT = (1 x 10-5) (1500m) 20 = 0.3 m

57 59. Calculate the energy to convert 50g of ice at 00C to water at 500C
m = 50 g heat = melt + raise temp Q = [80 (m) + mc ΔT] = m( )(1) c =1 50[ ] = 6500 calories

58 60. In one cycle, an engine burning a mixture of air and alcohol absorbs 525 J and expels 415 J. What is the engine’s efficiency? Qhot = 525 J Qcold = 415 J Although not asked, we need the Work = = 110 J efficiency = Work/Qhot = 110/525 = .21

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