1. Ch 11

2. Mass density ρ is the mass m divided by the volume v. ρ = m/v kg/m 3

3. The pressure P exerted by a fluid is the magnitude F of the force acting perpendicular to a surface divided by the area A over which the force acts. P = F/A

4. The unit is the N/m 2, a pascal (Pa). 10 5 Pa is one bar of pressure. Pressure is not a vector quantity. F refers only to the magnitude of the force. This force is always perpendicular to the surface.

5. If P 1 is the pressure at the top of a column of fluid, P 2 is the pressure at the bottom of a fluid, and h is the height of the column, then: P 2 = P 1 + ρgh

6. Ex. 5 - Blood in the arteries is flowing, but the effects of this flow can be ignored an the blood can be treated as a static fluid. Estimate the amount by which the blood pressure P 2 in the anterior tibial artery at the foot exceeds the blood pressure P 1 in the aorta at the heart when the body is (a) reclining horizontally and (b) standing (h = 1.35 m).

7. Pascal’s principle - Any change in the pressure applied to a completely enclosed fluid is transmitted undiminished to all parts of the fluid and the enclosing walls.

8. As long as the level at each end is at the same height, ρgh is zero. P 2 = P 1 + ρgh becomes P 2 = P 1.

9. If P 2 = P 1, then F 2 / A 2 = F 1 / A 1, and F 2 = F 1 (A 2 /A 1 ). If A 2 is larger than A 1, a large force F 2 can be produced with a small F 1. This is used in a hydraulic car lift.

10. The buoyant force is an upward force exerted by all fluids on objects submerged in them. This buoyant force is equal to the weight of the displaced fluid. This is Archimedes’ principle: Any fluid applies a buoyant force to an object that is partially or completely immersed in it; the magnitude of the buoyant force equals the weight of the fluid that the object displaces: F B = W fluid

11. Ex. 8 - A solid, square, pinewood raft measures 4.0 m on a side and is 0.30 m thick. (a) Determine whether the raft floats in water, and (b) if so, how much of the raft is beneath the surface. r pine is 550 kg/m 3.

12. This is the equation of continuity. ρ 1 A 1 v 1 = ρ 2 A 2 v 2 ρ = fluid density (kg/m 3 ) A = cross-sectional area of tube (m 2 ) v = fluid speed (m/s) The unit of mass flow rate is kg/s.

13. If we are dealing with an incompressible fluid, ρ 1 = ρ 2, and the equation of continuity reduces to A 1 v 1 = A 2 v 2. Av is the volume flow rate Q.

14. Ex. 11 - A garden hose has an unobstructed opening with a cross-sectional area of 2.85 x 10 -4 m 2, from which water fills a bucket in 30.0 s. The volume of the bucket is 8.00 x 10 -3 m 3. Find the speed of the water that leaves the hose through (a) the unobstructed opening and (b) an obstructed opening that has only half as much area.

15. Ex. 12 - In the carotid artery, blood flows three time faster through a partially blocked region than it does through an unobstructed region. Determine the ratio of the effective radii of the artery at the two places.

16. Bernoulli’s Principle - a moving fluid exerts less pressure than a stationary fluid. Bernoulli’s equation also includes elevation, the pressure is greater at a lower level than at higher elevation.

17. Bernoulli’s Equation: P 1 + ½ρv 1 2 + ρgy 1 = P 2 + ½ρv 2 2 + ρgy 2 y is the elevation. This applies to the steady, irrotational flow of a nonviscous, incompressible fluid.

18. P + ½ρv 2 + ρgy has a constant value at all points.

19. Ex. 14 - Because of an aneurysm, the cross-sectional area A 1 of the aorta increases to a value A 2 = 1.7A 1. The speed of the blood (ρ = 1060 kg/m 3 ) through a normal portion of the aorta is v 1 = 0.40 m/s. Assuming the aorta is horizontal, determine the amount by which the pressure P 2 in the enlarged region exceeds the pressure P 1 in the normal region.

20. Ex. 15 - A tank is open to the atmosphere at the top. A pipe allowing the fluid to exit is a height h below the top of the fluid. Find an expression for the speed of the liquid leaving the pipe. (This is called efflux speed).

21. Ch 12

22. Heat is energy that flows from a higher- temperature object to a lower-temperature object because of the difference in temperature.

23. A substance does not contain heat; the term heat refers only to the transfer of energy from one substance to another.

24. The heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount ∆T is: Q = cm∆T where c is the specific heat capacity of the substance.

25. 1 kcal = 4186 joules or 1 cal = 4.186 joules. This conversion factor is the mechanical equivalent of heat.

26. A calorimeter is an insulated container which can be used to measure heat loss or gain between materials contained within.

27. Ex.11 - A calorimeter cup is made from 0.15 kg of aluminum and contains 0.20 kg of water. Initially, the water and the cup have a common temp. of 18.0 °C. An unknown material (m = 0.040 kg) is heated to a temperature of 97.0 °C and then added to the water. The temp. of the water, cup, and unknown is 22.0 °C after thermal equilibrium. Find the specific heat capacity of the unknown material.

29. Ex.12 - Suppose you are cooking spaghetti for dinner, and the instructions say to boil the noodles in water fro ten minutes. To cook spaghetti in an open pot with the least amount of energy, should you turn up the burner to its fullest so the water vigorously boils, or should you turn down the burner so the water barely boils?

30. The latent heat of fusion L f refers to a solid to liquid change. The latent heat of vaporization L v refers to a liquid to gas change.The latent heat of sublimation L s refers to a solid to gas change.

31. Ex.13 - Ice at 0°C is placed in a Styrofoam cup containing 0.32 kg of lemonade at 27°C. The specific heat capacity of lemonade is virtually the same as water. After the ice and lemonade reach equilibrium, some ice still remains. Determine the mass of ice that has melted.

32. Ch 13

33. Heat Transfer Convection is the transfer of heat by the bulk movement of a fluid. Convection can be natural or forced.

34. Conduction is the transfer of heat through a material, any bulk motion of the material playing no role in the transfer. The free electrons in a metal allow heat energy to be transferred very easily.

35. Radiation is the transfer of heat by electromagnetic waves. All objects radiate energy in the form of electromagnetic waves, but the temperature must be over 1000 K for the light to be visible.

36. Ch 14

37. This number of particles in a mole is 6.022 x 10 23 and is known as Avogadro’s number N A. One mole of any substance contains Avogadro’s number of particles.

38. PV = nRT P is the absolute pressure V is the volume n is the number of moles T is the temperature in Kelvins R is the universal gas constant and has a value of 8.31 J/(molK)

39. Boyle’s law states that, if temperature and number of moles are held constant, the pressure and volume vary inversely; P i V i =P f V f.

41. The curve that passes through the initial and final points is called an isotherm. An isotherm at a different temperature would not intersect.

42. Charles law states that if the pressure and number of molecules are held constant; the volume is directly proportional to the temperature, V i /T i = V f /T f.

43. U = 3/2 nRT U is the internal energy of a monatomic ideal gas, n is the number of moles, R is the universal gas constant, T is the Kelvin temperature.

44. Ex 7. - Hydrogen is the most plentiful element in the universe. Why doesn’t our atmosphere contain hydrogen?

45. Ch 15

46. Thermodynamics is the study of the fundamental laws involving heat and work.

47. The zeroth law of thermodynamics - Two systems individually in equilibrium with a third system are in thermal equilibrium with each other.

48. The first law of thermodynamics The internal energy of a system changes from an initial value U i to a final value of U f due to heat Q and work W: ∆U = U f - U i = Q + W.

49. Q is positive when the system gains heat and negative when it loses heat. W is negative when work is done by the system and positive when work is done on the system.

50. Ex 1 - In situation A, a system gains 1500 J of heat from its surroundings, and 2200 J of work is done by the system on the surroundings. In part B, the system also gains 1500 J of heat, but 2200 J of work is done on the system by the surroundings. In each case determine the change in the internal energy of the system.

51. Ex 2 - The temperature of three moles of a monatomic ideal gas is reduced from T i = 540 K to T f = 350 K by two different methods. In the first method 5500 J of heat flows into the gas, while in the second method, 1500 J of heat flows into it. In each case find (a) the change in the internal energy of the gas and (b) the work done by the gas.

52. “An isobaric process is one that occurs at constant pressure. In an isobaric process, W = P∆V = P(V f - V i ). W is still negative for work done by a system (V f >V i ), and W is positive when work is done on a system (V f <V i ).”

53. An isochoric process is one that occurs at a constant volume. If heat is added and the volume stays constant, pressure increases, but no work is done since the walls do not move. The first law shows that the heat only changes the internal energy of the system: ∆U = Q + W = Q.

54. An isothermal process is one that takes place at constant temperature.

55. An adiabatic process is one that occurs without the transfer of heat. Q equals zero, so: ∆U = Q + W = W.

56. For the adiabatic expansion or compression of a monatomic ideal gas: W = 3/2 nR(T i - T f ). When the gas expands, W is negative and T i must be greater than T f. When the gas is compressed, W is positive and T i must be less than T f.

57. The second law of thermodynamics: The heat flow statement: Heat flows spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction.

58. A heat engine is a device that uses heat to perform work. The efficiency e of a heat engine is the ratio of work W done to the input heat Q H. Efficiency is often stated as a percentage.

59. For a heat engine the total of the work W done and the rejected heat Q C must equal the input heat Q H. Q H = W + Q C

60. The ratio of rejected heat Q C to input heat Q H is: Q C / Q H = T C / T H. T C and T H must be expressed in Kelvins.

61. The efficiency of a Carnot engine is equal to 1 - T C /T H.

62. Ex 7 - A heat engine has a hot reservoir at 298.2 K and a cold reservoir at 280.2 K. (a) Find the maximum possible efficiency for such an engine. (b) Determine the minimum input heat Q H that would be needed if a number of these heat engines were to produce an amount of work equal to the 9.3 x 10 19 J of energy that the US consumed in 1994.

63. The efficiency is low because the temperatures are so close to being the same. The efficiency increases as T C approaches absolute zero. Experiments have shown that it is not possible to cool a substance to 0 K; therefore a 100% efficient heat engine is not possible.

64. Refrigerators, air conditioners, and heat pumps use work to force heat to flow from the cold reservoir to the hot reservoir; this is called a refrigeration process. Q H = W + Q C

65. Any irreversible process increases the entropy of the universe. ∆S universe > 0 for an irreversible process. The entropy of the universe continually increases.

66. The second law of thermodynamics stated in terms of entropy - The total entropy of the universe does not change when a reversible process occurs (∆S universe = 0) and increases when an irreversible process occurs (∆S universe > 0).

67. An increase in entropy is associated with an increase in disorder.

68. The third law of thermodynamics - It is not possible to lower the temperature of any system to absolute zero in a finite number of steps.