1 1.4 Perform the following unit conversions: 0.05 ft 1 in. (a) 1L 61in. 1L 1ft (b) 1kJ 650 J 10 J 1Btu kJ Btu (c) 41 Btu/h 0.15 kw 1kW 1h 600 s ft lbf 1Btu ft lbf s (d) g 78 s 1kg 10 g 1lb kg 60 s 1min 50 lb min (e) 1lbf/in. 10 Pa lbf 04 kpa Pa 1kPa in. (f) m 55 h.808 ft 1m 1h 600 s ft 0.54 s (g) km 50 h 10 m 1km.808 ft 1m 1h 600 s ft s 1lbf 1ton (h) 8896 N 1ton N 000 lbf 1
2 1.5 Perform the following unit conversions: 1cm (a) 1 in in. 10 1m cm 10 1L - m L (b) 1kJ ft lbf kJ ft lbf 1kW (c) 100 hp kw 1.41hp lb 1h 1kg kg (d) h 600 s.046 lb s (e) lbf 9.9 in Pa 1lbf/in. 1 N/m 1Pa 10 1bar 5 N/m.07 bar (f) ft 500 min m 1ft 1min 60 s m 1.18 s (g) mile km/h km h 1mile/h h 000 lbf N (h) 1ton 8896 N 1ton 1lbf 1
3 1.6 Which of the following food items weighs approximately one newton? a. a grain of rice b. a small strawberry c. a medium-sized apple d. a large watermelon 1
4 1.7 A person whose mass is 150 lb weights lbf. Determine (a) the local acceleration of gravity, in ft/s, and (b) the person s mass, in lb, and weight, in lbf, if g =.174 ft/s. (a) F grav = mg F grav lbf.174 lb ft/s g = 0.97 ft/s m 150 lb 1lbf (b) Mass value remains the same. So ft 1lbf F grav = mg = (150 lb).174 s.174 lb ft/s = 150 lbf 1
5 1.8 The Phoenix with a mass of 50 kg was a spacecraft used for exploration of Mars. Determine the weight of the Phoenix, in N, (a) on the surface of Mars where the acceleration of gravity is.7 m/s and (b) on Earth where the acceleration of gravity is 9.81 m/s. KNOWN: Phoenix spacecraft has mass of 50 kg. FIND: (a) Weight of Phoenix on Mars, in N, and (b) weight of Phoenix on Earth, in N. SCHEMATIC AND GIVEN DATA: m = 50 kg g Mars =.7 m/s g Earth = 9.81 m/s ENGINEERING MODEL: 1. Acceleration of gravity is constant at the surface of both Mars and Earth. ANALYSIS: Weight is the force of gravity. Applying Newton s second law using the mass of the Phoenix and the local acceleration of gravity (a) On Mars, F = mg m 1 N F (50 kg).7 s 1kg m/s = N (b) On Earth, m 1 N F (50 kg) 9.81 s 1kg m/s = 4.5 N Although the mass of the Phoenix is constant, the weight of the Phoenix is less on Mars than on Earth since the acceleration due to gravity is less on Mars than on Earth. 1
8 1.11 At the grocery store you place a pumpkin with a mass of 1.5 lb on the produce spring scale. The spring in the scale operates such that for each 4.7 lbf applied, the spring elongates one inch. If local acceleration of gravity is. ft/s, what distance, in inches, did the spring elongate? KNOWN: Pumpkin placed on a spring scale causes the spring to elongate. FIND: Distance spring elongated, in inches. SCHEMATIC AND GIVEN DATA: m = 1.5 m x ENGINEERING MODEL: 1. Spring constant is 4.7 lbf/in.. Local acceleration of gravity is. ft/s. ANALYSIS: The force applied to the spring to cause it to elongate can be expressed as the spring constant, k, times the elongation, x. F = kx The applied force is due to the weight of the pumpkin, which can be expressed as the mass (m) of the pumpkin times acceleration of gravity, (g). F = Weight = mg = kx Solving for elongation, x, substituting values for pumpkin mass, acceleration of gravity, and spring constant, and applying the appropriate conversion factor yield ft 1.5 lb. mg s 1lbf x =.66 in. k lbf lb ft in. s
12 1.17 A communications satellite weighs 4400 N on Earth where g = 9.81 m/s. What is the weight of the satellite, in N, as it orbits Earth where the acceleration of gravity is 0.4 m/s? Express each weight in lbf. KNOWN: Weight of communications satellite on Earth. FIND: Determine weight of the satellite, in N, as it orbits Earth where the acceleration of gravity is 0.4 m/s. Express the satellite weight, in lbf, on Earth and in orbit. SCHEMATIC AND GIVEN DATA: W Sat(Earth) = 4400 N g Earth = 9.81 m/s g orbit = 0.4 m/s ENGINEERING MODEL: 1. Gravitational acceleration on Earth is constant at 9.81 m/s.. Gravitational acceleration at orbital altitude is constant at 0.4 m/s. ANALYSIS: Weight of the satellite is the force of gravity and varies with altitude. Mass of the satellite remains constant. Applying Newton s second law to solve for the mass of the satellite yields W = mg m = W/g On Earth, m = W Sat(Earth) /g Earth (4400 N) 1kg m/s m m 1 N 9.81 s = kg Solving for the satellite weight in orbit, W Sat(orbit) = mg orbit m 1 N W Sat(orbit) (448.5 kg) 0.4 s 1kg m/s = N Although the mass of the communications satellite is constant, the weight of the satellite is less at orbital altitude than on Earth since the acceleration due to gravity is less at orbital altitude than on Earth. 1
13 To determine the corresponding weights in lbf, apply the conversion factor, 1 lbf = N. 1lbf W Sat(Earth) (4400 N) N = 989. lbf 1lbf W Sat(orbit) (100.5 N) N =.6 lbf
16 1.1 A -lb sample of an unknown liquid occupies a volume of 6.6 in. For the liquid determine (a) the specific volume, in ft /lb, and (b) the density, in lb/ft. KNOWN: Volume and mass of an unknown liquid sample. FIND: Determine (a) the specific volume, in ft /lb, and (b) the density, in lb/ft. SCHEMATIC AND GIVEN DATA: m = lb V = 6.6 in. ENGINEERING MODEL: 1. The liquid can be treated as continuous. ANALYSIS: (a) The specific volume is volume per unit mass and can be determined from the total volume and the mass of the liquid V 6.6 in. v m lb 1ft 178 in. = ft /lb (b) Density is the reciprocal of specific volume. Thus, 1 1 = 55. lb/ft v ft lb
18 1. The specific volume of 5 kg of water vapor at 1.5 MPa, 440 o C is m /kg. Determine (a) the volume, in m, occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules. KNOWN: Mass, pressure, temperature, and specific volume of water vapor. FIND: Determine (a) the volume, in m, occupied by the water vapor, (b) the amount of water vapor present, in gram moles, and (c) the number of molecules. SCHEMATIC AND GIVEN DATA: m = 5 kg p = 1.5 MPa T = 440 o C v = m /kg ENGINEERING MODEL: 1. The water vapor is a closed system. ANALYSIS: (a) The specific volume is volume per unit mass. Thus, the volume occupied by the water vapor can be determined by multiplying its mass by its specific volume. m V mv (5 kg) = 1.08 m kg (b) Using molecular weight of water from Table A-1 and applying the appropriate relation to convert the water vapor mass to gram moles gives m n M 5 kg 1000 moles kg 1kmol 18.0 kmol = 77.5 moles (c) Using Avogadro s number to determine the number of molecules yields # Molecules Avogadro' s molecules Number # moles (77.5 moles) mole # Molecules = molecules
21 1.7 Three kg of gas in a piston-cylinder assembly undergo a process during which the relationship between pressure and specific volume is pv 0.5 = constant. The process begins with p 1 = 50 kpa and V 1 = 1.5 m and ends with p = 100 kpa. Determine the final specific volume, in m /kg. Plot the process on a graph of pressure versus specific volume. KNOWN: A gas of known mass undergoes a process from a known initial state to a specified final pressure. The pressure-specific volume relationship for the process is given. FIND: Determine the final specific volume and plot the process on a pressure versus specific volume graph. SCHEMATIC AND GIVEN DATA: State 1 State Gas Process 1 pv 0.5 = constant Gas m 1 = kg p 1 = 50 kpa V 1 = 1.5 m m = kg p = 100 kpa ENGINEERING MODEL: 1. The gas is a closed system.. The system undergoes a polytropic process in which pv 0.5 = constant. ANALYSIS: The final specific volume, v, can be determined from the polytropic process equation p 1 v = p v 0.5 Solving for v yields v = p1 1 v p Specific volume at the initial state, v 1, can be determined by dividing the volume at the initial state, V 1, by the mass, m, of the system v 1 = V 1 m 1.5 m kg = 0.5 m /kg 1
22 Pressure (kpa) Substituting values for pressures and specific volume yields v = m 50 kpa 0.5 kg 100 kpa =.15 m /kg The volume of the system increased while pressure decreased during the process. A plot of the process on a pressure versus specific volume graph is as follows: Pressure versus Specific Volume Specific Volume (m^/kg)
23 1.8 A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pv n = constant. The process begins with p 1 = 15 lbf/in., v 1 = 1.5 ft /lb and ends with p = 5 lbf/in., v = 0.5 ft /lb. Determine (a) the volume, in ft, occupied by the gas at states 1 and and (b) the value of n. (c) Sketch Process 1- on pressurevolume coordinates. KNOWN: Gas undergoes a process from a known initial pressure and specific volume to a known final pressure and specific volume. FIND: Determine (a) the volume, in ft, occupied by the gas at states 1 and and (b) the value of n. (c) Sketch Process 1- on pressure-volume coordinates. SCHEMATIC AND GIVEN DATA: State 1 State Gas m = 4 lb p 1 = 15 lbf/in. v 1 = 1.5 ft /lb Gas p = 5 lbf/in. v = 0.5 ft /lb ENGINEERING MODEL: 1. The gas is a closed system.. The relation between pressure and volume is pv n = constant during process 1-. ANALYSIS: (a) The specific volume is volume per unit mass. Thus, the volume occupied by the gas can be determined by multiplying its mass by its specific volume. For state 1 For state V = mv ft V 1 mv 1 (4 lb) 1.5 = 5 ft lb ft V mv (4 lb) 0.5 = ft lb 1
24 Pressure (lbf/in. ) (b) The value of n can be determined by substituting values into the relationship: p 1 (V 1 ) n = constant = p (V ) n Solving for n p1 p n V V 1 p 1 V ln nln p V1 p 15 lbf/in. ln 1 ln p 5 lbf/in. n = 1.8 V ln ft ln V1 5 ft (c) Process 1- is shown on pressure-volume coordinates below: 60 State Process State Volume (ft )
27 1.1 A gas contained within a piston-cylinder assembly undergoes four processes in series: Process 1-: Constant-pressure expansion at 1 bar from V 1 = 0.5 m to V = m Process -: Constant volume to bar Process -4: Constant-pressure compression to 1 m Process 4-1: Compression with pv 1 = constant Sketch the process in series on a p-v diagram labeled with pressure and volume values at each numbered state. p (bar) V (m ) 1
30 1.6 Liquid kerosene flows through a Venturi meter, as shown in Fig. P1.6. The pressure of the kerosene in the pipe supports columns of kerosene that differ in height by 1 cm. Determine the difference in pressure between points a and b, in kpa. Does the pressure increase or decrease as the kerosene flows from point a to point b as the pipe diameter decreases? The atmospheric pressure is 101 kpa, the specific volume of kerosene is m /kg, and the acceleration of gravity is g = 9.81 m/s. KNOWN: Kerosene flows through a Venturi meter. FIND: The pressure difference between points a and b, in kpa and whether pressure increases or decreases as the kerosene flows from point a to point b as the pipe diameter decreases. SCHEMATIC AND GIVEN DATA: p atm = 101 kpa g = 9.81 m/s L = 1 cm Kerosene v = kg/m a b ENGINEERING MODEL: 1. The kerosene is incompressible.. Atmospheric pressure is exerted at the open end of the fluid columns. ANALYSIS: Equation 1.11 applies to both columns of fluid (a and b). Let h b be the height of the fluid above point b. Then h b + L is the height of the fluid above point a. Applying Eq to each column yields p a = p atm + g(h b + L) = p atm + gh b + gl and p b = p atm + gh b Thus, the difference in pressure between point a and point b is p = p b p a = (p atm + gh b ) (p atm + gh b + gl) p = gl 1
31 Density of kerosene is the reciprocal of its specific volume Solving for the difference in pressure yields = 1/v = 1/0.001 m /kg = 80 kg/m kg m 1 N p (1 cm) m s kg m 1 s 1m 100 cm 1kPa N 1000 m = kpa Since points a and b are at the same elevation in the flow, the difference in pressure is indicated by the difference in height between the two columns. The negative sign indicates pressure decreases as the kerosene flows from point a to point b as the pipe diameter decreases.
32 1.7 Figure P1.7 shows a tank within a tank, each containing air. Pressure gage A, which indicates pressure inside tank A, is located inside tank B and reads 5 psig (vacuum). The U-tube manometer connected to tank B contains water with a column length of 10 in. Using data on the diagram, determine the absolute pressure of the air inside tank B and inside tank A, both in psia. The atmospheric pressure surrounding tank B is 14.7 psia. The acceleration of gravity is g =. ft/s. KNOWN: A tank within a tank, each containing air. FIND: Absolute pressure of air in tank B and in tank A, both in psia. SCHEMATIC AND GIVEN DATA: p atm = 14.7 psia Tank B L = 10 in. Tank A Gage A Water ( = 6.4 lb/ft ) g =. ft/s p gage, A = 5 psig (vacuum) ENGINEERING MODEL: 1. The gas is a closed system.. Atmospheric pressure is exerted at the open end of the manometer.. The manometer fluid is water with a density of 6.4 lb/ft. ANALYSIS: (a) Applying Eq p gas,b = p atm + gl where p atm is the local atmospheric pressure to tank B, is the density of the manometer fluid (water), g is the acceleration due to gravity, and L is the column length of the manometer fluid. Substituting values lbf lb ft 1lbf 1ft p gas,b (10 in.) = 15.1 lbf/in. lbm ft in. ft s. 178 in. s 1
33 Since the gage pressure of the air in tank A is a vacuum, Eq applies. p(vacuum) = p atm (absolute) p(absolute) The pressure of the gas in tank B is the local atmospheric pressure to tank A. Solving for p (absolute) and substituting values yield p(absolute) = p atm (absolute) p(vacuum) = 15.1 psia 5 psig = 10.1 psia
35 1.9 Show that a standard atmospheric pressure of 760 mmhg is equivalent to 101. kpa. The density of mercury is 1,590 kg/m and g = 9.81 m/s. KNOWN: Standard atmospheric pressure of 760 mmhg. FIND: Show that 760 mmhg is equivalent to 101. kpa. SCHEMATIC AND GIVEN DATA: Mercury vapor Hg = 1,590 kg/m L = 760 mm Mercury (Hg) ENGINEERING MODEL: 1. Local gravitational acceleration is 9.81 m/s.. Pressure of mercury vapor is much less than that of the atmosphere and can be neglected. ANALYSIS: Equation 1.1 applies. p atm = p vapor + Hg gl = Hg gl Neglecting the pressure of mercury vapor and applying appropriate conversion factors yield p atm kg m 1 N 1, (760 mm) m s kg m 1 s 1m 1000 mm 1kPa N 1000 m = 101. kpa
37 1.41 As shown in Figure P1.41, air is contained in a vertical piston-cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a pressure of 14.7 lbf/in. on top of the 6-in.-diameter piston. The absolute pressure of the air inside the cylinder is 16 lbf/in. The local acceleration of gravity is g =. ft/s. Determine (a) the mass of the piston, in lb, and (b) the gage pressure of the air in the cylinder, in psig. KNOWN: A piston-cylinder assembly contains air such that the piston is in static equilibrium. FIND: (a) The mass of the piston, in lb, and (b) the gage pressure of the air in the cylinder, in psig. SCHEMATIC AND GIVEN DATA: p atm = 14.7 lbf/in. g =. ft/s D piston = 6 in. Air p Air = 16 lbf/in. ENGINEERING MODEL: 1. The air is a closed system.. The piston is in static equilibrium.. Atmospheric pressure is exerted on the top of the piston. 4. Local gravitational acceleration is. ft/s. ANALYSIS: (a) Draw a free body diagram indicating all forces acting on the piston. Taking upward as the positive y-direction, the sum of the forces acting on the piston in the y-direction must equal zero for static equilibrium of the piston. Free Body Diagram F y 0 m pistong p Air A piston p atm A piston m piston g = 0 Solving for the mass of the piston, p atm A piston y mpiston pairapiston patmapiston g p Air A piston 1
38 mpiston pair patm g Apiston The area of the piston is determined from the piston diameter Apiston D (6 in.) = 8. in. 4 4 Substituting values and solving for the mass of the piston, m piston lbf 16 in. lbf in. in. ft. s lb ft. s 1lbf = 6.8 lb (b) Gage pressure of the air is given by Eq p(gage) = p(absolute) p atm (absolute) = 16.0 psia 14.7 psia = 1. psig
39 1.4 Air is contained in a vertical piston-cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a pressure of 101 kpa on top of the 0.5-meter-diameter piston. The gage pressure of the air inside the cylinder is 1. kpa. The local acceleration of gravity is g = 9.81 m/s. Subsequently, a weight is placed on top of the piston causing the piston to fall until reaching a new static equilibrium position. At this position, the gage pressure of the air inside the cylinder is.8 kpa. Determine (a) the mass of the piston, in kg, and (b) the mass of the added weight, in kg. KNOWN: A piston-cylinder assembly contains air such that the piston is in static equilibrium. Upon addition of a weight, the piston falls until reaching a new position of static equilibrium. FIND: (a) The mass of the piston, in kg, and (b) the mass of the added weight, in kg. SCHEMATIC AND GIVEN DATA: Part (a) p atm = 101 kpa g = 9.81 m/s Part (b) p atm = 101 kpa g = 9.81 m/s A piston = 0.5 m A piston = 0.5 m Weight Air p Air = 1. kpa (gage) Air p Air =.8 kpa (gage) ENGINEERING MODEL: 1. The air is a closed system.. The piston is in static equilibrium for both part (a) and part (b).. Atmospheric pressure is exerted on the top of the piston. 4. Local gravitational acceleration is 9.81 m/s. ANALYSIS: (a) Draw a free body diagram indicating all forces acting on the piston. Taking upward as the positive y-direction, the sum of the forces acting on the piston in the y-direction must equal zero for static equilibrium of the piston. 1
40 F y 0 Free Body Diagram m pistong p Air A piston p atm A piston m piston g = 0 Solving for the mass of the piston, p atm A piston y mpiston pairapiston patmapiston g p Air A piston mpiston pair patm g Apiston From Eq. 1.14, the quantity in parenthesis is the gage pressure of the air in the cylinder. Rewriting the equation above mpiston pair(gage) Apiston g Substituting values and solving for the mass of the piston, m piston N kg m kpa0.5 m m s m kPa 1 N s = 61. kg
41 (b) Draw a second free body diagram indicating all forces acting on the piston including the newly added weight expressed as the product of its mass and gravitational acceleration. Taking upward as the positive y-direction, the sum of the forces acting on the piston in the y-direction must equal zero for static equilibrium of the piston. Free Body Diagram F y 0 p Air A piston p atm A piston m piston g m weight g = 0 Solving for the mass of the weight, p atm A piston m pistong m weight g y mweight pairapiston patma piston g mpiston p Air A piston mweight pair patm Apiston g mpiston From Eq. 1.14, the quantity in parenthesis is the gage pressure of the air in the cylinder. Rewriting the equation above mweight pair(gage) Apiston g mpiston Substituting values and solving for the mass of the weight, m weight 1000 kg m 1.8 kpa0.5 m m s 61. kg m 9.81 s N 1kPa 1 N = 81.5 kg
45 1.46 As shown in Figure P1.46, an inclined manometer is used to measure the pressure of the gas within the reservoir. (a) Using data on the figure, determine the gas pressure, in lbf/in. (b) Express the pressure as a gage or a vacuum pressure, as appropriate, in lbf/in. (c) What advantage does an inclined manometer have over the U-tube manometer shown in Figure 1.7? KNOWN: A gas contained in a reservoir with inclined manometer attached. FIND: (a) Pressure of gas within the reservoir, in lbf/in. (b) Pressure expressed as gage or vacuum pressure, as appropriate, in lbf/in. (c) Advantage of inclined manometer over the U- tube manometer. SCHEMATIC AND GIVEN DATA: p atm = 14.7 lbf/in. g =. ft/s Gas a b 15 in. 40 o Oil ( = 54. lb/ft ) ENGINEERING MODEL: 1. The gas is a closed system.. Atmospheric pressure is exerted at the open end of the manometer.. The manometer fluid is oil with a density of 54. lb/ft. ANALYSIS: (a) Applying Eq p gas = p atm + gl where p atm is the local atmospheric pressure, is the density of the manometer fluid (oil), g is the acceleration due to gravity, and L is the vertical difference in liquid levels. Since level a is the same as level b, applying trigonometry to determine the vertical difference in liquid levels between level b and the liquid level at the free surface with the atmosphere yields Substituting values p gas = p atm + gl(sin 40 o ) 1
46 lbf lb ft 1lbf 1ft p gas (15 in.)(sin 40) = 15.0 lbf/in. in. ft s lbm ft. s 178 in. (b) Since the pressure of the gas is greater than atmospheric pressure, gage pressure is given by Eq p(gage) = p(absolute) p atm (absolute) = 15.0 psia 14.7 psia = 0. psig (c) The advantage of the inclined manometer is its easier readability since the surface of the liquid is wider than with a same diameter U-tube manometer. The scale on the inclined manometer is much more precise since more graduations are possible compared with the U-tube manometer.
47 Pressure (kpa) Substituting values for pressures and specific volume yields v = m 50 kpa 0.5 kg 100 kpa =.15 m /kg The volume of the system increased while pressure decreased during the process. A plot of the process on a pressure versus specific volume graph is as follows: Pressure versus Specific Volume Specific Volume (m^/kg)
51 1.5 Water in a swimming pool has a temperature of 4 o C. Express this temperature in K, o F, and o R. KNOWN: Water is at a specified temperature in o C. FIND: Equivalent temperature in K, o F, and o R. SCHEMATIC AND GIVEN DATA: T = 4 o C ANALYSIS: First convert temperature from o C to K by rearranging Eq to solve for temperature in K T water (K) = 4 o C = K T( o C) = T(K) 7.15 T(K) = T( o C) Next apply Eq to solve for temperature in o R T water ( o R) = (1.8)(97.15 K) = o R T( o R) = 1.8T(K) Finally, apply Eq to solve for temperatures in o F T water ( o F) = o R = 75. o F T( o F) = T( o R)
52 1.5 A cake recipe specifies an oven temperature of 50 o F. Express this temperature in o R, K, and o C. KNOWN: Oven temperature is specified in o F. FIND: Equivalent temperature in o R, K, and o C. SCHEMATIC AND GIVEN DATA: T = 50 o F ANALYSIS: First convert temperature from o F to o R using Eq to solve for temperature in o R T oven ( o R) = 50 o F = o R T( o F) = T( o R) T( o R) = T( o F) Next apply Eq to solve for temperature in K T oven (K) = o R/1.8 = K T( o R) = 1.8T(K) T(K) = T( o R)/1.8 Finally, apply Eq to solve for temperature in o C T oven ( o C) = K 7.15 = o C T( o C) = T(K) 7.15
54 1.56 Left for independent study using the Internet.
55 1.57 Air temperature rises from a morning low of 4 o F to an afternoon high of 70 o F. (a) Express these temperatures in o R, K, and o C. (b) Determine the temperature change in o F, o R, K, and o C from morning low to afternoon high. (c) What conclusion do you draw about temperature change for o F and o R scales? (d) What conclusion do you draw about temperature change for o C and K scales? KNOWN: Morning low temperature and afternoon high temperature, both in o F. FIND: (a) Express these temperatures in o R, K, and o C, (b) temperature change in o F, o R, K, and o C from morning low to afternoon high, (c) conclusion about temperature change for o F and o R scales, (d) conclusion about temperature change for o C and K scales. SCHEMATIC AND GIVEN DATA: T low = 4 o F T high = 70 o F ANALYSIS: (a) First convert temperatures from o F to o R using Eq to solve for temperatures in o R T low ( o R) = 4 o F = o R T high ( o R) = 70 o F = o R T( o F) = T( o R) T( o R) = T( o F) Next apply Eq to solve for temperature in K T low (K) = o R/1.8 = K T high (K) = o R/1.8 = 94.6 K T( o R) = 1.8T(K) T(K) = T( o R)/1.8 Finally, apply Eq to solve for temperature in o C T low ( o C) = K 7.15 = 5.56 o C T high ( o C) = 94.6 K 7.15 = 1.11 o C T( o C) = T(K) 7.15 (b) Temperature change, T, is T high T low. Calculating the differences yields T( o F) = 70 o F 4 o F = 8 o F 1
56 T( o R) = o R o R = 8 o R T(K) = 94.6 K K = K T( o C) = 1.11 o C 5.56 o C = o C (c) For o F and o R scales, the temperature change is the same since a Rankine degree and a Fahrenheit degree are the same temperature unit. (d) For o C and K scales, the temperature change is the same since a Kelvin degree and a Celsius degree are the same temperature unit.
57 1.58 Left for independent study using the Internet.
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