University Physics I - Classical Mechanics, 2019

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320 CHAPTER 13. THERMODYNAMICS 12. Microscopically, the entropy of a system is a measure of the range of distinct states available to its microscopic components (atoms or molecules) that are compatible with the set of macroscopic constraints that determine its thermal equilibrium state. More entropy means a greater range of possible “microstates.” 13. Entropy always increases in irreversible processes.
13.6. EXAMPLES 321 13.6 Examples 13.6.1 Calorimetry The specific heat of aluminum is 900 J/kg·K, and that of water is 4186 J·K. Suppose you drop a block of aluminum of mass 1 kg at a temperature of 80 ◦ C in a liter of water (which also has a mass of 1 kg) at a temperature of 20 ◦ C. What is the final temperature of the system, assuming no exchange of heat with the environment takes place? How much energy does the aluminum lose/the water gain? Solution Let us call T Al the initial temperature of the aluminum, T water the initial temperature of the water, and T f their final common temperature. The thermal energy given off by the aluminum equals ΔE Al = C Al (T f − T Al ) (this follows from the definition (13.1) of heat capacity; we could equally well call this quantity “the heat given off by the aluminum”). In the same way, the thermal energy change of the water (heat absorbed by the water) equals ΔE water = C water (T f − T water ). If the total system is closed, the sum of these two quantities, each with its appropriate sign, must be zero: This equation for T f has the solution 0=ΔE Al +ΔE water = C Al (T f − T Al )+C water (T f − T water ) (13.20) T f = C AlT Al + C water T water (13.21) C Al + C water As you can see, the result is a weighted average of the two starting temperatures, with the corresponding heat capacities as the weighting factors. The heat capacities C are equal to the given specific heats multiplied by the respective masses. In this case, the mass of aluminum and the mass of the water are the same, so they will cancel in the final result. Also, we can use the temperatures in degrees Celsius, instead of Kelvin. This is not immediately obvious from the final expression (13.21), but if you look at (13.20) you’ll see it involves only temperature differences, and those have the same value in the Kelvin and Celsius scales. Substituting the given values in (13.21), then, we get 900 × 80 + 4186 × 20 T f = =30.6 ◦ C (13.22) 900 + 4186 This is much closer to the initial temperature of the water, as expected, since it has the greater heat capacity. The amount of heat exchanged is C water (T f − T water ) = 4186 × (30.6 − 20) = 44, 440 J = 44.4 kJ (13.23)
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University Physics I: Classical Mec
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ii
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iv CONTENTS 1.6 Problems . . . . .
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vi CONTENTS 5.1 Conservative intera
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viii CONTENTS 7.7 Examples . . . .
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x CONTENTS 10.4 Examples . . . . .
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xii CONTENTS 13.2.1 Temperature and
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xiv CONTENTS they have read. Then,
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Chapter 1 Reference frames, displac
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1.2. POSITION, DISPLACEMENT, VELOCI
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1.3. REFERENCE FRAME CHANGES AND RE
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1.3. REFERENCE FRAME CHANGES AND RE
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1.4. IN SUMMARY 19 with a velocity
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1.5. EXAMPLES 21 1.5 Examples 1.5.1
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1.5. EXAMPLES 23 (twice what it was
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1.5. EXAMPLES 25 Note that I have d
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1.6. PROBLEMS 27 1.6 Problems Probl
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1.6. PROBLEMS 29 Problem 5 You are
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Chapter 2 Acceleration 2.1 The law
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2.1. THE LAW OF INERTIA 33 This is
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2.2. ACCELERATION 35 2.2 Accelerati
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2.2. ACCELERATION 37 called “conc
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2.2. ACCELERATION 39 2.2.2 Motion w
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2.2. ACCELERATION 41 2.2.3 Accelera
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2.3. FREE FALL 43 proportional to t
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2.4. IN SUMMARY 45 4. Changes in ve
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2.5. EXAMPLES 47 which the rocket r
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2.6. PROBLEMS 49 (a) How far did yo
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Chapter 3 Momentum and Inertia 3.1
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3.1. INERTIA 53 reliable, repeatabl
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3.1. INERTIA 55 3.1.2 Inertial mass
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3.2. MOMENTUM 57 the system is p i
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3.3. EXTENDED SYSTEMS AND CENTER OF
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3.4. IN SUMMARY 61 3.3.2 Recoil and
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3.5. EXAMPLES 63 3.5 Examples 3.5.1
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3.5. EXAMPLES 65 3.5.2 Collision in
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3.6. PROBLEMS 67 3.6 Problems Probl
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Chapter 4 Kinetic Energy 4.1 Kineti
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4.1. KINETIC ENERGY 71 v 2i =0,v 1f
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4.1. KINETIC ENERGY 73 special case
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4.1. KINETIC ENERGY 75 This immedia
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4.2. “CONVERTIBLE” AND “TRANS
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4.2. “CONVERTIBLE” AND “TRANS
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4.3. IN SUMMARY 81 7. Besides the c
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4.4. EXAMPLES 83 On the other hand,
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4.4. EXAMPLES 85 K cm ′ = 1 2 (m
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4.5. PROBLEMS 87 (a) What is the in
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Chapter 5 Interactions and energy 5
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5.1. CONSERVATIVE INTERACTIONS 91 T
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5.1. CONSERVATIVE INTERACTIONS 93 A
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5.1. CONSERVATIVE INTERACTIONS 95 d
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5.2. DISSIPATION OF ENERGY AND THER
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5.4. CONSERVATION OF ENERGY 99 leve
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5.5. IN SUMMARY 101 gravitational p
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5.6. EXAMPLES 103 5.6 Examples 5.6.
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5.6. EXAMPLES 105 problem involves
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5.7. ADVANCED TOPICS 107 where the
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5.8. PROBLEMS 109 5.8 Problems Prob
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5.8. PROBLEMS 111 here? Explain. (f
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Chapter 6 Interactions, part 2: For
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6.1. FORCE 115 however, somewhat mo
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6.2. FORCES AND POTENTIAL ENERGY 11
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6.2. FORCES AND POTENTIAL ENERGY 11
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6.3. FORCES NOT DERIVED FROM A POTE
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6.5. IN SUMMARY 127 F s,1 n a F s,1
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6.6. EXAMPLES 131 pushes on the bra
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6.6. EXAMPLES 133 This is a huge di
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6.7. PROBLEMS 135 Problem 6 Draw a
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Chapter 7 Impulse, Work and Power 7
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7.2. WORK ON A SINGLE PARTICLE 139
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7.3. THE “CENTER OF MASS WORK”
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7.4. WORK DONE ON A SYSTEM BY ALL T
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7.6. IN SUMMARY 151 which is equal
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7.7. EXAMPLES 155 To plot all this
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7.7. EXAMPLES 157 (a) The external
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7.8. PROBLEMS 159 Problem 5 A block
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Chapter 8 Motion in two dimensions
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8.1. DEALING WITH FORCES IN TWO DIM
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8.2. PROJECTILE MOTION 165 The over
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8.3. INCLINED PLANES 167 8.3 Inclin
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8.4. MOTION ON A CIRCLE (OR PART OF
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8.5. IN SUMMARY 175 As we shall see
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8.6. EXAMPLES 177 8.6 Examples You
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8.7. ADVANCED TOPICS 179 8.7 Advanc
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8.7. ADVANCED TOPICS 181 Now we jus
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8.7. ADVANCED TOPICS 183 Note that
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8.7. ADVANCED TOPICS 185 The cyan a
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8.8. PROBLEMS 187 (b) What is the w
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8.8. PROBLEMS 189 (e) The power is
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Chapter 9 Rotational dynamics Rotat
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9.2. ANGULAR MOMENTUM 193 9.2 Angul
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9.2. ANGULAR MOMENTUM 195 thing is
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9.3. THE CROSS PRODUCT AND ROTATION
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9.3. THE CROSS PRODUCT AND ROTATION
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9.4. TORQUE 201 momentum involves t
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9.4. TORQUE 203 wrenches). It is al
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9.5. STATICS 205 important in engin
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9.6. ROLLING MOTION 207 9.6 Rolling
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9.6. ROLLING MOTION 209 contact wit
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9.7. IN SUMMARY 211 (Note how I hav
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9.8. EXAMPLES 213 9.8 Examples This
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9.8. EXAMPLES 215 Solution The figu
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9.9. PROBLEMS 217 9.9 Problems Prob
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9.9. PROBLEMS 219 A 20-kg plank of
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Chapter 10 Gravity 10.1 The inverse
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10.1. THE INVERSE-SQUARE LAW 223 ne
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10.1. THE INVERSE-SQUARE LAW 225 it
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10.1. THE INVERSE-SQUARE LAW 227 in
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10.1. THE INVERSE-SQUARE LAW 229 an
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10.1. THE INVERSE-SQUARE LAW 231 el
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10.1. THE INVERSE-SQUARE LAW 233 10
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10.1. THE INVERSE-SQUARE LAW 235 Fr
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10.2. WEIGHT, ACCELERATION, AND THE
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10.2. WEIGHT, ACCELERATION, AND THE
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10.3. IN SUMMARY 241 shifted and/or
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10.4. EXAMPLES 243 10.4 Examples 10
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10.4. EXAMPLES 245 The diagram of t
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10.5. ADVANCED TOPICS 247 10.5 Adva
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10.6. PROBLEMS 249 10.6 Problems Pr
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10.6. PROBLEMS 251 an orbit around
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Chapter 11 Simple harmonic motion 1
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11.2. SIMPLE HARMONIC MOTION 255 is
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11.2. SIMPLE HARMONIC MOTION 257 Th
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11.2. SIMPLE HARMONIC MOTION 259 If
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11.2. SIMPLE HARMONIC MOTION 261 Th
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11.3. PENDULUMS 263 o θ l F t F G
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11.3. PENDULUMS 265 11.3.2 The “p
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11.4. IN SUMMARY 267 9. A simple pe
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University Physics I - Classical Mechanics, 2019

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