17 TEMPERATURE AND HEAT 17.1 IDENTIFY and SET UP: TF = 95 TC + 32° EXECUTE: (a) TF = (9/5)(−62.8) + 32 = −81.0°F (b) TF = (9/5)(56.7) + 32 = 134.1°F (c) TF = (9/5)(31.1) + 32 = 88.0°F 17.2 EVALUATE: Fahrenheit degrees are smaller than Celsius degrees, so it takes more F° than C° to express the difference of a temperature from the ice point IDENTIFY and SET UP: To convert a temperature between °C and K use TC = TK − 273.15 To convert from °F to °C, subtract 32° and multiply by 5/9 To convert from °C to °F, multiply by 9/5 and add 32° To convert a temperature difference, use that Celsius and Kelvin degrees are the same size and that F° = C° EXECUTE: (a) TC = TK − 273.15 = 310 − 273.15 = 36.9°C; TF = 95 TC + 32° = 95 (36.9°) + 32° = 98.4°F (b) TK = TC + 273.15 = 40 + 273.15 = 313 K; TF = 95 TC + 32° = 95 (40°) + 32° = 104°F (c) C° = K; C° = (7 C°)(9 F° /5 C°) = 13 F° (d) 4.0°C: TF = 95 TC + 32° = 95 (4.0°) + 32° = 39.2°F; TK = TC + 273.15 = 4.0 + 273.15 = 277 K −160°C: TF = 95 TC + 32° = 95 (−160°) + 32° = − 256°F; TK = TC + 273.15 = − 160 + 273.15 = 113 K (e) TC = 95 (TF − 32°) = 95 (105° − 32°) = 41°C; TK = TC + 273.15 = 41 + 273.15 = 314 K 17.3 EVALUATE: Celsius-Fahrenheit conversions not involve simple proportions due to the additive constant of 32°, but Celsius-Kelvin conversions require only simple addition/subtraction of 273.15 IDENTIFY: Convert ΔT between different scales SET UP: ΔT is the same on the Celsius and Kelvin scales 180 F° = 100 C°, so C° = 95 F° ⎛ C° ⎞ EXECUTE: (a) ΔT = 49.0 F° ΔT = (49.0 F°) ⎜ ⎟ = 27.2 C° ⎜ F° ⎟ ⎝5 ⎠ 17.4 ⎛ C° ⎞ (b) ΔT = −100 F° ΔT = (−100.0 F°) ⎜ ⎟ = −55.6 C° ⎜ F° ⎟ ⎝5 ⎠ EVALUATE: The magnitude of the temperature change is larger in F° than in C° IDENTIFY: Set TC = TF and TF = TK SET UP: TF = 95 TC + 32°C and TK = TC + 273.15 = 95 (TF − 32°) + 273.15 EXECUTE: (a) TF = TC = T gives T = 95 T + 32° and T = −40°; −40°C = −40°F ( ( ) (32°) + 273.15) = 575°; 575°F = 575 K (b) TF = TK = T gives T = 95 (T − 32°) + 273.15 and T = 94 − EVALUATE: Since TK = TC + 273.15 there is no temperature at which Celsius and Kelvin thermometers agree © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-1 17-2 17.5 Chapter 17 IDENTIFY: Convert ΔT in kelvins to C° and to F° SET UP: K = C° = 95 F° EXECUTE: (a) ΔTF = 95 ΔTC = 95 (−10.0 C°) = −18.0 F° (b) ΔTC = ΔTK = −10.0 C° 17.6 EVALUATE: Kelvin and Celsius degrees are the same size Fahrenheit degrees are smaller, so it takes more of them to express a given ΔT value IDENTIFY: Convert TK to TC and then convert TC to TF SET UP: TK = TC + 273.15 and TF = 95 TC + 32° EXECUTE: (a) TC = 400 − 273.15 = 127°C, TF = (9/5)(126.85) + 32 = 260°F (b) TC = 95 − 273.15 = −178°C, TF = (9/5)(−178.15) + 32 = −289°F (c) TC = 1.55 × 107 − 273.15 = 1.55 × 107°C, TF = (9/5)(1.55 × 107 ) + 32 = 2.79 × 107°F EVALUATE: All temperatures on the Kelvin scale are positive TC is negative if the temperature is below the freezing point of water 17.7 IDENTIFY: When the volume is constant, T2 p2 = , for T in kelvins T1 p1 SET UP: Ttriple = 273.16 K Figure 17.7 in the textbook gives that the temperature at which CO solidifies is TCO2 = 195 K ⎛T ⎞ ⎛ 195 K ⎞ p2 = p1 ⎜ ⎟ = (1.35 atm ) ⎜ ⎟ = 0.964 atm T ⎝ 273.16 K ⎠ ⎝ 1⎠ EVALUATE: The pressure decreases when T decreases IDENTIFY: Apply Eq (17.5) and solve for p SET UP: ptriple = 325 mm of mercury EXECUTE: 17.8 ⎛ 373.15 K ⎞ p = (325.0 mm of mercury) ⎜ ⎟ = 444 mm of mercury ⎝ 273.16 K ⎠ EVALUATE: mm of mercury is a unit of pressure Since Eq (17.5) involves a ratio of pressures, it is not necessary to convert the pressure to units of Pa IDENTIFY and SET UP: Fit the data to a straight line for p (T ) and use this equation to find T when p = EXECUTE: 17.9 EXECUTE: (a) If the pressure varies linearly with temperature, then p2 = p1 + γ (T2 − T1 ) γ= p2 − p1 6.50 × 104 Pa − 4.80 × 104 Pa = = 170.0 Pa/C° 100°C − 0.01°C T2 − T1 Apply p = p1 + γ (T − T1) with T1 = 0.01°C and p = to solve for T = p1 + γ (T − T1 ) T = T1 − p1 γ = 0.01°C − 4.80 × 104 Pa = −282°C 170 Pa/C° (b) Let T1 = 100°C and T2 = 0.01°C; use Eq (17.4) to calculate p2 Eq (17.4) says T2 /T1 = p2 /p1, where T is in kelvins ⎛T ⎞ ⎛ 0.01 + 273.15 ⎞ 4 p2 = p1 ⎜ ⎟ = 6.50 × 104 Pa ⎜ ⎟ = 4.76 × 10 Pa; this differs from the 4.80 × 10 Pa that was 100 273 15 T + ⎝ ⎠ ⎝ 1⎠ measured so Eq (17.4) is not precisely obeyed EVALUATE: The answer to part (a) is in reasonable agreement with the accepted value of −273°C 17.10 IDENTIFY: K = C° and C° = 95 F°, so K = 95 R ° SET UP: On the Kelvin scale, the triple point is 273.16 K EXECUTE: Ttriple = (9/5)273.16 K = 491.69°R © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17.11 17-3 EVALUATE: One could also look at Figure 17.7 in the textbook and note that the Fahrenheit scale extends from −460°F to + 32°F and conclude that the triple point is about 492°R IDENTIFY: ΔL = L0 α ΔT SET UP: For steel, α = 1.2 × 10−5 (C°) −1 EXECUTE: ΔL = (1.2 × 10−5 (C°)−1 )(1410 m)(18.0°C − (−5.0°C)) = +0.39 m 17.12 EVALUATE: The length increases when the temperature increases The fractional increase is very small, since αΔT is small IDENTIFY: Apply ΔL = α L0 ΔT and calculate ΔT Then T2 = T1 + ΔT , with T1 = 15.5°C SET UP: Table 17.1 gives α = 1.2 × 10−5 (C°) −1 for steel EXECUTE: ΔT = ΔL α L0 = 0.471 ft [1.2 × 10−5 (C°) −1 ][1671 ft] = 23.5 C° T2 = 15.5°C + 23.5 C° = 39.0°C EVALUATE: Since then the lengths enter in the ratio Δ L/L , we can leave the lengths in ft 17.13 IDENTIFY: Apply L = L (1 + α ΔT ) to the diameter D of the penny SET UP: K = C°, so we can use temperatures in °C EXECUTE: Death Valley: α D ΔT = (2.6 × 10−5 (C°)−1)(1.90 cm)(28.0 C°) = 1.4 × 10−3 cm, so the diameter is 1.9014 cm Greenland: α D ΔT = −3.6 × 10−3 cm, so the diameter is 1.8964 cm 17.14 EVALUATE: When T increases the diameter increases and when T decreases the diameter decreases IDENTIFY: Apply L = L (1 + α ΔT ) to the diameter d of the rivet SET UP: For aluminum, α = 2.4 × 10−5 (C°) −1 Let d be the diameter at –78.0°C and d be the diameter at 23.0°C EXECUTE: d = d + Δ d = d (1 + α ΔT ) = (0.4500 cm)(1 + (2.4 × 10−5 (C°) −1 )(23.0°C − [ −78.0°C]) 17.15 d = 0.4511 cm = 4.511 mm EVALUATE: We could have let d be the diameter at 23.0°C and d be the diameter at −78.0°C Then ΔT = −78.0°C − 23.0°C IDENTIFY: Find the change ΔL in the diameter of the lid The diameter of the lid expands according to Eq (17.6) SET UP: Assume iron has the same α as steel, so α = 1.2 × 10−5 (C°) −1 EXECUTE: Δ L = α L ΔT = (1.2 × 10−5 (C°)−1 )(725 mm)(30.0 C°) = 0.26 mm 17.16 EVALUATE: In Eq (17.6), ΔL has the same units as L IDENTIFY: ΔV = β V ΔT Use the diameter at −15°C to calculate the value of V0 at that temperature SET UP: For a hemisphere of radius R, the volume is V = 23 π R3 Table 17.2 gives β = 7.2 × 10−5 (C°) −1 for aluminum EXECUTE: V0 = 23 π R3 = 23 π (27.5 m)3 = 4.356 × 104 m3 ΔV = (7.2 × 1025 (C°) −1)(4.356 × 104 m3 )(35°C − [−15°C]) = 160 m3 EVALUATE: We could also calculate R = R0 (1 + α ΔT ) and calculate the new V from R The increase in volume is V − V0 , but we would have to be careful to avoid round-off errors when two large volumes of 17.17 nearly the same size are subtracted IDENTIFY: Apply ΔV = V0 β ΔT SET UP: For copper, β = 5.1 × 10−5 (C°)−1 ΔV/V0 = 0.150 × 10−2 EXECUTE: ΔT = ΔV/V0 0.150 × 10−2 = 29.4 C° Tf = Ti + ΔT = 49.4°C 5.1 × 10−5 (C°) −1 EVALUATE: The volume increases when the temperature increases β = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-4 Chapter 17 17.18 IDENTIFY: Apply ΔV = V β ΔT to the tank and to the ethanol SET UP: For ethanol, β e = 75 × 10−5 (C°) −1 For steel, βs = 3.6 × 10−5 (C°) −1 EXECUTE: The volume change for the tank is ΔVs = V β s ΔT = (2.80 m3 )(3.6 × 10−5 (C°) −1)(−14.0 C°) = −1.41 × 10−3 m3 = −1.41 L The volume change for the ethanol is ΔVe = V β e ΔT = (2.80 m3 )(75 × 10−5 (C°)−1 )(−14.0 C°) = −2.94 × 10−2 m3 = −29.4 L The empty volume in the tank is ΔVe − ΔVs = −29.4 L − (−1.4 L) = −28.0 L 28.0 L of ethanol can be added to the tank EVALUATE: Both volumes decrease But β e > βs , so the magnitude of the volume decrease for the 17.19 ethanol is greater than it is for the tank IDENTIFY: Apply ΔV = V0 β ΔT to the volume of the flask and to the mercury When heated, both the volume of the flask and the volume of the mercury increase SET UP: For mercury, β Hg = 18 × 10−5 (C°) −1 EXECUTE: 8.95 cm3 of mercury overflows, so ΔVHg − ΔVglass = 8.95 cm3 EXECUTE: ΔVHg = V0 β Hg ΔT = (1000.00 cm3 )(18 × 10 −5 (C°) −1 )(55.0 C°) = 9.9 cm3 17.20 ΔVglass 0.95 cm3 = 1.7 × 10−5 (C°) −1 (1000.00 cm3 )(55.0 C°) EVALUATE: The coefficient of volume expansion for the mercury is larger than for glass When they are heated, both the volume of the mercury and the inside volume of the flask increase But the increase for the mercury is greater and it no longer all fits inside the flask IDENTIFY: Apply Δ L = L α ΔT to each linear dimension of the surface ΔVglass = ΔVHg − 8.95 cm3 = 0.95 cm3 β glass = V0ΔT = SET UP: The area can be written as A = aL1L2 , where a is a constant that depends on the shape of the surface For example, if the object is a sphere, a = 4π and L1 = L2 = r If the object is a cube, a = and L1 = L2 = L, the length of one side of the cube For aluminum, α = 2.4 × 10−5 (C°) −1 EXECUTE: (a) A0 = aL01L02 L1 = L01 (1 + α ΔT ) L2 = L02 (1 + α ΔT ) A = aL1L2 = aL01L02 (1 + α ΔT )2 = A0 (1 + 2α ΔT + [α ΔT ]2 ) α ΔT is very small, so [α ΔT ] can be neglected and A = A0 (1 + 2α ΔT ) ΔA = A − A0 = (2α ) A0 ΔT (b) Δ A = (2α ) A0 ΔT = (2)(2.4 × 10−5 (C°)−1 )(π (0.275 m) )(12.5 C°) = 1.4 × 10−4 m 17.21 EVALUATE: The derivation assumes the object expands uniformly in all directions IDENTIFY and SET UP: Apply the result of Exercise 17.20a to calculate ΔA for the plate, and then A = A0 + ΔA EXECUTE: (a) A0 = π r02 = π (1.350 cm/2) = 1.431 cm (b) Exercise 17.20 says ΔA = 2α A0 ΔT , so ΔA = 2(1.2 × 10−5 C°−1 )(1.431 cm )(175°C − 25°C) = 5.15 × 10−3 cm A = A0 + ΔA = 1.436 cm 17.22 EVALUATE: A hole in a flat metal plate expands when the metal is heated just as a piece of metal the same size as the hole would expand IDENTIFY: Apply ΔL = L0 α ΔT to the diameter DST of the steel cylinder and the diameter DBR of the brass piston SET UP: For brass, α BR = 2.0 × 10−5 (C°) −1 For steel, αST = 1.2 × 10−5 (C°) −1 EXECUTE: (a) No, the brass expands more than the steel (b) Call D0 the inside diameter of the steel cylinder at 20°C At 150°C, DST = DBR D0 + Δ DST = 25.000 cm + Δ DBR This gives D0 + αST D0 ΔT = 25.000 cm + α BR (25.000 cm)ΔT © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat D0 = 17.23 17-5 25.000 cm(1 + α BR ΔT ) (25.000 cm)[1 + (2.0 × 10−5 (C°)−1 )(130 C°)] = = 25.026 cm + α ST ΔT + (1.2 × 10−5 (C°) −1)(130 C°) EVALUATE: The space inside the steel cylinder expands just like a solid piece of steel of the same size IDENTIFY and SET UP: For part (a), apply Eq (17.6) to the linear expansion of the wire For part (b), apply Eq (17.12) and calculate F/A EXECUTE: (a) ΔL = α L ΔT α= ΔL 1.9 × 10−2 m = = 3.2 × 10−5 (C°) −1 L0ΔT (1.50 m)(420°C − 20°C) (b) Eq (17.12): stress F/A = −Yα ΔT ΔT = 20°C − 420°C = −400 C° ( ΔT always means final temperature minus initial temperature) 17.24 F/A = −(2.0 × 1011 Pa)(3.2 × 10−5 (C°)−1 )(−400 C°) = +2.6 × 109 Pa EVALUATE: F/A is positive means that the stress is a tensile (stretching) stress The answer to part (a) is consistent with the values of α for metals in Table 17.1 The tensile stress for this modest temperature decrease is huge IDENTIFY: Apply Eq (17.12) and solve for F SET UP: For brass, Y = 0.9 × 1011 Pa and α = 2.0 × 10−5 (C°) −1 EXECUTE: F = −Yα ΔT A = −(0.9 × 1011 Pa)(2.0 × 10−5 (C°) −1)(−110 C°)(2.01 × 10−4 m ) = 4.0 × 104 N 17.25 EVALUATE: A large force is required ΔT is negative and a positive tensile force is required IDENTIFY: Apply Δ L = L0 α ΔT and stress = F/A = −Y α ΔT SET UP: For steel, α = 1.2 × 10−5 (C°) −1 and Y = 2.0 × 1011 Pa EXECUTE: (a) Δ L = L0 α ΔT = (12.0 m)(1.2 × 10−5 (C°) −1)(35.0 C°) = 5.0 mm (b) stress = −Y α ΔT = −(2.0 × 1011 Pa)(1.2 × 10−5 (C°) −1 )(35.0 C°) = −8.4 × 107 Pa The minus sign means 17.26 the stress is compressive EVALUATE: Commonly occurring temperature changes result in very small fractional changes in length but very large stresses if the length change is prevented from occurring IDENTIFY: The heat required is Q = mcΔT P = 200 W = 200 J/s, which is energy divided by time SET UP: For water, c = 4.19 × 103 J/kg ⋅ K EXECUTE: (a) Q = mcΔT = (0.320 kg)(4.19 × 103 J/kg ⋅ K)(60.0 C°) = 8.04 × 104 J 8.04 × 104 J = 402 s = 6.7 200.0 J/s EVALUATE: 0.320 kg of water has volume 0.320 L The time we calculated in part (b) is consistent with our everyday experience IDENTIFY and SET UP: Apply Eq (17.13) to the kettle and water EXECUTE: kettle Q = mcΔT , c = 910 J/kg ⋅ K (from Table 17.3) (b) t = 17.27 Q = (1.50 kg)(910 J/kg ⋅ K)(85.0°C − 20.0°C) = 8.873 × 104 J water Q = mcΔT , c = 4190 J/kg ⋅ K (from Table 17.3) Q = (1.80 kg)(4190 J/kg ⋅ K)(85.0°C − 20.0°C) = 4.902 × 105 J 17.28 Total Q = 8.873 × 104 J + 4.902 × 105 J = 5.79 × 105 J EVALUATE: Water has a much larger specific heat capacity than aluminum, so most of the heat goes into raising the temperature of the water IDENTIFY and SET UP: Use Eq (17.13) EXECUTE: (a) Q = mcΔT © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-6 Chapter 17 m = 12 (1.3 × 10−3 kg) = 0.65 × 10−3 kg Q = (0.65 × 10−3 kg)(1020 J/kg ⋅ K)(37°C − ( −20°C)) = 38 J (b) 20 breaths/min (60 min/1 h) = 1200 breaths/h 17.29 So Q = (1200)(38 J) = 4.6 × 104 J EVALUATE: The heat loss rate is Q/t = 13 W IDENTIFY: Apply Q = mcΔT m = w/g SET UP: The temperature change is ΔT = 18.0 K EXECUTE: c = 17.30 17.31 Q gQ (9.80 m/s )(1.25 × 104 J) = = = 240 J/kg ⋅ K (28.4 N)(18.0 K) mΔT wΔT EVALUATE: The value for c is similar to that for silver in Table 17.3, so it is a reasonable result IDENTIFY: The heat input increases the temperature of 2.5 gal/min of water from 10°C to 49°C SET UP: 1.00 L of water has a mass of 1.00 kg, so 9.46 L/min = (9.46 L/min)(1.00 kg/L)(1 min/60 s) = 0.158 kg/s For water, c = 4190 J/kg ⋅ C° EXECUTE: Q = mcΔT so H = (Q/t ) = ( m/t )c ΔT Putting in the numbers gives H = (0.158 kg/s)(4190 J/kg ⋅ C°)(49°C − 10°C) = 2.6 × 104 W = 26 kW EVALUATE: The power requirement is large, the equivalent of 260 100-watt light bulbs, but this large power is needed only for short periods of time The rest of the time, the unit uses no energy, unlike a conventional water heater which continues to replace lost heat even when hot water is not needed IDENTIFY: Apply Q = mcΔT to find the heat that would raise the temperature of the student’s body C° SET UP: W = J/s EXECUTE: Find Q to raise the body temperature from 37°C to 44°C Q = mcΔT = (70 kg)(3480 J/kg ⋅ K)(7 C°) = 1.7 × 106 J 1.7 × 106 J = 1400 s = 23 1200 J/s EVALUATE: Heat removal mechanisms are essential to the well-being of a person IDENTIFY and SET UP: Set the change in gravitational potential energy equal to the quantity of heat added to the water EXECUTE: The change in mechanical energy equals the decrease in gravitational potential energy, ΔU = − mgh; | ΔU | = mgh Q = | ΔU | = mgh implies mcΔT = mgh t= 17.32 ΔT = gh/c = (9.80 m/s )(225 m)/(4190 J/kg ⋅ K) = 0.526 K = 0.526 C° 17.33 EVALUATE: Note that the answer is independent of the mass of the object Note also the small change in temperature that corresponds to this large change in height! IDENTIFY: The work done by friction is the loss of mechanical energy The heat input for a temperature change is Q = mcΔT SET UP: The crate loses potential energy mgh, with h = (8.00 m)sin 36.9°, and gains kinetic energy mv 2 EXECUTE: (a) W f = − mgh + 12 mv22 = −(35.0 kg)((9.80 m/s )(8.00 m)sin 36.9° + 12 (2.50 m/s) ) = −1.54 × 103 J (b) Using the results of part (a) for Q gives ΔT = (1.54 × 103 J)/((35.0 kg)(3650 J/kg ⋅ K)) = 1.21 × 10−2 C° 17.34 EVALUATE: The temperature rise is very small IDENTIFY: The work done by the brakes equals the initial kinetic energy of the train Use the volume of the air to calculate its mass Use Q = mcΔT applied to the air to calculate ΔT for the air SET UP: K = 12 mv m = ρV © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17-7 EXECUTE: The initial kinetic energy of the train is K = 12 (25,000 kg)(15.5 m/s) = 3.00 × 106 J Therefore, Q for the air is 3.00 × 106 J m = ρV = (1.20 kg/m3 )(65.0 m)(20.0 m)(12.0 m) = 1.87 × 104 kg Q 3.00 × 106 J = = 0.157 C° mc (1.87 × 104 kg)(1020 J/kg ⋅ K) EVALUATE: The mass of air in the station is comparable to the mass of the train and the temperature rise is small IDENTIFY: Set K = 12 mv equal to Q = mcΔT for the nail and solve for ΔT Q = mcΔT gives ΔT = 17.35 SET UP: For aluminum, c = 0.91 × 103 J/kg ⋅ K EXECUTE: The kinetic energy of the hammer before it strikes the nail is K = 12 mv = 12 (1.80 kg)(7.80 m/s) = 54.8 J Each strike of the hammer transfers 0.60(54.8 J) = 32.9 J, Q 329 J = = 45.2 C° mc (8.00 × 10−3 kg)(0.91 × 103 J/kg ⋅ K) EVALUATE: This agrees with our experience that hammered nails get noticeably warmer IDENTIFY and SET UP: Use the power and time to calculate the heat input Q and then use Eq (17.13) to calculate c (a) EXECUTE: P = Q/t , so the total heat transferred to the liquid is Q = Pt = (65.0 W)(120 s) = 7800 J and with 10 strikes Q = 329 J Q = mcΔT and ΔT = 17.36 Q 7800 K = = 2.51 × 103 J/kg ⋅ K mΔT 0.780 kg(22.54°C − 18.55°C) (b) EVALUATE: Then the actual Q transferred to the liquid is less than 7800 J so the actual c is less than our calculated value; our result in part (a) is an overestimate IDENTIFY: Some of the kinetic energy of the bullet is transformed through friction into heat, which raises the temperature of the water in the tank SET UP: Set the loss of kinetic energy of the bullet equal to the heat energy Q transferred to the water Q = mcΔT From Table 17.3, the specific heat of water is 4.19 × 103 J/kg ⋅ C° Then Q = mcΔT gives c = 17.37 SOLVE: The kinetic energy lost by the bullet is Ki − K f = 12 m(vi2 − vf2 ) = 12 (15.0 × 10−3 kg)[(865 m/s)2 − (534 m/s)2 ] = 3.47 × 103 J, so for the water Q 3.47 × 103 J = = 0.0613 C° mc (13.5 kg)(4.19 × 103 J/kg ⋅ C°) EVALUATE: The heat energy required to change the temperature of ordinary-size objects is very large compared to the typical kinetic energies of moving objects IDENTIFY: The latent heat of fusion Lf is defined by Q = mLf for the solid → liquid phase transition For a temperature change, Q = mcΔT SET UP: At t = the sample is at its melting point and at t = 2.5 all the sample has melted EXECUTE: (a) It takes 1.5 for all the sample to melt once its melting point is reached and the heat input during this time interval is (1.5 min)(10.0 × 103 J/min) = 1.50 × 104 J Q = mLf Q = 3.47 × 103 J Q = mcΔT gives ΔT = 17.38 Q 1.50 × 104 J = = 3.00 × 104 J/kg m 0.500 kg (b) The liquid’s temperature rises 30 C° in 1.5 Q = mcΔT Lf = cliquid = Q 1.50 × 104 J = = 1.00 × 103 J/kg ⋅ K mΔT (0.500 kg)(30 C°) Q 1.00 × 104 J = = 1.33 × 103 J/kg ⋅ K mΔT (0.500 kg)(15 C°) EVALUATE: The specific heat capacities for the liquid and solid states are different The values of c and Lf that we calculated are within the range of values in Tables 17.3 and 17.4 The solid’s temperature rises 15 C° in 1.0 csolid = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-8 Chapter 17 17.39 IDENTIFY and SET UP: Heat comes out of the metal and into the water The final temperature is in the range < T < 100°C, so there are no phase changes Qsystem = (a) EXECUTE: Qwater + Qmetal = mwater cwater ΔTwater + mmetalcmetalΔTmetal = (1.00 kg)(4190 J/kg ⋅ K)(2.0 C°) + (0.500 kg)(cmetal )(−78.0 C°) = cmetal = 215 J/kg ⋅ K (b) EVALUATE: Water has a larger specific heat capacity so stores more heat per degree of temperature change (c) If some heat went into the styrofoam then Qmetal should actually be larger than in part (a), so the true cmetal is larger than we calculated; the value we calculated would be smaller than the true value 17.40 IDENTIFY: The heat that comes out of the person goes into the ice-water bath and causes some of the ice to melt SET UP: Normal body temperature is 98.6°F = 37.0°C, so for the person ΔT = −5 C° The ice-water bath stays at 0°C A mass m of ice melts and Qice = mLf From Table 17.4, for water Lf = 334 × 103 J/kg EXECUTE: Qperson = mcΔT = (70.0 kg)(3480 J/kg ⋅ C°)( −5.0 C°) = −1.22 × 106 J Therefore, the amount of heat that goes into the ice is 1.22 × 106 J mice Lf = 1.22 × 106 J and mice = 17.41 1.22 × 106 J = 3.7 kg 334 × 103 J/kg EVALUATE: If less ice than this is used, all the ice melts and the temperature of the water in the bath rises above 0°C IDENTIFY: The heat lost by the cooling copper is absorbed by the water and the pot, which increases their temperatures SET UP: For copper, cc = 390 J/kg ⋅ K For iron, ci = 470 J/kg ⋅ K For water, cw = 4.19 × 103 J/kg ⋅ K EXECUTE: For the copper pot, Qc = mccc ΔTc = (0.500 kg)(390 J/kg ⋅ K)(T − 20.0°C) = (195 J/K)T − 3900 J For the block of iron, Qi = mi ci ΔTi = (0.250 kg)(470 J/kg ⋅ K)(T − 85.0°C) = (117.5 J/K)T − 9988 J For the water, Qw = mw cw ΔTw = (0.170 kg)(4190 J/kg ⋅ K)(T − 20.0°C) = (712.3 J/K)T − 1.425 × 104 J ΣQ = gives 2.814 × 104 J = 27.5°C 1025 J/K EVALUATE: The basic principle behind this problem is conservation of energy: no energy is lost; it is only transferred IDENTIFY: The energy generated in the body is used to evaporate water, which prevents the body from overheating SET UP: Energy is (power)(time); calculate the heat energy Q produced in one hour The mass m of (195 J/K)T − 3900 J + (117.5 J/K)T − 9988 J + (712.3 J/K)T − 1.425 × 104 J T = 17.42 water that vaporizes is related to Q by Q = mLv 1.0 kg of water has a volume of 1.0 L EXECUTE: (a) Q = (0.80)(500 W)(3600 s) = 1.44 × 106 J The mass of water that evaporates each hour is m= Q 1.44 × 106 J = = 0.60 kg Lv 2.42 × 106 J/kg 0.60 L/h = 0.80 bottles/h 0.750 L/bottle EVALUATE: It is not unreasonable to drink 8/10 of a bottle of water per hour during vigorous exercise IDENTIFY: If it cannot be gotten rid of in some way, the metabolic energy transformed to heat will increase the temperature of the body SET UP: From Problem 17.42, Q = 1.44 × 106 J and m = 70 kg Q = mcΔT Convert the temperature (b) (0.60 kg/h)(1.0 L/kg) = 0.60 L/h The number of bottles of water is 17.43 change in C° to F° using that F° = C° Q 1.44 × 106 J EXECUTE: (a) Q = mcΔT so ΔT = = = 5.9 C° mc (70 kg)(3500 J/kg ⋅ C°) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17.44 17-9 ⎛ F° ⎞ (b) ΔT = (5.9°C) ⎜ ⎟ = 10.6°F T = 98.6°F + 10.6 F° = 109°F ⎝ C° ⎠ EVALUATE: A temperature this high can cause heat stroke and be lethal IDENTIFY: By energy conservation, the heat lost by the water is gained by the ice This heat must first increase the temperature of the ice from −40.0°C to the melting point of 0.00°C, then melt the ice, and finally increase its temperature to 20.0°C The target variable is the mass of the water m SET UP: Qice = micecice ΔTice + mice Lf + micecw ΔTmelted ice and Qwater = mcw ΔTw EXECUTE: Using Qice = micecice ΔTice + mice Lf + micecw ΔTmelted ice , with the values given in the table in the text, we have Qice = (0.200 kg)[2100 J/(kg ⋅ C°)](40.0C°) + (0.200 kg)(3.34 × 105 J/kg) + (0.200 kg)[4190 J/(kg ⋅ C°)](20.0C°) = 1.004 × 105 J Qwater = mcw ΔTw = m[4190 J/(kg ⋅ C°)](20.0C° − 80.0C°) = −(251,400 J/kg) m Qice + Qwater = gives 1.004 × 105 J = (251,400 J/kg)m m = 0.399 kg 17.45 EVALUATE: There is about twice as much water as ice because the water must provide the heat not only to melt the ice but also to increase its temperature IDENTIFY: By energy conservation, the heat lost by the copper is gained by the ice This heat must first increase the temperature of the ice from −20.0°C to the melting point of 0.00°C, then melt some of the ice At the final thermal equilibrium state, there is ice and water, so the temperature must be 0.00°C The target variable is the initial temperature of the copper SET UP: For temperature changes, Q = mcΔT and for a phase change from solid to liquid Q = mLF EXECUTE: For the ice, Qice = (2.00 kg)[2100 J/(kg ⋅ C°)](20.0C°) + (0.80 kg)(3.34 × 105 J/kg) = 3.512 × 105 J For the copper, using the specific heat from the table in the text gives Qcopper = (6.00 kg)[390 J/(kg ⋅ C°)](0°C − T ) = −(2.34 × 103 J/C°)T Setting the sum of the two heats equal to zero gives 3.512 × 105 J = (2.34 × 103 J/C°)T , which gives T = 150°C 17.46 EVALUATE: Since the copper has a smaller specific heat than that of ice, it must have been quite hot initially to provide the amount of heat needed IDENTIFY: Apply Q = mcΔT to each object The net heat flow Qsystem for the system (man, soft drink) is zero SET UP: The mass of 1.00 L of water is 1.00 kg Let the man be designated by the subscript m and the “‘water” by w T is the final equilibrium temperature cw = 4190 J/kg ⋅ K ΔTK = ΔTC EXECUTE: (a) Qsystem = gives mmcm ΔTm + mw cw ΔTw = mmcm (T − Tm ) + mw cw (T − Tw ) = mmcm (Tm − T ) = mw cw (T − Tw ) Solving for T, T = T= 17.47 mmcmTm + mw cwTw mmcm + mw cw (70.0 kg)(3480 J/kg ⋅ K)(37.0°C) + (0.355 kg)(4190 J/kg ⋅ C°)(12.0°C) = 36.85°C (70.0 kg)(3480 J/kg ⋅ C°) + (0.355 kg)(4190 J/kg ⋅ C°) (b) It is possible a sensitive digital thermometer could measure this change since they can read to 1°C It is best to refrain from drinking cold fluids prior to orally measuring a body temperature due to cooling of the mouth EVALUATE: Heat comes out of the body and its temperature falls Heat goes into the soft drink and its temperature rises IDENTIFY: For the man’s body, Q = mcΔT SET UP: From Exercise 17.46, ΔT = 0.15 C° when the body returns to 37.0°C Q mcΔT EXECUTE: The rate of heat loss is Q/t = and t = mcΔT (Q/t ) t t t= (70.355 kg)(3480 J/kg ⋅ C°)(0.15 C°) = 0.00525 d = 7.6 minutes 7.00 × 106 J/day EVALUATE: Even if all the BMR energy stays in the body, it takes the body several minutes to return to its normal temperature © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-10 17.48 Chapter 17 IDENTIFY: For a temperature change Q = mcΔT and for the liquid to solid phase change Q = −mLf SET UP: For water, c = 4.19 × 103 J/kg ⋅ K and Lf = 3.34 × 105 J/kg EXECUTE: Q = mcΔT − mLf = (0.350 kg)([4.19 × 103 J/kg ⋅ K][−18.0 C°] − 3.34 × 105 J/kg) = −1.43 × 105 J The minus sign says 1.43 × 105 J must be removed from the water ⎛ cal ⎞ (1.43 × 105 J) ⎜ ⎟ = 3.42 × 10 cal = 34.2 kcal 186 J ⎝ ⎠ ⎛ Btu ⎞ (1.43 × 105 J) ⎜ ⎟ = 136 Btu ⎝ 1055 J ⎠ EVALUATE: Q < when heat comes out of an object The equation Q = mcΔT puts in the correct sign automatically, from the sign of ΔT = Tf − Ti But in Q = ± L we must select the correct sign 17.49 IDENTIFY and SET UP: Use Eq (17.13) for the temperature changes and Eq (17.20) for the phase changes EXECUTE: Heat must be added to the following: ice at −10.0°C → ice at 0°C Qice = mcice ΔT = (12.0 × 10−3 kg)(2100 J/kg ⋅ K)(0°C − ( −10.0°C)) = 252 J phase transition ice (0°C) → liquid water (0°C)(melting) Qmelt = + mLf = (12.0 × 10−3 kg)(334 × 103 J/kg) = 4.008 × 103 J water at 0°C (from melted ice) → water at 100°C Qwater = mcwater ΔT = (12.0 × 10−3 kg)(4190 J/kg ⋅ K)(100°C − 0°C) = 5.028 × 103 J phase transition water (100°C) → steam (100°C)(boiling) Qboil = + mL v = (12.0 × 10−3 kg)(2256 × 103 J/kg) = 2.707 × 104 J The total Q is Q = 252 J + 4.008 × 103 J + 5.028 × 103 J + 2.707 × 104 J = 3.64 × 104 J (3.64 × 104 J)(1 cal/4.186 J) = 8.70 × 103 cal (3.64 × 104 J)(1 Btu/1055 J) = 34.5 Btu 17.50 EVALUATE: Q is positive and heat must be added to the material Note that more heat is needed for the liquid to gas phase change than for the temperature changes IDENTIFY: Q = mcΔT for a temperature change and Q = + mL f for the solid to liquid phase transition The ice starts to melt when its temperature reaches 0.0°C The system stays at 0.00°C until all the ice has melted SET UP: For ice, c = 2.10 × 103 J/kg ⋅ K For water, L f = 3.34 × 105 J/kg EXECUTE: (a) Q to raise the temperature of ice to 0.00°C: Q = mc ΔT = (0.550 kg)(2.10 × 103 J/kg ⋅ K)(15.0 C°) = 1.73 × 104 J t = 1.73 × 104 J = 21.7 800.0 J/min (b) To melt all the ice requires Q = mL f = (0.550 kg)(3.34 × 105 J/kg) = 1.84 × 105 J 1.84 × 105 J = 230 The total time after the start of the heating is 252 800.0 J/min (c) A graph of T versus t is sketched in Figure 17.50 EVALUATE: It takes much longer for the ice to melt than it takes the ice to reach the melting point t= Figure 17.50 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17-21 EXECUTE: In deriving Eq (17.12), it was assumed that ΔL = 0; if this is not the case when there are both ⎛ F ⎞ thermal and tensile stresses, Eq (17.12) becomes ΔL = L0 ⎜ α ΔT + ⎟ (See Problem 17.85.) For the AY ⎠ ⎝ situation in this problem, there are two length changes which must sum to zero, and so Eq (17.12) may be ⎛ ⎛ F ⎞ F ⎞ extended to two materials a and b in the form L0a ⎜ α a ΔT + ⎟ + L0b ⎜ α b ΔT + ⎟ = Note that in AY AY a ⎠ b⎠ ⎝ ⎝ the above, ΔT , F and A are the same for the two rods Solving for the stress F /A, F α L + α b L0b = a 0a ΔT Putting in the numbers gives A L0a /Ya + L0b /Yb F (1.2 × 10−5 (C°) −1 )(0.450 m) + (2.4 × 10−5 (C°) −1 )(0.250 m) =− (60.0 C°) = −1.2 × 108 Pa A (0.450 m)/(20 × 1010 Pa) + (0.250 m)/(7 × 1010 Pa) EVALUATE: F/A is negative and the stress is compressive If the steel rod was considered alone and its length was held fixed, the stress would be −Ysteelα steel ΔT = −1.4 × 108 Pa For the aluminum rod alone the stress would be −Yaluminumα aluminum ΔT = −1.0 × 108 Pa The stress for the combined rod is the average of 17.89 these two values (a) IDENTIFY and SET UP: The diameter of the ring undergoes linear expansion (increases with T) just like a solid steel disk of the same diameter as the hole in the ring Heat the ring to make its diameter equal to 2.5020 in ΔL 0.0020 in EXECUTE: ΔL = α L0ΔT so ΔT = = = 66.7 C° L0α (2.5000 in.)(1.2 × 10−5 (C°) −1) T = T0 + ΔT = 20.0°C + 66.7 C° = 87°C (b) IDENTIFY and SET UP: Apply the linear expansion equation to the diameter of the brass shaft and to the diameter of the hole in the steel ring EXECUTE: L = L0 (1 + α ΔT ) Want Ls (steel) = Lb (brass) for the same ΔT for both materials: L0s (1 + α s ΔT ) = L0b (1 + α b ΔT ) so L0s + L0sα s ΔT = L0b + L0bα b ΔT ΔT = L0b − L0s 2.5020 in − 2.5000 in = L0sα s − L0bα b (2.5000 in.)(1.2 × 10−5 (C°) −1 ) − (2.5050 in.)(2.0 × 10−5 (C°)−1 ) 0.0020 C° = −100 C° 3.00 × 10−5 − 5.00 × 10−5 T = T0 + ΔT = 20.0°C − 100 C° = −80°C ΔT = EVALUATE: Both diameters decrease when the temperature is lowered but the diameter of the brass shaft decreases more since α b > α s ; |ΔLb |−|ΔLs | = 0.0020 in 17.90 IDENTIFY: Follow the derivation of Eq (17.12) SET UP: For steel, the bulk modulus is B = 1.6 × 1011 Pa and the volume expansion coefficient is β = 3.6 × 10−5 K −1 EXECUTE: (a) The change in volume due to the temperature increase is βV ΔT , and the change in V volume due to the pressure increase is − Δp Setting the net change equal to zero, B Δp β V ΔT = V , or Δp = Bβ ΔT B (b) From the above, Δp = (1.6 × 1011 Pa)(3.6 × 10−5 K −1)(15.0 K) = 8.6 × 107 Pa EVALUATE: Δp in part (b) is about 850 atm A small temperature increase corresponds to a very large pressure increase © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-22 17.91 Chapter 17 IDENTIFY: Apply Eq (11.14) to the volume increase of the liquid due to the pressure decrease Eq (17.8) gives the volume decrease of the cylinder and liquid when they are cooled Can think of the liquid expanding when the pressure is reduced and then contracting to the new volume of the cylinder when the temperature is reduced SET UP: Let β1 and β m be the coefficients of volume expansion for the liquid and for the metal Let ΔT be the (negative) change in temperature when the system is cooled to the new temperature EXECUTE: Change in volume of cylinder when cool: ΔVm = β mV0ΔT (negative) Change in volume of liquid when cool: ΔV1 = β1V0ΔT (negative) The difference ΔV1 − ΔVm must be equal to the negative volume change due to the increase in pressure, which is −ΔpV0 /B = − k ΔpV0 Thus ΔV1 − ΔVm = − k ΔpV0 ΔT = − ΔT = − k Δp β1 − β m (8.50 × 10−10 Pa −1 )(50.0 atm)(1.013 × 105 Pa/1 atm) 4.80 × 10−4 K −1 − 3.90 × 10−5 K −1 T = T0 + ΔT = 30.0°C − 9.8 C° = 20.2°C 17.92 = −9.8 C° EVALUATE: A modest temperature change produces the same volume change as a large change in pressure; B >> β for the liquid IDENTIFY: Qsystem = Assume that the normal melting point of iron is above 745°C so the iron initially is solid SET UP: For water, c = 4190 J/kg ⋅ K and Lv = 2256 × 103 J/kg For solid iron, c = 470 J/kg ⋅ K EXECUTE: The heat released when the iron slug cools to 100°C is Q = mcΔT = (0.1000 kg)(470 J/kg ⋅ K)(645 K) = 3.03 × 104 J The heat absorbed when the temperature of the water is raised to 100°C is Q = mcΔT = (0.0850 kg)(4190 J/kg ⋅ K)(80.0 K) = 2.85 × 104 J This is less than the heat released from the iron and 3.03 × 104 J − 2.85 × 104 J = 1.81 × 103 J of heat is available for converting some of the liquid water at 100°C to vapor The mass m of water that boils is 1.81 × 103 J m= = 8.01 × 10−4 kg = 0.801 g 2256 × 103 J/kg (a) The final temperature is 100°C (b) There is 85.0 g − 0.801 g = 84.2 g of liquid water remaining, so the final mass of the iron and remaining water is 184.2 g EVALUATE: If we ignore the phase change of the water and write miron ciron (T − 745°C) + mwater cwater (T − 20.0°C) = 0, when we solve for T we will get a value slightly 17.93 larger than 100°C That result is unphysical and tells us that some of the water changes phase (a) IDENTIFY: Calculate K/Q We don’t know the mass m of the spacecraft, but it divides out of the ratio SET UP: The kinetic energy is K = 12 mv The heat required to raise its temperature by 600 C° (but not to melt it) is Q = mcΔT (7700 m/s) K 12 mv v2 = = = = 54.3 Q mcΔT 2cΔT 2(910 J/kg ⋅ K)(600 C°) (b) EVALUATE: The heat generated when friction work (due to friction force exerted by the air) removes the kinetic energy of the spacecraft during reentry is very large, and could melt the spacecraft Manned space vehicles must have heat shields made of very high melting temperature materials, and reentry must be made slowly IDENTIFY: The rate at which thermal energy is being generated equals the rate at which the net torque due to the rope is doing work The energy input associated with a temperature change is Q = mcΔT SET UP: The rate at which work is being done is P = τω For iron, c = 470 J/kg ⋅ K C° = K EXECUTE: The ratio is 17.94 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17.95 17-23 EXECUTE: (a) The net torque that the rope exerts on the capstan, and hence the net torque that the capstan exerts on the rope, is the difference between the forces of the ends of the rope times the radius of the capstan The capstan is doing work on the rope at a rate 2π rad 2π rad P = τω = Fnet r = (520 N)(5.0 × 10−2 m) = 182 W, or 180 W to two figures A larger number T (0.90 s) of turns might increase the force, but for given forces, the torque is independent of the number of turns ΔT Q/t P (182 W) (b) = = = = 0.064 C°/s t mc mc (6.00 kg)(470 J/kg ⋅ K) EVALUATE: The rate of temperature rise is proportional to the difference in tension between the ends of the rope and to the rate at which the capstan is rotating IDENTIFY and SET UP: To calculate Q, use Eq (17.18) in the form dQ = nC dT and integrate, using C (T ) given in the problem Cav is obtained from Eq (17.19) using the finite temperature range instead of an infinitesimal dT EXECUTE: (a) dQ = nCdT T2 T2 T2 T1 T1 T1 Q = n Ñ C dT = n Ñ k (T 3/Q3 )dT = ( nk/Q3 ) Ñ T dt = ( nk/Q3 ) Q= nk 4Q3 (T24 − T14 ) = (b) Cav = (1.50 mol)(1940 J/mol ⋅ K) 4(281 K)3 ( T T2 T1 ) ((40.0 K) − (10.0 K) ) = 83.6 J ΔQ 83.6 J ⎛ ⎞ = ⎜ ⎟ = 1.86 J/mol ⋅ K n ΔT 1.50 mol ⎝ 40.0 K − 10.0 K ⎠ (c) C = k (T/Q)3 = (1940 J/mol ⋅ K)(40.0 K/ 281 K)3 = 5.60 J/mol ⋅ K 17.96 EVALUATE: C is increasing with T, so C at the upper end of the temperature integral is larger than its average value over the interval IDENTIFY: For a temperature change, Q = mcΔT , and for the liquid → solid phase change, Q = − mL f SET UP: The volume Vw of the water determines its mass mw = ρ wVw For water, ρ w = 1000 kg/m3 , c = 4190 J/kg ⋅ K and Lf = 334 × 103 J/kg EXECUTE: Set the heat energy that flows into the water equal to the final gravitational potential energy Lf ρ wVw + cw ρ wVw ΔT = mgh Solving for h gives h= 17.97 (1000 kg/m3 )(1.9 × 0.80 × 0.160 m3 )[334 × 103 J/kg + (4190 J/kg ⋅ K)(37 C°)] (70 kg)(9.8 m/s ) h = 1.73 × 105 m = 173 km EVALUATE: The heat associated with temperature and phase changes corresponds to a very large amount of mechanical energy IDENTIFY: Apply Q = mcΔT to the air in the room SET UP: The mass of air in the room is m = ρV = (1.20 kg/m3 )(3200 m3 ) = 3840 kg W = J/s EXECUTE: (a) Q = (3000 s)(90 students)(100 J/s ⋅ student) = 2.70 × 107 J (b) Q = mcΔT ΔT = Q 2.70 × 107 J = = 6.89 C° mc (3840 kg)(1020 J/kg ⋅ K) ⎛ 280 W ⎞ (c) ΔT = (6.89 C°) ⎜ ⎟ = 19.3 C° ⎝ 100 W ⎠ EVALUATE: In the absence of a cooling mechanism for the air, the air temperature would rise significantly 17.98 T2 IDENTIFY: dQ = nCdT so for the temperature change T1 → T2 , Q = n Ñ C (T ) dT T1 SET UP: ∫ dT = T and ∫ TdT = 12 T Express T1 and T2 in kelvins: T1 = 300 K, T2 = 500 K © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-24 Chapter 17 EXECUTE: Denoting C by C = a + bT , a and b independent of temperature, integration gives b Q = n( a (T2 − T1) + (T22 − T12 )) Q = (3.00 mol)[(29.5 J/mol ⋅ K)(500 K − 300 K) + (4.10 × 10−3 J/mol ⋅ K )((500 K)2 − (300 K)2 )] Q = 1.97 × 104 J EVALUATE: If C is assumed to have the constant value 29.5 J/mol ⋅ K, then Q = 1.77 × 104 J for this temperature change At T1 = 300 K, C = 32.0 J/mol ⋅ K and at T2 = 500 K, C = 33.6 J/mol ⋅ K The average value of C is 32.8 J/mol ⋅ K If C is assumed to be constant and to have this average value, then Q = 1.97 × 104 J, which is equal to the correct value 17.99 IDENTIFY: Use Q = mLf to find the heat that goes into the ice to melt it This amount of heat must be conducted through the walls of the box; Q = Ht Assume the surfaces of the styrofoam have temperatures of 5.00°C and 21.0°C SET UP: For water Lf = 334 × 103 J/kg For styrofoam k = 0.01 W/m ⋅ K One week is 6.048 × 105 s The surface area of the box is 4(0.500 m)(0.800 m) + 2(0.500 m) = 2.10 m EXECUTE: Q = mLf = (24.0 kg)(334 × 103 J/kg) = 8.016 × 106 J H = kA TH − TC Q = Ht gives L tkA(TH − TC ) (6.048 × 105 s)(0.01 W/m ⋅ K)(2.10 m )(21.0°C − 5.00°C) = = 2.5 cm Q 8.016 × 106 J EVALUATE: We have assumed that the liquid water that is produced by melting the ice remains in thermal equilibrium with the ice so has a temperature of 0°C The interior of the box and the ice are not in thermal equilibrium, since they have different temperatures IDENTIFY: For a temperature change Q = mcΔT For the vapor → liquid phase transition, Q = − mLv L= 17.100 SET UP: For water, c = 4190 J/kg ⋅ K and Lv = 2256 × 103 J/kg EXECUTE: The requirement that the heat supplied in each case is the same gives mw cw ΔTw = ms (cw ΔTs + Lv ), where ΔTw = 42.0 K and ΔTs = 65.0 K The ratio of the masses is ms cw ΔTw (4190 J/kg ⋅ K)(42.0 K) = = = 0.0696, mw cw ΔTs + Lv (4190 J/kg ⋅ K)(65.0 K) + 2256 × 103 J/kg so 0.0696 kg of steam supplies the same heat as 1.00 kg of water 17.101 EVALUATE: Note the heat capacity of water is used to find the heat lost by the condensed steam, since the phase transition produces liquid water at an initial temperature of 100°C (a) IDENTIFY and SET UP: Assume that all the ice melts and that all the steam condenses If we calculate a final temperature T that is outside the range 0°C to 100°C then we know that this assumption is incorrect Calculate Q for each piece of the system and then set the total Qsystem = EXECUTE: copper can (changes temperature from 0.0° to T; no phase change) Qcan = mcΔT = (0.446 kg)(390 J/kg ⋅ K)(T − 0.0°C) = (173.9 J/K)T ice (melting phase change and then the water produced warms to T) Qice = + mLf + mcΔT = (0.0950 kg)(334 × 103 J/kg) + (0.0950 kg)(4190 J/kg ⋅ K)(T − 0.0°C) Qice = 3.173 × 104 J + (398.0 J/K)T steam (condenses to liquid and then water produced cools to T) Qsteam = −mLv + mcΔT = −(0.0350 kg)(2256 × 103 J/kg) + (0.0350 kg)(4190 J/kg ⋅ K)(T − 100.0°C) Qsteam = −7.896 × 104 J + (146.6 J/K)T − 1.466 × 104 J = −9.362 × 104 J + (146.6 J/K)T Qsystem = implies Qcan + Qice + Qsteam = (173.9 J/K)T + 3.173 × 104 J + (398.0 J/K)T − 9.362 × 104 J + (146.6 J/K)T = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17-25 (718.5 J/K)T = 6.189 × 104 J 6.189 × 104 J = 86.1°C 718.5 J/K EVALUATE: This is between 0°C and 100°C so our assumptions about the phase changes being complete were correct (b) No ice, no steam and 0.0950 kg + 0.0350 kg = 0.130 kg of liquid water T= 17.102 IDENTIFY: The final amount of ice is less than the initial mass of water, so water remains and the final temperature is 0°C The ice added warms to 0°C and heat comes out of water to convert that water to ice Conservation of energy says Qi + Qw = 0, where Qi and Qw are the heat flows for the ice that is added and for the water that freezes SET UP: Let mi be the mass of ice that is added and mw is the mass of water that freezes The mass of ice increases by 0.418 kg, so mi + mw = 0.418 kg For water, L f = 334 × 103 J/kg and for ice ci = 2100 J/kg ⋅ K Heat comes out of the water when it freezes, so Qw = −mLf EXECUTE: Qi + Qw = gives mici (15.0 C°) + (− mw Lf ) = 0, mw = 0.418 kg − mi , so mici (15.0 C°) + ( −0.418 kg + mi ) L f = (0.418 kg) Lf (0.418 kg)(334 × 103 J/kg) = = 0.382 kg 0.382 kg of ice was added ci (15.0 C°) + Lf (2100 J/kg ⋅ K)(15.0 K) + 334 × 103 J/kg EVALUATE: The mass of water that froze when the ice at −15.0C° was added was 0.868 kg − 0.450 kg − 0.382 kg = 0.036 kg mi = 17.103 IDENTIFY and SET UP: Heat comes out of the steam when it changes phase and heat goes into the water and causes its temperature to rise Qsystem = First determine what phases are present after the system has come to a uniform final temperature (a) EXECUTE: Heat that must be removed from steam if all of it condenses is Q = −mLv = −(0.0400 kg)(2256 × 103 J/kg) = −9.02 × 104 J Heat absorbed by the water if it heats all the way to the boiling point of 100°C: Q = mcΔT = (0.200 kg)(4190 J/kg ⋅ K)(50.0 C°) = 4.19 × 104 J EVALUATE: The water can’t absorb enough heat for all the steam to condense Steam is left and the final temperature then must be 100°C (b) EXECUTE: Mass of steam that condenses is m = Q/Lv = 4.19 × 104 J/2256 × 103 J/kg = 0.0186 kg Thus there is 0.0400 kg − 0.0186 kg = 0.0214 kg of steam left The amount of liquid water is 0.0186 kg + 0.200 kg = 0.219 kg 17.104 IDENTIFY: Heat is conducted out of the body At steady state, the rate of heat flow is the same in both layers (fat and fur) SET UP: Let the temperature of the fat-air boundary be T A section of the two layers is sketched in Figure 17.104 A Kelvin degree is the same size as a Celsius degree, so W/m ⋅ K and W/m ⋅ C° are equivalent units At steady state the heat current through each layer is equal to 50 W The area of each layer is A = 4π r , with r = 0.75 m Figure 17.104 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-26 Chapter 17 EXECUTE: (a) Apply H = kA T = TH − TH − TC to the fat layer and solve for TC = T For the fat layer TH = 31°C L HL (50 W)(4.0 × 10−2 m) = 31°C − = 31°C − 1.4°C = 29.6°C kA (0.20 W/m ⋅ K)(4π )(0.75 m) (b) Apply H = kA TH − TC to the air layer and solve for L = Lair For the air layer TH = T = 29.6°C and L kA(TH − TC ) (0.024 W/m ⋅ K)(4π )(0.75 m) (29.6°C − 2.7°C) = = 9.1 cm H 50 W EVALUATE: The thermal conductivity of air is much lass than the thermal conductivity of fat, so the temperature gradient for the air must be much larger to achieve the same heat current So, most of the temperature difference is across the air layer IDENTIFY: Heat Ql comes out of the lead when it solidifies and the solid lead cools to Tf If mass ms of TC = 2.7°C L = 17.105 steam is produced, the final temperature is Tf = 100°C and the heat that goes into the water is Qw = mw cw (25.0 C°) + ms Lv,w , where mw = 0.5000 kg Conservation of energy says Ql + Qw = Solve for ms The mass that remains is 1.250 kg + 0.5000 kg − ms SET UP: For lead, Lf,l = 24.5 × 103 J/kg, cl = 130 J/kg ⋅ K and the normal melting point of lead is 327.3°C For water, cw = 4190 J/kg ⋅ K and Lv,w = 2256 × 103 J/kg EXECUTE: Ql + Qw = − ml Lf,l + mlcl (−227.3 C°) + mw cw (25.0 C°) + ms Lv,w = ms = ms = ms = ml Lf,l + mlcl ( +227.3 C°) − mw cw (25.0 C°) Lv,w + (1.250 kg)(24.5 × 103 J/kg) + (1.250 kg)(130 J/kg ⋅ K)(227.3 K) − (0.5000 kg)(4190 J/kg ⋅ K)(25.0 K) 2256 × 103 J/kg 1.519 × 104 J = 0.0067 kg The mass of water and lead that remains is 1.743 kg 2256 × 103 J/kg EVALUATE: The magnitude of heat that comes out of the lead when it goes from liquid at 327.3°C to solid at 100.0°C is 6.76 × 104 J The heat that goes into the water to warm it to 100°C is 5.24 × 104 J The 17.106 additional heat that goes into the water, 6.76 × 104 J − 5.24 × 104 J = 1.52 × 104 J converts 0.0067 kg of water at 100°C to steam ΔT IDENTIFY: Apply H = kA and solve for k L SET UP: H equals the power input required to maintain a constant interior temperature EXECUTE: k = H 17.107 L (3.9 × 10−2 m) = (180 W) = 5.0 × 10−2 W/m ⋅ K AΔT (2.18 m )(65.0 K) EVALUATE: Our result is consistent with the values for insulating solids in Table 17.5 ΔT IDENTIFY: Apply H = kA L SET UP: For the glass use L = 12.45 cm, to account for the thermal resistance of the air films on either side of the glass 28.0 C° ⎛ ⎞ EXECUTE: (a) H = (0.120 W/m ⋅ K) (2.00 × 0.95 m ) ⎜ ⎟ = 93.9 W −2 −2 × + × 10 m 10 m ⎝ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat (b) The heat flow through the wood part of the door is reduced by a factor of − 17.108 17.109 17-27 (0.50) = 0.868, so (2.00 × 0.95) it becomes 81.5 W The heat flow through the glass is 28.0 C° 81.5 + 45.0 ⎛ ⎞ H glass = (0.80 W/m ⋅ K)(0.50 m) ⎜ = 1.35 ⎟ = 45.0 W, and so the ratio is −2 93.9 ⎝ 12.45 × 10 m ⎠ EVALUATE: The single-pane window produces a significant increase in heat loss through the door (See Problem 17.109) IDENTIFY: Apply Eq (17.23) HR1 HR2 SET UP: Let ΔT1 = be the temperature difference across the wood and let ΔT2 = be the A A temperature difference across the insulation The temperature difference across the combination is HR ΔT = ΔT1 + ΔT2 The effective thermal resistance R of the combination is defined by ΔT = A H H EXECUTE: ΔT = ΔT1 + ΔT2 gives ( R1 + R2 ) = R, and R = R1 + R2 A A EVALUATE: A good insulator has a large value of R R for the combination is larger than the R for any one of the layers IDENTIFY and SET UP: Use H written in terms of the thermal resistance R: H = AΔT/R, where R = L/k and R = R1 + R2 + … (additive) EXECUTE: single pane Rs = Rglass + Rfilm , where Rfilm = 0.15 m ⋅ K/W is the combined thermal resistance of the air films on the room and outdoor surfaces of the window Rglass = L/k = (4.2 × 10−3 m)/(0.80 W/m ⋅ K) = 0.00525 m ⋅ K/W Thus Rs = 0.00525 m ⋅ K/W + 0.15 m ⋅ K/W = 0.1553 m ⋅ K/W double pane Rd = Rglass + Rair + Rfilm , where Rair is the thermal resistance of the air space between the panes Rair = L/k = (7.0 × 10−3 m)/(0.024 W/m ⋅ K) = 0.2917 m ⋅ K/W Thus Rd = 2(0.00525 m ⋅ K/W) + 0.2917 m ⋅ K/W + 0.15 m ⋅ K/W = 0.4522 m ⋅ K/W H s = AΔT/Rs , H d = AΔT/Rd , so H s /H d = Rd /Rs (since A and ΔT are same for both) H s /H d = (0.4522 m ⋅ K/W)/(0.1553 m ⋅ K/W) = 2.9 17.110 EVALUATE: The heat loss is about a factor of less for the double-pane window The increase in R for a double-pane is due mostly to the thermal resistance of the air space between the panes kAΔT IDENTIFY: Apply H = to each rod Conservation of energy requires that the heat current through L the copper equals the sum of the heat currents through the brass and the steel SET UP: Denote the quantities for copper, brass and steel by 1, and 3, respectively, and denote the temperature at the junction by T0 EXECUTE: (a) H1 = H + H Using Eq (17.21) and dividing by the common area gives (k1/L1 ) k1 k k (100°C) Substitution (100°C − T0 ) = T0 + T0 Solving for T0 gives T0 = (k1/L1) + (k2 /L2 ) + ( k3/L3 ) L1 L2 L3 of numerical values gives T0 = 78.4°C (b) Using H = kA ΔT for each rod, with ΔT1 = 21.6 C°, ΔT2 = ΔT3 = 78.4 C° gives L H1 = 12.8 W, H = 9.50 W and H = 3.30 W EVALUATE: In part (b), H1 is seen to be the sum of H and H © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-28 17.111 Chapter 17 (a) EXECUTE: Heat must be conducted from the water to cool it to 0°C and to cause the phase transition The entire volume of water is not at the phase transition temperature, just the upper surface that is in contact with the ice sheet (b) IDENTIFY: The heat that must leave the water in order for it to freeze must be conducted through the layer of ice that has already been formed SET UP: Consider a section of ice that has area A At time t let the thickness be h Consider a short time interval t to t + dt Let the thickness that freezes in this time be dh The mass of the section that freezes in the time interval dt is dm = ρ dV = ρ A dh The heat that must be conducted away from this mass of water to freeze it is dQ = dmLf = ( ρ ALf ) dh H = dQ/dt = kA(ΔT/h), so the heat dQ conducted in time dt ⎛ T −T ⎞ throughout the thickness h that is already there is dQ = kA ⎜ H C ⎟ dt Solve for dh in terms of dt and ⎝ h ⎠ integrate to get an expression relating h and t EXECUTE: Equate these expressions for dQ ⎛ T −T ⎞ ρ ALf dh = kA ⎜ H C ⎟ dt ⎝ h ⎠ ⎛ k (TH − TC ) ⎞ h dh = ⎜ ⎟ dt ρ Lf ⎝ ⎠ Integrate from t = to time t At t = the thickness h is zero h t Ñ 0h dh = [k (TH − TC )/ρLf ]Ñ 0dt h2 = k (TH − TC ) 2k (TH − TC ) t and h = t ρ Lf ρ Lf t The thickness after time t is proportional to (c) The expression in part (b) gives t = h ρ Lf (0.25 m) (920 kg/m3 )(334 × 103 J/kg) = = 6.0 × 105 s 2k (TH − TC ) 2(1.6 W/m ⋅ K)(0°C − (−10°C)) t = 170 h (d) Find t for h = 40 m t is proportional to h , so t = (40 m/0.25 m) (6.00 × 105 s) = 1.5 × 1010 s This is 17.112 about 500 years With our current climate this will not happen EVALUATE: As the ice sheet gets thicker, the rate of heat conduction through it decreases Part (d) shows that it takes a very long time for a moderately deep lake to totally freeze IDENTIFY: Apply Eq (17.22) at each end of the short element In part (b) use the fact that the net heat current into the element provides the Q for the temperature increase, according to Q = mcΔT SET UP: dT/dx is the temperature gradient EXECUTE: (a) H = (380 W/m ⋅ K)(2.50 × 10−4 m )(140 C°/m) = 13.3 W Q ΔT ΔT = mc , where = 0.250 C°/s t t t dT ρ cΔx ⎛ ΔT ⎞ = + ⎜ ⎟ dx k ⎝ t ⎠ (b) Denoting the two ends of the element as and 2, H − H1 = kA kA dT dx dT dx 2 − kA dT dT ⎛ ΔT ⎞ = mc ⎜ ⎟ The mass m is ρ AΔx, so dx t dx ⎝ ⎠ = 140 C°/m + (1.00 × 10 kg/m )(520 J/kg ⋅ K)(1.00 × 10−2 m)(0.250 C°/s) = 174 C°/m 380 W/m ⋅ K EVALUATE: At steady-state the temperature of the short element is no longer changing and H1 = H 17.113 IDENTIFY: The rate of heat conduction through the walls is 1.25 kW Use the concept of thermal resistance and the fact that when insulating materials are in layers, the R values are additive SET UP: The total area of the four walls is 2(3.50 m)(2.50 m) + 2(3.00 m)(2.50 m) = 32.5 m © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat EXECUTE: H = A Rw = 17-29 A(TH − TC ) (32.5 m )(17.0 K) TH − TC = = 0.442 m ⋅ K/W For the wood, gives R = H R 1.25 × 103 W L 1.80 × 10−2 m = = 0.300 m ⋅ K/W For the insulating material, Rin = R − Rw = 0.142 m ⋅ K/W k 0.060 W/m ⋅ K Lin L 1.50 × 10−2 m = 0.106 W/m ⋅ K and kin = in = Rin 0.142 m ⋅ K/W kin EVALUATE: The thermal conductivity of the insulating material is larger than that of the wood, the thickness of the insulating material is less than that of the wood, and the thermal resistance of the wood is about three times that of the insulating material IDENTIFY: I1r12 = I 2r22 Apply H = Aeσ T (Eq 17.25) to the sun Rin = 17.114 SET UP: I1 = 1.50 × 103 W/m when r = 1.50 × 1011 m EXECUTE: (a) The energy flux at the surface of the sun is ⎛ 1.50 × 1011 m ⎞ I = (1.50 × 103 W/m ) ⎜ = 6.97 × 107 W/m ⎜ 6.96 × 108 m ⎟⎟ ⎝ ⎠ 1 ⎤4 ⎡ H ⎤ ⎡ 6.97 × 10 W/m = (b) Solving Eq (17.25) with e = 1, T = ⎢ ⎢ ⎥ = 5920 K ⎥ − ⎣Aσ⎦ ⎢⎣ 5.67 × 10 W/m ⋅ K ⎥⎦ EVALUATE: The total power output of the sun is P = 4π r22 I = × 1026 W 17.115 IDENTIFY and SET UP: Use Eq (17.26) to find the net heat current into the can due to radiation Use Q = Ht to find the heat that goes into the liquid helium, set this equal to mL and solve for the mass m of helium that changes phase EXECUTE: Calculate the net rate of radiation of heat from the can H net = Aeσ (T − Ts4 ) The surface area of the cylindrical can is A = 2π rh + 2π r (See Figure 17.115.) Figure 17.115 A = 2π r (h + r ) = 2π (0.045 m)(0.250 m + 0.045 m) = 0.08341 m H net = (0.08341 m )(0.200)(5.67 × 10−8 W/m ⋅ K )((4.22 K)4 − (77.3 K) ) H net = −0.0338 W (the minus sign says that the net heat current is into the can) The heat that is put into the can by radiation in one hour is Q = −( H net )t = (0.0338 W)(3600 s) = 121.7 J This heat boils a mass m Q 121.7 J = = 5.82 × 10−3 kg = 5.82 g Lf 2.09 × 104 J/kg EVALUATE: In the expression for the net heat current into the can the temperature of the surroundings is raised to the fourth power The rate at which the helium boils away increases by about a factor of (293/77) = 210 if the walls surrounding the can are at room temperature rather than at the temperature of of helium according to the equation Q = mLf , so m = 17.116 the liquid nitrogen IDENTIFY: The nonmechanical part of the basal metabolic rate (i.e., the heat) leaves the body by radiation from the surface SET UP: In the radiation equation, H net = Aeσ (T − Ts4 ), the temperatures must be in kelvins; e = 1.0, T = 30°C = 303 K, and Ts = 18°C = 291 K Call the basal metabolic rate BMR © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-30 Chapter 17 EXECUTE: (a) H net = Aeσ (T − Ts ) H net = (2.0 m )(1.0)(5.67 × 10−8 W/m ⋅ K )([303 K]4 − [291 K]4 ) = 140 W (b) (0.80)BMR = 140 W, so BMR = 180 W 17.117 EVALUATE: If the emissivity of the skin were less than 1.0, the body would radiate less so the BMR would have to be lower than we found in (b) IDENTIFY: The jogger radiates heat but the air radiates heat back into the jogger SET UP: The emissivity of a human body is taken to be 1.0 In the equation for the radiation heat current, H net = Aeσ (T − Ts ), the temperatures must be in kelvins EXECUTE: (a) Pjog = (0.80)(1300 W) = 1.04 × 103 J/s (b) H net = Aeσ (T − Ts ), which gives H net = (1.85 m )(1.00)(5.67 × 10−8 W/m ⋅ K )([306 K]4 − [313 K]4 ) = − 87.1 W The person gains 87.1 J of heat each second by radiation (c) The total excess heat per second is 1040 J/s + 87 J/s = 1130 J/s (d) In = 60 s, the runner must dispose of (60 s)(1130 J/s) = 6.78 × 104 J If this much heat goes to evaporate water, the mass m of water that evaporates in one minute is given by Q = mL v , so Q 6.78 × 104 J = = 0.028 kg = 28 g Lv 2.42 × 106 J/kg (e) In a half-hour, or 30 minutes, the runner loses (30 min)(0.028 kg/min) = 0.84 kg The runner must m= 0.84 L = 1.1 bottles 0.750 L/bottle EVALUATE: The person gains heat by radiation since the air temperature is greater than his skin temperature IDENTIFY: The heat generated will remain in the runner’s body, which will increase his body temperature SET UP: Problem 17.117 calculates that the net rate of heat input to the person is 1130 W Q = mcΔT F° = C° drink 0.84 L, which is 17.118 EXECUTE: (a) Q = Pt = (1130 W)(1800 s) = 2.03 × 106 J Q = mcΔT so Q 2.03 × 106 J = = 8.6 C° mc (68 kg)(3480 J/kg ⋅ C°) (b) ΔT = (8.6 C°)(9 F°/5 C°) = 15.5 F° T = 98.6°F + 15.5 F° = 114°F ΔT = 17.119 EVALUATE: This body temperature is lethal IDENTIFY: For the water, Q = mcΔT SET UP: For water, c = 4190 J/kg ⋅ K EXECUTE: (a) At steady state, the input power all goes into heating the water, so P = Q mcΔT and = t t Pt (1800 W)(60 s/min) = = 51.6 K, and the output temperature is cm (4190 J/kg ⋅ K)(0.500 kg/min) 18.0°C + 51.6 C° = 69.6°C ΔT = 17.120 EVALUATE: (b) At steady state, the temperature of the apparatus is constant and the apparatus will neither remove heat from nor add heat to the water IDENTIFY: For the air the heat input is related to the temperature change by Q = mcΔT SET UP: The rate P at which heat energy is generated is related to the rate P0 at which food energy is consumed by the hamster by P = 0.10 P0 EXECUTE: (a) The heat generated by the hamster is the heat added to the box; Q ΔT P = = mc = (1.20 kg/m3 )(0.0500 m3 )(1020 J/kg ⋅ K)(1.60 C° /h) = 97.9 J/h t t © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17-31 (b) Taking the efficiency into account, the mass M of seed that must be eaten in time t is M P0 P/(10%) 979 J/h = = = = 40.8 g/h t Lc Lc 24 J/g 17.121 EVALUATE: This is about 1.5 ounces of seed consumed in one hour IDENTIFY: Heat Qi goes into the ice when it warms to 0°C, melts, and the resulting water warms to the final temperature Tf Heat Qow comes out of the ocean water when it cools to Tf Conservation of energy gives Qi + Qow = SET UP: For ice, ci = 2100 J/kg ⋅ K For water, Lf = 334 × 103 J/kg and cw = 4190 J/kg ⋅ K Let m be the total mass of the water on the earth’s surface So mi = 0.0175m and mow = 0.975m EXECUTE: Qi + Qow = gives mici (30 C°) + mi Lf + micwTf + mow cw (Tf − 5.00°C) = Tf = − mici (30 C°) − mi Lf + mow cw (5.00 C°) (mi + mow )cw Tf = −(0.0175m)(2100 J/kg ⋅ K)(30 K) − (0.0175m)(334 × 103 J/kg) + (0.975m)(4190 J/kg ⋅ K)(5.00 K) (0.0175m + 0.975m)(4190 J/kg ⋅ K) Tf = 17.122 1.348 × 104 J/kg = 3.24°C The temperature decrease is 1.76 C° 4.159 × 103 J/kg ⋅ K EVALUATE: The mass of ice in the icecaps is much less than the mass of the water in the oceans, but much more heat is required to change the phase of kg of ice than to change the temperature of kg of water C°, so the lowering of the temperature of the oceans would be appreciable IDENTIFY: The oceans take time to increase (or decrease) in temperature because they contain a large mass of water which has a high specific heat SET UP: The radius of the earth is RE = 6.38 × 106 m Since an ocean depth of 100 m is much less than the radius of the earth, we can calculate the volume of the water to this depth as the surface area of the oceans times 100 m The total surface area of the earth is Aearth = 4π RE The density of seawater is 1.03 × 103 kg/m3 (Table12.1) EXECUTE: The surface area of the oceans is (2/3) Aearth = (2/3)(4π )(6.38 × 106 m)2 = 3.41 × 1014 m The total rate of solar energy incident on the oceans is (1050 W/m )(3.41 × 1014 m ) = 3.58 × 1017 W 12 hours per day for 30 days is (12)(30)(3600) s = 1.30 × 106 s, so the total solar energy input to the oceans in one month is (1.30 × 106 s)(3.58 × 1017 W) = 4.65 × 1023 J The volume of the seawater absorbing this energy is (100 m)(3.41 × 1014 m ) = 3.41 × 1016 m3 The mass of this water is m = ρV = (1.03 × 103 kg/m3 )(3.41 × 1016 m3 ) = 3.51 × 1019 kg Q = mcΔT , so Q 4.65 × 1023 J = = 3.4 C° mc (3.51 × 1019 kg)(3890 J/kg ⋅ C°) EVALUATE: A temperature rise of 3.4 C° is significant The solar energy input is a very large number, but so is the total mass of the top 100 m of seawater in the oceans dT is the temperature gradient at IDENTIFY: Apply Eq (17.22) to different points along the rod, where dx each point SET UP: For copper, k = 385 W/m ⋅ K EXECUTE: (a) The initial temperature distribution, T = (100°C)sin π x/L, is shown in Figure 17.123a ΔT = 17.123 (b) After a very long time, no heat will flow, and the entire rod will be at a uniform temperature which must be that of the ends, 0°C (c) The temperature distribution at successively greater times T1 < T2 < T3 is sketched in Figure 17.123b © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-32 Chapter 17 (d) dT = (100°C)(π /L)cos π x/L At the ends, x = and x = L, the cosine is ±1 and the temperature dx gradient is ± (100°C)(π /0.100 m) = ±3.14 × 103 C°/m (e) Taking the phrase “into the rod” to mean an absolute value, the heat current will be k A dT = (385.0 W/m ⋅ K)(1.00 × 10−4 m )(3.14 × 103 C°/m) = 121 W dx (f) Either by evaluating dT at the center of the rod, where π x/L = π /2 and cos(π /2) = 0, or by checking dx the figure in part (a), the temperature gradient is zero, and no heat flows through the center; this is consistent with the symmetry of the situation There will not be any heat current at the center of the rod at any later time k (385 W/m ⋅ K) (g) = = 1.1 × 10−4 m /s ρ c (8.9 × 103 kg/m3 )(390 J/kg ⋅ K) (h) Although there is no net heat current, the temperature of the center of the rod is decreasing; by considering the heat current at points just to either side of the center, where there is a non-zero temperature gradient, there must be a net flow of heat out of the region around the center Specifically, ⎛ ∂T ⎞ ∂T ∂T ∂ 2T ⎟ = kA Δ x, from H (( L/2) + Δ x) − H (( L/2) − Δ x) = ρ A(Δ x )c = kA ⎜ − ⎜ ∂x ( L/ + Δx ) ∂x ( L/2 − Δx ) ⎟ ∂t ∂x ⎝ ⎠ which the Heat Equation, ∂T k ∂ 2T = is obtained At the center of the rod, ∂t ρ c ∂x 2 ⎞ ∂ 2T = −(100 C°)(π /L) , and so ∂T = −(1.11 × 10−4 m /s)(100 C°) ⎛ π ⎜ ⎟ = −10.9 C°/s, or ∂t ∂x ⎝ 0.100 m ⎠ −11 C°/s to two figures (i) 100 C° = 9.17 s 10.9 C°/s (j) Decrease (that is, become less negative), since as T decreases, ∂ 2T ∂x decreases This is consistent with the graphs, which correspond to equal time intervals (k) At the point halfway between the end and the center, at any given time ∂ T2 is a factor of ∂x sin(π /4) = 1/ less than at the center, and so the initial rate of change of temperature is −7.71 C° /s EVALUATE: A plot of temperature as a function of both position and time for ≤ t ≤ 50 s is shown in Figure 17.123c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17-33 Figure 17.123 17.124 IDENTIFY: Apply Eq (17.21) For a spherical or cylindrical surface, the area A in Eq (17.21) is not constant, and the material must be considered to consist of shells with thickness dr and a temperature difference between the inside and outside of the shell dT The heat current will be a constant, and must be found by integrating a differential equation SET UP: The surface area of a sphere is 4π r The surface area of the curved side of a cylinder is 2π rl ln(1 + ε ) ≈ ε when ε (a) Equation (17.21) becomes H = k (4π r ) appropriate limits, dT H dr or = k dT Integrating both sides between the dr 4π r H ⎛1 1⎞ ⎜ − ⎟ = k (T2 − T1 ) In this case the “appropriate limits” have been chosen so that 4π ⎝ a b ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 17-34 Chapter 17 if the inner temperature T2 is at the higher temperature T1, the heat flows outward; that is, dT < dr k 4π ab(T2 − T1) Solving for the heat current, H = b−a (b) The rate of change of temperature with radius is of the form dT = B2 , with B a constant Integrating dr r ⎛ 1⎞ ⎛1 1⎞ from r = a to r and from r = a to r = b gives T ( r ) − T2 = B ⎜ − ⎟ and T1 − T2 = B ⎜ − ⎟ Using the a r ⎝ ⎠ ⎝a b⎠ ⎛ r − a ⎞⎛ b ⎞ second of these to eliminate B and solving for T(r) gives T ( r ) = T2 − (T2 − T1 ) ⎜ ⎟⎜ ⎟ There are, of ⎝ b − a ⎠⎝ r ⎠ course, many equivalent forms As a check, note that at r = a, T = T2 and at r = b, T = T1 dT H or = kLdT , which integrates, 2π r dr 2π k L(T2 − T1) H ln(b/a ) = kL(T2 − T1 ), or H = with the same condition on the limits, to 2π ln(b/a) (c) As in part (a), the expression for the heat current is H = k (2π rL) ln( r/a ) ln(b/a) EVALUATE: (e) For the sphere: Let b − a = l , and approximate b ~ a, with a the common radius Then the (d) A method similar to that used in part (b) gives T ( r ) = T2 + (T1 − T2 ) 17.125 surface area of the sphere is A = 4π a , and the expression for H is that of Eq (17.21) (with l instead of L, which has another use in this problem) For the cylinder: with the same notation, consider l⎞ l ⎛b⎞ ⎛ ln ⎜ ⎟ = ln ⎜1 + ⎟ ~ , where the approximation for ln(1 + ε ) for small ε has been used The expression a a ⎝ ⎠ ⎝ ⎠ a for H then reduces to k (2π La )(ΔT/l ), which is Eq (17.21) with A = 2π La IDENTIFY: From the result of Problem 17.124, the heat current through each of the jackets is related to the temperature difference by H = 2π lk ΔT , where l is the length of the cylinder and b and a are the inner ln(b/a ) and outer radii of the cylinder SET UP: Let the temperature across the cork be ΔT1 and the temperature across the styrofoam be ΔT2 , with similar notation for the thermal conductivities and heat currents EXECUTE: (a) ΔT1 + ΔT2 = ΔT = 125 C° Setting H1 = H = H and canceling the common factors, −1 ⎛ k ln1.5 ⎞ ΔT1k1 ΔT2k2 = Eliminating ΔT2 and solving for ΔT1 gives ΔT1 = ΔT ⎜1 + ⎟ Substitution of ln ln1.5 ⎝ k2 ln ⎠ numerical values gives ΔT1 = 37 C°, and the temperature at the radius where the layers meet is 140°C − 37°C = 103°C (b) Substitution of this value for ΔT1 into the above expression for H1 = H gives 2π (2.00 m)(0.0400 W/m ⋅ K) (37 C°) = 27 W ln 2π (2.00 m)(0.0100 W/m ⋅ K) EVALUATE: ΔT = 103°C − 15°C = 88 C° H = (88 C°) = 27 W This is the ln(6.00/4.00) H= 17.126 same as H1, as it should be IDENTIFY: Apply the concept of thermal expansion In part (b) the object can be treated as a simple pendulum SET UP: For steel α = 1.2 × 10−5 (C°) −1 yr = 86,400 s EXECUTE: (a) In hot weather, the moment of inertia I and the length d in Eq (14.39) will both increase by the same factor, and so the period will be longer and the clock will run slow (lose time) Similarly, the clock will run fast (gain time) in cold weather © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Temperature and Heat 17.127 17-35 (b) ΔL = αΔT = (1.2 × 10−5 (C°) −1 )(10.0 C°) = 1.2 × 10−4 L0 (c) To avoid possible confusion, denote the pendulum period by τ For this problem, Δτ = ΔL = 6.0 × 10−5 so in one day the clock will gain (86,400 s)(6.0 × 10−5 ) = 5.2 s τ L Δτ Δτ 1.0 s (d) = αΔT = gives ΔT = 2[(1.2 × 10−5 (C°) −1)(86,400)]−1 = 1.9 C° T must be τ τ 86,400 s controlled to within 1.9 C° EVALUATE: In part (d) the answer does not depend on the period of the pendulum It depends only on the fractional change in the period IDENTIFY: The rate in (iv) is given by Eq (17.26), with T = 309 K and Ts = 320 K The heat absorbed in the evaporation of water is Q = mL m = ρ V EXECUTE: (a) The rates are: (i) 280 W, (ii) (54 J/h ⋅ C° ⋅ m )(1.5 m )(11 C°)/(3600 s/h) = 0.248 W, SET UP: m = ρV , so (iii) (1400 W/m )(1.5 m ) = 2.10 × 103 W, (iv) (5.67 × 10−8 W/m ⋅ K )(1.5 m )((320 K)4 − (309 K)4 ) = 116 W The total is 2.50 kW, with the largest portion due to radiation from the sun P 2.50 × 103 W = = 1.03 × 10−6 m3/s This is equal to 3.72 L/h ρ Lv (1000 kg/m3 )(2.42 × 106 J/kg ⋅ K) (c) Redoing the above calculations with e = and the decreased area gives a power of 945 W and a corresponding evaporation rate of 1.4 L/h Wearing reflective clothing helps a good deal Large areas of loose-weave clothing also facilitate evaporation EVALUATE: The radiant energy from the sun absorbed by the area covered by clothing is assumed to be zero, since e ≈ for the clothing and the clothing reflects almost all the radiant energy incident on it For the same reason, the exposed skin area is the area used in Eq (17.26) (b) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... 17.23 17-5 25.000 cm(1 + α BR ΔT ) (25.000 cm)[1 + (2.0 × 10−5 (C°)−1 ) (130 C°)] = = 25.026 cm + α ST ΔT + (1.2 × 10−5 (C°) −1) (130 C°) EVALUATE: The space inside the steel cylinder expands just like... 10−3 kg)[(865 m/s)2 − (534 m/s)2 ] = 3.47 × 103 J, so for the water Q 3.47 × 103 J = = 0.0 613 C° mc (13. 5 kg)(4.19 × 103 J/kg ⋅ C°) EVALUATE: The heat energy required to change the temperature... kelvins EXECUTE: (a) Pjog = (0.80) (130 0 W) = 1.04 × 103 J/s (b) H net = Aeσ (T − Ts ), which gives H net = (1.85 m )(1.00)(5.67 × 10−8 W/m ⋅ K )([306 K]4 − [ 313 K]4 ) = − 87.1 W The person gains