Answers to Even-Numbered Conceptual Questions 2. If the temperature of the air in a house is increased, and the amount of air in the house remains constant, it
follows from the ideal gas law that the pressure will increase as well. 4. Yes. If the pressure and volume are changed in such a way that their product remains the same, it follows
from the ideal gas law that the temperature of the gas will remain the same. If the temperature of the gas is the same, the average kinetic energy of its molecules will not change.
6. No. The temperature at which water boils on a mountaintop is less than its boiling temperature at sea level due to the low atmospheric pressure on the mountain. Therefore, if the stove is barely able to boil water on the mountain, it will not be able to boil it at sea level, where the required temperature is greater.
8. If we look at the phase diagram in Figure 17-16, we can see that in order to move upward in the graph from the sublimation curve to the fusion curve, the pressure acting on the system must be increased.
10. No. Water is at 0˚C whenever it is in equilibrium with ice. The ice cube thrown into the pool will soon melt, however, showing that the ice cube–pool system is not in equilibrium. As a result, there is no reason to expect that the water in the pool is at 0˚C.
Solutions to Problems and Conceptual Exercises
2. Picture the Problem: We are to compare a mole of He and a mole of O2.
Strategy: Use the concept of a mole to answer the conceptual question. Solution: By the definition of a mole, the number of molecules in one mole of He is equal to the number of molecules
in one mole of O2. However, each molecule of O2 has two atoms, but the He molecules are monatomic, so the number of atoms in one mole of helium is less than the number of atoms in one mole of oxygen.
Insight: A mole is similar to a dozen; the words always refer to the same number, 6.02×1023 and 12, respectively.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
4. Picture the Problem: Two containers hold ideal gases at the same temperature. Container A has twice the volume and
half the number of molecules as container B. Strategy: Use the ideal gas model to answer the conceptual question. Solution: From the given information we know that VA = 2VB and nA = nB/2. The pressure of an ideal gas is given by
P = nRT/V. If PB = nBRT/VB it follows that PA = (nB/2)RT/(2VB) = PB/4, or A B 1 4 .P P =
Insight: A mole is similar to a dozen; the words always refer to the same number, 6.02×1023 and 12, respectively.
6. Picture the Problem: A person inhales and holds her breath. The volume of the air in her lungs expands as the temperature increases.
Strategy: Use equation 17-8 to solve for the final volume and then subtract the initial volume to find the increase in volume.
Solution: 1. Use equation 17-8 to find the volume ratio:
( )i f f f
i f i i
273.15 37 K 1.14
273.15 KV V V TT T V T
+= ⇒ = = =
3. Subtract the initial volume: ( )f i i i1.14 0.14 4.1 L 0.57 LV V V V VΔ = − = − = =
Insight: Another approach would be to calculate the final volume (4.7 L) and then subtract the initial volume (4.1 L) but the 0.6 L result has only one significant digit due to the rules of subtraction.
8. Picture the Problem: An automobile tire has a fixed volume and temperature. Adding air to the tire increases the tire pressure.
Strategy: Solve the ideal gas law, equation 17-5, for the number of moles before and after air is added. Subtract these two values to find the amount of added air.
Solution: 1. Solve equation 17-5 for the initial number of moles:
( )( )( ) ( )
3 3i
i
212 10 Pa 0.0185 m1.605 mol
8.31 J/ mol K 294 KPVnRT
×= = =
⋅⎡ ⎤⎣ ⎦
2. Solve for the final number of moles:
( )( )( ) ( )
3 3f
f
252 10 Pa 0.0185 m1.908 mol
8.31 J/ mol K 294 KPVnRT
×= = =
⋅⎡ ⎤⎣ ⎦
3. Subtract to find the air added to the tire: f i 1.908 mol –1.605 mol 0.303 moln n nΔ = − = =
Insight: The pressure is linearly proportional to the number of moles. Therefore the fractional increase in pressure is equal to the fractional increase in moles.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
10. Picture the Problem: A compressed air tank holds a quantity of air at a fixed pressure, volume, and temperature. This same quantity of air is released to atmospheric conditions where it occupies a larger volume.
Strategy: Since the number of moles remains constant, it follows from the ideal gas law that the ratio PV T is also constant. Use this relation to solve for the final volume.
Solution: 1. (a) Set the initial condi- tion equal to final condition:
i i f f
i f
PV PVT T
=
2. (b) Solve for the final volume: ( )3 3i f
f if i
880 kPa 303 K0.500 m 4.63 m101 kPa 285 K
P TV VP T
= = =
Insight: The final volume is more than nine times the initial volume. Therefore, as the gas expands, its density decreases by more than a factor of nine.
Strategy: Solve the ideal gas law (equation 17-2) for pressure.
Insight: This is 100 times lower than the lowest pressures achieved in a laboratory (10-13 Pa).
12. Picture the Problem: The pressure, volume, and number of moles for each of four ideal gases are given.
Strategy: Solve the ideal gas law (equation 17-5) for temperature to determine the ranking of these gases.
Solution: 1. Find TA and TB:
( )( )( )( )
3 3A A
AA
100 10 Pa 1 m1200 K,
10 mol 8.31 J/mol/KP VTn R
×= = =
( )( )( )( )
3 3B B
BB
200 10 Pa 2 m2400 K
20 mol 8.31 J/mol/KPVTn R
×= = =
2. Find TC and TD:
( )( )( )( )
3 3C C
CC
50 10 Pa 1 m120 K,
50 mol 8.31 J/mol/KPV
Tn R
×= = = ( )( )
( )( )
3 3D D
DD
50 10 Pa 4 m4800 K
5 mol 8.31 J/mol/KP VTn R
×= = =
3. By comparing their values we arrive at the ranking TC < TA < TB < TD.
Insight: If the pressure and volume are each doubled, the number of moles must be quadrupled in order to keep the temperature of an ideal gas the same.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
14. Picture the Problem: A spherical balloon is filled with helium gas at a specified temperature and pressure. The
number of atoms within the balloon is doubled, while the temperature and pressure are held constant. This increases the volume, and therefore the radius.
Strategy: Use the ideal gas law, equation 17-2, to calculate the number of atoms within the balloon. The volume of a sphere is related to the radius by 34
3 .V rπ= Doubling the number of atoms at constant temperature and pressure will double the volume. Set the final volume equal to twice the initial volume and solve for the final radius in terms of the initial radius.
Solution: 1. (a) Solve the ideal gas law for the number of atoms: PVPV NkT N
kT= ⇒ =
2. Write the volume in terms of the radius and insert the given values:
( ) ( ) ( )( )( )
33 54 43 3 24
23
2.4 10 Pa 0.25 m3.9 10 atoms
1.38 10 J/K 273.15 18 K
P rN
kT
π π−
×= = = ×
× +
3. (b) Set the final volume equal to twice the initial volume: ( )
2 1
3 34 42 13 3
2
2
V V
r rπ π
=
=
4. Solve for the final radius: 1/32 1 12 1.26r r r= =
The radius increases by a multiplicative factor of 1.26. Insight: Doubling the volume does not double the radius. To double the radius, the volume would need to increase by a
factor of 8.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
16. Picture the Problem: A known mass of an ideal gas fills a flask at a given pressure and temperature.
Strategy: Use the ideal gas law to calculate the number of moles in the gas. Calculate the molar mass by dividing the mass of the gas by the number of moles.
Solution: 1. Solve the ideal gas law, equation 17-5, for the number of moles:
( )( )( )
3 3 32
153 10 Pa 515 cm 1 m 2.945 10 mol100 cm8.31J/ mol K 322 K
PVnRT
−× ⎛ ⎞= = = ×⎜ ⎟⋅⎡ ⎤ ⎝ ⎠⎣ ⎦
2. Divide the mass by the number of moles:
2
0.460 g 15.6 g mol2.945 10 mol
M−
= =×
Insight: This gas is significantly lighter than air (Mair = 29 g/mole). It might be a mixture of CH3 (M = 15 g/mole) and CH4 (M = 16 g/mole).
18. Picture the Problem: The air inside a hot air balloon is warmer than the air outside the balloon. This produces a lower
air density within the balloon as compared with the cold air outside the balloon.
Strategy: Consider a given mass of air inside the balloon and an identical mass outside the balloon. The differing temperatures cause the masses to occupy different volumes. Calculate the ratio of the two densities, where equation 16-1 relates the density to the volume and equation 17-8 relates the volumes to the temperature.
Solution: 1. Write the ratio of densities in terms of masses and volumes and set the masses equal:
outin in in
out out out in
Vm Vm V V
ρρ
= =
2. Use equation 17-8 to eliminate the volumes in terms of temperatures:
( )( )
outin
out in
273.15 20.3 K0.8328
273.15 79.2 KTT
ρρ
+= = =
+
Insight: The lower density of the balloon allows it to rise because the buoyant force is greater than the weight. The balloon-to-air density ratio is inversely proportional to the temperature ratio. Therefore, increasing the temperature inside the balloon will cause it to rise, and lowering the temperature will cause it to descend.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
20. Picture the Problem: An ideal gas is held at constant temperature inside a flask. The pressure on the gas is increased by adding additional weight to the piston, which causes a decrease in the volume of the gas.
Strategy: Use equation 17-7 to solve for the final pressure in terms of the initial pressure and initial and final volume of the gas. Eliminate the volumes from the equation by writing them in terms of the radius and height of the gas, using 2V r hπ= .
Solution: 1. Write the volumes in eq. 17-7 in terms of the radius and heights: ( ) ( )
i i f f
2 2i i f f
PV PV
P r h P r hπ π
=
=
2. Solve for the final pressure: ( )( ) 5i i
ff
137 kPa 23.4 cm1.6 10 Pa
20.0 cmPhPh
= = = ×
Insight: The ratio of final to initial pressure is inversely proportional to the ratio of final height to initial height. Increasing the pressure proportionately decreases the height of the piston.
22. Picture the Problem: Air is a mixture of mostly of oxygen (O2) and nitrogen (N2) molecules. The oxygen molecules are more massive than the nitrogen molecules.
Strategy: Use the principles of gas kinetic theory and the expression rms 3v kT m= to answer the conceptual question.
Solution: 1. (a) The expression shows the rms speed is inversely proportional to the square root of the mass. We conclude that the rms speed of the more massive 2O molecules is less than the rms speed of the 2N molecules.
2. (b) The best explanation is III. The temperature is the same for both molecules, and hence their average kinetic energies are equal. As a result, the more massive oxygen molecules have lower speeds. Statement I is false (see the Insight statement), and statement II is false because equal temperature implies equal average kinetic energy, not speed.
Insight: Statement I is partly true because the O2 molecules do have a greater momentum by the factor 2 2O NM m due
to the difference in masses, but the O2 molecular speed is actually lower than that of the N2 molecules. 24. Picture the Problem: A piston held at the temperature T contains a gas mixture with molecules of three different types;
A, B, and C. The corresponding molecular masses are C B Am m m> > .
Strategy: Use the principles of gas kinetic theory and the expression rms 3v kT m= to determine the rankings.
Solution: 1. (a) Kinetic energy depends only on the absolute temperature of the gas, so Kav, A = Kav, B = Kav, C.
2. (b) The expression shows the rms speed is inversely proportional to the square root of the mass. We conclude that the ranking of the rms speeds is vrms, C < vrms, B < vrms, A.
Insight: In general, for a gas mixture in thermal equilibrium the more massive molecules have the smaller speed.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
26. Picture the Problem: Two ideal gases, oxygen and hydrogen, are comprised of molecules traveling with the same rms
speed. Since the gas molecules have different masses, the gases will have different temperatures.
Strategy: Use equation 17-13 to write the rms speed in terms of the mass and temperature of the gas. Set the rms speeds of the two gases equal and solve the resulting ratios of the temperatures to the masses of the two gas molecules.
Solution: 1. Set the rms speeds equal to each other using equation 17-13:
2 2
2 2 2 2
2 2 2 2
rms, O rms, H
O H O H
O H O H
3 3
v v
kT kT T Tm m m m
=
= ⇒ =
2. Square both sides and solve for TH2. Use a periodic table (Appendix E) to find the masses:
( )( )
2
2 2
2
HH O
O
2 1.01 g/mol313 K 19.8 K
2 16.00 g/molm
T Tm
⎡ ⎤= = =⎢ ⎥
⎢ ⎥⎣ ⎦
Insight: The hydrogen must be at 20 K, (−253°C), an extremely cold temperature, in order to have the same rms speed as oxygen at approximately room temperature.
28. Picture the Problem: An ideal gas is confined in a chamber. The gas expands to twice the original volume at constant
temperature. Strategy: Use the ideal gas law, equation 17-5, to determine the pressure and equation 17-12 to calculate the average
2. (b) Insert the given data into equation 17-12 for the kinetic energy: ( )( )–23 –203 3
av 2 2 1.38 10 J/K 273.15 295 K 1.18 10 JK kT= = × + = ×
3. (c) Pressure decreases by a factor of 2 because pressure is inversely proportional to volume. The average kinetic energy of a molecule remains the same, because it depends only on temperature.
Insight: The average kinetic energy of the gas molecules depends on the temperature, not on the pressure, volume, or number of moles present in the gas. As the gas expands, the number of collisions with the walls decreases, causing a decrease in pressure.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
30. Picture the Problem: An ideal gas is kept at constant volume and pressure while the number of molecules is doubled.
Strategy: Use the ideal gas law, equation 17-2, to determine the relationship between final and initial temperatures. Then using the final temperature and equation 17-13, calculate the final rms speed in terms of the initial rms speed.
Solution: 1. (a) By doubling the number of molecules N while holding the volume V and pressure P constant, the temperature T PV Nk= must decrease. We conclude that the rms speed rms 3v kT m= will decrease.
2. (b) Write the ideal gas law where pressure and volume are constant:
constantPV NkT= =
3. Set f i2N N= and solve for fT : 1i f f i22 NkT NkT T T= ⇒ =
4. Write equation 17-13 for the final rms speed, in terms of the initial temperature:
f i irms,f
3 33 12 2
kT T kTkvm m m
= = =
5. Use equation 17-13 to write the final rms speed in terms of the initial rms speed:
( )rms, f rms, i1 1 1300 m/s 920 m/s2 2
v v= = =
Insight: If the temperature remained constant as the number of molecules doubled, then twice as many molecules would collide with each wall in any given time. This would effectively double the pressure. To keep the pressure constant, the molecules must slow down so that the walls continue to receive the same impulse.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
35. Picture the Problem: A brick has faces with the following dimensions: face 1 is 1 cm by 2 cm; face 2 is 2 cm by 3 cm; face 3 is 1 cm by 3 cm.
Strategy: Use the concept of elastic deformation and (equation 17-17) to answer the question.
Solution: For a given force F, the change in length ΔL is proportional to the initial length, L0, and inversely proportional to the cross-sectional area A. Therefore, the change in dimensions of the brick will be least when it is placed on face 2. This corresponds to smallest height (L0 = 1 cm) and the largest cross-sectional area (A = 2 cm × 3 cm = 6 cm2).
Insight: The largest would occur when the brick is placed on face 1, which corresponds to the largest height (L0 = 3 cm) and the smallest cross-sectional area (A = 1 cm × 2 cm = 2 cm2).
Picture the Problem: An ideal gas is contained within a fixed volume. Increasing the rms speed of the gas molecules increases the gas temperature and pressure.
Strategy: Write the temperature as a function of the rms speed using equation 17-13. Then calculate the temperature when the rms speed is 1% higher than the initial rms speed. From the new temperature calculate the percent increase. Rearrange the ideal gas law to write an equation relating pressures and temperatures when volume is held constant. Determine the percent increase in pressure from this equation:
Solution: 1. (a) Solve equation 17-13 for T:
2rms
3v M
TR
=
2. Increase rmsv by 1% and solve for fT : ( ) ( )
2 22rms rms
f
1.011.01 1.0201
3 3v M v M
T TR R
= = =
3. Calculate the percent increase: f 1.0201 1 0.0201 100% 2.01%T TT−
= − = × =
4. (b) Write the ideal gas law as a ratio of final to initial states:
f f f f PV nRT P TPV nRT P T
= ⇒ =
5. Insert the final temperature: f f 1.0201 1.0201
P T TP T T= = =
6. Solve for the percent increase in pressure: 1.0201 1 0.0201 100% 2.01%− = × =
Insight: Since the temperature is proportional to the square of the rms speed, a small increase in speed corresponds to about double the increase in temperature. Since the temperature and pressure are proportional to each other, the pressure will increase by the same percentage as the temperature.
34. Picture the Problem: A spherical flask contains an ideal gas at a fixed temperature. The gas molecules collide with the walls of the container producing an outward net force.
Strategy: Solve the ideal gas law, equation 17-2, for the pressure per olecule. Since pressure is the force per unit area, calculate the force by multiplying the pressure per molecule by the surface area of the sphere. To find the surface area of the sphere, solve the spherical volume equation for the radius and insert the result into the equation for the surface area of a sphere.
Solution: 1. Solve the ideal gas law for pressure per atom and then multiply by surface area:
P kT PA F kTAPV NkTN V N N V
= ⇒ = ⇒ = =
2. Solve the volume of a sphere for the radius:
1 334 3
3 4VV r rππ
⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠
3. Insert the radius into the surface area equation: ( )
2 31 32 2 334 4 36
4VA r Vπ π ππ
⎛ ⎞= = =⎜ ⎟⎝ ⎠
3 cm 2 cm
1 cm
face 2
face 3
face 1
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
4. Insert the surface area into the force equation: ( )
( ) ( )( )
( )( )3
1 3 –231 3 2 3 –19
1 31 m
1000 L
36 1.38 10 J/K 293K36 2.8 10 N
0.350 L
F kT VN V
ππ
×= = = ×
⎡ ⎤⎣ ⎦
Insight: The total outward force on the flask is 13 kN (1.3 tons!). This can be found by multiplying the force per molecule by the number of molecules ( 224.5 10N = × ).
36. Picture the Problem: A hollow cylindrical rod (rod 1) and a solid cylindrical rod (rod 2) are made of the same material. The two rods have the same length and the same outer radius, and the same compressional force is applied to each rod.
Strategy: Use the concept of elastic deformation and ( )0F Y L L A= Δ (equation 17-17) to answer the question.
Solution: 1. (a) The change in length 0L F L YAΔ = is inversely proportional to the area. The hollow cylindrical rod has a smaller effective cross-sectional area than does the solid rod. We conclude that the change in length of rod 1 is greater than the change in length of rod 2.
2. (b) The best explanation is I. The solid rod has the greater effective cross-sectional area, since the empty part of the hollow rod doesn’t resist compression. Therefore, the solid rod has the smaller change in length. Statement II erroneously ignores the hollow part of rod 1, and statement III is false.
Insight: In some cases the use of hollow structural members, such as the tubes used to create a bicycle frame, can be adequately strong and much lighter than the corresponding solid structural members. An example in nature is the hollow bones that are characteristic of birds.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
38. Picture the Problem: When a force of 25 N is applied to a relaxed bicep muscle, it stretches by 2.5 cm.
Strategy: Solve equation 17-17 for Young’s modulus of the bicep.
Solution: 1. Write equation 17-17:
0
LF Y AL
⎛ ⎞Δ= ⎜ ⎟
⎝ ⎠
2. Solve for Young’s Modulus:
( )( )4 20
22 1 m100 cm
25N 24 cm 5.1 10 N m2.5 cm47 cm
LFYA L⎛ ⎞ ⎛ ⎞= = = ×⎜ ⎟⎜ ⎟Δ ⎝ ⎠⎝ ⎠
Insight: Young’s Modulus for the bicep is six orders of magnitude smaller than the moduli given in Table 17-1. The force required to stretch a metal by the same distance as the bicep would be about a million times stronger than the force on the bicep.
40. Picture the Problem: A copper sphere is submerged to the bottom of the Marianas Trench. The increased pressure
compresses the ball.
Strategy: Solve equation 17-19 for the change in volume. Calculate the initial volume of the sphere from the diameter.
Solution: 1. Solve equation 17-19 for :VΔ 0V P
VBΔ
Δ = −
2. Write the volume in terms of the radius:
( )
( ) ( )
34 13 2
3 8 54 13 2
10 2
6 3
0.15 m 1.10 10 Pa –1.01 10 Pa
14 10 N/m 1.4 10 m
D PV
B
V
π
π
−
⎡ ⎤ Δ⎣ ⎦Δ = −
⎡ ⎤− × × ×⎣ ⎦=×
Δ = − ×
Insight: The volume decreases by 0.08%, which is not a noticeable change. At the bottom of the trench, the diameter of the ball is 14.996 cm, a change of 4/100 of a millimeter.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
42. Picture the Problem: When placed under tension, a given length of steel wire stretches. The amount of stretch can be
related to the diameter of the wire.
Strategy: Replace the area in equation 17-17 with 214A Dπ= , the cross-sectional area of a cylinder. Solve the resulting
equation for the diameter of the cylinder.
Solution: 1. (a) Substitute 214A Dπ=
into equation 17-17:
2
0 0 4L L DF Y A YL L
π⎛ ⎞ ⎛ ⎞⎛ ⎞Δ Δ= =⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠ ⎝ ⎠
2. Solve for the diameter: ( )
( )0
10 2
4 360 N4 4.7 m 3.1 mm0.0011 m20 10 N/m
LFDY Lπ π
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟Δ × ⎝ ⎠⎝ ⎠
3. (b) The diameter should be increased because a wire’s cross-sectional area and its elongation are inversely related.
Insight: To produce an elongation one half as long ( 0.055 cmLΔ = ) the diameter would need to be 4.4 mm, which is a factor of 2 thicker than the diameter in the problem.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
44. Picture the Problem: Two rods of differing materials but
identical dimensions are placed end to end under a force of 8400 N. The force compresses each rod.
Strategy: The compressive force is felt equally by each rod. Therefore, use equation 17-17 to calculate the change in length of each rod separately. Sum the changes in length of both rods to determine to total change in length.
Solution: 1. (a) Solve equation 17-17 for the change in length:
20
20
4
4L FL DF Y L
L Y Dπ
π
⎛ ⎞⎛ ⎞Δ= ⇒ Δ =⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
2. Sum the changes in length for each rod and combine like terms:
0 0 0total Al Br 2 2 2
Al BrAl Br
4 4 4 1 1L F L F L FL L L
Y YY D Y D Dπ π π
⎛ ⎞Δ = Δ +Δ = + = +⎜ ⎟
⎝ ⎠
3. Solve numerically: ( )( )
( )total 2 10 2 10 2
4 0.55 m 8400 N 1 1 0.52 mm6.9 10 N/m 9.0 10 N/m0.017 m
Lπ
⎛ ⎞Δ = + =⎜ ⎟× ×⎝ ⎠
4. (b) Aluminum will have the larger change in length because it has the smaller Young’s modulus. Insight: The aluminum rod’s change in length is 0.295 mm and the brass rod’s change in length is 0.226 mm. As
predicted, aluminum has the greater change in length.
46. Picture the Problem: Water, initially at the freezing point, freezes to ice.
Strategy: Examine the properties of water at the melting point. During a phase change temperature remains constant. Adding heat during the phase change converts ice to water. Extracting heat converts water to ice. Unlike most materials, water expands as it freezes.
Solution: The answer is (d). Removing heat from the water will cause it to freeze.
Insight: For most materials at the freezing point, answer (c) would also have been correct, but water expands as it freezes.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
48. Picture the Problem: The liquid vapor curve shows the
relationship between the vapor pressure and boiling point temperature of water.
Strategy: Examine the graph to find at what temperature water boils when the vapor pressure is 1.5 kPa:
Solution: The temperature is about 13°C.
Insight: Note from the graph that the boiling temperature increases as the vapor pressure increases.
50. Picture the Problem: The vapor pressure curve shows the
relationship between the vapor pressure and boiling point temperature of carbon dioxide.
Strategy: Examine the graph to find at what temperature carbon dioxide boils when the vapor pressure is 1.5 MPa:
Solution: 1. (a) It boils at about –28°C.
2. (b) The graph indicates that an increase in the boiling pressure results in an increase in the boiling temperature.
Insight: When the boiling pressure is 2.0 MPa, a higher pressure than in the problem, the associated boiling temperature is about −20°C, which is higher than the problem’s boiling temperature.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
52. Picture the Problem: The phase diagram for carbon dioxide
shows the pressures and temperatures for which carbon dioxide is in the solid, liquid, and vapor phases. The lines separating the phases are the sublimation, fusion, and vapor pressure curves.
Strategy: Examine the graph to determine the phase at the given temperatures and pressures. Find the boiling pressure at 20°C.
Solution: 1. (a) It is in the gas phase.
2. (b) It is in the solid phase.
3. (c) The boiling pressure is 5707 kPa.
Insight: Note that CO2 does not have a liquid phase at atmospheric pressure.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
54. Picture the Problem: The phase diagram for water shows the
pressures and temperatures for which water is in the solid, liquid, and vapor phases. The sample of water begins at atmospheric pressure just above the melting temperature.
Strategy: Examine the phase diagram. Locate the melting point at atmospheric pressure and note that the initial state is just to the right of the melting point. Observe the phase change that occurs if the temperature increases. Observe what phase change would occur if the pressure decreases.
Solution: 1. (a) As the temperature increases, the liquid turns to gas.
2. (b) As the pressure decreases, the liquid first turns to solid. Then the solid sublimates to gas.
Insight: One of the unusual characteristics of water is that it can pass from liquid to solid to gas at constant temperature as the pressure decreases.
56. Picture the Problem: Heat is removed from four liquids that are at their freezing temperature, and they each solidify
completely. The amount of heat that must be removed, Q, and the mass, m, of each of the liquids is given. Strategy: Use fL Q m= (equation 17-20) to determine the ranking of the latent heats of fusion.
Solution: 1. Find LfA and LfB: 5 5A B
fA fBA B
33,500 J 166,000 J= = 3.35 10 J/kg, and = = 3.32 10 J/kg0.100 kg 0.500 kg
Q QL Lm m
= × = ×
2. Find LfC and LfD: 4 5C D
fC fDC D
31,500 J 5,400 J= = 1.26 10 J/kg, and = = 1.08 10 J/kg0.250 kg 0.0500 kg
Q QL Lm m
= × = ×
3. By comparing the values of the latent heats we arrive at the ranking C < D < B < A.
Insight: A large latent heat of fusion corresponds to a material that is difficult to melt, requiring a large amount of heat Q to melt a small amount of mass m. Such a material typically has very strong attractive forces between its molecules.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
initially at −15°C, the heat first increases the temperature to the melting point, then melts the ice, and finally raises the temperature of the melted water to 15°C.
Strategy: Set the total heat equal to the sum of the heat needed to (i) raise the ice to the melting point, (ii) melt the ice, and (iii) increase the water to the final temperature. Solve the resulting equation for the mass.
Solution: 1. Sum the heats using equations 16-13 and 17-20: ( ) ( )
( ) ( )
i ii iii
ice f water1 2
ice f water1 2
Q Q Q Qmc T mL mc T
Q m c T L c T
= + +
= Δ + + Δ
⎡ ⎤= Δ + + Δ⎣ ⎦
2. Solve for the mass:
( ) ( )
( ) ( ) ( ) ( )
ice f water1 25
4
9.5 10 J2090 J/ kg C° 15C° 33.5 10 J/kg 4186 J/ kg C° 15C°
2.2 kg
Qmc T L c T
m
=Δ + + Δ
×=
⋅ + × + ⋅⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
=
Insight: From the mass we can calculate the amount of heat used in each of the three processes: i 69.4 kJ, Q = ii 742 kJ, Q = and iii 139 kJ. Q = Most of the heat is needed to melt the ice.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
60. Picture the Problem: As a specified amount of heat is added to ice initially at −5.0°C, the heat first increases the
temperature to the melting point, then melts the ice, and finally raises the water to its final temperature.
Strategy: Use equation 16-13 to calculate the amount of heat necessary to raise the temperature to the melting point. Use equation 17-20 to calculate the amount of heat necessary to melt the ice. Subtract these two amounts of heat from the total heat available. Insert the remainder of the heat into equation 16-13 to calculate the final temperature of the water.
Solution: 1. (a) Calculate the heat necessary to raise the ice to the melting point:
( ) ( ) 41 ice 1.1 kg 2090 J/ kg C° 5.0 C 1.15 10 JQ mc T= Δ = ⋅ ° = ×⎡ ⎤⎣ ⎦
2. Calculate the heat necessary to melt the ice: ( )4 52 f 1.1 kg 33.5 10 J/kg 3.685 10 JQ mL= = × = ×
3. Subtract the heats from the total heat: 3 total 1 25 4 5 55.2 10 J 1.15 10 J 3.685 10 J 1.40 10 J
Q Q Q Q= − −
= × − × − × = ×
4. Solve equation 16-13 for fT :
( )
3 water f5
3f
water
( 0 C)
1.40 10 J 30 C1.1 kg 4186 J/ kg C°
Q mc TQ
Tmc
= − °
×= = = °
⋅⎡ ⎤⎣ ⎦
5. No ice remains because f 0 C.T > °
6. (b) The mass must double also. All the temperature changes, specific heat values, and the latent heat value remain the same, so doubling the amount of heat added requires doubling the mass because they are proportional.
Insight: If the heat added to the ice had been only 53.0 10 J,× the result of step 3 above would have been a negative number (not possible). This would mean that only part of the ice melted. The final temperature would be 0°C, with 0.86 kg of water and 0.24 kg of ice remaining in the system.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
kilogram of water initially as ice at −20°C until the water has been converted to steam at 120°C.
Strategy: Use equations 16-13 and 17-20 to calculate the heat added to raise the temperature to points A, B, C, and D on the graph. Calculate the slope between points B and C by dividing the temperature difference by the heat added, (i.e., rise over run).
Solution: 1. (a) Calculate heat added to increase the temperature to the melting point:
( ) ( )A
41.000 kg 2090 J/ kg C° 20.0 C 4.18 10 J
Q mc T= Δ
= ⋅ ° = ×⎡ ⎤⎣ ⎦
2. Add the heat needed to melt the ice:
( )
4B f
4 4 5
4.18 10 J
4.18 10 J 1.000 kg 33.5 10 J/kg 3.77 10 J
Q mL= × +
= × + × = ×
3. Add the heat needed to raise the temperature to the boiling point:
( ) ( )
4C
4
5
37.7 10 J
37.7 10 J 1.000 kg 4186 J/ kg C° 100 C
7.96 10 J
Q mc T= × + Δ
= × + ⋅ °⎡ ⎤⎣ ⎦
= ×
4. Add the heat needed to boil off the water:
( )
5D v
5 5 6
7.96 10 J
7.96 10 J 1.000 kg 22.6 10 J/kg 3.06 10 J
Q mL= × +
= × + × = ×
5. (b) Calculate the slope of the line: 4C B
5C B
100 Cslope 2.39 10 C J4.186 10 J
T TQ Q
−− °= = = × ° /
− ×
6. Calculate water1 c to show it equals the slope:
( )( )4
water
1 1 2.39 10 kg C° /J4186 J/ kg C°c
−= = × ⋅⋅⎡ ⎤⎣ ⎦
Insight: In this problem the slope is equal to the inverse of the specific heat, because the graph represents 1.00-kilogram of water. In general, the slope is the inverse of the heat capacity (C = mc).
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
64. Picture the Problem: As heat is added to 0.550 kg of
ice that is initially at −20°C, the temperature increases linearly until the ice reaches the melting point. The temperature remains constant as the ice melts. After the ice is melted, the temperature again rises linearly to the boiling point as heat is added.
Strategy: Calculate the slope between points A and B, by dividing the temperature difference by the heat added, that is, rise over run. Do the same for the slope between points C and D.
Solution: 1. (a) Calculate the heat added between points A and B: ( ) ( ) 4
AB 0.550 kg 2090 J/ kg K 20.0 C 2.30 10 JQ mc T= Δ = ⋅ ° = ×⎡ ⎤⎣ ⎦
2. Calculate the slope:
( ) 4B A4
AB
0 C 20.0 Cslope 8.70 10 C J
2.30 10 JT TQ
−° − − °−= = = × ° /
×
3. Show that 1/mc is equal to the slope:
( )41 1 8.70 10 C /J slope
(0.550 kg) 2090 J/ kg Kmc−= = × ° =
⋅⎡ ⎤⎣ ⎦
4. (b) Calculate the heat added between points C and D:
( ) ( ) 4CD 0.550 kg 4186 J/ kg K 20.0 C 4.60 10 JQ mc T= Δ = ⋅ ° = ×⎡ ⎤⎣ ⎦
5. Calculate the slope between C and D: 4D C
4CD
20.0 C 0 Cslope 4.34 10 C /J4.60 10 J
T TQ
−− ° − °= = = × °
×
6. Show that 1/mc is equal to the slope:
( )41 1 4.34 10 C /J slope
0.550 kg 4186 J/ kg Kmc−= = × ° =
⋅⎡ ⎤⎣ ⎦
Insight: Since the specific heat of water is about double the specific heat of ice, the slope between C and D is about one-half the slope between points A and B.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
66. Picture the Problem: Ice covers a rectangular windshield. Heat is added to the ice until it melts.
Strategy: Calculate the mass of the ice on the windshield by multiplying the density of the ice by its volume, which is its area A multiplied by its thickness d. Use equations 16-13 and 17-20 to calculate the heat necessary to melt the ice.
Solution: 1. Use equation 15-1, setting V Ad= and solving for m: ( )( )( )3 2
ice ice 917 kg/m 1.6 m 0.0058 m 8.5 kgm V Adρ ρ= = = =
2. Add the heat to increase the tem- perature to the heat to melt the ice, and solve for the total heat:
( )( ) ( )( )
ice f ice f
48.5 kg 2090 J kg K 2.0 C 33.5 10 J/kg 2.9 MJ
Q mc T mL m c T L= Δ + = Δ +
⎡ ⎤= ⋅ ° + × =⎣ ⎦
Insight: A typical 300-watt auto defroster would take over 2.5 hours to completely melt the ice.
68. Picture the Problem: A cube of very cold aluminum is placed into a container of water. Heat transfers from the water
to the aluminum until the two are in equilibrium. Strategy: Assume that the final temperature is f 0 C,T = ° with part of the water frozen into ice. Use equation 16-13 to
calculate the amount of heat the aluminum absorbs as it heated up to 0°C Subtract from that heat the amount of heat the water gives off as it cools to 0°C. Use equation 17-20 to calculate the mass of ice that freezes as the aluminum absorbs the remaining heat.
Solution: 1. Calculate the heat gained by the aluminum:
( )( ) ( )
A1 Al Al f Al,i
0.155 kg 653 J/ kg K 0 C– –196 C 19,838 J
Q m c T T= −
= ⋅ ° ° =⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
2. Calculate the heat given off by water as it cools to freezing: ( ) ( )
w w w
0.0800 kg 4186 J/ kg K 15.0 C 5023 J
Q m c T= Δ
= ⋅ ° =⎡ ⎤⎣ ⎦
3. Find the difference in heats: f Al w 19,838 J 5023 J 14,815 JQ Q Q= − = − =
69. Picture the Problem: A hot iron block is immersed in cool water. The iron loses heat to the water until the two come to equilibrium.
Strategy: Assume that the equilibrium temperature is greater than 100°C and that all of the water has boiled off to steam. Use equations 16-13 and 17-20 to find the heat gained by the water gainedQ in coming to the final temperature. Use equation 16-13 to find the heat lost by the iron. Set the two equations equal and solve for the final temperature.
Solution: 1. (a) Calculate the heat needed to boil the water:
( ) ( )
( ) ( ){ }w w water w,i w w water
5w
100 C – 100 – 20 C
0.040 kg 4186 J/ kg K 80 C° 22.6 10 J/kg 103.8 kJ
v vQ m c T m L m c L
Q
= ° + = ° +⎡ ⎤⎣ ⎦
= ⋅ + × =⎡ ⎤⎣ ⎦
2. Calculate gained :Q ( )gained w w steam f 100°CQ Q m c T= + −
3. Calculate the heat lost by the iron: Fe Fe Fe Fe,i f( )Q m c T T= −
4. Set the heat given off equal to the heat absorbed: ( ) ( )
Fe gained
Fe Fe Fe,i f w w steam f 100 C
Q Q
m c T T Q m c T
=
− = + − °
5. Solve for the final temperature: ( )
( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( ) ( )
w steam Fe Fe Fe,i wf
w steam Fe Fe
100 C
0.040 kg 2010 J/ kg K 100°C
0.825 kg 560 J/ kg K 352°C 103.8 kJ123°C
0.040 kg 2010 J/ kg K 0.825 kg 560 J/ kg K
m c m c T QT
m c m c° + −
=+
⎧ ⎫⋅⎡ ⎤⎪ ⎣ ⎦ ⎪⎨ ⎬
+ ⋅ −⎡ ⎤⎪ ⎪⎣ ⎦⎩ ⎭= =⋅ + ⋅⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
Insight: The final temperature is greater than 100°C, so it agrees with our initial assumption. If the calculated final temperature had been less than 100°C, we would have to change our strategy to accommodate a final condition in which only part of the water boiled off.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
4. Divide the heat by the latent heat of fusion to calculate the mass of the water that freezes to ice:
ff 4
f
14,815 J 44.2 g33.5 10 J/kg
QmL
= = =×
Insight: Since the mass of water that is frozen is greater than zero but less than the total mass of the water ( 0 44.2 g < 80.0 g< ), our assumption that the final temperature would be zero and some, but not all, of the water would be frozen to ice, was correct. If the calculations gave a mass less than zero, we would have to reevaluate our strategy to calculate a final temperature greater than zero. If the calculation gave us a mass greater than the total mass of the water, we would reevaluate the strategy to find a temperature less than zero with all of the water frozen to ice.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
70. Picture the Problem: An ice cube is placed into an aluminum cup filled with water. Heat flows out of the water and
aluminum cup and into the ice until the water, ice, and cup are in thermal equilibrium. Strategy: Assume that the final temperature is greater than 0°C. In this state all of the ice will have melted and the
resulting water will have heated to the final temperature. Use equation 16-13 and equation 17-20 to calculate the heat gained by the ice gainedQ as its temperature rises to the equilibrium temperature. Set that heat equal to the heat lost by the cup and water lostQ as they cool to the equilibrium temperature and solve for the final temperature.
Solution: 1. (a) Find gainedQ : ( )gained ice f ice w f 0.0 CQ m L m c T= + − °
2. Find lostQ : ( ) ( ) ( )( )lost c Al f w w f c Al w w f23 C 23 C 23 CQ m c T m c T m c m c T= ° − + ° − = + ° −
3. Set the heat absorbed equal to the heat released: ( )( )
gained lost
ice f ice w f c Al w w f23 C
Q Q
m L m c T m c m c T
=
+ = + ° −
4. Solve for the final temperature:
( )( )( )
( )( )( ) ( ){ }( )
( ) ( ) ( )
ice f c Al w wf
w ice w c Al
4
23 C
0.035 kg 33.5 10 J/kg
0.062 kg 900 J/ kg K 0.110 kg 4186 J/ kg K 23 C0.22 C
4186 J/ kg K 0.035 kg 0.110 kg 0.062 kg 900 J/ kg K
m L m c m cT
c m m m c− + + °
=+ +
⎛ ⎞− × +⎜ ⎟⎜ ⎟⋅ + ⋅ °⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⎝ ⎠= = °
⋅ + + ⋅⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦
5. (b) The equilibrium temperature with silver is less than with aluminum. In fact, not all the ice will melt. Less heat loss is required to lower the temperature of silver because it has a smaller specific heat.
Insight: When the aluminum is replaced by silver, the equilibrium temperature is 0°C, with 32.6 grams of the ice melted and 2.4 grams of ice remaining.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
72. Picture the Problem: Ice is dropped into some water. Heat transfers from the water and melts the ice until the water
and ice are in thermal equilibrium. Strategy: Assume that the equilibrium temperature is 0°C and that not all of the ice has melted. Using this assumption
and equation 16-13, calculate the heat lost by the water as it cools to 0°C. Set that heat equal to the latent heat of fusion of the ice, equation 17-20, to calculate the mass of ice that melts. If that mass is less than the total mass of ice, then our assumption is correct. To calculate the minimum temperature for which all of the ice will melt, set the latent heat of the melting ice equal to the heat lost by the water and solve for the initial temperature.
Solution: 1. (a) Calculate the heat lost by the water: ( ) ( ) 40.33 kg 4186 J/ kg K 14C° 1.93 10 J Q mc T= Δ = ⋅ = ×⎡ ⎤⎣ ⎦
2. Calculate the mass of the melted ice:
4
4f
1.93 10 J 0.058 kg33.5 10 J/kg
QmL
×= = =
×
3. Calculate the remaining ice: f = 0.075 kg 0.058 kg = 0.017 kgm −
4. (b) Set the latent heat required to melt the ice equal to the heat lost by the water: ( )w ice0°Cm c T m L− =
5. Solve for the initial temperature of the water:
( )( )( ) ( )
4ice
w
0.075 kg 33.5 10 J/kg18 C
0.33 kg 4186 J/ kg Km L
Tm c
×= = = °
⋅⎡ ⎤⎣ ⎦
Insight: A small amount of ice can cool a large drink because the latent heat of fusion of water is quite large. 74. Picture the Problem: Plastic bubble wrap is used as a protective packing material.
Strategy: The bubble wrap protects the contents of a package by cushioning them with small pockets of air. The pressure in the air pockets should be sufficiently large to provide good cushioning.
Solution: The bubble wrap is more effective on a warm day because the air pressure within the bubbles will be greater, leading to more effective cushioning. At low temperature, the bubbles are almost flat.
Insight: If the bubble wrap becomes too warm, however, the plastic shells will become too soft and fragile, and the plastic will melt at sufficiently high temperatures.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
76. Picture the Problem: The ratio of oxygen to nitrogen is measured at different altitudes in the atmosphere.
Strategy: The maximum altitude that can be achieved by a moving object can be determined from the conservation of energy: 2 21
2 2 .mgh mv h v g= ⇒ = Note that it is independent of mass, but depends on the speed. Use this principle together with the kinetic theory of gases to determine the change in the ratio of O2 to N2 with increasing altitude.
Solution: In the gas mixture of air the O2 and N2 molecules have the same average kinetic energy, but O2 molecules have a smaller average speed because they are more massive. Therefore, O2 molecules cannot reach as high an altitude
2 2h v g= as can the lighter and faster-moving N2 molecules. We expect that the ratio of oxygen to nitrogen in the atmosphere to will decrease as you go up in altitude.
Insight: Thermal convection and gas diffusion (a process that results from collisions between molecules) tend to mix the atmosphere and diminish this expected change in atmospheric composition with altitude. Measurements indicate the atmospheric composition is essentially uniform from the surface to an altitude of 80 km.
78. Picture the Problem: The absolute temperature of an ideal gas is doubled from 100 K to 200 K.
Strategy: Use the principles of the Kelvin temperature scale and the kinetic theory of gases to answer the question.
Solution: 1. (a) If the absolute temperature of an ideal gas is doubled, the average kinetic energy of its molecules doubles as well. Recall that kinetic energy depends on speed squared. It follows, then, that the average speed of the molecules increases by a factor that is less than 2; in fact, the speed increases by a factor of 2 .
2. (b) The best explanation is I. Doubling the Kelvin temperature doubles the average kinetic energy, but this implies an increase in the average speed by a factor of 2 1.414= …, which is less than two. Statement II confuses the volume V in the ideal gas law with the average speed vav of the molecules, and statement III is false.
Insight: The average speed of the molecules will double if the temperature is changed from 100 K to 400 K.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
80. Picture the Problem: A new laptop design removes heat by vaporizing methanol.
Strategy: Use equation 17-20 to calculate the latent heat of methanol.
Solution: Calculate the latent heat: 65100 J 1.1 10 J/kg
0.0046 kgQLm
= = = ×
Insight: The latent heat of water is about twice that of methanol. However, the boiling point of methanol is only 65°C, making it a better choice to keep the laptop components cool.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
82. Picture the Problem: A sealed container of fixed volume contains hydrogen and oxygen molecules. The molecules
react to convert all of hydrogen and oxygen into water molecules.
Strategy: Use the molar masses of hydrogen and oxygen to determine the number of moles of each that are in the container. Summing these values gives the total number of moles. Use this number and the ideal gas law to calculate the volume of the vessel. During the reaction, three molecules (two hydrogen and one oxygen) interact to create two water molecules. This decreases the number of moles present in the gas to two-thirds the initial number totaln . Use the new number of moles
2H On in the ideal gas law to calculate the new gas pressure.
Solution: 1. (a) Calculate the number of moles of hydrogen and oxygen gas: ( )
( )
2
2
2
2
2
2
HH
H
OO
O
8.06 g 4.00 mol2 1.00794 g/mol
64.0 g 2.00 mol2 15.9994 g/mol
mn
M
mn
M
= = =
= = =
2. Sum to find the total moles of gas: 2 2total H O 4.00 mol 2.00 mol 6.00 moln n n= + = + =
3. Solve the equation 17-5 for the volume:
( ) ( )
total
35
6.00 mol 8.31 J/ mol K 273.15 125 K0.196 m
1.013 10 Pa
n RTV
P=
⋅ +⎡ ⎤⎣ ⎦= =×
4. (b) Calculate 2H O
n : ( )2
2 2H O total3 3 6.00 mol 4.00 moln n= = =
5. Calculate the final pressure:
( ) ( )
2H O
3
4.00 mol 8.31 J/ mol K 273.15 125 K67.5 kPa
0.196 m
n RTP
V=
⋅ +⎡ ⎤⎣ ⎦= =
Insight: Since the number of moles decreases to 2/3 of its initial value, and all other terms remain the same, the ideal gas law stipulates that the pressure also decreases to 2/3 of its initial value.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
84. Picture the Problem: Peter pulls a fish out of the water from the
top of a pier. The sketches show free-body diagrams for the fish when it is not accelerating and when it is accelerating upward.
Strategy: Use equation 17-17 to calculate the fractional increase in length of the fishing line. Use Newton’s Second Law to determine the tension in the line.
Solution: 1. (a) Solve Newton’s Second Law for the forces on the fish with no acceleration:
1
1
0F T mgT mg= − =
=∑
2. Solve equation 17-17 for the fractional increase in length:
( )( )( ) ( )
2
29 20
4.8 kg 9.81 m/s0.040
5.1 10 N/m .00027 mL F mgL YA YA π
Δ= = = =
×
3. (b) Constant speed implies zero acceleration and zero net force. Therefore, 0/ 0.040 .L LΔ =
4. (c) Solve Newton’s Second Law for the forces on the fish with upward acceleration:
2
2 ( )
F maT mg ma
T m a g
=
− =
= +
∑
5. (d) Insert the tension to solve for the fractional change in length:
( ) ( )( )( ) ( )
2
29 20
4.8 kg 1.5 9.81 m/s0.046
5.1 10 N/m .00027 m
m a gLL YA π
++Δ= = =
×
Insight: Upward acceleration requires more force than the constant speed case, so the line stretches more.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
86. Picture the Problem: A steel ball is attached to the end of an aluminum wire
and swung in a vertical circle. The resulting tension in the wire stretches the wire. Free-body diagrams for the ball at the bottom and top are shown in the figure.
Strategy: Solve Newton’s Second Law for the tension in the wire. Insert the resulting tension in equation 17-17 to solve for the change in length of the wire.
Solution: 1. (a) Write Newton’s Second Law at the top of the circle and solve for T:
( )cp
cp
F T mg ma
T m a g
= + =
= −
∑
2. Substitute m Vρ= and 2
cp :a v r= 2vT V gr
ρ⎛ ⎞
= −⎜ ⎟⎝ ⎠
3. Insert numerical values: ( ) ( ) ( )2
33 27.8 m/s7860 kg/m 0.064 cm 9.81 m/s 69.5 N
6 0.82mT π ⎡ ⎤⎡ ⎤= − =⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
4. Solve equation 17-17 for :LΔ ( )( )
( ) ( )0
210 2 312
0.82 m 69.46 N0.17 mm
6.9 10 N/m 2.5 10 m
L TL
YA π −Δ = = =
× × ×
5. (b) Solve Newton’s Second Law for the tension at the bottom of the circle:
( )
( ) ( ) ( )
cp
cp
233 29.3m/s
7860kg/m 0.064cm 9.81m/s 124 N6 0.82m
F T mg ma
T m a g
π
= − =
= +
⎡ ⎤⎡ ⎤= + =⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
∑
6. Solve equation 17-17 for :LΔ ( )( )
( ) ( )0
210 2 312
0.82 m 124.4 N0.30 mm
6.9 10 N/m 2.5 10 m
L TL
YA π −Δ = = =
× × ×
Insight: The wire stretches more at the bottom of the circle, where the tension is much greater.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
88. Picture the Problem: Five molecules move with known speeds. Strategy: Sum the five speeds and divide by five to calculate the average speed. Sum the squares of the speeds and
divide by five to calculate the average of the squares. Finally, take the square root of the average of the squares to calculate the rms speed.
Solution: 1. (a) Sum the speeds and divide by 5:
( )av
221 301 412 44.0 182 m/s232 m s
5v
+ + + += =
2. (b) Since in general, rms av ,v v> ( )2av
v will be greater than ( )2avv .
3. (c) Average the squares of the speeds: ( )
( ) ( ) ( ) ( ) ( )( )2 2 2 2 2 2 2
2
av
4 2 2
221 301 412 44.0 182 m /s
56.88 10 m s
v+ + + +
=
= ×
4. Compare the average of the squares with the square of the average: ( )24 2 2 4 2 26.88 10 m /s 232 m/s 5.38 10 m /s× > = ×
5. (d) Take the square root of the average of the squares: ( )2 4 2 2
rms av6.88 10 m /s 262 m/sv v= = × =
6. Compare the rms speed with the average speed: rms av rms av262 m/s > 232 m s v v v v= = ∴ >
Insight: The rms speed is always greater than or equal to the average speed, because the average of the squares is always equal to or larger than the square of the averages.
90. Picture the Problem: Water within a pipe or rock cavity expands as it freezes. To maintain the same volume, the pipe
or rock must exert an increased pressure on the water. Strategy: Use equation 17-19 to calculate the pressure that must be exerted onto the water ice to compress it back to its
original volume. The initial volume of the compression is the original volume plus the expanded volume. Solution: 1. Calculate the relative
volume compression from the relative volume expansion: 0
– 0.0905 –0.083001 0.0905
VVΔ
= =+
2. Use equation 17-19 to calculate the increase in pressure: ( )( )10 2 8 2
0
0.80 10 N/m 0.08300 6.64 10 N/mVP BV
⎛ ⎞ΔΔ = − = − × − = ×⎜ ⎟
⎝ ⎠
3. Solve for the final pressure: 5 8 8f i 1.01 10 Pa 6.64 10 Pa 6.6 10 PaP P P= +Δ = × + × = ×
Insight: Most copper and lead pipes cannot withstand a pressure this large, and they burst.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
92. Picture the Problem: Ice inside a Styrofoam container melts as heat flows into the container from the surrounding air.
Strategy: Use equation 16-16 to calculate the rate at which heat flows into the cooler. Use equation 17-17 to calculate the amount of heat necessary to melt the ice. Divide the heat by the heat flow rate to calculate the time necessary to melt the ice.
Solution: 1. (a) Solve equation 16-16 for Q t :
( ) ( ) ( )( )
2 21 C 0.0 C0.030 W/ m C° 1.5 m 25 W
0.038 m
Q Tk At L
Δ⎛ ⎞= ⎜ ⎟⎝ ⎠
° − °= ⋅ =⎡ ⎤⎣ ⎦
2. (b) Calculate heat necessary to melt ice: ( )4 6melt f 5.1 kg 33.5 10 J/kg 1.71 10 JQ mL= = × = ×
3. Divide Qmelt by the flow rate to find t: ( )
( )
64melt
1.71 10 J hr6.87 10 s 19 hrs24.9 W 3600 s
QtQ t
× ⎛ ⎞= = = × =⎜ ⎟⎝ ⎠
Insight: The ice will melt at a much faster rate if the cooler is periodically opened and warm items are placed in the cooler while cold items are removed.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
94. Picture the Problem: Heat conducts through a copper rod from the base to the top.
The heat melts a small amount of ice at the top of the rod, as shown in the figure.
Strategy: Set the heat necessary to melt the ice equal to the heat conducted through the copper tube using equation 16-16. Solve the resulting equation for the time necessary to melt the ice.
Solution: 1. Set equation 16-16 equal to the heat needed to melt the ice and solve for the time. Write the cross-sectional area of the rod in terms of its diameter:
f
f
f2
4
TQ k A t mLL
mL Ltk A TmL Lk d Tπ
Δ⎛ ⎞= =⎜ ⎟⎝ ⎠
=Δ
=Δ
2. Simplify and insert numerical values: ( )( )( )
( ) ( ) ( )
4f
2 2
4 0.025 kg 33.5 10 J/kg 0.37 m415 s
390 W/ m K 0.075 m 120 KmL Lt
k d Tπ π
×= = =
Δ ⋅⎡ ⎤⎣ ⎦
Insight: Copper is an excellent conductor of heat, so very little time is required to melt the ice. If the copper tube were replaced by glass (a poor conductor) it would take almost 2 hours to melt the ice.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
96. Picture the Problem: A bathysphere, a hollow spherical chamber, is submerged into the ocean.
Strategy: Use the ideal gas law (equation 17-5) to find the number of moles of air molecules in the bathysphere.
Solution: Solve equation 17-5 for n: ( ) ( )
( )( )
35 4 13 21.01 10 Pa 4.75 ft 0.305 m/ft
65.2 mol8.31 J/mol K 297 K
PVnRT
π× × ×= = =
⋅
Insight: If a diver breathes 0.5 mol of air per minute, the two divers breathed 1.0 mol of air per minute. The air in the bathysphere would therefore be breathed at least once every hour. That is why the divers brought extra oxygen and chemicals to absorb the carbon dioxide.
98. Picture the Problem: While the bathysphere is submerged in the water it is supported by a cable to
the surface. The cable stretches due to the tension caused by the weight and buoyant forces.
Strategy: Sum the vertical forces on the bathysphere to determine the tension in the cable, where the buoyant force is given by equation 15-9. Then use equation 17-17 to find the stretch distance.
Solution: 1. (a) Set the sum of the vertical forces equal to zero:
0yF T B mgT mg B= + − =
= −
∑
2. Calculate the buoyant force:
( )( ) ( )33 2 4 1231000 kg/m 9.81 m/s 4.75 ft 3.281 ft/m
15.6 kN
B gV
B
ρ
π
=
=
=
3. Solve for the tension: ( )212,700 kg 9.81 m/s 15.6 kN 109 kNT mg B= − = − =
4. Solve equation 17-17 for the change in the cable’s length:
( )( )
2
0 0
302 10 2 2
109 10 N (923 m)47 cm
20 10 N/m (0.0185 m)
L LF Y A Y rL L
FLL
Y r
π
π π
⎛ ⎞ ⎛ ⎞Δ Δ= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
×Δ = = =
×
Insight: Without the buoyancy, the tension in the cable would be 15.6 kN larger, or 125 kN. This force would result in a stretch of 54 cm. The buoyancy reduces the stretch of the cable.
Chapter 17: Phases and Phase Changes James S. Walker, Physics, 4th Edition
100. Picture the Problem: An ice cube is dropped into a container of lemonade initially at 1°C. Heat from the lemonade
melts the ice cube until the lemonade and ice cube are in thermal equilibrium.
Strategy: Use equation 16-13 to calculate the amount of heat lost by the lemonade as it cools to the freezing point of water. Use this heat and equation 17-17 to calculate the mass of the ice cube that melts. To calculate the minimum initial temperature of the lemonade, set the heat lost by the lemonade equal to the heat necessary to melt the ice and solve for the initial temperature.
Solution: 1. (a) Calculate the heat lost by the lemonade:
( )( ) ( ) ( )
lem lem i 0°C
2.00 kg 4186 J/ kg K 1 0 C 8370 J
Q m c T= −
= ⋅ − ° =⎡ ⎤⎣ ⎦
2. Solve 17-17 for the melted mass of the ice cube:
lemmelted 4
8372 J 0.025 kg33.5 10 J/kg
Qm
L= = =
×
3. Compare the melted mass with the total mass of the ice cube:
melted ice0.025 kg 0.045 kgm m= < = so not all of the ice has melted and the final temperature is T = 0°C.
4. (b) Set the heat to melt the ice cube equal to the heat lost by the lemonade: ( )lem i ice0 °Cm c T m L− =
5. Solve for the initial temperature: ( )
( )
4ice
ilem
0.045 kg 33.5 10 J/kg1.8°C
2.00 kg 4186 J/ kg Km L
Tm c
×= = =
⋅⎡ ⎤⎣ ⎦
Insight: For initial lemonade temperatures less than 1.8°C, some but not all of the ice will melt. The final temperature will therefore be 0°C. For initial lemonade temperatures greater then 1.8°C all of the ice will melt and the final temperature will be between the initial temperature and 0°C.
