+whw6: chapter 16: 11,19,42,55, chapter 17: 17,28,72 ...erbion.com › index_files › phys111 ›...
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Home Work Solutions PHYS111.02 SFSU Eradat [WH6 CHAPTERS 16-‐18] 1 +WHW6: Chapter 16: 11,19,42,55, Chapter 17: 17,28,72, Chapter 18: 27,38,51 Chapter 16 11. Picture the Problem: Our sketch shows the pressure
in the gas thermometer as a function of the temperature. Note that at T = 100°C the pressure is 227 mmHg and the pressure extrapolates to zero at the temperature −273.15°C. Assuming ideal behavior we want to find the temperature associated with a pressure of 162 mmHg.
Strategy: We assume that the pressure lies on a straight line. Using the known pressures at 100°C and −273.15°C, calculate the rate at which pressure increases as a function of temperature. Calculate the temperature at 162 mmHg using this rate.
Solution: 1. Divide the change in pressure by the change in temperature to obtain the rate:
rate = 227 mmHg − 0100°C − −273.15°C( ) = 0.60833 mmHg/C°
2. Solve the rate equation for the temperature:
rate =P − P0
T − T0
⇒ T = T0 +P − P0
rate
3. Insert given data:
T = 100°C + 162 mmHg − 227 mmHg
0.60833 mmHg/C°= − 6.85°C
Insight: The gas thermometer is first placed in the boiling water to calibrate it, because water boils at a known temperature. Once the thermometer is calibrated, the pressure variation can accurately give temperatures over a wide range.
19. Picture the Problem: A steel bar has a larger diameter than that
of an aluminum ring that you would like to slip over the bar. Strategy: Use equation 14-6 to calculate the temperature at which the
ring’s inner diameter will equal the diameter of the bar. The coefficient of thermal expansion is given in Table 16-1.
Solution: 1. (a) The ring should be heated. Imagine that the ring is cut and “unrolled.” It would be a rectangle. If the rectangle is heated, it will expand along its length and width. Its length is the circumference of the ring. Since the length of the rectangle increases, the circumference of the circle increases, and therefore, so does its diameter.
2. (b) Solve equation 16-‐4 for the change in temperature:
Δd = α dΔT ⇒ ΔT = T − T0 =Δdαd
3. Solve for the final temperature:
T = T0 +Δdαd0
= 10.00 °C + 4.040 cm – 4.000 cm
2.4 ×10−5 C°( )−1( ) 4.000 cm( )= 430°C
Insight: When the aluminum is heated to 430°C it will slip over the steel rod. As it cools back down it will
shrink to form a tight bond with the steel. 42. Picture the Problem: Heat transfers from a hot horseshoe to the cold water. This decreases the
temperature of the horseshoe and increases the temperature of the water until the water and horseshoe are at the same equilibrium temperature.
2
Strategy: Since only two objects are transferring heat, use equation 16-‐15 to calculate the equilibrium temperature. To determine which object would cause a larger final temperature, you should compare the heat capacities heats of the two objects. The object with the higher heat capacity will have more heat to transfer to the water, causing the final temperature to be greater.
Solution: 1. (a) Insert given data into equation 16-‐15:
T =mhchTh + mwcwTw
mhch + mwcw
=0.50 kg 448 J/ kg ⋅K( )⎡⎣ ⎤⎦ 450 °C + 25 kg 4186 J/ kg ⋅K( )⎡⎣ ⎤⎦ 23 °C
0.50 kg( ) 448 J/ kg ⋅K( )⎡⎣ ⎤⎦ + 25 kg( ) 4186 J/ kg ⋅K( )⎡⎣ ⎤⎦T = 24 °C
2. (b) Write the heat capacities of the lead and iron:
CPb = 1 kg ×128J/ kg ⋅K( )⎡⎣ ⎤⎦ = 128 J/K
CFe = 0.50 kg × 448 J/ kg ⋅K( )⎡⎣ ⎤⎦ = 224 J/K
3. Compare the heat capacities: Since the heat capacity of the lead is less than the heat capacity of the iron, the final temperature will be less than 24°C.
Insight: Even though the lead had twice the mass of the iron, the specific heat of lead is small enough that the heat capacity of the iron was larger than the heat capacity of the lead.
55. Picture the Problem: Two metal bars, of equal lengths, are
connected in parallel. Heat transfers across both bars. The diameter of the lead rod is known. We want to know what diameter of the copper rod is necessary for the rate of heat transfer through both rods to be 33.2 J/s.
Strategy: Set the total rate of heat transfer equal to the sum of the heat transfers through each rod. Using equation 16-‐16, solve for the diameter of the copper rod.
Solution: 1. Sum the heat transfer rates of the two rods, using equation 16-‐16:
Qtotal
t=
QCu
t+
QPb
t= kCu ACu
ΔTL
⎛⎝⎜
⎞⎠⎟+ kPb APb
ΔTL
⎛⎝⎜
⎞⎠⎟
= kCu
π4
dCu2⎛
⎝⎜⎞⎠⎟+ kPb
π4
dPb2⎛
⎝⎜⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
ΔTL
⎛⎝⎜
⎞⎠⎟= kCudCu
2 + kPbdPb2⎡⎣ ⎤⎦
πΔT4L
⎛⎝⎜
⎞⎠⎟
2. Solve for the diameter of the copper rod:
dCu =
Qtotal
t⎛
⎝⎜⎞
⎠⎟4LπΔT
⎛⎝⎜
⎞⎠⎟− kPbdPb
2⎡
⎣⎢⎢
⎤
⎦⎥⎥
kCu
=
33.2 J/s( ) 4 0.650 m( )π 112 − 21( ) °C − 34.3W/ m ⋅K( )⎡⎣ ⎤⎦ 0.0276 m( )2
395W/ m ⋅K( ) = 2.64 cm
Insight: The diameters of both rods are about the same. However, because the thermal conductivity of
the copper is over 10 times the thermal conductivity of lead, over 90% of the heat passes through the copper.
Home Work Solutions PHYS111.02 SFSU Eradat [WH6 CHAPTERS 16-‐18] 3 Chapter 17 17. Picture the Problem: The Martian atmosphere can be considered as an ideal gas with a given
temperature and pressure.
Strategy: Use the ideal gas law (equation 17-‐2) to solve for the number of molecules per volume. Compare the results for the Martian atmosphere and Earth’s atmosphere at standard temperature and pressure.
Solution: 1. (a) Convert the Martian temperature to Kelvin:
Tc =59
Tf − 32( ) = 59
− 64 − 32( ) = −53.3°C
Tk = 273.15− 53.3 K( ) = 219.8 K
2. Solve the ideal gas law for number of molecules per volume:
NV
= PkT
= 0.92 ×103 Pa1.38 ×10−23J/K( ) 219.8 K( ) = 3.0 ×1023 molecules m3
3. (b) The number of molecules per volume on Earth is greater than that on Mars. The temperatures of Earth and Mars have the same order of magnitude, but the pressure on Earth is far greater than that on Mars.
4. (c) Solve for the number of molecules per volume on Earth:
NV
= PkT
= 1.01×105 Pa1.38 ×10–23J/K( ) 273.15 K( ) = 2.68 ×1025 molecules m3
Insight: The density of molecules on Earth is about 100 times the density of molecules on Mars. When landing a rover on Mars, the scarce atmosphere results in a higher terminal velocity, even with a parachute.
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 kinetic energy.
Solution: 1. (a) Solve equation 17-‐5 for P:
P = nRTV
=3 mol 8.31J/ mol ⋅K( )⎡⎣ ⎤⎦ 273.15+ 295 K( )
0.0035 m3 = 4.0 ×106 Pa
2. (b) Insert the given data into equation 17-‐12 for the kinetic energy:
Kav = 3
2 kT = 32 1.38 ×10–23J/K( ) 273.15+ 295 K( ) = 1.18 ×10–20 J
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.
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
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temperature. Solution: 1. (a) Calculate the heat lost by the
water: Q = mcΔT = 0.33 kg 4186 J/ kg ⋅K( )⎡⎣ ⎤⎦ 14C°( ) = 1.93×104 J
2. Calculate the mass of the melted ice: m = Q
Lf
= 1.93×104 J33.5×104 J/kg
= 0.058 kg
3. Calculate the remaining ice: mf = 0.075 kg − 0.058 kg = 0.017 kg
4. (b) Set the latent heat required to melt the ice equal to the heat lost by the water:
mwc T − 0°C( ) = mice L
5. Solve for the initial temperature of the water:
T =mice Lmwc
=0.075 kg( ) 33.5×104 J/kg( )0.33 kg( ) 4186 J/ kg ⋅K( )⎡⎣ ⎤⎦
= 18°C
Insight: A small amount of ice can cool a large drink because the latent heat of fusion of water is quite large.
Chapter 18 27. Picture the Problem: An ideal gas completes a three-‐process
cycle.
Strategy: Use the ideal gas law to calculate the temperature at each state. Solve the first law of thermodynamics for the heat absorbed in each process.
Solution: 1. (a) Solve the ideal gas law for temperature:
T = PVnR
2. Solve for TA :
TA =150 ×103 Pa( ) 1.00 m3( )
67.5 mol( ) 8.31 J mol ⋅K( )= 267 K
3. Solve for TB :
TB =
50 ×103 Pa( ) 4.00 m3( )67.5 mol( ) 8.31 J mol ⋅K( ) = 357 K
4. Solve for TC :
TC =
50 ×103 Pa( ) 1.00 m3( )67.5 mol( ) 8.31 J mol ⋅K( ) = 89.1 K
5. (b) A → B :The temperature rises and the gas does work, so heat enters the system. B→ C : The temperature drops and work is done on the gas, so heat leaves the system. C → A : The temperature rises and no work is done on or by the gas, so heat enters the system.
6. (c) Solve the first law for Q: Q = ΔU +W = 32 nRΔT +W
7. Insert values for A → B :
Q = 32 67.5 mol( ) 8.31 J mol ⋅K( ) 357 K − 267 K( )+ 1
2 150 kPa + 50 kPa( ) 3.00 m3( ) = 376 kJ
8. Insert values for B→ C :
Q = 32 67.5 mol( ) 8.31 J mol ⋅K( ) 89.1 K − 357 K( )+ 50 kPa −3.00 m3( ) = −375 kJ
Home Work Solutions PHYS111.02 SFSU Eradat [WH6 CHAPTERS 16-‐18] 5 9. Insert values for C → A :
Q = 3
2 67.5 mol( ) 8.31 J mol ⋅K( ) 267 K − 89.1 K( ) + 0 = 150 kJ
Insight: The net work done during this cycle is equal to the net heat absorbed, or W = 150 kJ. Although the sum of the heats in part (d) appears to be 151 kJ, that is an artifact of significant digits and rounding issues, it’s exactly 150 kJ.
38. Picture the Problem: A monatomic ideal gas is expanded at
constant pressure by a fixed change in volume. This process is done twice with differing initial volumes.
Strategy: Use the area under the PV plot to calculate the work done in each process.
Solution: 1. (a) Write the work as area under the PV plot: W = PΔV
2. Solve numerically: W = 160 ×103 Pa( ) 8600 cm3 − 5400 cm3( ) 1×10−6 m3
cm3
⎛
⎝⎜⎞
⎠⎟= 0.51 kJ
3. (b) Work is directly proportional to the change in volume. Therefore, the work done by the gas in the second expansion is equal to that done in the first expansion.
4. (c) Solve numerically:
W = 160 ×103 Pa( ) 5400 cm3 − 2200 cm3( ) 1×10−6 m3
cm3
⎛
⎝⎜⎞
⎠⎟= 0.51 kJ
Insight: Because the work is proportional to the change in volume, increasing the volume by 3200 cm3
from any initial volume will produce the same amount of work. 51. Picture the Problem: The efficiency of a Carnot engine is increased by lowering the cold temperature
reservoir. Strategy: Use equation 18-‐13 to solve for the temperature of the cold temperature reservoir using the
initial efficiency, and then the new efficiency. Solution: 1. (a) Solve equation 18-‐13 for Tc :
Tc = Th 1− emax( ) = 545 K 1− 0.300( ) = 382 K
2. (b) The efficiency of a heat engine increases as the difference in temperature of the hot and cold reservoirs increases. Therefore, the temperature of the low temperature reservoir must be decreased.
3. (c) Insert the new efficiency: Tc = 545 K 1− 0.400( ) = 327 K
Insight: The efficiency of a Carnot engine is increased when the temperature difference between the two reservoirs increases. When the hot temperature reservoir is fixed, the efficiency can be increased by lowering the temperature of the cold reservoir.
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