# Physics 202 at Portland State 2014

start

## Announcements

past announcements have moved to sidebar and can be found here.

### Final exam solutions

Are added to D2L under “Sample and Past Exams and Solutions” — Nicholas Kuzma 2014/03/24 17:14

### End-of-class Email

Dear Class,

The first term of my engagement as PSU faculty has almost come to a close - the final grades are due on Tuesday, and our grader is working away full steam trying to finish by Monday.

Here is a number of things I want you to take a look at by the end of the business day on Monday though:

1. I just posted all the grades (except for the final exam) on D2L:
• extra credit (= Quizzes + Workshop/Paper + Wiki Edit/Course improvement bonus) and the last homework.
• Quizzes include Ameena's survey (as quiz 6), you get 5 points if you submitted anything to her. The total quiz grade is $\frac{\Sigma\,\text{best 5 quizzes}}{5}$. (Out of 5 maximum total).
• Workshop or paper are added to that, also out of 5 maximum total.
• Wiki edits, or emails to me with suggestions for course improvement/feedback/constructive physics questions are an additional bonus out of 5 points total.
• Theoretically, one can have up to 15 points extra credit.
• Please check these grades and, if you have evidence that some points weren't added correctly (such as 0 for work that you submitted, or maybe some wiki edits that I missed), - please send me an email by 4 pm Monday. Please include specific details that I can use to look for missing credit (i.e. the date of the edit or the topic, etc). Please, at this point don't send general inquiries such as “is there something else I can get extra credit for?” etc. The time window for these is now closed.
2. Because of a miscommunication with Prof. Widenhorn, I didn't realize this department has a long history of doing the CLASS survey after each term, not to be confused with Ameena's workshop survey. I really apologize to you guys for not catching it earlier (this was supposed to be the extra credit quiz 6 survey!), and, humbled by my mistake, I am asking you to please go and fill out this survey as well. You can do it during the break as well, I'll finalize the data by the end of the first week of class next term.
• This is the email from Prof. Widenhorn that I overlooked, please click on the link and fill it out in your spare time:
• The PSU physics department and I are committed to collecting real data about how effective our teaching is. With this in mind we survey student attitude throughout the year….
• The survey is about your personal beliefs, there is no right or wrong answer and neither I nor any other instructor will know how you responded.
• DO NOT fill it out on a smart phone they often fail to submit properly. If you take it twice we will only keep your first response unless it is incomplete. The entire survey should take 20-30 minutes and must be completed in one sitting. Please respond to each question quickly according to your first impressions and/or gut feeling. Remember to respond how you believe, don’t even think about how others might respond.
3. Corrections (please read if you are taking MCAT or any other test with a physics component in the future, or if you are planning to be an engineer):
• There was a genuine confusion (on my part) between the terms “Coefficient of performance (COP)” and “efficiency”. When I studied physics, there only was one term to this effect, basically the ratio between the “useful output” and “the cost-determining input” of any thermodynamic engine or appliance.
• For engines, this is $\frac{W}{Q_h}$, for coolers such as fridges and air conditioners, this is $\frac{Q_c}{W}$, and for heat pumps this is $\frac{Q_h}{W}$.
• What I didn't realize is that nowadays in this country, “Coefficient of performance (COP)” is reserved just for coolers and heat pumps (see Coefficient of performance), with two different definitions depending on whether it is a cooler or a heat pump
• “efficiency” is mostly reserved just for engines: Thermal efficiency, but can also be loosely applied to anything that consumes heat (for example, in a co-gen plant, the input is the $Q_h$, but the useful output is the sum $W+Q_c$). So, for engines and any other heat-driven processes efficiency cannot exceed 1 because no other source of heat is considered, and some of that energy is invariably wasted and doesn't contribute to the output. Again, efficiency is $\frac{\text{output}}{\text{input}}$ ratio, where input is the heat extracted from (the only) source of heat such as fuel, nuclear reactor, or geothermal source. Output is work and/or heat released into the useful application.
• Please accept my apologies and make a note of this correction for your test-preparation notes. This Wiki has been corrected accordingly.
4. Final grades are due Tuesday by 5pm, please check at that point. I am hoping to submit Monday night / Tuesday morning, in case there are glitches with the submission system and I need to call IT department for help.
5. Finally, my heart-felt thanks to all of you for making it not only an enjoyable class, but also for contributing to each other's, as well as to my own education: I learned a lot from you guys, and I want to thank you for this amazing experience!

Best of luck, and hopefully I'll see you in my Physics 203 class! Those who are excited about it and want to have an early start, please read chapters 13 and 14 in our textbook. I'll post the syllabus and get the new wiki going soon… Please google “Kuzma PDX” to go to my homepage and check whether there is a link to the new wiki. — Nicholas Kuzma 2014/03/22 17:29

### Workshop survey

survey_workshop_form.pdf - Deadline for extra credit: Tuesday March 18, 11:59pm PDT survey_workshop_form.doc - a Word document for those who cannot edit/save the PDF

#### Survey checklist

• right-clicking on it and selecting “Save (link) as”
• or left-clicking on it, and clicking a pop-up “Save” button in your viewer plug-in (PDF only)
2. open it on your computer
3. fill it out
4. quit and save the file
5. email it as an attachment to Ameena ( aka6 [at] pdx . edu )
6. Those of you who wish to get extra Quiz credit, please sign your email to her with your full name.

##### A warning for those using Google Chrome Browser (at least on a mac)

Be sure to check the PDF before emailing it to see if it is blank. In google chrome's PDF plug-in for mac, it will sometimes just save a blank form. In order to preserve your data, you can alternatively follow this procedure:

1. Hit the print icon.
2. When the print menu launches, click the “open PDF in Preview” link near the bottom.
3. Preview should open with your document and include the data you entered.
4. Select “File” pull down, then “Save…” and name the file something, being sure to include the .pdf extension in the name and save.
5. Then email the pdf with your data to Ameena at the address listed above.

## Lectures and Notes

Use WolframAlpha for calculations

## Ch.15: Fluids (Feb 25)

### Examples Ch.15

#### Problem 12.1

Find the mass of a bucket (3 US gallons) of mercury. The density of Hg is $\rho_\text{Hg}=13.5\,$g/cm3.

Steps:

1. Convert units to SI:
• V = 3 US gal = 3 $\cdot$ 231 inch3 = 3 $\cdot$ 3.7854 L = 11.356 L $\approx$ 0.011 m3
• $\rho_\text{Hg} \approx 13.53\,\frac{\text g}{{\text{cm}}^3}$ $=13530\,\frac{\text kg}{{\text{m}}^3}$
2. Find the mass from the definition of density $\rho=\frac{m}{V}$:
• $m=\rho_\text{Hg}\!\cdot\!V=0.011\,{\text{m}}^3\times 13530\,\frac{\text kg}{{\text{m}}^3}=153\,$kg
3. Try to make sense of the answer:
• This is about 300 pounds!!! Mercury is heavy!
• A joke in the old good cold war research labs was to tell a new student to “go get a bucket of mercury”.
• Don't try that at home! Mercury is “quite toxic” and the guy in the video is quite foolish.

#### Problem 12.2

You and your bike weigh 170 pounds. When you sit on it, each wheel makes an 8 cm2 imprint on the pavement. Find the pressure in the tires.

Steps:

1. Convert units to SI:
• 170 pounds = 77.1 kg
• 8 cm2 = 0.0008 m2
2. Write the balance of forces on the flat portion of the tires in contact with the road
• $F_\text{support}+F_\text{atm}=F_\text{pressure}\;\;\Rightarrow\;\;mg+p_AA=pA$
3. Solve for absolute pressure in the tires
• $p=p_A+\frac{mg}{A}\approx 101\,{\text{kPa}}+\frac{77.1\,{\text{kg}}\,\cdot\,9.8\frac{\text m}{{\text s}^2}}{2\,\cdot\,0.0008\,{\text m}^2}=p_A+473\,{\text{kPa}}$ $=p_A+69\,{\text{psi}}$
4. Try to make sense of the answer:
• Sounds about right for a non-professional bike like mine
• Do you multiply the given surface area of the tires by two since there are two tires?
• Yes!
• Also, what happens to the $p_A$ in the equation? Is 69 psi the final answer?
• Technically, the final answer should be 574 kPa. But when you are looking at your bicycle pressure gauge, it gives the pressure in psig units, where “g” stands for “gauge”. That means the gauge reading is relative to 1 atm (i.e. 0 psig = 1 atm). Essentially, your gauge will read the psi pressure difference between the actual tire pressure and $p_A=1\,$atm. That is, it will read 69 psi (more accurately, 69 psig).

#### Problem 12.3

A partially submerged cylindrical tank is pulled out of the lake with the closed end up and the open end still under water. The level of the water inside the tank is 2 m above the lake water level. Find the air pressure in the air bubble trapped in the tank. Can the water level in the tank be 15 m above the lake surface?

Steps:

1. Write down the air pressure at the lake surface (assume atmospheric pressure)
• $p_A=101325\,$Pa
2. Calculate the pressure difference due to 2 m column of water
• $\left(p_A-p\right)=\rho gh = 19.6\,$kPa
3. Figure out the air pressure at the water surface in the tank
• $p=p_A-\rho gh=81.7\,$kPa
4. Do the same calculation for $h$ = 15 m
• $-50\,$kPa ???

#### Problem 12.4

A solid object made from a pure element weighs 4.413% less when submerged in pure water at room temperature (compared to its weight in air). What material is it made from?

Steps (as discovered by Archimedes):

1. Assign symbols to the quantities of interest:
• Density of the unknown material $\rho_x$
• Mass of the unknown object $m_x$
• Volume of the unknown object $V_x$
2. Look up the necessary constants:
• Density of water at $20^\circ:\;\;\;\rho_\text{wat}=0.99820\!\frac{\text g}{\text{cm}^3}=998.20\!\frac{\text kg}{\text{m}^3}$
• Density of air at $20^\circ:\;\;\;\rho_\text{air}=1.2041\!\frac{\text kg}{\text{m}^3}$
3. Put together the equations for the observables:
• Weight of the object in air: $\;\;\;\;\;F_\text{in air}\,\;=\;m_xg \!-\! \rho_\text{air}V_xg\,\;=\;\rho_xV_xg\!-\!\rho_\text{air}V_xg$ $=\; (\rho_x\!-\!\rho_\text{air})V_xg$
• Weight of the object in water: $\;\;F_\text{in wat}\;=\;m_xg \!-\! \rho_\text{wat}\!V_xg\;=\;\rho_xV_xg\!-\!\rho_\text{wat}V_xg$ $=\; (\rho_x\!-\!\rho_\text{wat})V_xg$
• The difference of these two weights is a known percentage of the weight in air (the weight in water is lower): $0.04413=\frac{F_\text{in air}-F_\text{in wat}}{F_\text{in air}}$
4. Plug in the values and solve the equations:
• $0.04413=\frac{F_\text{in air}-F_\text{in wat}}{F_\text{in air}}=\frac{(\rho_x-\rho_\text{air})V_xg-(\rho_x-\rho_\text{wat})V_xg}{(\rho_x-\rho_\text{air})V_xg}$ $=\frac{(\rho_x-\rho_\text{air})-(\rho_x-\rho_\text{wat})}{\rho_x-\rho_\text{air}}=\frac{\rho_\text{wat}-\rho_\text{air}}{\rho_x-\rho_\text{air}}\;$ $\Rightarrow$ $\;\rho_x=\rho_\text{air}\!+\!\frac{\rho_\text{wat}-\rho_\text{air}}{0.04413}=22593\!\frac{\text kg}{\text{m}^3}=22.59\!\frac{\text g}{\text{cm}^3}$
5. Look up the unknown element in the density of elements table:
• Osmium density is the highest of all elements at 22.59 g/cm3, with iridium being a close second at 22.56 g/cm3
6. Try to make sense of the answer:
• At the time of Archimedes, gold had the highest density of all known materials, and was also the most expensive metal. Therefore, if the density determined by this method was less than that of pure gold, the unknown material must have been gold diluted with some other, less expensive and less dense, metal such as copper or silver.

#### Problem 12.5

An injection of 2 mL of vaccine takes 10 s. The syringe inner diameter is 5 mm, the needle is 5 cm long, with a 0.2$\,$mm inner diameter. Find the force required to perform the injection, assuming the viscosity of the vaccine is equal to that of water (0.001 N$\cdot$s/m2). If the vaccine is squirted into the air straight up (to remove the air bubble), how high will it go? Assume the length of the syringe (not including the needle) can be calculated from the volume of the vaccine and the syringe cross-section.

Steps:

1. Assign symbols to the quantities of interest, convert units to SI:
• Volume $\;\;V=2\,{\text{mL}}=2\times 10^{-6}\,{\text{m}}^3$
• Syringe parameters:
• Cross-section $\;\;A_1=\pi\!\cdot\!\left(\frac{d_1}{2}\right)^2=\pi\cdot (0.0025\,{\text{m}})^2$ $\approx 2\times 10^{-5}\,{\text{m}}^2=0.2\,{\text{cm}}^2$
• Length of vaccine bolus inside the syringe initially $\;\;\Delta x_1=\frac{V}{A_1}=10.2\,{\text{cm}}=0.102\,{\text{m}}$
• Needle parameters:
• Cross-section $\;\;A_2=\pi\!\cdot\!\left(\frac{d_1}{2}\right)^2=\pi\cdot (0.0001\,{\text{mm}})^2$ $\approx 3.1\times 10^{-8}\,{\text{m}}^2=3.1\times 10^{-4}\,{\text{cm}}^2$
• Length $\;\;l_2=0.05\,$m
2. Look up the necessary constants:
• Viscosity of water at $20^\circ:\;\;\;\eta_\text{wat}=1.002\times 10^{-3}\!\frac{{\text N}\cdot{\text s}}{\text{m}^3}=1.002\times 10^{-3}\,{\text{Pa}}\!\cdot\!{\text{s}}$
• Density of water at $20^\circ:\;\;\;\;\rho_\text{wat}=0.99820\!\frac{\text g}{\text{cm}^3}=998.20\!\frac{\text kg}{\text{m}^3}$
3. To be able to use Poiseuille equation, need to find velocity:
• Velocity of vaccine in the syringe: $v_1=\frac{\Delta x_1}{\Delta t}\approx\frac{10\,{\text{cm}}}{10\,{\text{s}}}= 1\,$cm/s $=0.01\,$m/s
• Continuity equation along the path from the syringe into the needle: $\;\;A_1\!\cdot\!v_1=A_2\!\cdot\!v_2$
• Velocity of vaccine in the needle: $v_2=v_1\!\cdot\!\frac{A_1}{A_2}\approx 6.4\,$m/s
4. Write down Poiseuille equation and plug the values to find the pressure difference (the velocity is the average across $A_2$):
• $\Delta p=p\!-\!p_A=8\pi\,\eta_\text{wat}\frac{v_2l_2}{A_2}$ $=8\cdot 3.14159\cdot 10^{-3}\,{\text{Pa}}\!\cdot\!{\text s}\cdot\frac{6.4\,{\text{m/s}}\cdot 0.05\,{\text m}}{3.1\times 10^{-8}\,{\text m}^2}\approx 2.6\times 10^5\,$Pa $\approx 2.5\,$atm.
5. Find the force on the plunger from the balance of forces on it $\;\;F+p_A\!\cdot\!A_1=p\!\cdot\!A_1$:
• $F=p\!\cdot\!A_1\!-p_A\!\cdot\!A_1=\left(p\!-\!p_A\right)\!\cdot\!A_1\approx 2.6\times 10^5\,$Pa$\,\cdot\,2\times\!10^{-5}\,$m$^2=5.02\,$N
6. Try to make sense of the answer:
• This is equivalent to approximately 1.1 pounds of force, not that effortless after all for one thumb!
7. Find the height of the squirted liquid from the Bernoulli equation $\;\frac{1}{2}\rho v_2^2=\rho gh$:
• $h=\frac{1}{2}\frac{v_2^2}{g}=\frac{(6.4\,{\text{m/s}})^2}{2\,\cdot\,9.8\,{\text{m/s}}^2}\approx 2.1\,$m
• (Note: you can also use Torricelli's Law and solve for height)
• Should the velocity used in part 4 be 6.4 m/s and not 0.01 m/s?
• You are absolutely right, it was a cut-and-paste typo. The final answer is correct though. I fixed the number in part 4. — Nicholas Kuzma 2014/03/15 00:02

#### Problem 12.6

A river boat with a blunt bow, moving with a velocity $v=18.52\,\frac{\text{km}}{\text{hr}}$ (10 knots), “piles up” water at the bow. Find the height of the resulting wake at the bow. Steps:

1. Convert all units to SI and look up constants:
• $v=18.52\,\frac{\text{km}}{\text{hr}}$ $=18520\,\frac{\text m}{\text{hr}}$ $=\frac{18520}{3600}\,\frac{\text m}{\text s}$ $\approx 5.144\,\frac{\text m}{\text s}$
• $\rho_\text{wat}=1000\,\frac{\text{kg}}{ {\text m}^3}$
• $g=9.8\,\frac{\text m}{ {\text s}^2}$
2. In the boat's frame of reference, pressure at the top of the wake is $p_\text{top}=p_A$, and the velocity relative to the boat is zero. In the same frame, the velocity of far-away water is $-v$.
3. We can write Bernoulli's equation in this frame, comparing the top of the wake (subscript 2) to the water surface far away (subscript 1):
• $p_1+\rho g h_1+\frac{1}{2}\rho v_1^2=p_2+\rho g h_2+\frac{1}{2}\rho v_2^2$
• $p_A+0+\frac{1}{2}\rho\!\cdot\!(-v)^2=p_A+\rho g\Delta h+0$
4. Solve the above to find the height of the wake $\Delta h$:
• $\frac{1}{2}\rho v^2=\rho g\Delta h$
• $h=\frac{1}{2}\frac{v^2}{g}=\frac{\left(5.144\,\frac{\text m}{\text s}\right)^2}{2\,\cdot\,9.8\,\frac{\text m}{ {\text s}^2}}$ $=1.35\,{\text m}$
5. Try to make sense of the answer:
• Ten knots is pretty fast, and the bow in most boats is made sharp (not blunt) on purpose, to minimize the height of the wake and the drag force of the water on the boat.

## Ch.16: Temperature and heat (Feb 27)

### Concepts Ch.16

• Linear expansion
• Volume expansion

### Material properties Ch. 16

From: Walker, James S. “Ch. 16 Temperature and Heat.” Physics. Custom ed. Vol. 2. Upper Saddle River, NJ: Pearson/Prentice Hall, 2004

 Coefficients of Thermal Expansion Near 20$\,^\circ$C Specific Heats at Atmospheric Pressure Thermal Conductivities Table 16-1 Table 16-2 Table 16-3 Substance Coefficient of linear expansion $\alpha$ $\;\left[K^{-1}\right]$ Substance Specific Heat $c$ $\;\left[\frac{\text J}{{\text{kg}}\cdot{\text K}}\right]$ Substance Thermal Conductivity $k$ $\;\left[\frac{\text W}{{\text m}\cdot{\text K}}\right]$ Lead $2.9\times 10^{-5}$ Water 4186 Silver 417 Aluminum $2.4\times 10^{-5}$ Ice 2090 Copper 395 Brass $1.9\times 10^{-5}$ Steam 2010 Gold 291 Copper $1.7\times 10^{-5}$ Beryllium 1820 Aluminum 217 Iron (steel) $1.2\times 10^{-5}$ Air 1004 Steel (low carbon) 67 Concrete $1.2\times 10^{-5}$ Aluminum 900 Lead 34 Window glass $1.1\times 10^{-5}$ Glass 837 Stainless steel alloy 302 16 Pyrex glass $3.3\times 10^{-6}$ Silicon 703 Ice 1.6 Quartz $5.0\times 10^{-7}$ Iron (steel) 448 Concrete 1.3 Substance Coefficient of volume expansion $\beta$ $\;\left[K^{-1}\right]$ Copper 387 Glass 0.84 Silver 234 Water 0.60 Ether $1.51\times 10^{-3}$ Gold 129 Asbestos 0.25 Carbon tetrachloride $1.18\times 10^{-3}$ Lead 128 Wood 0.10 Alcohol $1.01\times 10^{-3}$ Wool 0.040 Gasoline $9.5\times 10^{-4}$ Air 0.0234 Olive oil $6.8\times 10^{-4}$ Water $2.1\times 10^{-4}$ Mercury $1.8\times 10^{-4}$

### Equation Sheet Ch.16

##### Please update if you have an equation not included
1. Temperature in Fahrenheit = 1.8 (Temperature in Celsius) + 32
• $T_F = 1.8\cdot T_C + 32$
• Example (“warm & cosy” room temperature): $\;T_C=25\,^\circ{\text{C}}\;\;\Rightarrow\;\;T_F=1.8\cdot 25+32=77\,^\circ$F
• Example (“energy-saver” room temperature): $\;T_C=20\,^\circ{\text{C}}\;\;\Rightarrow\;\;T_F=1.8\cdot 20+32=68\,^\circ$F
2. Temperature in Celsius = (5/9) (Temperature in Fahrenheit $- 32$)
• $T_C =\frac{5}{9}\!\cdot \left(T_F - 32\right)$
• Example (body temperature): $\;T_F=98.6\,^\circ{\text{F}}\;\;\Rightarrow\;\;T_C=\frac{5}{9}\!\cdot(98.6-32)=37.0\,^\circ$C
3. Temperature in Kelvin = Temperature in Celsius + 273.15
• $T_K = T_C + 273.15$
• Example (boiling point of liquid nitrogen at $p_A$: $\;T_C=-195.7\,^\circ{\text{C}}\;\;\Rightarrow\;\;T_K=-195.7+273.15=77.45\,$K
4. Linear Expansion = Coefficient of linear expansion * initial length * change in temperature
• $\Delta L=\alpha\!\cdot\!L_0\!\cdot\!\Delta T$
5. Area expansion = 2 * Coefficient of linear expansion * initial area * change in temperature
• $\Delta A\approx 2\alpha\!\cdot A_0\!\cdot\!\Delta T$
6. Volume expansion = Coefficient of volume expansion * initial volume * change in temperature $\approx$ 3 * Coefficient of linear expansion * initial volume * change in temperature
• $\Delta V=\beta\!\cdot\!V_0\!\cdot\!\Delta T\approx 3\alpha\!\cdot V_0\!\cdot\!\Delta T$
7. Energy (and heat) unit conversions:
• 1 calorie = 4.186 J (chemistry)
• 1 Calorie = 1 kilocalorie = 4186 J (nutrition labels)
• 1 Btu = 0.252 kcal = 1055 J (heating bills)
8. $Q$ = heat = energy transferred due to temperature differences or friction
9. Heat Capacity = heat / change in temperature
• $C = \frac{Q}{ΔT} = c\!\cdot\!m\;\;$ (see next item)
10. Specific heat = Q / (mass * change in temperature)
• $c = \frac{Q}{m\Delta T}=\frac{C}{m}$
11. Heat transfer by conduction = Coefficient of thermal conductivity * Area * (Temperature difference betw. sides / Thickness) * time interval
• $Q = k\!\cdot\!A\frac{T_\text{hot}-T_\text{cold}}{L}\Delta t$
12. Rate of heat transfer (in watts) = Coefficient of thermal conductivity * Area * (Temperature difference betw. sides / Thickness)
• $\frac{\Delta Q}{\Delta t} = k\!\cdot\!A\frac{T_\text{hot}-T_\text{cold}}{L}$
13. Radiated Power = emissivity * (Stefan-Boltzmann constant) * area * (temperature to the 4)
• $P = e\!\cdot\!\sigma\!\cdot\!A\!\cdot\!T^4$
• Stefan-Boltzmann constant $\;\;\sigma= 5.67\times 10^{-8}\,\frac{\text{W}}{{\text m}^2\cdot{\text K}^4}$
• Emissivity $\;e\approx 1$ for black, dark-colored objects; $\;e\approx 0$ for shiny, mirror-like objects
14. Net Radiated Power = emissivity * (Stefan-Boltzmann constant) * area * (temperature (object) to the 4 - temperature (surroundings) to the 4)
• $P_\text{net} = e\!\cdot\!\sigma\!\cdot\!A\!\cdot\!\left(T_\text{obj}^4-T_\text{surr}^4\right)$
15. Notation:
• A temperature $T$ of five degrees is 5 $^\circ$C (five degrees Celsius)
• A temperature change $\Delta T$ of five degrees is 5 C$^\circ$ (five Celsius degrees)
• Temperature values and differences in Kelvin units are using the same notation: $T$ = 5 K and $\Delta T$ = 5 K
• Conversion of temperature differences is much simpler (compared to values): $\Delta T_K=\Delta T_C=\frac{5}{9}\Delta T_F$

#### Homework Questions Ch.16

##### Problem 16.41

To determine the specific heat of an object, a student heats it to 100$\,^\circ$C in boiling water. She then places the 98.3 g object in a 193 g aluminum calorimeter containing 147 g of water. The aluminum and water are initially at a temperature of 20.0$\,^\circ$C, and are thermally insulated from their surroundings.

If the final temperature is 23.7$\,^\circ$C, what is the specific heat of the object?

Hint: Solve it by balancing the heat released from cooling of the object (100$\,^\circ$C $\rightarrow$ 23.7$\,^\circ$C) and the heat absorbed by the aluminum body + the water of the calorimeter (20.0$\,^\circ$C $\rightarrow$ 23.7$\,^\circ$C).

### Examples Ch.16

#### Problem 13.1

At what temperature Celsius and Fahrenheit scales are equal?

Steps:

1. Write down the conversion equation for $T_F$ (temperature value in $^\circ$F units) and $T_C$ (value in $^\circ$C units):
• $T_F=\frac{9}{5}\!\cdot\!T_C$ $+32$
2. Equate this to the Celsius value and solve for $T_C$:
• $T_C=T_F$ $=\frac{9}{5}\!\cdot\!T_C+32\;\;\Rightarrow\;\;T_C-\frac{9}{5}\!\cdot\!T_C=32$ $\;\Rightarrow\;\;T_C\!\cdot\!\left(1-\frac{9}{5}\right)=32$
• $T_C=\frac{32}{1-\frac{9}{5}}=\frac{\;32}{-4/5}=-40\,^\circ{\text C}=-40\,^\circ{\text F}$
3. Make sense of the answer:
• Check out the picture to the right

#### Problem 13.2

Contracting window frame

A 6 x 6 ft window is encased in an aluminum frame. Find the minimal gap on each side at room temperature (20$\,^\circ$C) to avoid cracking at $-40\,$degrees.

1. Look up the necessary constants:
• $\alpha_\text{Al}= 24\times 10^{-6}\,{\text K}^{-1}$
• $\alpha_\text{glass}= 11\times 10^{-6}\,{\text K}^{-1}$
2. Calculate $\Delta L$ for each material:
• $L = 6\,$ft = 1.83 m
• $\Delta L_\text{glass} = \alpha_\text{glass}\!\!\cdot\!L\!\cdot\!\Delta T$ $= 11\times 10^{-6}\,{\text K}^{-1}\cdot 1.83\,\text{m}\cdot(-40 - 20)\,{\text C}^\circ$ $= -1.2\times 10^{-3}\,{\text m} = -1.2\,$mm (minus indicates shrinkage)
• $\Delta L_\text{Al} = \alpha_\text{Al}\!\!\cdot\!L\cdot\!\Delta T$ $= 24\times 10^{-6}\,{\text K}^{-1}\cdot 1.83\,\text{m}\cdot(-40 - 20)\,{\text C}^\circ$ $= -2.6\times 10^{-3}\,{\text m} = -2.6\,$mm (greater shrinkage)
3. Total gap needed:
• $\Delta L_\text{glass} - \Delta L_\text{Al} = 1.4\,$mm
• $\frac{1.4\,{\text{mm}}}{2} = 0.7\,$mm needed on each side of the square frame

Note that the temperature difference $\Delta T$ is the same in K units as in C$^\circ$ (difference) units.

#### Problem 13.3

Gasoline expansion

How much do you gain by filling up a 15 gal tank in the winter at 20$\,^\circ$F and using it in summer at 90$\,^\circ$F?

1. Look up the volume expansion coefficient
• $\beta_\text{gasoline} = 9.5\times 10^{-4}\,{\text K}^{-1}$
• Cost of gasoline 3.29 $/gal 2. Calculate •$\Delta T = 70\,{\text F}^\circ = \frac{5}{9}\cdot 70 = 39\,{\text C}^\circ = 39\,$K. (Note, conversion of temperature differences does not require subtracting 32) •$\Delta V = \beta\cdot V\cdot \Delta T = 9.5\times 10^{-4}\,{\text K}^{-1}\cdot 15\,{\text{gal}}\cdot 39\,{\text K}= 0.554\,$gal • Cost gain 0.554 US gallons times 3.29$/gal = 1.82 US dollars
1. Why is V not 15 gal? Solving for V with the $\Delta V$ of 0.554 gal yields 4.98 gal for V. I don't understand why.
• Not sure what equation you are trying to solve… V is 15 gal! I made the calculation above more explicit, please check again! — Nicholas Kuzma 2014/03/16 20:23

#### Problem 13.4

Calories from chocolate

How many steps up the stairs are necessary to burn off all the calories from 1/4 of the bar of chocolate pictured to the right? Assume the human muscle is 100% efficient in converting the calories to mechanical energy, each step is 15 cm high, and your mass is 70 kg.

Steps (literally and figuratively):

1. Convert all units to SI
• $\Delta h=15\,{\text{cm}}$ $=0.15\,{\text m}$
• Energy equivalent of the whole 100-gram bar
• $230\,{\text{Cal}}\times 2.5\,{\text{servings}}$ $=575\,{\text{kcal}}$ $=2.4\times 10^6\,{\text J}$
• Energy equivalent of the quarter-bar (25 g):
• $\frac{1}{4}\cdot 575\,{\text{kcal}}$ $=6\times 10^5\,{\text J}$
2. The energy needed to lift oneself by one step is
• $mgh=70\,{\text{kg}}\cdot 9.8\,\frac{\text m}{ {\text s}^2}\cdot 0.15\,{\text m}$ $=103\,{\text J}$
3. The ratio of the energy equivalent of 1/4 bar of chocolate to that needed to climb one step is
• $\frac{\frac{1}{4}\,{\text{bar choc}}}{mgh}$ $=5841$ steps
1 bar of “Green & Black's” = Mt. Hood

The cost of OHSU sky-tram going up is about the price of a bar of “Green & Black's” - not a bad deal! (Going down is free, by the way)

#### Problem 13.5

A sleeping person “burns” about 70 W of energy just to keep warm during sleep. Assuming the thermostat in the bedroom is set to 72$\,^\circ$F and that the area of the upward-facing side of the body is 0.5 m2,

1. Find the thermal conductivity of the blanket, which is half-inch thick.
2. How many more kilocalories are burned during 6 hours of sleep if the thermostat is set to 68$\,^\circ$F instead?

Steps:

1. Convert all units to SI

#### Problem 13.6

How hot a black electric stove needs to be to radiate 1 kW. The diameter of the hot plate is 6 inches.

Steps:

1. Convert all units to SI

## Ch.17: Phases and Phase Changes (3/4,6)

### Concepts Ch.17

• Phase equilibrium

### Constants Ch.17

#### Properties of water

These values depend somewhat on temperature and pressure. See properties of water for more details.

Property Symbol Value Unit
Latent heat of water vaporization $L_v({\text H}_2{\text O})$ $22.6\times 10^5$ $\frac{\text J}{\text{kg}}$
Specific heat of water $c_\text{wat}$ 4186 $\frac{\text J}{{\text{kg}}\cdot{\text K}}$
Latent heat of ice fusion $L_f({\text H}_2{\text O})$ $33.5\times 10^4$ $\frac{\text J}{\text{kg}}$
Specific heat of ice $c_\text{ice}$ 2090 $\frac{\text J}{{\text{kg}}\cdot{\text K}}$

#### Alternate Gas Constant Values

Value Units
8.314 m3 Pa K-1 mol-1
0.08206 L atm mol-1 K-1
8.206 x 10-5 m3 atm mol-1 K-1
62.364 L torr mol-1 K-1
0.062364 m3 torr mol-1 K-1

### Units Ch.17

#### Units of P, V, and T

Factor Variable Units
Pressure P Atm, Torr, Pa, mmHg
Volume V L, m3
Moles n or $\nu$ mol
Temperature T K

#### Pressure Units

See the full table in this Wikipedia article

Unit Symbol Value of $p_A$ in these units Value of 1 Pa in these units
Atmosphere atm 1 atm $9.86923\times 10^{-6}\,$atm
Millimeter of Mercury mmHg 760 mmHg 0.0075 mmHg
Torr torr 760 torr 0.0075 torr
Pascal (SI Unit) Pa 101325 Pa 1 Pa
Kilopascal kPa 101.325 kPa 0.001 kPa
Pound per square inch psi, psia (“absolute”) 14.696 psia $14.5\times 10^{-5}\,$psi

### Equation Sheet Ch.17

1. Equation of State for an Ideal Gas in terms of the total number of molecules:
• $PV=Nk_BT$
• $k_B= 1.38\times 10^{-23}\,\frac{\text J}{\text K}\;\;$ (Boltzmann’s constant)
• $N=$ number of molecules
• $T$ is temperature in Kelvin
2. Equation of State for an Ideal Gas in terms of moles:
• $PV= nRT\;\;$ or $\;\;PV=\nu RT\;\;$, where $n$ (or $\nu$) is the number of moles:
• $n=\nu=\frac{N}{N_A}\;\;$, where $N_A$ is Avogadro constant
• $R= 8.31\!\frac{\text J}{{\text{mol}}\cdot{\text K}}$
3. Pressure in the Kinetic Theory of Gases, where the subscript av denotes an average quantity per molecule
• $P= \frac{1}{3}\left(\frac{N}{V}\right)\,2K_\text{av} =\frac{2}{3} \frac{N}{V}\left( \frac{1}{2}m_0v^2\right)_\text{av}\;\;,$ where $m_0$ is the mass of each molecule in kg
4. Average kinetic energy per molecule and Temperature of any ideal gas
• $K_\text{av}=\left(\frac{1}{2}m_0v^2\right)_\text{av}= \frac{3}{2}\!k_BT$
5. RMS Speed of a Gas Molecule
• $v_\text{rms}= \sqrt{(v^2)_\text{av}}= \sqrt{\frac{3k_BT}{m_0}} = \sqrt{\frac{3RT}{M}}\;\;,$ where $M$ is the molar mass of the gas in $\frac{\text{kg}}{\text{mol}}$
• SI units: $\frac{\text m}{\text s}$
6. Internal Energy of a monoatomic Ideal Gas
• $U=NK_\text{av}=\frac{3}{2}\!Nk_BT = \frac{3}{2}\!nRT= \frac{3}{2}\!\nu RT$
7. Definition of Latent Heat, L
• $L=\frac{Q}{m}\;,$ where $Q$ is the heat absorbed or released during phase transition, and $m$ is the total mass of the material
• SI units: $\frac{\text J}{\text{kg}}$
8. Changing the Length of a Solid under stress
• $F = Y\left(\!\frac{\Delta L}{L_0}\!\right)A$
9. Shear Deformation
• $F= S\left(\!\frac{\Delta x}{L_0}\!\right)A$

### Homework Questions Ch.17

#### Prob. 17.72

A xxx-kg ice cube at 0.0 oC is dropped into a Styrofoam cup holding yyy kg of water at zzz oC. (a) Find the final temperature of the system and (b) the amount of ice (if any) remaining. Assume the cup and the surroundings can be ignored. (c) Find the initial temperature of the water that would be enough to just barely melt all of the ice.

• I keep calculating the mass of the ice remaining to be 0.075 kg, but it says it is wrong. I calculated this using $Q=mc\Delta T= 0.09\,{\text{kg}}\cdot 4186\,\frac{\text{J}}{{\text{kg}}\cdot{\text{K}}}\cdot 13\,{\text{C}}^\circ=4897\,$J. Then I divide $\frac{Q}{L}=\frac{4897}{33.5\times 10^4} = 0.01455\,$kg…which is the amount of ice melted, so I subtract it from the initial mass…. $0.09-0.01455=0.075\,$kg
• I think you may be confusing the initial masses of the ice and the water. When you initially calculate the heat $Q$ released from cooling of the water from $13\,^\circ{\text{C}}$ to $0\,^\circ{\text{C}}$, you should use the initial mass of the water, not 0.09 kg. — Nicholas Kuzma 2014/03/05 22:56
• In my case mass of the ice $=8.0\times10^{−2}\,$kg, the mass of water = 0.35 kg, and initial water temperature = 13 $^\circ$C. I have calculated 4.33 $^\circ$C should be the final temperature and have asked a few different people and they seem to agree with that. however the answer says it is 0 $^\circ$C. The only reasoning i can find for this would be that because it takes more heat energy to turn the Ice into water than it does to cool the water to zero then the system should be at zero degrees with more ice forming than originally was there however i can not seem to get my calculations to reflect this. I was just curious if it was me who has fumbled somewhere I have tried calculating this in about every possible way i can imagine and can not seem to get zero degrees. I am just curious if I am missing something.
• Compare the latent heat necessary to melt all of the ice to the heat released when the water cools down from 13 $^\circ$C to 0 $^\circ$C. Is there enough heat to melt all of the ice? If not, some ice will not melt, and you will have a mixture of water and ice in your cup. Such mixture is characterized by equilibrium between two phases of water, and must be at the melting point of water, i.e. 0 $^\circ$C. However, your idea that the amount of ice will increase is not correct: some ice has to melt to cool the (separate) amount of water from 13$^\circ$ to zero. — Nicholas Kuzma 2014/03/07 16:11

### Quiz questions Ch.17

#### Quiz 5

Find the contents of the insulated container into which 1 ton of steam at 100 $^\circ$C and 1 ton of ice at 0 $^\circ$C were inserted.

Steps:

• First, figure out how much heat it takes
1. to melt the ice:
• $Q_m=L_fm=33.5\times 10^4\!\frac{\text J}{\text{kg}}\cdot 10^3\,{\text{kg}}=335\times 10^6\,$J
2. to condense the steam:
• $Q_c=L_vm=22.6\times 10^5\!\frac{\text J}{\text{kg}}\cdot 10^3\,{\text{kg}}=2.26\times 10^9\,$J
3. to change the temperature of 1 ton of water by 100 $^\circ$C:
• $Q_{100}=c_\text{wat}m\Delta T=4186\!\frac{\text J}{{\text{kg}}\cdot{\text K}}\cdot 10^3\,{\text{kg}}\cdot 100\,{\text K}$ $=419\times 10^6\,$J
• Will the heat consumed by the melting be sufficient to condense all the steam and cool down the resulting water to 0 $^\circ$C?
• Obviously, not! ($Q_m<Q_c+Q_{100}$)
• That means the resulting temperature must be above 0 $^\circ$C.
• Will the heat released by the condensation be sufficient to melt all the ice and warm up the resulting water to 100 $^\circ$C?
• Apparently, yes. ($Q_c>Q_m+Q_{100}$)
• That means, all the ice will melt, the resulting water will warm up to 100 $^\circ$C, and there still will be some steam left.
• The correct answer then is, steam+water at 100 $^\circ$C.
• If neither heat were sufficient to bring the other substance through the phase transition and all the way to the first substance's temperature, then the answer would be: water at some intermediate temperature.
• It would be closer to 100 $^\circ$C, because $L_vm>L_fm$, so it takes a fraction of the steam to melt all of the ice. The remaining fraction of the steam would then go to warm up the water produced by the ice to some higher temperature $T$. When the steam is done condensing, the water produced by the steam is still at 100 $^\circ$C. So the average (mass-weighted) temperature of the water would be above 50 $^\circ$C at that point, and will remain the same as the temperatures equilibrate.

### Examples Ch.17

#### Problem 14.1

1. How many molecules are there at 1 atmosphere in 1 L of air at $T_1=0\,^\circ$C and $T_2=25\,^\circ$C?
1. Write out what you have and see if you need to convert anything to SI units
• $V=1$ L (not SI)$=0.001\,{\text m}^3$
• $T_1=0\,^\circ$C (not SI)$\,=0\,^\circ$C$\,+\,273=273\,$K
• $T_2=25\,^\circ$C (not SI)$\,=25\,^\circ$C$\,+\,273=298\,$K
• $N_A = 6.022\times 10^{23}\,$mol (we need this because we are looking for the number of molecules!)
• $k_B = 1.38\times 10^{-23}\,\frac{\text J}{\text K}$ (Boltzmann’s constant)
• $P = 1.01\times 10^5\,$Pa (atmospheric pressure)
2. Determine the equation you need to use
• $PV =Nk_BT$
3. Solve for the number of molecules ($N$)
• $N = \frac{PV}{k_BT}$
4. Plug in the values!
• Solve separately for each temperature (press Control and + or Command and + to zoom in):
• $N_1 = \frac{1.01\times 10^5\,{\text{Pa}}\,\cdot\,0.001\,{\text m}^3}{1.38\times 10^{-23}\frac{\text J}{\text K}\,\cdot\,273\,{\text K}} = 2.687\times 10^{22}$ molecules
• $N_2 = \frac{1.01\times 10^5\,{\text{Pa}}\,\cdot\,0.001\,{\text m}^3}{1.38\times 10^{-23}\frac{\text J}{\text K}\,\cdot\,298\,{\text K}} = 2.461\times 10^{22}$ molecules
2. Express these numbers in moles, by dividing them by the Avogadro number (which is the number of molecules in 1 mol):
• $n_1 = 2.687\times 10^{22}\,{\text{molecules}}\times\frac{1\,{\text{mol}}}{6.022\times 10^{23}\,{\text{molecules}}}$ $= 0.0446\,{\text{mol}}$
• $n_2 = 2.461\times 10^{22}\,{\text{molecules}}\times\frac{1\,{\text{mol}}}{6.022\times 10^{23}\,{\text{molecules}}}$ $= 0.0409\,{\text{mol}}$
3. How many liters would each mole take up at these temperatures?
• The number of moles in a liter and the number of liters taken by a mole are (mathematically) inverse of each other:
• At $T_1=0\,^\circ$C, 1 mol takes up $V_1=1\,{\text{mol}}\cdot\frac{1}{0.0446\frac{\text{mol}}{\text L}}$ = 22.4 L.
• At $T_2=25\,^\circ$C, 1 mol takes up $V_2=1\,{\text{mol}}\cdot\frac{1}{0.0409\frac{\text{mol}}{\text L}}$ = 24.5 L.

#### Problem 14.2

3 moles of xenon are heated to 400 K at atmospheric pressure. Find volume.

1. Write out what you have and see if you need to convert anything to SI units:
• $n = 3\,$mol Xe (number of moles, NOT molecules) $T = 400$ K
• $P = 1.01\times 10^5$ Pa (atmospheric pressure)
• $R = 8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}$ (ideal gas constant)
2. Determine the equation you need to use
• $PV = nRT$
3. Solve for volume
• $V = \frac{nRT}{P}$
4. Plug in the values!
• $V = \frac{3\,{\text{mol}}\,\cdot\,8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}\,\cdot\,400\,{\text K}}{1.01\times 10^5\,{\text{Pa}}}\approx 0.098\,{\text m}^3=98\,$L

#### Problem 14.3

1 m3 of steam at 100$\,^\circ$C is compressed to 0.7 m3 at constant temperature. Find pressure and the amount of water (liquid) in the chamber.

1. If not for the condensation, compressing at constant temperature would increase the pressure of the ideal gas ($PV={\text{const}}$) and thus drive the gas into the “forbidden” liquid zone on the phase diagram.
• This tells us that condensation will occur, and throughout the process steam will be converting to liquid water
• Therefore the pressure will remain constant, dictated by the water-steam equilibrium $P_\text{eq}(T)$ curve at constant temperature
• $P=1\,$atm
2. Find the number of moles of steam initially:

#### Problem 14.4

Find the amount of heat released by condensing 1 oz of liquid water from steam.

Steps:

1. Convert all units to SI

#### Problem 14.5

100% efficient 700-watt microwave oven is trying to warm up 1 pound of ice, initially at $-4\,^\circ$C.

1. How long does it take before it starts melting?
2. How long does it take to finish melting?
3. How long does it take to bring the resulting water to a boil?
4. How long does it take to boil off all the water?

Steps:

1. Convert all units to SI

#### Problem 14.6

Find the spring constant of a steel cable 1 cm in diameter, and 100 m in length. How much will this cable stretch when lifting a load of 1 metric ton? Assume the Young's modulus of steel to be $\Upsilon_\text{steel}=20\times 10^{10}\,\frac{\text N}{{\text{m}}^2}$.

Steps:

1. Convert all units to SI and find cross-sectional area:
• $d=0.01\,$m
• $A=\pi\left(\frac{d}{2}\right)\approx 7.85\times 10^{-5}\,$m2 [check: 1 cm2=0.0001 m2, the area of a circle is about 3/4 of the corresponding square]
2. Equate the expressions for force in terms of the spring constant $k$ and the Young's modulus to find $k$:
• $|F|=k\,\Delta L=\Upsilon_\text{steel}\left(\frac{\Delta L}{L_0}\right)A$
• $k=\frac{\Upsilon}{L_0}\cdot A=\frac{20\times 10^{10}\,\frac{\text N}{{\text{m}}^2}}{100\,{\text m}}\cdot 7.85\times 10^{-5}\,{\text m}^2$ $=157\,\frac{\text{kN}}{\text m}=1.57\times 10^5\,\frac{\text{N}}{\text m}$
3. Find elongation:
• $\Delta L=\frac{|F|}{k}=\frac{1000\,{\text{kg}}\,\cdot\,9.8\!\frac{\text m}{{\text s}^2}}{157\times 10^3\!\frac{\text N}{\text m}}$ $=0.062\,{\text m}=6.2\,{\text{cm}}\approx 2.5\,$inches.

## Ch.18: The Laws of Thermodynamics

### Examples Ch.18

#### Problem 15.1

20 L of xenon at 25$\,^\circ$C and 1 atm of pressure are compressed to 19 L

1. isothermally
3. isobarically

Find work done by the compressor and the amount of heat released in each case

Steps:

1. Convert units:
• $T_1=25\,^\circ$C = 298.15 K
• $P_1=1\,$atm = 101325 Pa
• $V_1=20\,$L = 0.020 m3
• $V_2=19\,$L = 0.019 m3
2. Find the number of moles:
• $n=\nu=\frac{PV}{RT}=0.82$ mol
3. Calculate work on the gas for each process (hint: work on the gas is the negative of the work by the gas, $W_\text{on gas}=-W_\text{by gas}$):
1. isothermal, $T={\text{const}}$
• $PV=nRT={\text{const}}$
• $U={\text{const}}$
• $\Delta U=0$
• $W_\text{on gas}\approx-P\!\cdot\!\Delta V\approx -P_\text{av}\Delta V=103.99\,$J. This is the approximate method, works well when change in volume is small: $\Delta V\ll V$
• $P_1=101325\,$Pa
• $P_2=\frac{P_1V_1}{V_2}=106657\,$Pa
• $P_\text{av}=\frac{P_1+P_2}{2}=103991\,$Pa
• $\Delta V=V_2-V_1=-0.001\,{\text m}^3$
• $W_\text{on gas}=-nRT_1\!\ln\left(\!\frac{V_2}{V_1}\!\right)=-nRT_1\ln\frac{19}{20}$ $=P_1\!V_1\!\ln\frac{20}{19}=103.95\,$J. This is the exact method, works well for any change in volume
• Because the change in the internal energy is zero ($\Delta U=0$), the heat released is equal to the work done on gas: $Q_\text{out}=W_\text{on gas}\!-\Delta U\,=\,W_\text{on gas}=103.95\,$J
2. adiabatic, $Q=0$
• $PV^\gamma=\,$const, where $\gamma=\frac{c_p}{c_v}=\frac{5/2\;R}{3/2\;R}=\frac{5}{3}\approx 1.67$ for xenon, a monoatomic gas
• $P_1V_1^{1.67}=P_2V_2^{1.67}\;$ $\Rightarrow\;\;P_2=P_1\frac{V_1^{1.67}}{V_2^{1.67}}=101325\,{\text{Pa}}\cdot\left(\frac{20\,{\text L}}{19\,{\text L}}\right)^{1.67}$ $=110370\,$Pa
• $T_2=\frac{P_2V_2}{nR}=\frac{110370\,{\text{Pa}}\,\cdot\,0.019\,{\text m}^3}{0.82\,{\text{mol}}\,\cdot\,8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}}=308.5\,$K
• $W_\text{on gas}\approx -P\Delta V\approx P_\text{av}\!\cdot(V_2-V_1)$ $=-\frac{1}{2}(P_1+P_2)\cdot(V_2-V_1)$ $=0.5\cdot 211695\,{\text{Pa}}\cdot 1\times 10^{-3}\,{\text m}^3$ $=105.85\,$J
• The above is an approximate method, assuming the change in volume is much smaller than the volume
• The exact method is to equate the work to the change in internal energy:
• $W_\text{on gas}=\Delta U-Q_\text{out}=\Delta U=c_vn\Delta T$ $=\frac{3}{2}R\,n\,(T_2-T_1)$ $=1.5\cdot 8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}\cdot 0.82\,{\text{mol}}\cdot(308.5-298.2)\,{\text K}$ $=105.8\,$J
• $Q_\text{out}=0$
3. isobaric, $P={\text{const}}$
• Because the pressure stays constant in an isobaric process, $P_1=P_2=P_\text{av}$ and the $P\Delta V$ formula for work is exact:
• $W_\text{on gas}=-P\Delta V$ $=101325\,{\text{Pa}}\cdot 1\times 10^{-3}\,{\text m}^3=101.3\,$J
• $Q_\text{out}=-c_p\,n\,(T_2-T_1)=-\frac{5}{2}R\,n\,\left(\frac{PV_2}{nR}-\frac{PV_1}{nR}\right)$ $=-\frac{5}{2}P(V_2-V_1)=2.5\cdot 101325\,{\text{Pa}}\cdot 1\times 10^{-3}\,{\text m}^3$ $=253.3\,$J

#### Problem 15.2

The cycle 1-2-3-4 performed on a monoatomic gas consists of:

• $1\rightarrow 2$: isochoric heating at $V_1=V_2=2\,{\text m}^3$
• $2\rightarrow 3$: isobaric expansion at $P_2=P_3=5\,$atm
• $3\rightarrow 4$: isochoric cooling at $V_3=V_4=6\,{\text m}^3$
• $4\rightarrow 1$: isobaric compression at $P_4=P_1=2\,$atm

Find the net work by the gas in the cycle and the efficiency of the heat engine based on it, if the lowest temperature in the cycle is 298 K.

Steps:

1. Convert the units to SI:
• $P_2=P_3=5\,{\text{atm}}=506.6\times 10^3\,$Pa
• $P_1=P_4=2\,{\text{atm}}=202.7\times 10^3\,$Pa
2. Since the cycle has a rectangular shape on a $P\!-\!V$ diagram, the net work can be easily calculated as the area inside the rectangle:
• $W_\text{net by gas}=(P_2-P_1)(V_3-V_2)$ $=(506.6-202.7)\times 10^3\,{\text{Pa}}\cdot(6-2)\,{\text m}^3$ $=1.22\times 10^6\,{\text J}=1.22\,$MJ.
3. To find efficiency$\,=\frac{W_\text{net by gas}}{Q_h}$, we need to find the heat supplied to the gas during heating segments
• The temperature $T\sim PV$, therefore it (as well as the internal energy) is increasing during $1\rightarrow 2$ and $2\rightarrow 3$ legs of the cycle.
• Since both $c_v$ and $c_p$ are positive, in these processes the heat going into the gas has the same sign as the temperature change: positive when the temperature increases and negative (i.e. out of the gas) when the temperature decreases.
• $Q_h=Q_\text{in}=c_vn\,(T_2-T_1)+c_pn\,(T_3-T_2)$ $=\frac{3}{2}R\,n\,(T_2-T_1)+\frac{5}{2}R\,n\,(T_3-T_2)$ $=R\,n\,\left(\frac{3}{2}(T_2-T_1)+\frac{5}{2}(T_3-T_2)\right)$
• The lowest temperature in the cycle is at the point with the lowest $PV$ product, i.e. point 1.
• $T_1=298\,$K
• $T_2=T_1\frac{P_2}{P_1}=298\,{\text K}\cdot\frac{5\,{\text{atm}}}{2\,{\text{atm}}}=745\,$K (in an isochoric process, because V = const, $T\sim P$)
• $T_3=T_2\frac{V_3}{V_2}=745\,{\text K}\cdot\frac{6\,{\text m}^3}{2\,{\text m}^3}=2235\,$K (in an isobaric process, because P = const, $T\sim V$)
• $n=\frac{P_1V_1}{RT_1}=\frac{202.7\times 10^3\,{\text{Pa}}\,\cdot\,2\,{\text m}^3}{8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}\,\cdot\,298\,{\text K}}=164\,$mol
• Finally, can plug these temperature and mole values into the formula for heat and find $Q_h$:
• $Q_h=8.31\frac{\text J}{ {\text{mol}}\cdot{\text K}}\cdot 164\,{\text{mol}}\big(\frac{3}{2}\cdot(745\!-\!298)\,{\text K}$ $+\frac{5}{2}\cdot(2235\!-\!745)\,{\text K}\big)$ $=6\times 10^6\,{\text J}=6\,$MJ.
• Now we can find the efficiency (Note, this is not a Carnot cycle, and we cannot use the $1-\frac{T_c}{T_h}$ formula!):
• ${\text{efficiency}}=\frac{W_\text{net by gas}}{Q_h}=\frac{1.22\,{\text{MJ}}}{6\,{\text{MJ}}}=0.204=20.4$%.

#### Problem 15.3

Gasoline in a certain engine is combusted at 1000 K, and the exhaust comes out at 400 K. Assuming the laws governing thermodynamic closed cycles apply here, what is the maximum efficiency that this engine can have?

Steps:

1. The maximum efficiency is achieved in the Carnot cycle with $T_h=1000\,$K and $T_c=400\,$K.
• ${\text{eff}}_\text{max}=1-\frac{T_c}{T_h}=1-0.4=0.6=60$%.

#### Problem 15.4

An ideal heat pump operating on a Carnot cycle in reverse consumes 1 kW from the electrical grid. How much heat per unit time does it supply to keep the inside of the house warm (at 25$\,^\circ$C) during a frigid winter day at 0$\,^\circ$C outside?

Steps:

1. Convert all the units to SI:
• $T_h=298\,$K
• $T_c=273\,$K
2. For a Carnot cycle, no matter straight or reverse, $\frac{|Q_c|}{|Q_h|}=\frac{T_c}{T_h}$. Also, $P_\text{electr}=\frac{W_\text{net on gas}}{t}=\frac{|Q_h|}{t}-\frac{|Q_c|}{t}$
3. Solving for the heat per unit time $|Q_h|/t$,
• $P_\text{heat}=\frac{|Q_h|}{t}=\frac{P_\text{electr}}{\frac{T_h}{T_c}-1}=11\,$kW

## Exam 3 review

Can be found here or on the sidebar

### Final Exam Equation Sheets

One can be found here : ph202_final_equation_sheet.pdf

Another compilation can be found here : notes-15-18.pdf

1. Submerged volume of a solid floating in a fluid: $V_\text{subm}=V_\text{solid}\frac{\rho_\text{solid}}{\rho_\text{fluid}}$
• this is only applicable to chunks of solids (e.g. ice, wood) without cavities, and is not applicable to vessels such as ships (because there is some air below the water line, contributing to $V_\text{subm}$ but not part of the vessel material).
• For floating vessels, use the more general formula $\rho_\text{fluid}\!\cdot\!V_\text{subm}=m_\text{vessel}$, true for anything afloat larger than a few cm (when the surface tension can be neglected).
2. Coefficients of linear and volume expansions are related by an approximate formula (which works well when they are much less than 1):
• $\beta \approx 3\alpha$
3. Conduction of heat:
• $\Delta T$ is not the temperature change in time, rather it is the temperature difference between the two surfaces of the material
4. Internal energy of an ideal gas and the $U=\frac{3}{2}nRT$ formula:
• $\frac{3}{2}nRT$ is the internal energy of a monoatomic ideal gas, which is equal to the internal kinetic energy of the molecules of any ideal gas.
• Non-monoatomic (polyatomic) ideal gases such as O2, N2, and CO2, have an additional internal energy due to rotations, and at some higher temperatures, vibrations of these molecules.
5. Molar specific heats:
• $c_v=\frac{3}{2}\!\cdot\!R\;\;$ and $\;\;c_p=\frac{5}{2}\!\cdot\!R\;\;$ are only true for monoatomic ideal gases, such as He, Ne, Ar, Kr, Xe, Ra, or their mixtures. For all other gases, these constants are higher because of contributions from rotational and vibrational degrees of freedom
6. Efficiency of the heat engine:
• $\frac{Q_h-Q_c}{Q_h}=1-\frac{T_c}{T_h}$ is only true for Carnot engine, where the cycle consists of alternating isotherms and adiabats

Nicholas Kuzma 2014/03/16 14:26