Contemporary Challenges: Fall-2023
Homework 3 (SOLUTION): Due 11 Friday 10/20
- Three ideas for the term project
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Read the description of the term project on the class website at “Introduction to term project”. Identify three (3) subjects that you find interesting/intriguing (for example, solar energy, exoplanets, ...). Within each subject, pose a question that might have an interesting quantitative answer: “Since it requires energy to make a solar panel, how long does it take to recoup that energy?”, “How far away could we see an Earth-like planet orbiting a Sun-like star?” ... You should turn in 3 different subjects and 3 different quantitative questions (quantitative means “quantities that can be calculated and/or measured”)
Let your mind wander as broadly as possible. Subjects and questions are not restricted to the topics taught in PH315. During this exploratory stage, be bold and daring; you are not committing yourself to solve all 3 questions. To spark your imagination, there is a list of ideas on the class website. The instructor will read your ideas and give you feedback. Whenever possible, the feedback will point you towards a coarse-grained model that is helpful for answering your question. Use the feedback to help decide which question you will develop further (or whether you need to go back to the drawing board).
- Multiplicity of an ideal gas
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- (T3M.7) The multiplicity of an ideal monatomic gas with \(N\) atoms, internal energy \(U\), and volume \(V\) turns out to be roughly
\begin{align}
\Omega (U, V, N) = CV^NU^{\left(\frac{3 N}{2}\right)}
\end{align}
where \(C\) is a constant that depends on \(N\) alone. Use this expression, together with the fundamental definition of temperature, and they fundamental definition of entropy, to find \(U\) as a function of \(N\) and \(T\) for an ideal gas.
Based on \(\Omega\) for an ideal gas, find the relationship between \(U\) and \(T\). From class we know \(S = k_\text{B} \ln \Omega\) and \(1/T = dS/dU\). Therefore, \begin{align} \frac{1}{T} &= k_\text{B} \frac{d}{d U} \ln \left(C V^N U^{3N/2}\right)\\ &= k_\text{B} \frac{d}{d U} \left[\ln C + N \ln V + \frac{3 N}{2} \ln U\right]\\ &= k_\text{B} \frac{3 N}{2}\frac{1}{U} \end{align} So, rearranging the equation, \(U = \frac{3}{2} N k_\text{B} T\).
- (From the GRE PhysicsSubject GR0177, given in 2001)
Note 1: The irreversibility of this process tells you that entropy must go (up or down?).
Note 2: The gas constant, \(R\), is equal to Avagadro's number times \(k_B\).
Since the process is irreversible, the total change in entropy (for the whole system) must be positive. This eliminates option (D) and (E) in the multiple choice question.
I cannot describe this process as a simple “flow of heat between objects”, so I cannot use the relationship \(\Delta S = Q/T\).
The volume doubles while \(N\) and \(U\) stay constant (there is no change in the internal energy of the gas). I'll put these quantitative statements into the expression for total entropy of a monatomic ideal gas. \begin{align} S_\text{initial} = k_\text{B} \ln\left(C V^N U^{3N/s}\right) \quad \quad \quad S_\text{final} = k_\text{B} \ln\left(C (2V)^N U^{3N/s}\right)\\ \frac{\Delta S}{k_\text{B}} = \ln C + N \ln (2V) + \frac{3N}{2} \ln U - \ln C - N \ln V - \frac{3 N}{2} \ln U \end{align} \begin{align} \Delta S &= k_\text{B} N \left(\ln(2V) - \ln V\right)\\ &= k_\text{B} N \ln 2 \end{align} This is option (B) in the GRE question.
- (T3M.7) The multiplicity of an ideal monatomic gas with \(N\) atoms, internal energy \(U\), and volume \(V\) turns out to be roughly
\begin{align}
\Omega (U, V, N) = CV^NU^{\left(\frac{3 N}{2}\right)}
\end{align}
where \(C\) is a constant that depends on \(N\) alone. Use this expression, together with the fundamental definition of temperature, and they fundamental definition of entropy, to find \(U\) as a function of \(N\) and \(T\) for an ideal gas.
- Heat Pump
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The diagram shows a machine (the white circle) that moves energy from a cold reservoir to a hot reservoir. We will consider whether a machine like this is useful for heating a family home in the winter when the temperature inside the family home is \(T_\text{H}\), and the temperature outside the house is \(T_\text{C}\). To quantify the performance of this machine, I'm interested inthe ratio \(Q_\text{H}/W\), where \(Q_{\text{H}}\) is the heat energy entering the house, and \(W\) is the net energy input in the form of work. (\(W\) is the energy I need to buy from the electricity company to run an electric motor). Starting from the 1\(^{\text{st}}\) and 2\(^\text{nd}\) laws of thermodynamics, find the maximum possible value of \(Q_\text{H}/W\). This maximum value of \(Q_\text{H}/W\) will depend solely on the ratio of temperatures \(T_\text{H}\) and \(T_\text{C}\).
We are tring to maximize the heat flowing into the high temperature reservoir for a given amount of work input. By the first law of thermodynamics we know that
\[Q_{\text{C}}+W = Q_{\text{H}}\]
If I fix W, then increasing \(Q_{\text{C}}\) will increase \(Q_\text{H}\).
By the second law of thermodynamics we know \begin{align} -\frac{Q_{\text{C}}}{T_{\text{C}}} + \frac{Q_{\text{H}}}{T_{\text{H}}} \geq 0 \end{align}
If we use the beggest \(Q_{\text{C}}\) then \begin{align} \frac{Q_{\text{C}}}{T_{\text{C}}} = \frac{Q_{\text{H}}}{T_{\text{H}}}\\ \rightarrow \frac{Q_{\text{C}}}{Q_{\text{H}}} = \frac{T_{\text{C}}}{T_{\text{H}}} \end{align}
We can combine the two equations to get the efficiency: \begin{align*} \epsilon = \frac{Q_\text{H}}{W} &= \frac{Q_\text{H}}{Q_\text{H}-Q_\text{C}}\\[6pt] \frac{1}{\epsilon} &= \frac{Q_\text{H}-Q_\text{C}}{Q_\text{H}} \\[6pt] &= 1- \frac{Q_\text{C}}{Q_\text{H}} \\[6pt] &= 1- \frac{T_\text{C}}{T_\text{H}} \\[6pt] &= \frac{T_{\text{H}}-T_{\text{C}}}{T_\text{H}} \\[6pt] \epsilon &= \frac{T_\text{H}}{T_{\text{H}}-T_{\text{C}}} \end{align*}
To understand this answer, I can consider a few special cases. If \(T_\text{H} = T_{\text{C}}\), then no work is needed. If \(T_{\text{C}}\rightarrow 0\), then \(Q_\text{H} = W\). All the heat must come from work.
I'm also noticing that the difference in temperature between the reservoirs is going to be smaller than the temperature of the hot reservoirs, so I'm expecting efficiencies that are bigger than 1.
Sensemaking: Choose realistic values of \(T_\text{H}\) and \(T_\text{C}\) to describe a family home on a snowy day. Based on your temperature estimates, what is the maximum possible value of \(Q_\text{H}/W\)?
For a snowy day, let \(T_{\text{C}} = 270\) K and \(T_\text{H} = 290\) K.
Substituting that into my equation, I get \(Q_\text{H}/W = 290/20 \approx 15\).
That sounds great! How well do real heat pumps perform? See wikipedia: “Heat Pump”, subsection “performance considerations.” It says that \(Q_\text{H}/W \approx 3.2-4.5\) for a unit you can install at your house.
- Thermal energy in the earth's atmosphere
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Thermal energy is stored in all materials on Earth, including the air, water and rocks. The air is composed mostly of diatomic molecules such as N2 and O2.
Use Google to look up the mass of the earth's atmosphere. Now, exercise some skepticism and make sure that Goggle's answer is consistent with other facts about the earth: Air pressure at sea level is about 100 kPa and the radius of the earth is about 6400 km. The air pressure at sea level (force per unit area) is caused by the downward force of gravity acting on the atmosphere directly above a unit area. The thickness of the atmosphere is much much less than the radius of the earth. Give your argument supporting or refuting the internet's value for the mass of the earth's atmosphere.
Google says the mass of the Earth's atmosphere is \(5\times10^{18}\) kg. I want to check this. First, consider a column of air.
Every square meter of earth's surface has 10,000 kg of air directly above it. Radius of the Earth is \(\approx\) 6400 km. \begin{align} m_\text{atm} &= \left(4 \pi r_\text{earth}^2\right)(m_\text{sqr})\\ &= 4 \pi (6.4\times10^6 \text{ m})^2 (10^4 \text{ kg/m}^2) \\ &= 5\times 10^{18} \text{ kg} \end{align}
We know that between 1955 and 2010, the temperature of the top 2000 meters of the ocean rose by about 0.05 C. Given this fact, assess the validity of the following statement:
“If the same amount of heat that has gone into the top 2000 meters of the ocean between 1955-2010 had gone into the lower 10 km of the atmosphere, then the atmosphere would have warmed by about 20°C (36°F).”
Is this statement reasonable, or ridiculous? Show your calculations that support your conclusion. Your starting assumptions will include the specific heat capacity of air and water, and a reasonable guess regarding the fraction of the earth's surface that is covered with ocean.
I need to calculate the mass of water in top 2000 m of ocean. \begin{align} \text{Surface area of earth} &= 4 \pi R_\text{earth}^2\\ &=4 \pi (6.4\times10^{6})^2\\ &= 5.14\times10^{14} \text{ m}^2\\ \\ \text{Volume of water} &= \left(\frac{2}{3}\right)(5\times10^{14})(2\times10^3) \text{ m}^3\\ &\text{The 2/3 is because 2/3 of the planet is ocean}\\ &= 7\times10^{17}\text{ m}^3\\ \\ \text{Mass of water} &= (10^3 \text{ kg/m}^3)(7\times10^{17} \text{ m}^3) = 7\times10^{20} \text{ kg} \end{align} Now we can find the heat required to raise the ocean temperature by 0.05° C. \begin{align} Q &= m c_\text{p,water} \Delta T \\ &= [7\times10^{20} \text{ kg}][4.2 \text{ kJ}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}][5\times10^{-2} \text{ K}]\\ &= 10^{20}\text{ kJ} \end{align} Heat required to raise air temperature by 20°C. \begin{align} Q &= m c_\text{p,air} \Delta T \\ &= (5\times10^{18} \text{ kg})(1 \text{ kJ}\cdot\text{kg}^{-1}\cdot\text{K}^{-1})(20 \text{ K})\\ &= 10^{20}\text{ kJ} \end{align}
Note that I've used the heat capacity of air at constant pressure (\(c_\text{p,air} \approx 1 \text{ kJ}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}\)), which is greater than the heat capacity of air at constant volume (\(c_\text{v,air} \approx 0.7 \text{ kJ}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}\)). The reason for this difference is the following. If we heat a gas while keeping the pressure of the gas constant, then some heat energy must be turned into work - expanding the volume of the gas. Thus, less heat energy goes into raising the internal energy of the gas.
When answering this question, \(c_\text{p,air}\) is the correct choice. I won't dock points if you used \(c_\text{v,air}\). For example, if you estimated the heat capacity of air using the equipartition theorem, you should have found a reasonable value for \(c_\text{v,air}\).
The statement appears to be true, the same amount of heat is required to either raise the ocean temp by 0.05°C or to raise the air temp by 20°C.