1.1: Thermal Equilibrium This chapter introduces the several basic definitions of temperature, starting with the operational (“temperature is what you measure with a thermometer”) and ending with a qualitative theoretical definition (“Temperature is a measure of the tendency of an object to spontaneously give up energy to its surroundings. When two objects are in thermal contact, the one that tends to spontaneously lose energy is at the higher temperature.”) Some time is spent on what exactly thermal contact entails–that the objects are in some way able to exchange energy, and that this energy is called heat. When the two objects have been in contact long enough, we learn that they are considered to be in thermal equilibrium, and that the time required to come to thermal equilibrium is the relaxation time. Curiously, Schroeder doesn’t say in the definition that the two objects are in thermal equilibrium when they’ve reached the same temperature. This is probably because of the way he defined relaxation time, which gets a footnote that some authors consider the relaxation time as the time required for the temperature difference between the two objects to decrease by a factor of e. The most interesting part of the section is an analogy of thermal equilibrium to other kinds of equilibrium. Schroeder gives the example that if you pour cold cream into a hot cup of coffee, the relaxation time of the cream/coffee system is very short, but the relaxation time for the coffee to come to thermal equilibrium with the room is still very long. But he notes that there are more kinds of equilibrium on display in the example; the cream in the coffee reaches diffusive equilibrium when it is blended with the coffee such that its molecules have no greater tendency to move in one direction than another. Also defined is mechanical equilibrium, when large-scale motions can take place, but no longer do. Whereas the exchanged quantity to reach thermal equilibrium is energy, the exchanged quantity for diffusive equilibrium is particles, and for mechanical equilibrium, volume. The rest of the section describes how thermometers work, and introduces the concepts of absolute zero and the kelvin scale. These things are sufficiently basic that I’m not going to synopsize them.
I don’t think I’ll always work every problem in a chapter, but to start out I’m going to.
Problem 1.1: The Fahrenheit temperature scale is defined so that ice melts at 32°F and water boils at 212°F. (a) Derive the formulas for converting from Fahrenheit to Celsius and back. (b) What is absolute zero on the Fahrenheit scale?
These are two linear scales, meaning that every degree F is the same size as every other degree F, and equivalently for every degree C. But F degrees and C degrees are different sizes, and the scales don’t start in the same place. So first I’ll figure out how do describe the size of a degree F in terms of a degree C, or the number of degrees F per degrees C. The magnitude of the temperature change between where water freezes and where it boils is the same regardless of scale, so let F be the size of a degree Fahrenheit and C be the size of a degree Celsius. Then,
( 212F - 32F ) = (100C - 0C).
Solving this equation gives us
F = \frac { 100 }{ 180 } C = \frac { 5 }{ 9 } C.
So one degree F is only five ninths the size of a degree C. So if we have a number of degrees Celsius, we have to divide by five ninths (same as multiplying by nine fifths) to get the equivalent number of degrees Fahrenheit. But there’s still the issue that the scales don’t start in the same place. When we’re at 0° C, we’re still at 32° F. So we need to add a constant factor of 32 to the equation above to adjust for the different zero-point, giving us a temperature conversion equation
{ T }_{ F }=\frac { 9 }{ 5 } { T }_{ C }+32 and { T }_{ C }=\frac { 5 }{ 9 } ({ T }_{ F }-32). (Solution to part a.)
For part b, we know from the chapter that absolute zero is -273°C, so
{ T }_{ F }=\frac { 9 }{ 5 } (-273)+32=-459. (Solution to part b.)
Problem 1.2: The Rankine temperature scale (abbreviated °R) uses the same size degrees as Fahrenheit, but measured up from absolute zero like kelvin (so Rankine is to Fahrenheit as kelvin is to Celsius). Find the conversion formula between Rankine and Fahrenheit, and also between Rankine and kelvin. What is room temperature on the Rankine scale?
Degrees Fahrenheit and Degrees Rankine are the same size, the scales just start in a different place. We know from the previous problem what absolute zero in °F is, so
{ T }_{ R }={ T }_{ F }+459.
Rankine and kelvin start in the same place, but have different degree sizes. The sizes are the same as for Fahrenheit and Celsius respectively, so we already know that one degree Rankine is five ninths the size of one kelvin. (I’ll note here that, for no reason I’m aware of, it’s considered improper to say “degree kelvin.”) So again we divide our kelvins by five ninths to get the number of degrees Rankine, giving us a temperature conversion equation
{ T }_{ R }=\frac { 9 }{ 5 } { T }_{ K }.
Room temperature is defined to be 300 K, so by the equation above, room temperature is also 540°R.
Problem 1.3: Determine the kelvin temperature for each of the following: (a) human body temperature; (b) the boiling point of water (at standard pressure of 1 atm); (c) the coldest day you can remember; (d) the boiling point of liquid nitrogen (-196°C); (e) the melting point of lead (327°C).
(a) 98.6°F = 37°C = 310 K. (b) 100°C = 373 K. (c) -10°F ≈ -23.3°C = 249.6 K. (d) -196°C = 77 K. (e) 327°C = 600 K.
Problem 1.4: Does it ever make sense to say that one object is “twice as hot” as another? Does it matter whether one is referring to Celsius or kelvin temperatures? Explain.
The book hasn’t explicitly defined “heat” yet, so I’m going to interpret this problem to be asking just about temperature scales. Without a defined scale, saying “twice as hot” is meaningless. Twice as many degrees Celsius is a much greater difference in temperature than twice as many degrees Fahrenheit. Also, suppose our first object is at 0°C. Within the Celsius scale, you can’t define what temperature an object “twice as hot” would be. On the kelvin scale, excepting the special case of absolute zero, the phrase “twice as hot” does always have a mathematically coherent definition. (I’m not sure that it has a physically coherent definition, but the book hasn’t gotten to specific enough definitions for me to make that argument.)
Problem 1.5: When you’re sick with a fever and you take your temperature with a thermometer, approximately what is the relaxation time?
I’m not sure what the intention with this problem is. Actually calculating this would both require information that hasn’t been introduced yet, and defining a specific type of thermometer. But I don’t see anything in the chapter to provide a basis for approximation. From anecdotal experience, the relaxation time (i.e. how long it takes for an oral thermometer to go from room temperature to the temperature of the inside of my mouth) is on the order of one minute.
Problem 1.6: Give an example to illustrate why you cannot accurately judge the temperature of an object by how hot or cold it feels to the touch.
The key here is judging accurately. Imagine that you touch an object that is at exactly your body temperature. Your body and the object are in thermal equilibrium, so there will be no transfer of energy, and you will feel no temperature difference. So you know that it’s the same temperature as you are, but what temperature is that? You can’t find out by touching yourself, as you are always in thermal equilibrium with yourself. You can estimate your own temperature, but that wouldn’t be accurate. Since body temperature is variable, it’s impossible to know where the “feels hot, feels cold” scale starts without recourse to a thermometer. (There’s also the issue that beyond certain temperature ranges the tissues of your body with do the sensing will be damaged, making you unable to feel temperature at all.)
Problem 1.7: When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume change increases by one part in 5500. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β:
\beta =\frac { { \Delta V }/{ V } }{ \Delta T }
(where V is volume, T is temperature, and the Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β = 1/5500 K^-1 = 1.81 x 10 ^-4 K^-1. (The exact value varies with temperature, but between 0°C and 200°C the variation is less than 1%.)
(a) Get a mercury thermometer,
No.
estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.
Screw getting an thermometer. Let’s just assert that the bulb at the bottom has volume V of one cubic centimeter. Let’s also assert, because I’m making things up and I can, that the degree markings are a height h ofone millimeter apart. The amount of mercury that rises up the tube radius r with a 1 K increase in temperature, then, is
\beta (V)(\Delta T)=\pi { r }^{ 2 }h
Solve for the radius of the cylinder,
r=\sqrt { \frac { \beta (V)(\Delta T)}{\pi h}} =0.02. So the radius is 0.002 cm. (Frustratingly, I haven’t been able to make the unit symbols parse in the LaTeX plugin, but the dimensional analysis checks out.
(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10^-4 K^-1 at 100°C, but decreases as the temperature is lowered until it becomes zero at 4°C. Below 4°C it is slightly negative, reaching a value of -0.68×10^-4 K^-1 at 0°C. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.
Okay, this one is interesting. Imagine a lake getting colder and colder. The positive thermal expansion coefficient means that, above 4°C, the colder the water is the denser it is, so the coldest water will sink to the bottom. (This matches my intuition that the bottom of a swimming pool is colder than the surface.) Let this behavior continue until, hypothetically, we have an entire lake in which all the water is at 4°C. Then let some of the water in that lake get colder. Because the thermal expansion coefficient turns negative now, the colder water will be less dense, and will start rising to the top. That means that by the time the first water gets down to 0°C and freezes, it will be at the surface, and the water below will be warmer. So lakes freeze from the surface down. If the thermal expansion was always positive, I would expect lakes to freeze from the bottom up.
Problem 1.8: For a solid, we also define the linear thermal expansion coefficient, α, as the fractional increase in length per degree:
\alpha =\frac { { \Delta L }/{ L } }{ \Delta T } .
(a) For steel, α is 1.1 x 10^-5 K^-1. Estimate the total variation in the length of a 1-km steel bridge between a cold winter night and a hot summer day.
Let’s say that a hot summer day is 43°C, and a cold winter night is 0°C. For an L of 1km, we have
\alpha (L)(\Delta T)=\alpha (1)(43)=0.00047
in units of kilometers, which is just under half a meter.
(b) The dial thermometer in Figure 1.2 uses a coiled metal strip made of two different metals laminated together. Explain how this works.
The two different metals have different thermal expansion coefficients. Assume that they are the same length when laminated together at some thermal equilibrium. When the temperature changes, the strips will not be able to expand linearly because they are attached to each other and changing by different lengths. As such, there will be a lateral displacement; the laminated strip will curl. This curling is proportional to the change in temperature.
(c) Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions: \beta ={\alpha}_{ x }+{\alpha}_{y}+{\alpha}_{z}. (So for an isotropic solid, which expands the same in all directions, β=3α.)
Recall that \Delta V={ V }_{ final }-{ V }_{ initial }. If {V}_{initial}=xyz, then it follows that
{ V }_{ final }=(x+\Delta x)(y+\Delta y)(z+\Delta z)\\ =xyz+\Delta xyz+x\Delta yz+xy\Delta z+x\Delta y\Delta z+\Delta xy\Delta z+\Delta x\Delta yz+\Delta x\Delta y\Delta z\\ \approx xyz+\Delta xyz+x\Delta yz+xy\Delta z
That last step works because all of the terms with multiple deltas are so tiny compared to the other terms that they are physically negligible, so I can drop them. This gives me
{\Delta V=V }_{ final }-{ V }_{ initial }=xyz+\Delta xyz+x\Delta yz+xy\Delta z-xyz\\ =\Delta xyz+x\Delta yz+xy\Delta z
Which I can plug into the equation for β to get
\beta =\frac { { \Delta V }/{ V } }{ \Delta T } =\frac { { (\Delta xyz+x\Delta yz+xy\Delta z) }/{ (xyz) } }{ \Delta T } =\frac { ({ \Delta x }/{ x })+({ \Delta y }/{ y })+({ \Delta z }/{ z }) }{ \Delta T }\\ ={ \alpha}_{ x }+{ \alpha}_{ y }+{ \alpha}_{ z }
thus proving the theorem.
To find an equation for the shape of that curve, we look at the equipartition of energy theorem, U = \frac {f}{2}NkT, where f is the number of degrees of freedom per molecule. The infinitesimal change in energy along the curve is then given by dU=\frac {f}{2}NkdT. If we assume the compression is quasistatic, then from the equation for work we know that dU=-PdV. (We can say this because we previously established that in adiabatic compression the entire change in energy comes from work.) This gives \frac {f}{2}NkdT=-PdV. Now you can plug in the ideal gas law for P and do some canceling to get \frac {f}{2}\frac {dT}{T}=-\frac {dV}{V} \rightarrow \frac {f}{2}\ln {\frac {{T}_{f}}{{T}_{i}}}=-\ln{\frac{{V}_{f}}{{V}_{i}}}. This ends up simplifying down to V{T}^{{f}/{2}}=C for some constant C. If you are looking for pressure instead of temperature, you can use the ideal gas law to rewrite this {V}^{\gamma}P=D for some constant D, where \gamma = {(f+2)}/{f} and is called the adiabatic constant.
Bubble A undergoes adiabatic compression (in this case, expansion), while bubble B undergoes isothermal compression. Initially, the bubbles are identical, so their pressures and volumes are equal. For isothermal compression, {P}_{B}{V}_{B}=C where C is a constant, and for adiabatic compression {P}_{A}{V}_{A}^{\gamma}=C. The constants C must be the same, because the bubbles are initially identical. Since \gamma > 1, {V}_{A}<{V}_{B}. Thus, the bubble that undergoes isothermal compression, bubble B, is larger when it reaches the surface. This can be seen by visual inspection of a graph depicting an isotherm and an adiabat. Notice that for the same initial conditions, the volume of the isotherm rises faster than that of the adiabat.
