Today we will be melting ice using our microwave ovens. The purpose of this is
to examine how energy affects matter. If we were in Weniger we would transfer our
energy using a power supply and resistor, and you would be able to use a couple
of multimeters to measure voltage and current and quantitatively determine how
much energy was being dissipated per second. Instead we will use a microwave
oven, and assume that its power is constant so we can treat time as a measure of
the energy transfered.
In Math Bits, you have learned that the way amount internal energy changes
relates to the work done:
\begin{align}
dU &= F_L dx_L + F_R dx_R
\end{align}
You made small changes in \(dx_L\) and \(dx_R\) and determined from that how much
the energy changed.
Today we are gong to examine energy transfer in a backwards manner.
When we transfer energy to something by heating, it's hard to measure the “thing
we changed,” which was entropy. It is, however, possible in some cases
to measure the amount
of energy transfered by heating, and from that we can work backwards to find out
how much the entropy changed.
The amount of energy transfered into a system by heating is generally written as
\(Q\).* An infinitesimal amount of energy
transfered by heating is called \({\mathit{\unicode{273}}} Q\). Recall that \({\mathit{\unicode{273}}} \) indicates an
inexact differential, which you can think of as a “small chunk” that is not
the change of something. \({\mathit{\unicode{273}}} Q\) is nota small change in the
amount of energy transfered by heating, but rather is a small amount of energy
transfered by heating.
Heat here is analogous to left work \(F_Ldx_L\), which is
also an amount
of energy transfered to a system. So you might wonder what the “thing changing”
is, which is analogous to \(dx_L\). A natural guess might be temperature, since you
know that has something to do with heat, but we can see that temperature can't be
the “thing that changes” when you heat something, because you can transfer
energy by heating without changing the temperature.
Latent heat
A phase transition is when a material changes state of matter, as in melting or
boiling. At most phase transitions (technically, abrupt phase transitions
as you will learnin the Capstone), the temperature remains constant while the
material is changing from one state to the other. So you know that as long as
you have ice and water coexisting in equilibrium at one atmosphere of pressure,
the temperature must be \(0^\circ\)C. Similarly, as long as water is boiling at
one atmosphere of pressure, the temperature must be \(100^\circ\)C. In both of
these cases, you can transfer energy to the system (as we will) by heating
without changing the temperature! This relates to why I keep awkwardly
saying
“transfer energy to a system by heating” rather than just “heating a system”
which means the same thing. We have deeply ingrained the idea that “heating”
is synonymous with “raising the temperature,” which does not align with the
physics meaning.
So now let me define the latent heat. The latent heat is the amount
of energy that must be transfered to a material by heating in order to change
it from one phase to another. The latent heat of fusion is the amount
of energy required to melt a solid, and the latent heat of vaporization
is the amount of energy required to turn a liquid into a gas. We will be
measuring both of these for water.
A question you may ask is whether the latent heat is extensive or intensive.
Technically the latent heat is extensive, since if you have more material
then more energy is required to melt/boil it. However, when you hear latent heat
quoted, it is almost always the specific latent heat,
which is the energy
transfer by heating required per unit of mass. It can be confusing that people
use the same words to refer to both quantities. Fortunately, dimensional checking
can always give you a way to verify which is being referred to. If \(L\) is an
energy per mass, then it must be the specific latent heat, while if it is an
energy, then it must be the latent heat.
Heat capacity and specific heat
The heat capacity is the amount of energy transfer required per
temperature to raise the temperature of a system. If we hold the pressure fixed
(as in our experiment) we can write this as:
\begin{align}
{\mathit{\unicode{273}}} Q &= C_p dT
\end{align}
where \(C_p\) is the heat capacity at fixed pressure.
You might think to rewrite this expression as a derivative, but we can't
do that since the energy transfered by heating is not a state function.
Note that the heat capacity, like the latent heat, is an extensive quantity.
The specific heat is the the heat capacity per unit mass, which is an
intensive quantity that we can consider a property of a material independently
of the quantity of that material.
I'll just mention as an aside that the term “heat capacity” is another one of
those unfortunate phrases that reflect the inaccurate idea that heat is a
property of a system.
Entropy
Finally, we can get to entropy. Entropy is the “thing that changes” when you
transfer energy by heating. I'll just give this away:
\begin{align}
{\mathit{\unicode{273}}} Q &= TdS
\end{align}
where this equation is only true if you make the change quasistatically
(see another lecture). This allows us to find the change in entropy if we know
how much energy was transfered by heating, and the temperature in the process.
\begin{align}
\Delta S &= \int \frac1T {\mathit{\unicode{273}}} Q
\end{align}
where again, we need to know the temperature as we add heat.
Students work out heat and work for rectangular paths on \(pV\) and \(TS\) plots. This gives with computing heat and work, applying the First Law, and recognizing that internal energy is a state function, which cannot change after a cyclic process.
Stefan-Boltzmannblackbody radiationThermal and Statistical Physics 2020
A black (nonreflective) sheet of metal at high
temperature \(T_h\) is parallel to a cold black sheet of metal at temperature
\(T_c\). Each sheet has an area \(A\) which is much greater than the distance between them. The sheets are in vacuum, so energy can only be transferred by radiation.
Solve for the net power transferred between the two sheets.
A third black metal sheet is
inserted between the other two and is allowed to come to a steady
state temperature \(T_m\). Find the temperature of the middle sheet,
and solve for the new net power transferred between the hot and cold sheets.
This is the principle of the heat shield, and is part of how the James Web telescope shield works.
Optional: Find the power through an \(N\)-layer sandwich.
phase transformationClausius-ClapeyronThermal and Statistical Physics 2020
Consider a phase
transformation between either solid or liquid and gas. Assume that the
volume of the gas is way bigger than that of the liquid or
solid, such that \(\Delta V \approx V_g\). Furthermore, assume that
the ideal gas law applies to the gas phase. Note: this problem
is solved in the textbook, in the section on the Clausius-Clapeyron
equation.
Solve for \(\frac{dp}{dT}\) in terms of the pressure of the vapor
and the latent heat \(L\) and the temperature.
Assume further that the latent heat is roughly independent of
temperature. Integrate to find the vapor pressure itself as a
function of temperature (and of course, the latent heat).
Plot your data I
Plot the temperature versus total energy
added to the system (which you can call \(Q\)). To do this, you will
need to integrate the power. Discuss this curve and any interesting
features you notice on it.
Plot your data II
Plot the heat capacity versus
temperature. This will be a bit trickier. You can find the heat
capacity from the previous plot by looking at the slope.
\begin{align}
C_p &= \left(\frac{\partial Q}{\partial T}\right)_p
\end{align}
This is what is called the heat capacity, which is the amount
of energy needed to change the temperature by a given amount. The
\(p\) subscript means that your measurement was made at constant
pressure. This heat capacity is actually the total heat capacity of
everything you put in the calorimeter, which includes the resistor
and thermometer.
Specific heat From your plot of \(C_p(T)\), work out the
heat capacity per unit mass of water. You may assume the effect of
the resistor and thermometer are negligible. How does your answer
compare with the prediction of the Dulong-Petit law?
Latent heat of fusion
What did the temperature do while the ice was melting? How much
energy was required to melt the ice in your calorimeter? How much
energy was required per unit mass? per molecule?
Entropy of fusion The change in entropy is easy to measure for a
reversible isothermal process (such as the slow melting of ice),
it is just
\begin{align}
\Delta S &= \frac{Q}{T}
\end{align}
where \(Q\) is the energy thermally added to the system and \(T\) is
the temperature in Kelvin. What is was change in the entropy of
the ice you melted? What was the change in entropy per
molecule? What was the change in entropy per molecule divided
by Boltzmann's constant?
Entropy for a temperature change Choose two temperatures
that your water reached (after the ice melted), and find the change
in the entropy of your water. This change is given by
\begin{align}
\Delta S &= \int \frac{{\mathit{\unicode{273}}} Q}{T} \\
&= \int_{t_i}^{t_f} \frac{P(t)}{T(t)}dt
\end{align}
where \(P(t)\) is the heater power as a function of time and \(T(t)\) is
the temperature, also as a function of time.
In this remote-friendly activity, students use a microwave oven (and optionally a thermometer) to measure the latent heat of melting for water (and optionally the heat capacity). From these they compute changes in entropy. See also Ice Calorimetry Lab.
VaporizationHeatThermal and Statistical Physics 2020
The pressure
of water vapor over ice is 518 Pa at \(-2^\circ\text{C}\). The vapor
pressure of water at its triple point is 611 Pa, at
0.01\(^\circ\text{C}\) (see
Estimate in \(\text{J
mol}^{-1}\) the heat of vaporization of ice just under freezing. How
does this compare with the heat of vaporization of water?
The students will set up a Styrofoam cup with heating element and a thermometer in it. They will measure the temperature as a function of time, and thus the energy transferred from the power supply, from which they compute changes in entropy.
Consider the bottle in a bottle problem in a previous problem set, summarized here.
A small bottle of
helium is placed inside a large bottle, which otherwise contains
vacuum. The inner bottle contains a slow leak, so that the helium
leaks into the outer bottle. The inner bottle contains one tenth
the volume of the outer bottle, which is insulated.
The volume of the small bottle is 0.001 m23 and the volume of the big bottle is 0.01 m3. The initial state of the gas in the small bottle was \(p=106\) Pa and its temperature \(T=300\) K. Approximate the helium gas as an ideal gas of equations of state \(pV=Nk_BT\) and \(U=\frac32 Nk_BT\).
How many molecules of gas does the large bottle contain? What is the final temperature of the gas?
Compute the integral \(\int \frac{{\mathit{\unicode{273}}} Q}{T}\) and the change of entropy \(\Delta S\) between the initial state (gas in the small bottle) and the final state (gas leaked in the big bottle).
The isothermal
compressibility is defined as
\begin{equation}
K_{T}=-\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_{T}
\end{equation}
\(K_T\) is be found by measuring the fractional change in volume when
the the pressure is slightly changed with the temperature held
constant. In contrast, the adiabatic compressibility is defined as
\begin{equation}
K_{S}=-\frac{1}{V} \left(\frac{\partial V}{\partial p}\right)_{S}
\end{equation}
and is measured by making a slight change in pressure without
allowing for any heat transfer. This is the compressibility, for
instance, that would directly affect the speed of sound. Show that
\begin{equation}
\frac{K_{T}}{K_{S}} = \frac{C_{p}}{C_{V}}
\end{equation}
Where the heat capacities at constant pressure and volume are given
by
\begin{align}
C_{p} &= T \left(\frac{\partial S}{\partial T}\right)_{p} \\
C_{V} &= T \left(\frac{\partial S}{\partial T}\right)_{V}
\end{align}
Clausius-ClapeyronThermal and Statistical Physics 2020
Calculate based on the Clausius-Clapeyron equation the value of
\(\frac{dT}{dp}\) near \(p=1\text{atm}\) for the liquid-vapor
equilibrium of water. The heat of vaporization at
\(100^\circ\text{C}\) is \(2260\text{ J g}^{-1}\). Express the result
in kelvin/atm.
