This short lecture introduces the ideas required for Ice Calorimetry Lab or Microwave oven Ice Calorimetry Lab.
This is part of a pandemic activity. Not sure whether I might use some of these changes when we are teaching in person again.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 not a 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.
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.
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.
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.
group Small Group Activity
30 min.
assignment Homework
Solve for the net power transferred between the two sheets.
Optional: Find the power through an \(N\)-layer sandwich.
assignment Homework
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).
assignment Homework
biotech Experiment
60 min.
heat entropy water ice thermodynamics
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.assignment Homework
group Small Group Activity
60 min.
assignment Homework
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 m^{23} and the volume of the big bottle is 0.01 m^{3}. 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).
assignment Homework
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}
assignment Homework