Chapter 20

Heat

For a long time, scientists did not have a clear idea of what heat, or, more precisely, heat transfer really was. In fact, the most common belief was that heat was carried by some sort of intangible, mysterious fluid, called either phlogiston or caloric. The eventual realization that heat transfer was in reality yet another form of energy transfer was a major milestone, that effectively opened up the road to the industrial revolution, the development of steam engines, the science of Thermodynamics, etc. Interpreting heat as a form of energy transfer closed the loop in most mechanical processes, showing that, whenever mechanical energy is lost (e.g. due to friction) in any given process, an equivalent amount of energy is generated in the form of thermal energy, and this in turn gives to the Principle of Energy Conservation a much much broader scope. Moreover, it was possible to introduce the concept of internal energy, i.e. energy carried by the elementary constituents (e.g. atoms or molecules) of any given substance.

Internal Energy : in the simplest case of a monoatomic gas (e.g. noble gases) internal energy can only be present in the form of linear kinetic energy of the gas atoms. In general, internal energy will also include rotational kinetic energy and electrostatic potential energy (for completeness, one could also list nuclear potential energy as contributing to internal energy, but this energy plays no role in ordinary thermodynamical processes).

In general, heat can be thought of as a transfer of internal energy from one object to another. This is equivalent to the transfer of mechanical energy as mediated by work. In general, internal energy is related to temperature, and a process of heat transfer is accompanied by a change of temperature, but, as we will see, there are some important exceptions.

When studying heat transfers, it is more straightforward to deal with the unit originally introduced to describe such phenomena, i.e. the calorie. 1 calorie is defined as the amount of heat required to increase by one degree Celsius 1 gram of water (the exact defintion would state ....to bring 1 g of water from 14.5 to 15.5 degrees). A related unit is the Calorie (notice the capital C), which is equal to 1000 calories, therefore it is the amount of heat required to increase by one degree the temperature of 1 kilogram of water ( or, to a good approximation, to increase the temperature of 100 g of water by 10 degrees, etc.).

In the next step, we define the specific heat of any substance as the amount of heat (in Calories) required to increase by 1⁰C 1 kg of the substance. From these definitions it follows immediately that the heat capacity of water, measured in Cal/(kg ⁰C), is 1. The heat capacity of water is among the highest of most materials, it takes lots of heat to warm up water and conversely, hot or warm water can store and release lots of heat. From these definitions, it follows immediately that the quantity Q of heat involved in changing the temperature of a body of mass m by an amount $\Delta T$ is given by

$Q = m c \Delta T$

where c is the specific heat of the particular substance

When, through the work of scientists like Benjamin Thompson Count Rumford (a rare case of an American emigrating to Europe) and James P. Joule, it was realized that heat was another form of energy transfer, it was possible to obtain the exact correspondence between the traditional units for measuring heat and the standard units for energy. And the answer was:

1 cal = 4.186 joules, or 1 Cal = 4186 joules

This result is extremely useful, since it allows us to work out how much weight lifting we need to do in order to burn off a 100 Calorie snack ...

Phase Changes
Adding or removing heat to/from a substance will change its temperature up to a certain point : we all know what happens to water when we heat it or cool it too much. The most important thing to remember when dealing with phase changes is that, throughout the change the temperature remains constant. When we say that water freezes at 0⁰C, we mean that, if we have some water and lower its temperature, when it reaches 0 it will stay at that value until all the water has frozen, and only then removing more heat will further cool the sample. Conversely, if we start with very cold ice and warm it up, it will start melting when it reaches 0⁰, and the ice/water mixture will remain at that temperature until all the ice has melted. The same effect occurs when water boils away, i.e it turns into its gaseous phase, at 100⁰C (but remember: the boiling point of a liquid depends on the pressure to which it is subjected, in Denver water boils below 100⁰C).

The amount of heat that is involved in the phase change (the latent heat of fusion or of vaporization) is considerable: the latent heat of fusion for water is 79.7 cal/g (or Cal/kg), which means for instance that it takes the same amount of heat to melt 1 kg of ice originally at 0⁰C as to increase its temperature from 0 to $\approx$ 80⁰ !!!. Similarly, the latent heat of evaporation for water is 539 cal/g .

Remember that the process goes both way: if e.g. lots of heat is required to vaporize liquid water at 100⁰, condensing steam will release an equivalent amount of heat: a burn can be much more severe when caused by steam rather than water at 100⁰C.

Energy Conservation : 1st Law of Thermodynamics

To start, we note that, if a system has some internal energy then (almost by definition) it is capable of doing work. Recalling that work is Force x displacement, we realize that an expanding gas does work; the force at play comes from the gas pressure, and the displacement can be that of a piston, or of the molecules of another gas, or the walls of a stretchable container, etc. (exception: a gas expanding into a vacuum does not displace anything therefore it does no work). Choosing the simple case of a gas in a cylindrical container with a movable piston of area A

dW = F dx = PA dx = P dV

$W~=~\int_{V1}^{V2}P dV$

notice that in general the work done does depend on the path followed to go from V1 to V2 . Examples : start from (P1,V1) and reach (P2,V2) and compare compare the work done

• either when the process occurs at constant temperature (therefore PV = nRT =costant)
• or when first allowing to expand at constant pressure and then reducing the pressure (e.g. by reducing the temperature) while keeping the volume constant

In spite of the fact that work depends upon the path followed, we can still introduce the idea of energy conservation in a thermodynamical process. Let us first clarify a confusing issue concerning the signs of the quantities involved : we are interested in finding the change in the internal energy of our system $\Delta E_{int}=E^{f}_{int}-E^{i}_{int}$ . A change in internal energy can occur

• by transfering some amount of heat Q in or out of the system
• or when some work W is done onto or by the system

It is rather intuitive that E_int will increase $(\Delta E_{int}\gt)$ if we

• either transfer heat into the system, or
• do some work onto the the system

For both of these cases Q and W will then be positive. Conversely, E_int will decrease $(\Delta E_{int}<0)$ if we

• transfer heat out of the system, or
• the system itself does some work

In these cases, Q and W will be negative. Having clarified this, we can write the condition of energy conservation as

$\Delta E_{int}~=~Q~+~W$

with the understanding that both Q and W can be positive or negative, according to the intuitive rules discussed above.

Note: "my" sign convention is different from the one given in the book, but, in my opinion, it is less confusing. Using my convention, when facing a particular problem and needing to assign signs to W and Q, you should ask yourself (separately for Q and W) : is this process going to increase or decrease the internal energy? and assign the signs accordingly. Notice also that in "my" convention, Q and $\Delta E$ have the same sign as the book, while W has the opposite sign.

Problem 20.40

A few special cases:

• adiabatic = no heat exchange with the environment:
  $Q=0\rightarrow \Delta E = W$
• isovolumetric or isochoric : $W = 0 \rightarrow \Delta E = Q$
• isobaric : $W = \int P dV = P\Delta V$
• isothermic : PV = nRT = constant, $W = \int P dV = nRT\int dV/V = nRT\ln (V_{f}/V_{i})$

Heat Transmission

Heat Transmission con occur via Convection, Conduction or Radiation. Convection is the dominant process in fluids (liquids and gases), Conduction acts mostly in solids (but also in liquids and, to some extent, in gases). Heat transfer by radiation can take place through a vacuum , as well as through any medium transparent to radiation.

Convection

The mechanism for convection is the density difference in regions of a fluid at different temperatures, and the consequent action of gravity. Take a fluid at a constant temperature throughout and heat some region of it. In the warmer region density will decrease and the fluid in that region will rise (according to Archimedes principle). Simultaneously, cooler fluid will move in to fill in the space vacated by the warmer fluid. The effect of such convective currents is the transport of thermal energy throughout the fluid.

Typical instances of convection are e.g. water being heated in a pan or the formation of cumulus clouds. In several applications, one can also have forced convection, e.g. when using a fan or a pump (to be correct, the radiator in you car should not be called Radiator but "Convector"). Although conceptually easy to understand, it is not easy to describe convection in a quantitative way.

Conduction

While in convection heat is carried by an actual displacement of matter within the fluid, in conduction only heat is transferred, without any transportation of matter taking place. To interpret the phenomenon of heat conduction at the atomic level, we should be aware that temperature of a body is related to the average kinetic energy of its molecules. In the contact region between a colder and a warmer body, collisions among molecules will have the effect of transfering some of the (thermal) energy from the higher to the lower temperature. But to analyze heat conduction quantitatively it is not necessary to think in terms of collisions among molecules. Instead, we can easily write the basic expression representing the amount of heat Q transmitted across a given thickness of material :

$Q = k A t \Delta T/L$

where A is the area of two surfaces which are at a temperature difference $\Delta T$, L is the material thickness and t is the elapsed time. k is a constant, typical of the material, which represents how well (or how badly) the material can transmit heat. According to the application, we will want to employ either very good conductors (e.g. in a frying pan) or good insulators (for the pan's handle). Remembering that Q is an energy, and that Power = Energy/time, the power transmitted by conduction is

$P = k A \Delta T/$L

In many applications, dealing mainly with construction and home insulation, the expression above is written as

$Q = A t\Delta T/R$

with R, the R-value, = L/k . R is useful to compare simultaneously insulating power and material thickness; obviously the larger is R, the better the insulation. It is also useful to remember that air (especially dry air) is a good insulator, while water is a good conductor (on a windless day, you could stand in 50⁰F air without too much discomfort, but you would quickly get very chilled if you were in water of the same temperature).

When dealing with conduction across layers of different material, the useful rule to remember is that, at equilibrium, the same amount of heat has to go through each material.

Radiation

Every object, no matter at how low a temperature (except for absolute zero) emits some amount of energy in the form of electromagnetic radiation, i.e. electromagnetic waves. We haven't yet studied electromagnetism, but for now let us just accept the idea that electromagnetic waves (corresponding to oscillations of the electromagnetic field) can manifest themselves in many different forms, ranging from, in increasing energy, radio waves, microwaves, infrared, light, ultraviolet, X-rays and gamma-rays. The member of this family usually associated with heat and temperature is the infrared radiation, since this type of radiation is felt by our bodies as heat, and the range of radiation emitted by bodies at ordinary temperatures is peaked mainly in the infrared region (but all other types of electromagnetic radiation do transfer energy and consequently increase the temperature of an object, you can get burnt equally well by a laser producing purely visible light or by UV rays).

If a body at a given temperature continuously emits radiation, that is if it emits some of its internal energy, then, if left to itself, it would gradually cool down (this is in fact what happens if we have a hot body in a cooler room). But, when in equilibrium with its surroundings, an object maintains its temperature; this must mean that the body, in addition to radiating away energy, must also be able to absorb it from its surroundings. And this is exactly what happens: when in thermal equilibrium with its surroundings a body at the same time absorbs and emits radiation at the rate of

P = Power = Energy/time = Q/t = $e\sigma T^{4}A$

where T is the absolute temperature (in kelvin), A is the area of the emitting/absorbing surface, $\sigma$is a universal constant, and e is a number between 0 and 1 that depends upon the properties of the emitting/absorbing surface.

It is important to understand the role and meaning of e : let's take a certain object, and "shine" it with a certain amount of electromagnetic radiation (e.g. light, or infrared, etc.). Experience tells us that different objects will absorb the radiation in different amounts : we define as emissivity e the ratio of energy absorbed by the body to the energy received by it. Obviously e will range between 0 and 1 (and what happens to the energy that is not absorbed? it is reflected).

Now you would be entitled to ask : why did I call this ratio emissivity and not, e.g. absorptivity ? The reason is that there is a one to one correspondance between energy absorbed and energy emitted : the rate of energy that a body absorbs is identical (for the same temperature) to the rate it will emit, i.e. the formula above applies equally well to emission and absorption. But do not be confused: the actual value of emitted energy depends upon the temperature of the emitting body, while that of absorbed energy depends upon the "temperature" of the energy reaching the body. In formula

$P_{net}~=~e\sigma A(T^{4}_{body}-T^{4}_{environment})$