Thermodynamics I: Air Conditioners and Refrigerators

Refrigerators and air conditioners operate the same way because they have the same basic function: to keep some object colder than its surrounding environment. For the refrigerator, the object is the volume full of food and the environment is the warmer house. For the air conditioner, the object is the inside of the house and the environment is the warm outdoors.

It is necessary for some system (the refrigerant) to operate continuously, i.e. not simply cool the object once, because heat (thermal energy) will always leak into the object from its warmer surroundings. One hopes that the object is well insulated, so that the rate that heat leaks in (J/s = Watt) is small, but it is never zero.

One might ask why not simply “destroy” the thermal energy that leaks in? From our previous classes, the answer should be obvious – energy is conserved and cannot be destroyed: if you want to remove it from an object, you have to put it somewhere else. While the answer my be obvious to us, it wasn’t clear until the middle of the 19th century. Since atoms were not understood, neither was thermal energy, and heat was thought to be a substance (called caloric) that flowed from hot to cold objects.

[Aside: The amount of caloric was measured in “calories”: one calorie was the amount of caloric needed to increase the temperature of 1 gram of water by 1 oC. Since we now recognize heat as a form of energy transfer, we can relate the calorie to the Joule: 1 calorie = 4.19 J. One kcal, often written as “Calorie” (with upper case C) = 4190 J is the unit used for a food’s energy content.]

Following the text, we will call the energy inside a stationary object its internal energy, U. (It is mostly the thermal energy, but may also include changes in the material’s potential energy due to changes in electric, magnetic, and gravitational fields.)

The recognition that heat was a form of energy transfer led to the First Law of Thermodynamics (equivalent to a statement about energy conservation): The change in internal energy of an object is equal to the net heat that flows in minus the net work the object does:

ΔU = Q – W (First Law of Thermodynamics) The change in internal energy of an object is equal to the net heat that flows in minus the net work the object does:

ΔU = Q – W (First Law of Thermodynamics)

The words “net” are important. Both Q and W in this expression can be positive or negative. Q is positive if heat flows in and negative if heat flows out. W is positive if work is done by the object (e.g. if it expands, pushing a piston out) and negative if work is done on the object (e.g. if it is compressed). Although the above expression is a usual way of writing the First Law, perhaps a clearer statement would be:

ΔU = |Qin| – |Qout| – |Wby| + |Won| where |…| means absolute value.

I will use absolute value symbols in my equations to show the + and – signs of each term explicitly. Back to refrigerators and air conditioners: We’d want the refrigerant to operate in a “cycle”, i.e. after removing thermal energy from the object to be cooled, it somehow returns to its original condition (pressure, volume, temperature, charge, electric and magnetic fields, etc.) so that it can repeat the operation. Therefore, the change in the refrigerant’s internal energy after completing one cycle should be zero. Since it is absorbing heat from the object to be cooled, it needs to get rid of this energy, e.g. by exhausting heat (Qout) or doing work (Wby):

Here, Qc is the heat that flows out of the cold object in one cycle and either |Qout| = |Qc| (first scheme) or |Wby| = |Qc| (second scheme) because the refrigerant goes through a cycle and has no net change.

In fact, both of these schemes are impossible! The first scheme is impossible because for heat to flow from the object to the refrigerant, the refrigerant should be colder than the system, and for heat to flow from the refrigerant to the outside, the refrigerant must be hotter than the outside: i.e. Tobject ≥ Trefrigerant ≥ Toutside, but we know that the object is colder than the outside! The second scheme is impossible for a more subtle reason. Thermal energy is, as we have discussed, energy associated with disorder in the material. Work implies putting order into a system; for example, electrical work means separating positive and negative charges. So the second scheme implies removing disorder from the object to make something else (in the surroundings) more ordered. (The refrigerant itself does not change since it is going through a cycle.)

The Second Law of Thermodynamics, also based on our observations of nature can be stated as: No processes are possible if the net effect is to transform disorder into order. No processes are possible if the net effect is to transform disorder into order.

In physics, it is necessary to quantify this statement. There is a quantity called Entropy (S) that is a measure of the amount of disorder in a system in terms of the properties of all the particles (atoms, electrons, molecules, …) that make it up. While we won’t write down an explicit expression for entropy, we can restate the Second Law:.

The entropy of a thermally isolated system of objects (i.e. in which no energy flows in or out) never decreases: ΔS ≥ 0 for any process in an isolated system.

[For our refrigerator schemes, the isolated system is the object to be cooled plus the refrigerant plus the surroundings.] The entropy of a thermally isolated system of objects (i.e. in which no energy flows in or out) never decreases: ΔS ≥ 0 for any process in an isolated system.

The Second Law is unique in physics: it is the only law that distinguishes the past from the future. It states that the future cannot be more ordered than the past. All other physical laws are “time reversible”: e.g. they describe what quantities are conserved, or the magnitude of forces. Even Newton’s Second Law is time reversible: If a particle moving toward the right is increasing speed because of a force, it would be decreasing speed if moving toward the left for the same force. Therefore, if you were to take a movie of its motion and show the movie backwards, it would look OK.

However, for macroscopic systems, i.e. systems with lots of particles with thermal energy, changes in time are no longer reversible. If you watch a movie with macroscopic objects (including people), you can usually tell when it is being shown backwards. Actions seem awkward – people and things don’t move like that. You can tell because you instinctively know the Second law of Thermodynamics.

Here is a trivial example. Suppose you drop a cup and it shatters into lots of pieces. Although all other physical laws (e.g. conservation of energy) would allows the pieces of the cup to spontaneously reassemble themselves and jump up back into your hand, we know it will never happen. That is because the broken cup is more disordered, i.e. has more entropy, than the whole cup, so spontaneously reassembling and jumping back up would require a decrease in entropy and cannot occur because of the Second Law. Because the Second Law distinguishes the past from the future, we say that it determines the direction of the “arrow of time”. It governs the laws of cosmology (the history of the universe), biology, consciousness (we remember the past, not the future), and has deep philosophical consequences.

Cosmologically, it means that the universe (the only perfectly isolated system) has always been growing more disordered since the big bang. This may not be obvious as stars and planets form out of the very disordered interstellar medium. However, each star or planet is not itself an isolated system. For example, as a planet forms, its own entropy may decrease but the entropy of the surrounding material increases by a greater amount, so there is a net increase in entropy.

Similarly, as a biological organism (e.g. a person) develops, its own entropy may decrease but the waste products that are produced increase the total entropy. Similarly, it has been shown that evolution of more complicated organisms leads to a net increase in entropy.

The maximum disorder for a system with lots of objects is when they are all at the same temperature – i.e. all have the same average thermal energy/particle. That is where the universe is heading, with a time scale of trillions of years: it is called the “heat death of the universe”. We will see (next lecture) that if all objects are at the same temperature, no work is possible, and all changes stop. [So do your projects now!!] Although we have not given a general mathematical expression for the entropy of an object, there is actually an easy expression for the change in entropy of an object at a fixed (Kelvin) temperature T when heat flows in or out:

ΔS = (|Qin| – |Qout|) / T

[One of the ways heat can flow in or out of an object without its temperature changing is if the object is very large (i.e. large mass). Since ΔT = (change in internal energy) / [(mass) ⸱(specific heat)], ΔT →0 if mass →∞. A large object which can absorb or emit heat with negligible change in temperature is called a heat reservoir.]

Consider an isolated system comprised of two reservoirs.

Heat is flowing out of object 1, at temperature T1, into object 2, at temperature T2, so Q1 = -|Q | and Q2 = +|Q|. So for the whole system, ΔS = |Q|/T2 – |Q|/T1 = |Q| (1/T2 - 1/T1). Since ΔS ≥ 0, we must have T1 ≥ T2; that is heat cannot spontaneously (i.e. without something else going on) flow from cold to hot – we have proven our assertion from the last lecture. Now let’s return to our discussion of the refrigerator/air conditioner. This is what we’ve determined so far: We want a refrigerant, running a cycle, to remove heat from a cold object. We know it cannot simply exhaust heat into the surroundings, which are hotter than the cold object, and cannot get rid of this energy by doing work. There is a solution, however, that obeys both the first law (conservation of energy) and the second law (total entropy cannot decrease), and it is shown schematically here:

Heat Qc is absorbed by the refrigerant from the object to be cooled and heat Qh is exhausted to the hotter environment. During the cycle, however, the environment does work on the refrigerant.

Since the refrigerant goes through a cycle, there is no

change in its internal energy, so |Qc| + |Won| = |Qh| (i.e. energy in arrows = energy out arrow). Since the entropy of the refrigerant also does not change in a cycle, the total change in entropy of the universe is that of the object and surroundings, which we approximate as reservoirs at constant temperatures:

ΔS = |Qh|/Th – |Qc|/Tc = |Qc|(1/Th-1/Tc) + |Won|/Th, which can be ≥ 0 if Won is large enough. Now let’s look at this refrigerator/air conditioner from an “economic” point of view. Qc is what we want – to remove heat from the inside of the refrigerator (or house for an air conditioner). Won is the work done on the refrigerant – it is what we pay the utility company for. |Qh| = |Qc| + |Won| is the heat that comes out at the back of the refrigerator our air conditioner – it is “wasted heat.” So we always waste more heat than the heat removed for cooling. This is not an efficient device, but from the laws of thermodynamics, it’s the best we can do.

The coefficient of performance (COP) of the refrigerator is COP = |Qc| / |Won| = (what you want) /(what you pay for) In fact, the very best we can do is if ΔS = 0 (which is permitted, we just cannot allow the entropy to decrease ). One can approach this limit if there is no friction (which always generates extra heat) and if one allows all the steps to occur very slowly (if the work or heat flows are rapid, they inevitably result in messiness and increasing the disorder and entropy). Since ΔS = |Qh|/Th – |Qc|/Tc = |Qc|(1/Th-1/Tc) + |Won|/Th, the ideal refrigerator has |Qh|/Th = |Qc|/Tc, so the energy you pay for is |Won| = |Qc| (Th-Tc) / Tc: The smaller the temperature difference between the inside and outside the more efficient the refrigerator/air conditioner (i.e. the less work needs to be done/cycle and the less money it costs to run).

Then the coefficient performance of the ideal refrigerator/air conditioner is:

COPideal = Tc /(Th – Tc) A schematic of a practical refrigerator or air conditioner is shown here. As described in the text, the refrigerant evaporates and absorbs heat when in contact with the cold object. Work is done in the compressor, and then the refrigerant condenses into a liquid, exhausting heat, when in contact with the warm outside. To make this work practically, the refrigerant must be a material that readily transforms between liquid and gaseous state near room temperature. Chlorine contained in the traditional refrigerant material, freon, unfortunately can attack the earth’s ozone layer if it leaks into the atmosphere. It is being replaced with “hydrofluorocarbons” (hfc’s), which do not attack ozone (but can contribute to global warming if they leak out). A typical “coefficient of performance”,

COP = |Qc| / |Won| is ~ 6 for a good refrigerator and ~ 5 for an air conditioner. These are much less than the ideal values:

COPideal = Tc / (Th – Tc).

o o For a refrigerator, with Tc = 277 K (~ 40 F) and Th = 295K (~ 72 F), the ideal COP = 277/18 = 15.

o For an air conditioner, with Tc = 295 K (~ 72 F) and o Th = 306 K (~91 F) → COPideal = 295/11 = 27.

One reason that real systems are so inefficient compared to ideal systems is that an ideal system must operate at an impractically slow speed – we make compromises to get our cooling reasonably quickly. Exercises: 1. Drinking fountains that actively chill the water they serve can’t work without ventilation. They usually have louvers on their sides so that air can flow through them. Why do they need this airflow? 2. If you open the door of your refrigerator with the hope of cooling your room, you will find that the room’s temperature actually increases somewhat. Why doesn’t the refrigerator remove heat from the room? 3. The outdoor portion of a central air-conditioning unit has a fan that blows air across the condenser coils. If this fan breaks, why won’t the air conditioner cool the house properly? 7. A soda siphon carbonates water by injecting carbon dioxide gas into it. The gas comes compressed in a small steel container. As the gas leaves the container and pushes its way into the water, why does the container become cold? Problems: 2. While polishing a 1-kg brass statue, you do 760 J of work against sliding friction. Assuming that all the resulting heat flows into the statue, how much does its temperature rise? 3. You drop a lead ball on a cement floor from a height of 10 m. When the ball stops bouncing, how much will its temperature have risen? (Assume that the ball has a mass of 3 kg and that all the energy stays in the ball and not in the cement.) • What is the (approximate) change in entropy of the ball if its initial temperature = 295 K? 6. An ideally efficient freezer cools food to 260 K. If room temperature is 300 K, how much work does this freezer consume when removing 100 J of heat from the food? 7. An ideally efficient refrigerator removes 900 J of heat from food at 270 K. How much heat does it then deliver to the 300 K room air?