101 Heat Death

The Second Law of Thermodynamics Revisited

The introduced in the previous chapters states that thermal energy will always spontaneously transfer from higher temperature to lower temperature. In order to aid in our study of other thermodynamic processes, such as phase changes, a more general version of the Second Law can be stated in terms of  energy concentration and dispersion: any must move an isolated system toward a state of more uniform dispersion of energy throughout the system.  By we mean one for which energy does not leave or enter. For example, we know that higher means a greater average per molecule, so we can think of temperature as a measure of the of thermal energy in an object. If we consider a hot object in a cold room as our complete system, then the thermal energy in our system is not very well dispersed because it’s more concentrated in the hot object. The predicts that energy should move from the hot object to the cold environment to better disperse the energy, and that is what we observe. However, the heat transfer will only occur until the thermal energy is maximally dispersed, which corresponds to , and is indicated by the object and environment  having the same temperature.

Everyday Examples: Sweating, Dew, and Rain

When we sweat to exhaust by we aren’t actively grabbing the hottest water molecules, pulling then away from their neighbors, and throwing them into the gas phase. The evaporation happens spontaneously because stored in water molecules that are stuck together is relatively concentrated compared to thermal energy stored in water molecules zipping around in the air and free to disperse. The transfer of thermal energy to sweat (by ), followed by evaporation is a because it increases the dispersion of energy throughout the system made up of you, the sweat, and the surrounding air. Therefore, evaporation of sweat is a .

When the reaches 100% then has maximally dispersed the available . Any additional evaporation would begin to over-concentrate energy in the air and decrease the overall level of energy dispersion. Therefore, we don’t see evaporation occurring once 100% humidity is reached. In fact if the humidity gets pushed above 100% (by a drop in air without a loss of water vapor) then energy is over-concentrated in the air and thus increasing dispersion of energy requires that water molecules come out of the vapor phase and occurs spontaneously. When the liquid condenses on surfaces we call it , when the liquid condenses on particles in the air and falls to the ground we call it rain.

Reinforcement Exercise

You place -10  °C ice in warm room. Will thermal energy flow from room air to ice, or from ice to warm air? Explain in terms of and and the .

When the ice warms up to 0 °C it begins to melt instead of increase . Based on the , which of the following processes provides more ways of dispersing through the ice and air?

  1. Thermal energy is removed from the warm room and stored as kinetic energy (vibrations) in ice molecules, which increases the ice temperature.
  2. Thermal energy is removed from the warm room and stored as kinetic energy in vibrations and rotations of liquid water molecules, which requires melting the ice instead of raising its temperature.

Entropy

(S) is a measure of energy within a system. An increase in the entropy corresponds to an increase in dispersion of energy. A decrease in entropy would correspond to energy being less dispersed, or increasing energy . Therefore the can also be stated as:  a process will happen spontaneously if it increases the total entropy of an isolated system. The change in entropy for a constant-temperature process can be calculated from the heat transferred (Q) and the at which the transfer occurs (T) as:

(1)   \begin{equation*} \Delta S = Q/T \end{equation*}

Notice that we definitely need to use the absolute temperature scale when working with the change in equation, or else we might find ourselves attempting to divide by zero.  Let’s apply this equation and the entropy version of the to the second part of the previous Reinforcement Exercise.

Everyday Example: Entropy Change of Melting

If you place ice in a  warm room and leave it alone, it will melt. The ice would be a because it will happen all on it’s own, so we should find that melting increases the total entropy (ΔS > 0). Let’s check that out. We’ll keep it simple and calculate change in entropy of one kilogram of ice, which melts at 0 °C, or 273 K.

    \begin{equation*} \Delta S_{ice} = Q/T = \frac{mL_f}{T} = \frac{(1\,\bold{kg})(334,000\,\bold{J/kg})}{273 \,\bold{K}} = 1223\, \bold{J/K} \end{equation*}

Next we calculate the change in for the room. The same that went into melting the ice came out of the room, so the Q for the room is the same as for the ice, only negative. Let’s pick a typical room temperature of 20 °C for our example, which is 293 K:

    \begin{equation*} \Delta S_{room} = Q/T = \frac{-mL_f}{T} = \frac{-(1\,\bold{kg})(334,000\,\bold{J/kg})}{293 \,\bold{K}} = -1140\, \bold{J/K} \end{equation*}

Now we just add up the two changes to get the total:

    \begin{equation*} \Delta S_{T} =\Delta S_{ice} + \Delta S_{room} = 1223\, \bold{J/K} + (-1140)\, \bold{J/K} = 17\, \bold{J/K} \end{equation*}

Our entropy change is greater than zero, so the states that ice melting in warm room is a , which we observe it to be.

Maximum Efficiency

are inherently . For example, we could not reverse the spontaneous process of  transfer from your body to sweat and then to the environment by . Imagine trying to run around and grab all the water vapor molecules and shove them back into the liquid on our skin and then make those water molecules collide with skin molecules in just the right way to conduct thermal energy back into your body! Good luck. If the process could be reversed, the net change would be zero, but that is not a real possibility.  Any real process increases the entropy because irreversible at some level, meaning energy is further dispersed throughout the system without a realistic opportunity to put it back where it was. We have arrived at yet another version of the : any real process increases the total entropy of the universe. For example, even if you could reverse the process we described above, the you released during that running around would decrease the of energy in your body and disperse it throughout the room.The system, which includes you, would not cause have returned to the original conditions at all. In fact, the total entropy would have increased by even more than before you tried to reverse the process. For example, as you ran around, more sweat would have evaporated and you would have to chase those molecules down as well, and so on–you could never win! We can’t keep the of the universe from increasing.

Everyday Examples: Human Mechanical Efficiency

Muscle contraction relies on the release of chemical potential energy stored in ATP molecules. When that energy is released, the entropy of the molecules decreases, so the entropy of their environment must increase by more, and this is achieved because most of the energy released from the ATP molecules is degraded to heat and distributed to the environment. “The muscle cell moves toward chemical and electrical equilibrium as sodium and potassium ions pass through the cell membrane. In thermodynamics terms the cell has lost a marked degree of differentiation, […] and increased its degree of randomness as it approaches equilibrium.”  [1]

After contraction a muscle cell must be reorganized, which decreases entropy, but we know overall entropy must increase, so some energy must be dispersed to the environment to provide the necessary entropy increase.  That energy is “wasted” because it is not available to do work. Therefore, entropy and the 2nd Law of Thermodynamics limit the mechanical efficiency of the human body.

Everyday Example: Geothermal Heat Engine

Let’s imagine molten rock from the Earth’s mantle pushes partway through the Earth’s crust, and keeps a region of bedrock at a constant 300 °C temperature. If the rock was not too deep, we could install pipes in the rock, and then boil water by running it through the pipes. We would basically have a giant pressure cooker! Rather than cook food, we could release the pressurized steam to push on a piston or spin a turbine and produce mechanical work.  After releasing the pressure and getting some work out, we would be left with low pressure steam. We could condense the steam back to water by running it through pipes exposed to the 20 °C air above ground. Thermal energy would transfer from the steam to the air, the steam would condense into liquid water, and we could start again.

Any machine like ours that converts thermal energy into mechanical work is called a heat engine.  Your car is powered by an internal combustion heat engine.  Let’s see how Entropy and the 2nd Law of Thermodynamics affects the efficiency of our geothermal heat engine.

First we calculate the entropy change when 1000 J of thermal energy transfers out of the rock to the water to run the engine, remembering to convert the 300 °C rock temperature to Kelvin by adding 273 K:

    \begin{equation*} \Delta S_{rock} = \frac{Q_{rock}}{T_{rock}} = \frac{-1000\,\bold{J}}{573 \,\bold{K}}=-1.75\, \bold{J/K} \end{equation*}

If our engine is real, then so are its processes,  which means running the engine must increase the total entropy of the universe. We need to find out how much energy must be transferred from the low pressure steam into the air at 30 °C (293 K) in order for the entropy of the air to increase by at least as much as the rock entropy decreased (1.75 J/K). Rearranging the change in entropy equation:

    \begin{equation*} Q_{air} = \frac{T_{air}}{\Delta S_{air}} = \frac{293\,\bold{K}}{1.75 \,\bold{J/K}}=511\, \bold{J} \end{equation*}

Transferring the 511 J of energy into the air leaves only 489 J of the original 1000 J input energy available for doing work. Therefore the maximum possible efficiency of our engine, no matter how well designed and even after eliminating all mechanical inefficiencies would be:

    \begin{equation*} e = \frac{Work}{Input} =\frac{489\,\bold{J}}{1000\,\bold{J}} = 0.49 \end{equation*}

Multiplying by 100 would give us the efficiency as a percentage: 49 %. This is a maximum possible efficiency. Any engine we actually built would be less efficient.

The told us that you cannot build an engine that is more than 100 % efficient because energy cannot be created. Even worse, the 2nd Law tells us that even if we managed to eliminate all mechanical inefficiencies, such as friction, we still can’t get up to 100 % because all engines must exhaust some energy in order to increase entropy overall. The theoretical maximum efficiency, which is always less than 100 %, is known as the Carnot efficiency (ec) and depends only on the high and low operating temperatures (TH and TL), as we saw in the previous example.  All the work we did in the previous example can be short-cut by the equation for Carnot Efficiency (ec):

(2)   \begin{equation*} e_c = 1 - \frac{T_L}{T_H} \end{equation*}

The theoretical engine which could actually produce that theoretically maximum efficiency is known as the Carnot Engine. The operating principles of the Carnot Engine are well known, having been developed by Nicolas Léonard Sadi Carnot in 1824, but the engine cannot be realistically designed or built.[2]

Reinforcement Exercise: Carnot Efficiency

Check that the previous equation gives the correct maximum efficiency for our geothermal heat engine.

What is the maximum theoretical efficiency of our geothermal heat engine if the hot rock was actually at 550 C° instead of 300  ?

You will find that this engine is more efficient, just because the hot  operating temperature higher, even though it still works in the same way and nothing else has changed. The efficiency increased because the input energy started out more concentrated and less dispersed (measured as higher temperature), so less of that energy had to become dispersed in the environment in order to ensure that entropy increased. Energy that starts out concentrated is known as high quality energy.

Reinforcement Exercise:

A particular internal combustion engine in a car operates between the temperature of the hot combustion gasses immediately after combustion and the outside air temperature. These are typically 1000 K and 290 K.  What is the theoretical maximum efficiency of such engines?

Actual automobile engines have efficiencies that are about 75 % of the maximum theoretical efficiency. Roughly how efficient are actual automobile engines?

Heat Death

In addition to limiting our efficiency in doing mechanical work, the 2nd Law drives our bodies toward higher entropy, which means thermal equilibrium with the environment. Unless the environmental temperature happens to be near body temperature, reaching thermal equilibrium means death. Life also requires concentration of chemical energy, but due to the the 2nd Law we tend toward chemical equilibrium, which is not survivable. Concentrations of electrical energy drive your nervous system, but due to the 2nd Law we are constantly at risk of reaching an internal electrical equilibrium with no electrical activity. Life is a constant battle against various types of equilibrium that would correspond to maximum entropy, but also to death.  The work necessary to fight off our own entropy increase might be considered basic metabolism. Doing that work, and even taking in the energy required to do that work,  involves processes that provide even more opportunity for entropy increases in a seemingly viscous cycle. You can’t beat the 2nd Law of Thermodynamics! Even as we manage to prevent our own entropy increase we cause the entropy of the environment to increase by a greater amount than what we prevented in ourselves. In fact, a complete dispersion of energy, so that all matter is at equilibrium, no processes remain which would increase the entropy, and nothing really happens all, is one possible fate of the universe which has been dubbed heat death. Well, at least we don’t expect heat death of the universe to occur for at least 10100 years. [3]


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