Lecture 18 - Second Law of Thermodynamics

Introduction

The basic laws of physics are time reversible, meaning that they are unchanged under the change in the direction of time (in mathematical language, symmetry under time reversal $t\rightarrow -t$). For example, the elastic collsion of two particles obeys Newton's Laws, conservation of energy, conservation of momentum. If you record the whole collision on tape and play the video backward, the reversed process will obey the same laws. In other words, it appears that the fundamental laws of nature do not distinguish between future and past (or direction of the arrow of time).

However in our day to day experience, clearly future and past are very different. Here are some examples of such irreversible processes:

We only age in one direction in time (getting older).
A glass that falls and shatters into a thousand pieces do not spontaneously recombine back into an unbroken glass.
An explosion does not un-explode back into an intact bomb.
Heat always flows from hot objects to cold objects when left in contact.
A gas in a box always diffuses out to fill the room, but air in a room does not spontaneously flow into a small box leaving vacuum in the rest of the room.

The second law of thermodynamics is an empirical law that describes such irreversible process. There are many equivalent ways to state the second law. In this course we will state the law with the following simple (although not entirely unambiguous) statement via the change in entropy:

$$ \Delta S_{\text{closed system}} \geq 0 $$

It states that the entropy of any closed system (such as the universe as a whole) can never decrease. Roughly speaking, entropy is the disorderness (or messiness) of a system. Note the "$\geq$" as opposed to "$\gt$", which means entropy of a closed system could increase or stay the same according to the law. The term "closed system" refers to a system isolated from interactions with anything else.

Note that $\Delta S_{open system} \lt 0$ does not violate the second law. In other words, it is perfectly fine for a part of a close system to have a negative change in entropy (say $\Delta S_1 = -100J/K$), as long as that negative change is compensated by positive change elsewhere in the system (say $\Delta S_2 = +400J/K$).

The main deficiency in this way of formulating the Second Law this way is the ambiguity in the precise definition of the term entropy. Historically there were many different definitions to entropy, and papers are still being written these days on other possible definitions of entropy. For us we will be content with using the simple classical definition of entropy based on the idea of heat. We will explore the concept of entropy further in the next section.

Historically, there are popular (but equivalent) ways to state the second law:

It is impossible for any process to have as its sole result the transfer of heat from a cooler object to a hotter object.
It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending in the same state in which it began.

For this course, we will not explore this and other statements of the second law further, and will instead focus on the statment based on the concept of entropy.

Entropy

Roughly speaking, entropy is the disorderness (or the messiness) of a system. The second law of thermodynamics essentially states that whenever the entropy of a closed system changes, it is always in a way that will make the system more disordered.

The intuition behind this is that there are always vastly more ways to mess up than there are to keep a system neat and tidy. Think of a drawer of many red socks and a drawer of many black socks. There are many ways you can mix up the colored socks in the two drawers and end up with each having a mix of red and black socks, but only two configurations that are considered "organized" - all red socks in one drawer, and all black ones in the other drawer (2 configurations because you could exchange the two drawers and they would still be organized).

The basic equation for entropy change is: $$ \Delta S = \int_{\text{reversible}} \frac{dQ}{T} $$ The SI unit for entropy is $J/K$. The above integral is actually not so straight forward to use, because technically, it only gives you the correct change in entropy if the integration is carried out over a reversible process. For us, we will not worry about this subtlety and will specialize in only two simple special cases:

$$ \Delta S = \left\{ \begin{array}{lll} \frac{Q}{T} &&\text{ when } \Delta T = 0 \\ mc \ln(\frac{T_f}{T_i}) && \text{ when } \Delta T \neq 0 \end{array} \right. $$

Proof (click to show)

Proof of case 1 - no temperature change $\Delta T = 0$ (e.g. during a change of phase, or when changes in $T$ is negligible such as for a large object). Because $T$ is constant under this assumption, it can be taken outside the integral: $$ \begin{eqnarray} \Delta S &=& \int \frac{dQ}{T} \\ &=& \frac{1}{T}\int dQ \\ &=& \frac{Q}{T} \end{eqnarray} $$

Proof of case 2 - with temperature change $\Delta T \neq 0$. We will assume sample obeys the simple relation $Q = md\Delta T$, or its infinitesimal version $dQ = mc dT$: $$ \begin{eqnarray} \Delta S &=& \int_{T_i}^{T_f} \frac{dQ}{T} \\ &=& \int_{T_i}^{T_f} \frac{mc dT}{T} \\ &=& mc \int_{T_i}^{T_f} \frac{dT}{T} \\ &=& mc \ln(\frac{T_f}{T_i}) \end{eqnarray} $$

Try It Yourself (click to show)

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Heat Engine

A heat engine is any device that uses heat as a source of energy to do mechanical work. It involves at least three parts:

Hot reservoir (sources of energy)
Cold reservoir (absorbs waste heat)
Engine (generates work)

The meaning of the variables:

$|Q_H|$ is the heat extracted from the hot reservoir.
$W$ is the amount of energy converted into mechanical work by the engine.
$|Q_C|$ is the heat that is not used by the engine, released to the cold reservoir.

Technically $Q_C$ is actually negative because it is heat leaving the engine. However, when dealing with heat engines it is more intuitive to consider the absolute value (so you won't carelessly commit sign errors), as a result we write the heat as $|Q_H|$ and $|Q_C|$.

Your car engine is an example of a heat engine. The hot reservoir is the hot chamber of burning gasoline, where heat $|Q_H|$ is generated, of which part is converted into useful work $W$ pushing your car forward. The amount of heat that is not used $|Q_C|$ is lost (and wasted) to the atmosphere, which acts as the cold reservoir.

By conservation of energy (or using the first law of thermodynamics with $\Delta U = 0$):

$$ |Q_H| = W + |Q_C| $$

This states that energy flowing into the engine equals to the energy flowing out.

The efficieny of an engine is defined to be:

$$ e = \frac{W}{|Q_H|} $$

Using $|Q_H| = W + |Q_C| \Rightarrow W = |Q_H| - |Q_C|$, we get: $$ e = 1 - \frac{|Q_C|}{|Q_H|} $$

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Reversible Engine

The most efficient engines allowed by the laws of thermodynamics is a reversible engine. In a reversible engine, all the processes are reversible. For such engines, the efficiency depends only on the temperatures of the heat reservoirs:

$$ e_{\text{reversible}} = 1 - \frac{T_C}{T_H} $$

Here is another related equation that is useful for reversible engines:

$$ \frac{T_C}{T_H} = \frac{|Q_C|}{|Q_H|} \text{ (true only for reversible engines)} $$

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Notations

Name	Symbol	Unit	Meaning
Heat input	$\|Q_H\|$	$ J$	heat extracted from the hot reservoir
Heat wasted	$\|Q_C\|$	$ J $	heat released to the cold reservoir
Useful work	$W$	$J$	useful mechanical work performed by the engine
Efficiency	$e$	no unit	fraction of energy that is turned into useful work

PHYS 170

Introduction

Entropy

Proof (click to show)

Try It Yourself (click to show)

Try It Yourself (click to show)

Heat Engine

Try It Yourself (click to show)

Reversible Engine

Try It Yourself (click to show)

Notations