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# 物理代写|热力学代写Thermodynamics代考|ENES232 What Is the Second Law of Thermodynamics?

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## 物理代写|热力学代写Thermodynamics代考|What Is the Second Law of Thermodynamics?

The concepts of the Second Law of Thermodynamics and of entropy have their roots in the early nineteenth century, when engineers started to develop and to operate heat engines. One of the problems was that not all forms of energy are of equal “utility”. For example, electrical or mechanical work (from a resistance heater or a stirrer) can-without restriction be used to increase the internal energy of a water reservoir. It is not possible, however, to transform all the energy stored within a warm water reservoir into useful work. This statement holds for all processes. In a Clausius-Rankine process, as an example of a heat

engine discussed in Chapter 6 , the heat provided by the combustion of coal can only partially be converted into work. In heat engines, part of the energy provided in the form of heat always leaves the engine as “waste” heat at a low temperature.

This can be seen in Figure 3.1, which depicts the basic principle of a heat engine. $Q$ indicates the heat typically provided by a combustion process at a high temperature $T$, whereas $Q_{0}$ indicates the heat lost at a low temperature $T_{0}$ to the environment. Therefore, only the difference may be obtained as useful mechanical (shaft) work $W_{\mathrm{s}}$.
An energy balance shows
$$Q+W_{\mathrm{s}}+Q_{0}=0 \quad \text { or } \quad\left|W_{\mathrm{s}}\right|=Q-\left|Q_{0}\right| .$$
In Equation 3.1, $W_{\mathrm{s}}$ and $Q_{0}$ have negative values because they are leaving the system. In general, and if not otherwise specified, quantities added to the system have positive values while leaving quantities count negatively (see Questions 1.4.2 and 1.4.3).

The second form of Equation $3.1$ has been written using absolute values for $W_{\mathrm{s}}$ and $Q_{0}$ because we are normally interested in these rather than in the negative values. As a result, $Q$ represents the heat transferred to the system at high temperature, whereas $\left|W_{\mathrm{s}}\right|$ and $\left|Q_{0}\right|$ stand for the shaft work obtained and the heat leaving the system at a low temperature. It follows that a thermal efficiency factor of the machine, defined as the ratio between the net work output, and the heat input may be computed as follows (see Chapter 6)
$$\eta_{\text {th }} \equiv \frac{\left|W_{\mathrm{s}}\right|}{Q}=1-\frac{\left|Q_{0}\right|}{Q} .$$

## 物理代写|热力学代写Thermodynamics代考|How Did the Need for a New State Function Arise?

The big question in the early days of heat machines was how much heat had to be “wasted” as $\left|Q_{0}\right|$ because this would determine the amount of energy that could be retrieved as shaft work and also the efficiency of the machine: The lower this could be kept, the more efficient the process would be.

In a purely mathematical analysis of heat engines Sadi Carnot showed (Carnot 1824) that the fraction of wasted heat with respect to the heat provided must indeed reach a minimum and that the efficiency would reach a maximum value if the heat engine could be operated strictly reversibly. If there was a way to predict this “magical” heat minimum $Q_{0 \text { rev }}$ one could use the above equations to predict at least the theoretical maximum efficiency of heat engines. Unfortunately, this is not possible because heat is not a state function. Carnot has also shown that the efficiency must be exactly the same for any conceivable reversible heat engine operating between the same temperatures, and that they will be the more efficient the larger the temperature difference $\left(T-T_{0}\right)$.

Based on these results, the question arose whether a new state function could be found that would permit one to predict if not $Q_{0}$ then at least $Q_{0 \mathrm{rev}}$ in the case of reversible processes. Since the heat engine is a cyclic process (see Question 6.1), the amount of this state function in the engine cannot change over a complete cycle, because the cycle could only work repeatedly if the system returned to the exact same state after a complete cycle. The amount of this state function taken up would therefore have to be exactly equal to the amount released over a whole cycle.

A crucial element in the search for such a state function was probably the realization by Lord Kelvin around 1850 (see e.g. Levenspiel 1996), and in a preliminary form already by Carnot, that the amounts of heat reversibly exchanged with the environment are proportional to the absolute temperatures $T$ and $T_{0}$ :
$$\frac{\left|Q_{0 \text { rev }}\right|}{Q_{\text {rev }}}=\frac{T_{0}}{T} \quad \text { or } \quad \frac{\left|Q_{0 \text { rev }}\right|}{T_{0}}=\frac{Q_{\text {rev }}}{T} .$$
The last equation demonstrates that the same amount of $Q_{\mathrm{rev}} / T$ is released and taken up during a complete working cycle, so that the amount accumulated inside the engine stays constant. The function $Q_{\mathrm{rev}} / T$ can thus be considered as the desired state function. It can also be shown that the change of such a state function between two states of the system only depends on the two states and not on the path taken to bring the system from one state to the other, as is the case with any state function.

Clausius defined, around 1865, the new state function, which he named entropy, for the case of non-isothermal processes as
$$\mathrm{d} S \equiv \frac{\delta Q_{\mathrm{rev}}}{T} .$$
For isothermal heat transfer, this becomes
$$\Delta S \equiv \frac{Q_{\mathrm{rev}}}{T} .$$
These definitions are not just applicable to heat engines but are absolutely general. In Equations $3.4$ and $3.5$ and in the following equations, $\delta Q_{\text {rev }}$ and $Q_{\text {rev }}$ do not therefore only signify the heat provided (reversibly) to a heat engine at high temperature, but any amount of heat reversibly exchanged between any closed system and its environment.

## 物理代写|热力学代写Thermodynamics代考|What Is the Second Law of Thermodynamics?

$$Q+W_{\mathrm{s}}+Q_{0}=0 \quad \text { or } \quad\left|W_{\mathrm{s}}\right|=Q-\left|Q_{0}\right| .$$

$$\eta_{\mathrm{th}} \equiv \frac{\left|W_{\mathrm{s}}\right|}{Q}=1-\frac{\left|Q_{0}\right|}{Q} .$$

## 物理代写|热力学代写Thermodynamics代考|How Did the Need for a New State Function Arise?

$$\frac{\left|Q_{0 \mathrm{rev}}\right|}{Q_{\mathrm{rev}}}=\frac{T_{0}}{T} \quad \text { or } \quad \frac{\left|Q_{0 \mathrm{rev}}\right|}{T_{0}}=\frac{Q_{\mathrm{rev}}}{T} .$$

$$\mathrm{d} S \equiv \frac{\delta Q_{\mathrm{rev}}}{T} .$$

$$\Delta S \equiv \frac{Q_{\mathrm{rev}}}{T} .$$

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