12.2: Heat Engine

The magnetic-dipole system we are considering is shown in Figure 12.1, where there are two environments at different temperatures, and the interaction of each with the system can be controlled by having the barriers either present or not (shown in the Figure as present). Although Figure 12.1 shows two dipoles in the system, the analysis here works with only one dipole, or with more than two, so long as there are many fewer dipoles in the system than in either environment.

Now let us rewrite the formulas from Chapter 11 with the use of \(\beta\) replaced by temperature. Thus Equations 11.8 to 11.12 become

Table 12.1: Various temperatures of interest (bp = boiling point, mp = melting point)

	K	\(^{\circ}\)C	\(^{\circ}\)F	\(^{\circ}\)R	\(k_BT = \frac{1}{\beta}\) (J)	\(\beta (J^{−1})\)
Absolute Zero	0	-273.15	-459.67	0	0	\(\infty\)
Outer Space (approx)	2.7	-270	-455	4.9	3.73 × 10\(^{−23}\)	2.68 × 10\(^{22}\)
Liquid Helium bp	4.22	-268.93	-452.07	7.6	5.83 × 10\(^{−23}\)	1.72 × 10\(^{22}\)
Liquid Nitrogen bp	77.34	-195.81	-320.46	139.2	1.07 × 10\(^{−21}\)	9.36 × 10\{^20}\)
Water mp	273.15	0.00	32.00	491.67	3.73 × 10\(^{−21}\)	2.65 × 10\(^{20}\)
Room Temperature (approx)	290	17	62	520	4.00 × 10\(^{−21}\)	2.50 × 10\(^{20}\)
Water bp	373.15	100.00	212.00	671.67	5.15 × 10\(^{−21}\)	1.94 × 10\(^{20}\)

Figure 12.1: Dipole moment example. (Each dipole can be either up or down.)

\[\begin{align*}
1 &=\sum_{i} p_{i} \tag{12.4}\\
E &=\sum_{i} p_{i} E_{i} \tag{12.5}\\
S &=k_{B} \sum_{i} p_{i} \ln \left(\frac{1}{p_{i}}\right) \tag{12.6}\\
p_{i} &=e^{-\alpha} e^{-E_{i} / k_{B} T} \tag{12.7}\\
\alpha &=\ln \left(\sum_{i} e^{-E_{i} / k_{B} T}\right) \\
&=\frac{S}{k_{B}}-\frac{E}{k_{B} T} \tag{12.8}
\end{align*} \nonumber \]

The differential formulas from Chapter 11 for the case of the dipole model where each state has an energy proportional to H, Equations 11.30 to 11.36 become

\(\begin{align*}
0 &=\sum_{i} d p_{i} \tag{12.9}\\
d E &=\sum_{i} E_{i}(H) d p_{i}+\left(\frac{E}{H}\right) d H \tag{12.10}\\
T d S &=d E-\left(\frac{E}{H}\right) d H \tag{12.11}\\
d \alpha &=\left(\frac{E}{k_{B} T}\right)\left[\left(\frac{1}{T}\right) d T-\left(\frac{1}{H}\right) d H\right] \tag{12.12}\\
d p_{i} &=p_{i}\left[\frac{E_{i}(H)-E}{k_{B} T}\right]\left[\left(\frac{1}{T}\right) d T-\left(\frac{1}{H}\right) d H\right] \tag{12.13} \\
d E &=\left[\sum_{i} p_{i}\left(E_{i}(H)-E\right)^{2}\right]\left(\frac{1}{k_{B} T}\right)\left[\left(\frac{1}{T}\right) d T-\left(\frac{1}{H}\right) d H\right]+\left(\frac{E}{H}\right) d H \tag{12.14}\\
T d S &=\left[\sum_{i} p_{i}\left(E_{i}(H)-E\right)^{2}\right]\left(\frac{1}{k_{B} T}\right)\left[\left(\frac{1}{T}\right) d T-\left(\frac{1}{H}\right) d H\right] \tag{12.15}
\end{align*}\)

and the change in energy can be attributed to the effects of work \(dw\) and heat \(dq\)

\[\begin{align*}
d w&=\left(\frac{E}{H}\right) d H \tag{12.16}\\
d q&=\sum_{i} E_{i}(H) d p_{i} \\
&=T d S \tag{12.17}
\end{align*} \nonumber \]