19th Ave New York, NY 95822, USA

# 物理代写|热力学代写Thermodynamics代考|JNE239 Wiedemann-Franz’ Law

my-assignmentexpert™提供最专业的一站式服务：Essay代写，Dissertation代写，Assignment代写，Paper代写，Proposal代写，Proposal代写，Literature Review代写，Online Course，Exam代考等等。my-assignmentexpert™专注为留学生提供Essay代写服务，拥有各个专业的博硕教师团队帮您代写，免费修改及辅导，保证成果完成的效率和质量。同时有多家检测平台帐号，包括Turnitin高级账户，检测论文不会留痕，写好后检测修改，放心可靠，经得起任何考验！

## 物理代写|热力学代写Thermodynamics代考|Wiedemann-Franz’ Law

What if both $I_S, I_e, \Delta T$ and $V_{e m f}$ are $\neq 0$ ? Two distinct, irreversible processes occur simultaneously: heat conduction and Joule heating. We denote the length and the cross section of the wire with $l_{\text {wire }}$ and $A_{\text {wire }}$ respectively. Both $I_S$ and $I_e$ are time derivatives of physical quantities, hence both are eligible as thermodynamic currents. According to the notation above, we write $J_1=I_S$ and $J_2=I_e$. The corresponding thermodynamic forces are $X_1=\frac{\Delta T}{T}$ and $X_2=\frac{V}{T}$; both vanish at thermodynamic equilibrium. Our model of heat transport is linearized, as $\Delta T \ll T$. If $I_e$ then we define the thermal conductivity $\chi$ such that $Q=\chi \frac{A_{\text {wire }}}{l_{\text {wire }}} \Delta T^{16}$; thus, $I_s=L_{11} \frac{\Delta T}{T}$.
Since $\left(\frac{d S}{d t}\right){\text {produced in the wire }}>0, \chi>0$. In copper, the relationship between $V{e m f}$ and $I_e$ is linear. If $I_S=0$ then we define the electrical conductivity $\sigma_{\Omega}$ such that $V_{e m f}=R_{\Omega} I_e$ where the electrical resistance $R_{\Omega}$ is equal to $R_{\Omega}=\frac{l_{\text {wire }}}{A_{\text {wire }}} \sigma_{\Omega}$. Thus, $I_e=L_{22} \frac{V_{\text {emf }}}{T}$ where $L_{22}=T \sigma_{\Omega} \frac{A_{\text {wire }}}{l_{\text {wire }}}$. Since $\left(\frac{d S}{d t}\right){\text {produced in the wire }}>0, \sigma{\Omega}>0$.

In the general case, $J_i=L_{i j} X_j(i, j=1,2)$ with $L_{12}=T\left(\frac{\partial I_S}{\partial V_{e m f}}\right){\Delta T \rightarrow 0}$ and $L{21}=T\left(\frac{\partial I_e}{\partial \Delta T}\right){V{e m f} \rightarrow 0}$. In order to understand if $L_{12}=+L_{21}$ or $L_{12}=-L_{21}$, we investigate the behaviour of the quantities which $I_S$ and $I_e$ are time derivatives of under the transformation $\mathbf{v} \rightarrow-\mathbf{v}^{17}$

As for $I_e$, it is the time derivative of the electric charge, which does not depend on $\mathbf{v}$ and therefore does not change sign. As for $I_S$, it is equal to $-\frac{d S_1}{d t}$, where $S_1$ is the entropy of the reservoir at temperature $T_1=T$. When applied to this heat reservoir, ${ }^{18}$ the First Principle of thermodynamics reads $d E_1=T d S_1$, and since $T=$ const. we write $I_S=$ const. $\cdot \frac{d E_1}{d t}$. Energy is quadratic in $\mathbf{v}$; hence $E_1$ does not change sign. Since the electric charge and $E_1$ behave the same way under reversal of $\mathbf{v}$, we conclude that Onsager’s symmetry $L_{12}=L_{21}$ holds and $\left(\frac{\partial I_S}{\partial V_{e m f}}\right){\Delta T \rightarrow 0}=\left(\frac{\partial I_e}{\partial \Delta T}\right){V_{e m f} \rightarrow 0}$.
It is customary to define the ‘Seebeck coefficient’ $\varepsilon \equiv\left(\frac{I_S}{I_e}\right){\Delta T \rightarrow 0}$. Since $L{12}=$ $L_{21}$, the definitions of $X_1$ and $X_2$ imply $\varepsilon=\frac{L_{12}}{L_{22}}=\frac{L_{21}}{L_{22}}=-\left(\frac{V_{e m f}}{\Delta T}\right){I_e \rightarrow 0}$. Accordingly, applied e.m.f. induce differences of temperature, and applied differences of temperature induce differences of e.m.f.. In the following, we take advantage of the fact that $d L{i j}=0$, divide the relationships concerning $I_S$ and $I_e$ by the same constant factor $L_{22}$, and introduce the new thermodynamic fluxes $I_S^{\prime} \equiv \frac{I_S}{L_{22}}$ and $I_e^{\prime} \equiv \frac{I_e}{L_{22}}$. In contrast, we leave the thermodynamic forces unchanged. Onsager’s symmetry and the definition of $\varepsilon$ give $^{19}$ :
$$I_S^{\prime}=\left(\frac{\chi}{T \sigma_{\Omega}}\right)\left(\frac{\Delta T}{T}\right)+\varepsilon \frac{V_{e m f}}{T} \quad ; \quad I_e^{\prime}=\varepsilon \frac{\Delta T}{T}+\frac{V_{e m f}}{T}$$
Now, with this new, perfectly allowable set of thermodynamic fluxes the Onsager coefficient $L_{11}$ is equal to $L_{11}=\frac{\chi}{T \sigma_{\Omega}}$. Since $L_{11}$ is constant, we obtain WiedemannFranz’ law:
$$\chi \propto T \cdot \sigma_{\Omega}$$

## 物理代写|热力学代写Thermodynamics代考|Seebeck Effect

The relationships below hold at LTE. We denote with $\phi_g$ the gravitational potential, which affects chemical potentials as follows ${ }^{25}$
$$\mu_k \rightarrow \mu_k+m_k \phi_g \text { hence } \mu_k^0 \rightarrow \mu_k^0+\phi_g$$
Since $\phi_g$ depends neither on $p$, nor on $T$, nor on $c_k$, it leaves the corresponding derivatives of $h$ and $\mu_k^0$ unaffected. Moreover, ${ }^{26}$ the relationship $g=\mu_k^0 c_k$ implies that gravity makes us to replace $g$ as follows: $g=\mu_k^0 c_k \rightarrow \mu_k^0 c_k+\phi_g \sum_k c_k=g+\phi_g$. Analogously, $h \rightarrow h+\phi_g$ and $u \rightarrow u+\phi_g$, in agreement with the theorem of small corrections. In other words, Gibbs’ free energy per unit mass, enthalpy per unit mass and internal energy per unit mass transform like chemical potentials per unit mass. In particular, gravity leaves the relationship $h=u+\frac{p}{p}$ unaffected, because it adds the same term $\phi_g$ to $h$ and $u$. Finally, the relationship $h \rightarrow h+\phi_g$ implies $h_k \rightarrow h_k+\phi_g$ as no species is privileged. But we have shown that $h_k=\mu_k^0+T s_k$ and that $\mu_k^0 \rightarrow \mu_k^0+\phi_g$, hence $\phi_g$ leaves $s_k$ unaffected. In other words, gravity leaves entropy unaffected.

Term-by-term subtraction provides us with the net e.m.f.:
$$V_{A B}=V_{I-I I(A)}-V_{I-I I(B)}=\left(\varepsilon_A-\varepsilon_B\right)\left(T_I-T_{I I}\right)$$
For finite temperature jump, Seebeck’s law follows:
$$V_{A B}=\int_{T_I}^{T_{I I}} d T\left(\varepsilon_A-\varepsilon_B\right)$$

## 物理代写热力学代写THERMODYNAMICS代考|WIEDEMANNFRANZ’ LAW

$$I_S^{\prime}=\left(\frac{\chi}{T \sigma_{\Omega}}\right)\left(\frac{\Delta T}{T}\right)+\varepsilon \frac{V_{e m f}}{T} \quad ; \quad I_e^{\prime}=\varepsilon \frac{\Delta T}{T}+\frac{V_{e m f}}{T}$$

$$\chi \propto T \cdot \sigma_{\Omega}$$

## 物理代写热力学代写THERMODYNAMICS代考|SEEBECK EFFECT

$$\mu_k \rightarrow \mu_k+m_k \phi_g \text { hence } \mu_k^0 \rightarrow \mu_k^0+\phi_g$$

$g=\mu_k^0 c_k \rightarrow \mu_k^0 c_k+\phi_g \sum_k c_k=g+\phi_g$. 类似地， $h \rightarrow h+\phi_g$ 和 $u \rightarrow u+\phi_g$ ，与小修正定理一數。换句话说，每单位质量的吉布斯自由能、每单位质量的焓和 每单位质量的内能像每单位质量的化学势一样变换。特别是，重力离开了关系 $h=u+\frac{p}{p}$ 不受影响，因为它添加了相同的术语 $\phi_g$ 至 $h$ 和 $u$. 最后，关系 $h \rightarrow h+\phi_g$ 暗示 $h_k \rightarrow h_k+\phi_g$ 因为没有物种字有特权。但我们已经证明 $h_k=\mu_k^0+T s_k$ 然后 $\mu_k^0 \rightarrow \mu_k^0+\phi_g$ ，因此 $\phi_g$ 树叶 $s_k$ 不受影响。换句话说，重力不影响熵。

$$V_{A B}=V_{I-I I(A)}-V_{I-I I(B)}=\left(\varepsilon_A-\varepsilon_B\right)\left(T_I-T_{I I}\right)$$

$$V_{A B}=\int_{T_I}^{T_{I I}} d T\left(\varepsilon_A-\varepsilon_B\right)$$

## Matlab代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。