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物理学|PHYS1002 Physics 1 (Fundamentals)代写 monash代写

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This course is a real-time, online course where the instructor and students meet via web conferencing tools, at scheduled days and times. Instructors and students share information, ideas and learning experiences in a virtual course environment. Participation in synchronous courses requires students to have reliable, high-speed internet access, a computer (ideally with a webcam), and a headset with a microphone.

这是一份monash悉尼大学PHYS1002的成功案例

物理学|PHYS1002 Physics 1 (Fundamentals)代写 monash代写


问题 1.

Let $Q_{B} Q_{R}, T_{R}$ and $W$, respectively, denote, the heat drawn by the refrigerator from the body at temperature $T$, out of which $Q_{R}$ is rejected in the room at temperature $T_{R}$ and a work $W$ is performed on the refrigerator. Now,
$$
Q_{R}=Q_{B}+W \text { and } Q_{B}=Q_{R}-W
$$
And the Coefficient of performance of the refrigerator, $(\mathrm{COF})=\frac{W}{Q_{R}}=\frac{\left(T_{R}-T\right)}{T_{R}}$ or
$$
Q_{R}=\frac{W T_{R}}{\left(T_{R}-T\right)}
$$


证明 .


Substituting this value of $Q_{R}$ in Eq. S-7.10.1 gives:
$$
\begin{aligned}
&Q_{B}=Q_{R}-W=\frac{W T_{R}}{\left(T_{R}-T\right)}-W=W\left(\frac{T}{\left(T_{R}-T\right)}\right) \
&W=\left(\frac{\left(T_{R}-T\right)}{T}\right) Q_{B}
\end{aligned}
$$
shows that in order to draw a quantity of heat $Q_{B}$ from the body at temperature $T$ the amount of work required to be done is $W$ whichis equal to $\left(\frac{\left(T_{R}-T\right)}{T}\right) Q_{B}$.







英国论文代写Viking Essay为您提供作业代写代考服务

PHS2061 COURSE NOTES ：

The thermal capacity (at constant volume) is defined as $C_{V} \equiv\left(\frac{\partial U}{\partial T}\right){V}$, which in the present case is $K T^{3}$. Using the relation $\left(\frac{\partial S}{\partial T}\right){V}=\frac{1}{T}\left(\frac{\partial U}{\partial T}\right){V}=\frac{K T^{3}}{T}=K T^{2}$, and hence, $$ \Delta S=\int{T_{R}}^{0} K T^{2} d T=-\frac{K}{3} T_{R}^{3}
$$
Negative sign indicates that this is the decrease in the entropy.
Change in internal energy of the body may be calculated using the relation
$$
d U=C d T \quad \text { or } \quad \Delta U=\int_{T_{R}}^{0} K T^{3} d T=-\frac{K}{4} T_{R}^{4}
$$




















物理 Physics F303

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这是一份imperial帝国理工大学 F303作业代写的成功案例

物理 Physics F303
问题 1.

$$
\frac{1}{2} m\left(v_{x}-2 u\right)^{2}-\frac{1}{2} m v_{x}^{2}=-2 m u v_{x}
$$
where we have discarded the second order term in $u$. Therefore, the change in internal energy of the ideal gas due to the collision between this class of molecules with the infinitesimal area $d A$ is
$$
-2 \operatorname{m\eta ug}\left(v_{x}\right) v_{x}^{2} d v_{x} d A d t
$$



证明 .


$$
d E=-2 m \eta d V \int_{0}^{\infty} g\left(v_{x}\right) v_{x}^{2} d v_{x}
$$
Since $\int_{0}^{\infty} g\left(v_{x}\right) v_{x}^{2} d v_{x}=\frac{1}{2}\left\langle v_{x}^{2}\right\rangle$
$$
d E=-m \eta\left\langle v_{x}^{2}\right\rangle d V=-\frac{1}{3} m \eta\left\langle v^{2}\right\rangle d V
$$

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F303 COURSE NOTES ：

$$
d S-\frac{\delta Q}{T_{e x t}} \geq 0
$$
for the total entropy change of the universe to be non-negative. Applying the first law of thermodynamics $\delta Q=d U+\delta W_{b y}$, where $d U$ is the change in internal energy of the system and $\delta W_{b y}$ is the work done by the system,
$$
d U+\delta W_{b y}-T_{e x t} d S \leq 0
$$
If the work done by the system is only that against its external environment, $\delta W_{b y}=p_{e x t} d V$ where $d V$ is the change in volume of the system. Thus,
$$
d U+p_{e x t} d V-T_{e x t} d S \leq 0
$$







物理 Physics F300

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这是一份imperial帝国理工大学 F300作业代写的成功案例

物理 Physics F300
问题 1.

$$
\begin{aligned}
\rho v+\Delta \rho c &=0, \
\mu c v+c_{p} \Delta T &=0 .
\end{aligned}
$$
$\Delta T$ can be related to $\Delta \rho$ through the adiabatic condition. Since $p^{1-\gamma} T^{\gamma}=$ constant and $\rho \propto \frac{p}{T}$ by the ideal gas law,
$$
\rho^{1-\gamma} T=c
$$



证明 .


for some constant $c$. Taking the total derivative of the above,
$$
\begin{gathered}
(1-\gamma) \rho^{-\gamma} T d \rho+\rho^{1-\gamma} d T=0 \
d T=\frac{(\gamma-1) T}{\rho} d \rho .
\end{gathered}
$$
Since $\Delta \rho$ and $\Delta T$ are small,
$$
\Delta T \approx \frac{(\gamma-1) T}{\rho} \Delta \rho
$$
Substituting this expression for $\Delta T$ and summarizing our equations,
$$
\begin{aligned}
\mu c v+\frac{c_{p}(\gamma-1) T}{\rho} \Delta \rho &=0 \
\rho v+c \Delta \rho &=0
\end{aligned}
$$

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F300 COURSE NOTES ：

$$
\begin{gathered}
\mu c^{2}-c_{p}(\gamma-1) T=0 \
c=\sqrt{\frac{c_{p}(\gamma-1) T}{\mu}} .
\end{gathered}
$$
Notice that $c_{p}(\gamma-1)=c_{p} \cdot \frac{c_{p}-c_{v}}{c_{v}}=\frac{c_{p}}{c_{v}} \cdot R=\gamma R$ and $\frac{\mu}{R T}=\frac{\rho}{p}$ by the ideal gas law such that the above becomes
$$
c=\sqrt{\frac{\gamma p}{\rho}}
$$