# 物理代写|热力学代写thermodynamics代考|NEM2201

## 物理代写|热力学代写thermodynamics代考|THERMODYNAMIC OR Absolute Temperature SCale

To obtain a temperature scale, one simply measures the efficiency of a reversible engine (hence $Q_2 / Q_1$ ) between two chosen heat reservoirs. Different choices of a judicial system and various assignments of its temperature result in different temperature scales. As an example, for the Kelvin temperature scale, we keep the triple point of water as a standard reservoir and assign a temperature of $273.16 \mathrm{~K}$ for the triple point of water. By measuring the ratio of the heat transfer between a reservoir at temperature $T$ to a reservoir at the triple point of water where we have assigned a temperature of $273.16 \mathrm{~K}$, the temperature of the reservoir can be written as
$$T=273.16\left(\frac{Q}{Q_{t . p}}\right)$$
We have replaced $\theta$ by $T$ in the above for the absolute temperature. $Q$ is the heat transfer from reservoir at temperature $T$ while $Q_{t, p}$ is from the reservoir at the triple point temperature of $273.16 \mathrm{~K}$. The advantage of the thermodynamic temperature scale is that it is independent of the type of reversible engine and the working fluids used in it.

## 物理代写|热力学代写thermodynamics代考|CLAUSIUS INEQUALITY

Clausius inequality states that for a system undergoing a cycle and delivering work $W s$ while exchanging heat from various reservoirs at different temperatures, the sum of the heat transfer divided by the temperature of the reservoir over the cycle of operation is negative or zero, that is,
$$\sum_{\substack{\text { cycle; } \ i=1,2 \ldots}} \frac{Q_i}{T_i} \leq 0$$
The cyclic system with the interacting reservoirs at temperatures $T_1, T_2, \ldots, T_k$ is illustrated in Figure 4.9.

To prove Clausius inequality, we first arrange to have the system exchanging heat with only one heat reservoir at temperature $T_0$ using reversible engines and reversible heat pumps as indicated in Figure 4.10. Here, the total system within the boundary includes the original system $S$ exchanging heat with reservoirs at temperatures $T_1, T_2, \ldots, T_i, T_j$ and $T_k$ and doing work $W_S$ and the reservoirs at temperatures $T_1, T_2, \ldots, T_i, T_j$ and $T_k$ exchanging heat with only a single reservoir at temperature $T_0$ through reversible heat engines and heat pumps.

Applying the first law to the system within the boundary, we get the total work
$$W_T=\sum_{i=1,2,-} W_i+W_s=\sum_{\substack{\text { Cyclic: } \ i=1,2,-}} Q_{\mathrm{oi}}$$
Here, $\sum W_i$ is the sum of the work done by reversible engines and on the heat pumps (all $R_i$ ‘s viz., $R_1, R_2, \ldots, R_i, R_j . R_k$, ) and $W_S$ is the work done by the system. $\sum Q_{\mathrm{oi}}$ is the net heat exchanges with the reservoir at $T_0$.

For the reversible engines or the reversible heat pumps, we have from the equality of ratios of heat transfer and temperatures
$$\frac{Q_{\mathrm{oi}}}{Q_i}=\frac{T_0}{T_i}$$
where $Q_i$ represents the heat transfers between the system and the reversible engine or pump at temperature $T_i$ as shown in rigure 4.10. We get from the above equation
$$Q_{\mathrm{oi}}=T_0 \frac{Q_i}{T_i}$$

# 热力学代考

## 物理代写|热力学代写thermodynamics代考|THERMODYNAMIC OR Absolute Temperature SCale

$$T=273.16\left(\frac{Q}{Q_{t . p}}\right)$$

## 物理代写|热力学代写thermodynamics代考|CLAUSIUS INEQUALITY

$$\sum_{\text {cycle; } i=1,2 \ldots} \frac{Q_i}{T_i} \leq 0$$

$$W_T=\sum_{i=1,2,-} W_i+W_s=\sum_{\text {Cyclic: }} Q_{i=1,2,-} Q_{\mathrm{oi}}$$

$$\frac{Q_{\mathrm{oi}}}{Q_i}=\frac{T_0}{T_i}$$

$$Q_{\mathrm{oi}}=T_0 \frac{Q_i}{T_i}$$

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