The First Law: Energy Analysis of Closed Systems第一定律：封閉系統的能量分析

What you'll be able to do本章學習成果

• Start from conservation of energy and write the energy balance for a closed system.由能量守恆出發，寫出封閉系統的能量平衡式。
• Identify the parts of total energy ($U$, $KE$, $PE$) and the only two ways energy can cross a closed boundary (heat and work).辨識總能量的組成（$U$、$KE$、$PE$），以及能量穿越封閉邊界的僅有兩種方式（熱與功）。
• Move between the finite, differential and rate forms of the balance.在能量平衡的有限差分式、微分式與速率式之間轉換。
• Reduce the balance for adiabatic, steady-state and cyclic systems.將能量平衡化簡至絕熱、穩態與循環系統。
• Derive moving-boundary work $\int p\,dV$, polytropic work, and see why enthalpy appears at constant pressure.推導移動邊界功 $\int p\,dV$ 與多變功，並了解為何定壓下會出現焓。

Key equations重要公式

Notation符號說明

Conservation of energy for a closed system封閉系統的能量守恆

A first look: Joule's free-expansion experiment初探：焦耳自由膨脹實驗

Joule connected a tank of compressed air to an evacuated tank, both immersed in a water bath, then opened the valve. The air expanded into the vacuum.焦耳將一個裝有壓縮空氣的容器與一個抽真空的容器相連，兩者皆浸於水浴中，然後打開閥門，空氣膨脹進入真空。

• $W = 0$: the air pushed against nothing.$W = 0$：空氣沒有推動任何東西。
• $Q = 0$: the bath temperature did not change.$Q = 0$：水浴溫度不變。
• So $\Delta U = Q - W = 0$: the stored energy cannot change.故 $\Delta U = Q - W = 0$：儲存的能量不能改變。
• Yet $V$ and $p$ changed a lot while $T$ stayed the same, so $u$ depends on $T$ only (for an ideal gas).然而 $V$ 與 $p$ 大幅改變，$T$ 卻不變——因此（理想氣體的）$u$ 僅與 $T$ 有關。

Caveat: Joule's water bath was too large to detect small temperature changes. Precise later experiments (Joule–Thomson) show real gases do cool slightly on expansion, because their attractive forces are not zero. The effect fades as density drops, which is exactly the regime where the ideal-gas model applies.附註：焦耳的水浴熱容量過大，無法測出微小溫度變化。後來較精密的實驗（焦耳–湯姆森）顯示，真實氣體膨脹時確實會略微降溫，因為其分子間吸引力不為零。此效應隨密度降低而消失——正是理想氣體模型適用的範圍。

The key idea: energy can enter or leave only as heat or work. Set both to zero and the stored energy cannot change. The rest of this chapter turns that idea into a general law.關鍵概念：能量只能以熱或功的形式進出。兩者皆為零，儲存的能量就不能改變。本章其餘部分將此概念發展為一般定律。

• Closed system: a fixed quantity of matter. No mass crosses its boundary.封閉系統：固定量的物質，沒有質量穿越其邊界。
• The boundary may move or deform, as with a piston.邊界可以移動或變形，例如活塞。
• Energy is conserved: it can be neither created nor destroyed.能量守恆：既不能被創造，也不能被消滅。

So between state 1 and state 2, the system's energy can change only by what passes through its boundary:因此在狀態 1 與狀態 2 之間，系統能量的變化只能來自穿越邊界的能量：

This is bookkeeping, like a bank account: change in balance = deposits − withdrawals. Two questions remain:這是一種記帳關係，如同銀行帳戶：餘額變化 = 存入 − 提出。尚需回答兩個問題：

• What energy does the system store? → $E$, the left-hand side (next section).系統儲存什麼能量？→ $E$，等號左邊（下一節）。
• How can energy cross the boundary? → $E_{in} - E_{out}$, the right-hand side (the section after).能量如何穿越邊界？→ $E_{in} - E_{out}$，等號右邊（再下一節）。

What the system stores: total energy系統儲存的能量：總能量

Total energy has two parts:總能量分為兩部分：

• Macroscopic: motion and position of the system as a whole, $KE$ and $PE$.巨觀：系統整體的運動與位置——$KE$ 與 $PE$。
• Microscopic: energy of its molecules, the internal energy $U$.微觀：分子層級的能量——內能 $U$。

Kinetic energy $KE$動能 $KE$

Push a mass $m$ with a net force $F$ along a path $x$. Newton's second law gives $F = m\,d\mathcal{V}/dt$, where $\mathcal{V}$ is the speed of the mass, a scalar, written in script to distinguish it from volume $V$. The work done on the mass over a displacement $dx = \mathcal{V}\,dt$ is以淨力 $F$ 沿路徑 $x$ 推動質量 $m$。由牛頓第二定律 $F = m\,d\mathcal{V}/dt$（$\mathcal{V}$ 為質量的速率，屬純量；以草寫體區別於體積 $V$）。在位移 $dx = \mathcal{V}\,dt$ 上對質量所做的功為

Integrating from speed $\mathcal{V}_1$ to $\mathcal{V}_2$ gives the energy the motion stores:由速率 $\mathcal{V}_1$ 積分至 $\mathcal{V}_2$，即得運動所儲存的能量：

Potential energy $PE$位能 $PE$

Lift the mass slowly, with no acceleration, against gravity. The applied force just balances the weight, $F = mg$, and moves the mass through $dz$:緩慢（無加速度）地抵抗重力舉起質量。施力恰好平衡重量 $F = mg$，使其移動 $dz$：

Internal energy $U$內能 $U$

Everything else is internal energy:其餘全歸為內能：

• Molecular kinetic energy: translation, rotation, vibration.分子動能：移動、轉動、振動。
• Molecular potential energy: intermolecular forces, chemical bonds, nuclear binding.分子位能：分子間作用力、化學鍵、原子核結合。
• Depends only on the state ($T$, $p$, phase, …), not on the system's position or speed.僅與狀態有關（$T$、$p$、相態…），與系統的位置或速度無關。
• In practice, only the sensible and latent parts change in most engineering problems.實務上，多數工程問題中只有顯熱與潛熱部分會變化。

Adding the three parts, the change of stored energy between states 1 and 2 is將三者相加，狀態 1 至 2 之間的儲存能量變化為

How energy crosses the boundary for a closed system: heat and work能量如何穿越邊界：熱與功

In general, energy can cross a boundary in three ways:一般而言，能量可透過三種方式穿越邊界：

• Mass flow: carried by matter crossing the boundary. Not possible in a closed system.質量流：隨穿越邊界的物質攜帶。封閉系統不可能發生。
• Heat $Q$熱 $Q$
• Work $W$功 $W$

Why only two for a closed system? Energy crossing without mass is either driven by a temperature difference (heat) or it is not (work). The split is exhaustive by definition, so there is no third kind.封閉系統為何只有兩種？不靠質量而穿越邊界的能量，要嘛由溫差驅動（熱），要嘛不是（功）。此分類在定義上即已窮盡，不存在第三種。

Heat $Q$熱 $Q$

• Definition: energy transferred because of a temperature difference.定義：因溫差而傳遞的能量。
• Mechanisms: conduction, convection, radiation.方式：傳導、對流、輻射。
• Direction: hot → cold; stops when temperatures equalise.方向：高溫 → 低溫；溫度相等時停止。
• Adiabatic: a process with $Q = 0$, from insulation or no temperature difference.絕熱：$Q = 0$ 的過程（因隔熱或無溫差）。

A precise definition of heat. "Energy moved by a temperature difference" is intuitive, but it doesn't tell us how to measure $Q$. Joule's experiments give an operational definition using only work, which we can measure mechanically:熱的精確定義。「因溫差而移動的能量」雖直觀，卻未說明如何量測 $Q$。焦耳的實驗只用可由力學量測的功，給出可操作的定義：

• Adiabatic work between two given states is the same for every method: paddle wheel, electric heater or compression.兩給定狀態間的絕熱功，無論以何種方式皆相同：攪拌輪、電熱器或壓縮。
• Path-independent, so it measures a property change: $\Delta E = -W_{ad}$.與路徑無關，故量度一性質變化：$\Delta E = -W_{ad}$。
• For a non-adiabatic process the work differs. Heat is defined as the difference: $Q \equiv \Delta E + W$.非絕熱過程的功不同，熱即定義為其差：$Q \equiv \Delta E + W$。

Work $W$功 $W$

• Definition: energy transfer not driven by a temperature difference. In mechanics, $\int F\,dx$.定義：非由溫差驅動的能量傳遞。力學中為 $\int F\,dx$。
• Test: an interaction is work if its sole effect on the surroundings could be the raising of a weight. Nothing has to be lifted in practice; it only has to be possible in principle. Example: current from a battery-and-resistor system is electrical work, because the same current could instead drive an ideal motor winding a weight up a pulley, with nothing else changing outside the system. Heat fails this test: energy leaving as heat cannot be turned entirely into lifting a weight without other changes in the surroundings (the Second Law).判定法則：若某互動對外界的唯一效果可以是舉起一重物，即為功。實際上不必真的舉起，只需原理上可行。例：電池與電阻構成的系統輸出電流屬電功，因同一電流可改為驅動理想馬達，經滑輪舉起重物，而系統外別無其他變化。熱不符合此判定：以熱離開的能量無法在外界不留其他變化的情況下全部用於舉起重物（第二定律）。
• Common forms: moving boundary $p\,dV$, shaft $T\omega$, electrical $\mathcal{E} I$, spring $\tfrac12 k_s\left(x_2^2 - x_1^2\right)$.常見形式：移動邊界 $p\,dV$、軸功 $T\omega$、電功 $\mathcal{E} I$、彈簧功 $\tfrac12 k_s\left(x_2^2 - x_1^2\right)$。

Sign convention符號規定

This course uses the engineering convention, which mirrors a heat engine (heat in, work out):本課程採工程慣例，對應熱機的「熱進、功出」：

• $Q > 0$: heat enters the system. $Q < 0$: heat leaves.$Q > 0$：熱進入系統。$Q < 0$：熱離開。
• $W > 0$: work is done by the system on its surroundings. $W < 0$: work is done on the system.$W > 0$：系統對外界做功。$W < 0$：外界對系統做功。

So the net energy transferred into the system is因此進入系統的淨能量為

The closed-system energy balance封閉系統能量平衡

Finite (difference) form有限差分式

Substitute the stored energy from the left-hand side and the net transfer from the right-hand side into the conservation statement:將等號左邊的儲存能量與右邊的淨傳遞能量代入守恆式：

• $Q$ and $W$ are the totals transferred between states 1 and 2.$Q$、$W$ 為狀態 1 至 2 之間傳遞的總量。
• Per unit mass: $\Delta u + \Delta ke + \Delta pe = q - w$.每單位質量：$\Delta u + \Delta ke + \Delta pe = q - w$。
• Stationary system ($\Delta KE = \Delta PE = 0$, the usual case): $\Delta U = Q - W$.靜止系統（$\Delta KE = \Delta PE = 0$，最常見）：$\Delta U = Q - W$。

Differential form微分式

Shrink the interval to an infinitesimal change of state:將時段縮小至狀態僅有無限小變化：

• Property change → exact differential $dE$.性質變化 → 恰當微分 $dE$。
• Transfers → small path-dependent amounts $\delta Q$, $\delta W$.傳遞量 → 與路徑有關的微小量 $\delta Q$、$\delta W$。

Integrating it from 1 to 2 recovers the finite form: $\int_1^2 dE = E_2 - E_1$, while $\int_1^2 \delta Q = Q_{12}$ and $\int_1^2 \delta W = W_{12}$.由 1 積分至 2 即回到有限式：$\int_1^2 dE = E_2 - E_1$，而 $\int_1^2 \delta Q = Q_{12}$、$\int_1^2 \delta W = W_{12}$。

Rate form速率式

If the change happens over a time $dt$, divide the differential form by $dt$ and let $dt \to 0$. The heat-transfer rate $\dot Q = \delta Q/dt$ (W) and the power $\dot W = \delta W/dt$ (W) appear:若變化發生於時間 $dt$ 內，將微分式除以 $dt$ 並令 $dt \to 0$，即得熱傳率 $\dot Q = \delta Q/dt$（W）與功率 $\dot W = \delta W/dt$（W）：

Use the rate form when a problem gives power or heating in watts, or asks how long a process takes.使用速率式的時機：題目以瓦特給出功率或加熱率，或問過程需時多久。

Adiabatic systems絕熱系統

Put $Q = 0$ (or $\delta Q = 0$) into the balance:將 $Q = 0$（或 $\delta Q = 0$）代入能量平衡：

• All the work shows up as a change in stored energy.所有功皆轉為儲存能量的變化。
• Work in ($W < 0$) raises $U$: compression heats a gas.對系統做功（$W < 0$）使 $U$ 增加：壓縮使氣體升溫。
• Work out ($W > 0$) comes entirely from $U$: expansion cools a gas.對外做功（$W > 0$）完全取自 $U$：膨脹使氣體降溫。
• Path-independent: adiabatic work between two given states is fixed, the fact used to define heat.與路徑無關：兩給定狀態間的絕熱功為定值——這正是定義熱所用的事實。

Steady-state systems and cycles穩態系統與循環

Steady state穩態

• Steady state: no property changes with time, while heat and work keep flowing.穩態：所有性質皆不隨時間改變，但熱與功持續流動。
• Example: an electric motor at constant speed and temperature.例：以定速、定溫運轉的電動機。
• $E$ is constant, so $dE/dt = 0$:$E$ 為常數，故 $dE/dt = 0$：

Every watt of work out must be matched by a watt of heat in. For the motor, the electrical power in (negative $\dot W$) leaves as shaft power plus heat to the room.每一瓦的輸出功率必須由一瓦的輸入熱平衡。以電動機為例，輸入的電功率（負的 $\dot W$）以軸功率加上排至室內的熱離開。

Cycles循環

• Cycle: the system returns to its initial state.循環：系統回到初始狀態。
• Not steady at every instant, but over one full cycle $E_2 = E_1$, so $\oint dE = 0$.不必每一瞬間皆為穩態，但一個完整循環後 $E_2 = E_1$，故 $\oint dE = 0$。
• Integrate the differential form around the cycle:將微分式沿循環積分：

A power cycle takes in $Q_{in}$, rejects $Q_{out}$ and delivers $W_{cycle} = Q_{in} - Q_{out}$. Its thermal efficiency follows directly:動力循環吸收 $Q_{in}$、排出 $Q_{out}$，輸出 $W_{cycle} = Q_{in} - Q_{out}$。其熱效率直接得出：

Moving boundary: boundary work移動邊界：邊界功

The most common work in a closed system comes from its boundary moving, such as gas in a piston–cylinder. The gas exerts force $F = pA$ on a piston of area $A$. When the piston moves outward by $dx$, the gas does work封閉系統中最常見的功來自邊界移動——例如活塞–汽缸內的氣體。氣體對面積 $A$ 的活塞施力 $F = pA$。活塞向外移動 $dx$ 時，氣體做功

using $A\,dx = dV$.其中 $A\,dx = dV$。

• Condition: the process must be quasi-equilibrium, slow enough that $p$ is uniform throughout the gas.條件：過程須為準平衡——足夠慢，使氣體各處壓力均勻。

Integrating between states:在兩狀態間積分：

Substituting into the stationary energy balance gives a working equation for piston–cylinder problems:代入靜止系統的能量平衡，即得活塞–汽缸問題的實用式：

Constant pressure: why enthalpy appears定壓過程：為何出現焓

If the piston is free and loaded by a constant weight, $p$ is constant, so $W_{12} = p\,(V_2 - V_1)$. Put this into the balance and group terms by state:若活塞可自由移動並承受固定重量，$p$ 為常數，故 $W_{12} = p\,(V_2 - V_1)$。代入能量平衡並依狀態整理：

• Enthalpy: $H \equiv U + pV$. The combination appears so often it gets its own name.焓：$H \equiv U + pV$。此組合出現頻繁，因而另立名稱。
• At constant pressure, heat added = enthalpy change:定壓下，加入的熱 = 焓變化：

Types of processes過程的種類

From the energy law to processes由能量定律到過程

The balance $\Delta U = Q - W$ holds for every closed-system change, but on its own it cannot be solved:能量平衡 $\Delta U = Q - W$ 適用於任何封閉系統變化，但單憑它無法求解：

• $\Delta U$: fixed by the end states.$\Delta U$：由端點狀態決定。
• $W = \int p\,dV$: needs the whole path $p(V)$.$W = \int p\,dV$：需要完整路徑 $p(V)$。
• $Q$: whatever is left over.$Q$：剩下的差額。

So we must say how the system gets from state 1 to state 2.因此必須說明系統如何由狀態 1 到達狀態 2。

That description is a process:這段描述就是過程：

• Named by the one quantity held fixed: volume, pressure, temperature, or heat.命名依被固定的量：體積、壓力、溫度或熱。
• Supplies the missing path $p(V)$.提供缺少的路徑 $p(V)$。
• Usually zeroes one term of the balance.通常使能量平衡中某一項為零。
• Assumed below: stationary ideal gas, fixed mass $m$, $pV = mRT$, $\Delta U = m c_v \Delta T$.以下假設：靜止理想氣體，固定質量 $m$，$pV = mRT$，$\Delta U = m c_v \Delta T$。

1 · Isochoric (isometric, isovolumetric): $V$ = const1 · 等容（定容）過程：$V$ = 常數

A rigid tank. On a p–V diagram it is a vertical line. Since $dV = 0$ everywhere along the path:剛性容器。在 p–V 圖上為一條鉛直線。因沿途 $dV = 0$：

All the heat goes into internal energy. From $pV = mRT$ with $V$ fixed, $p_2/p_1 = T_2/T_1$.所有熱皆轉為內能。由 $pV = mRT$ 且 $V$ 固定，得 $p_2/p_1 = T_2/T_1$。

2 · Isobaric: $p$ = const2 · 等壓過程：$p$ = 常數

A freely sliding piston under a fixed load. It is a horizontal line. Take $p$ outside the integral and use $pV = mRT$:承受固定負載、可自由滑動的活塞。為一條水平線。將 $p$ 提出積分並利用 $pV = mRT$：

Substitute into the energy balance and use $c_p = c_v + R$:代入能量平衡並利用 $c_p = c_v + R$：

• Part of the heat raises $U$; the rest, $mR\,\Delta T$, pushes the piston. This is why $c_p > c_v$.部分熱使 $U$ 增加，其餘 $mR\,\Delta T$ 推動活塞——這就是 $c_p > c_v$ 的原因。
• $V_2/V_1 = T_2/T_1$.$V_2/V_1 = T_2/T_1$。

3 · Isothermal: $T$ = const3 · 等溫過程：$T$ = 常數

Slow expansion in good thermal contact with a reservoir. For an ideal gas $pV = mRT$ = const, a hyperbola. Substitute $p = mRT/V$ and integrate:在與熱庫良好接觸下緩慢膨脹。理想氣體 $pV = mRT$ = 常數，為一雙曲線。代入 $p = mRT/V$ 並積分：

Because $u$ depends only on $T$, $\Delta U = 0$ and the balance gives由於 $u$ 僅與 $T$ 有關，$\Delta U = 0$，能量平衡給出

Every joule of work done by the gas is replaced by a joule of heat from the reservoir.氣體所做的每一焦耳功，都由熱庫補充一焦耳的熱。

4 · Adiabatic (reversible): $Q = 0$4 · 絕熱（可逆）過程：$Q = 0$

An insulated cylinder, or a process too fast for heat to flow. Here the constraint is on $Q$, not on a property, so we must derive the path. With $\delta Q = 0$ the differential balance gives $dU = -\delta W$:隔熱汽缸，或進行過快而來不及傳熱的過程。此時限制的是 $Q$ 而非某一性質，因此必須推導路徑。由 $\delta Q = 0$，微分式能量平衡給出 $dU = -\delta W$：

With $k = c_p/c_v$, $R/c_v = (c_p - c_v)/c_v = k - 1$. Integrate, taking $k$ constant:令 $k = c_p/c_v$，則 $R/c_v = (c_p - c_v)/c_v = k - 1$。以 $k$ 為常數積分：

Replace $T$ with $pV/(mR)$ to get the path on a p–V diagram:以 $T = pV/(mR)$ 代換，得 p–V 圖上的路徑：

The work comes entirely from internal energy, and the gas cools as it expands:功完全取自內能，氣體膨脹時降溫：

The last step uses $m c_v T = pV/(k-1)$. The curve is steeper than the isotherm because the falling temperature drops the pressure further.最後一步利用 $m c_v T = pV/(k-1)$。此曲線比等溫線陡，因為溫度下降使壓力進一步降低。

5 · Polytropic: $pV^n$ = const5 · 多變過程：$pV^n$ = 常數

Real compressors and engines usually lose some heat, but not all of it, so they fall between isothermal and adiabatic. They are fitted with $pV^n = C$ for a measured exponent $n$. Write $p = C\,V^{-n}$ and integrate:實際的壓縮機與引擎通常會散失部分（但非全部）的熱，介於等溫與絕熱之間，以量測的指數 $n$ 配合 $pV^n = C$ 描述。寫成 $p = C\,V^{-n}$ 並積分：

This uses $C\,V_2^{\,1-n} = p_2V_2$ and $C\,V_1^{\,1-n} = p_1V_1$. At $n = 1$ the integrand is $C/V$, which gives the isothermal logarithm. The heat follows from $Q = \Delta U + W$ with $\Delta U = (p_2V_2 - p_1V_1)/(k-1)$:此處利用 $C\,V_2^{\,1-n} = p_2V_2$、$C\,V_1^{\,1-n} = p_1V_1$。$n = 1$ 時被積函數為 $C/V$，即得等溫對數式。熱量由 $Q = \Delta U + W$ 及 $\Delta U = (p_2V_2 - p_1V_1)/(k-1)$ 求得：

Every other process is a special case:其他過程皆為特例：

• $n = 0$ → isobaric ($p$ = const)$n = 0$ → 等壓（$p$ = 常數）
• $n = 1$ → isothermal ($T$ = const)$n = 1$ → 等溫（$T$ = 常數）
• $n = k$ → reversible adiabatic ($Q = 0$)$n = k$ → 可逆絕熱（$Q = 0$）
• $n \to \infty$ → isochoric ($V = C^{1/n}p^{-1/n} \to$ const)$n \to \infty$ → 等容（$V = C^{1/n}p^{-1/n} \to$ 常數）

Summary for an ideal gas理想氣體總整理

Now try them in the explorer below. Pick a process, set the volume ratio, and compare the shaded work areas.接著在下方探索器中實際操作：選擇過程、設定體積比，比較陰影所示的功。

p–V work explorerp–V 功探索器

The explorer works per unit mass of air, so it uses specific volume $v$ (m³/kg) and gives specific work $w$ and heat $q$ (kJ/kg). Pick a process, set the initial pressure and specific-volume ratio, and watch the work appear as the shaded area. Compare how much work the four processes deliver for the same volume change.探索器以每單位質量空氣計算，因此使用比容 $v$（m³/kg），並給出比功 $w$ 與比熱傳 $q$（kJ/kg）。選擇過程、設定初始壓力與比容比，觀察陰影面積所代表的功。比較四種過程在相同比容變化下分別做了多少功。

Specific heats from the energy balance由能量平衡得比熱

Problem statement問題敘述

A "specific heat" is the heat required per unit mass per unit temperature rise, $\delta q/dT$. Since $\delta q$ is path-dependent, this ratio has no single value: we must say which path. The energy balance tells us what it is for the two simplest paths.「比熱」為每單位質量、每升高單位溫度所需的熱，$\delta q/dT$。由於 $\delta q$ 與路徑有關，此比值沒有唯一值——必須指明路徑。能量平衡可告訴我們兩種最簡單路徑下的結果。

Case 1 — Heating at constant volume情況 1——定容加熱

We define the right-hand side as the constant-volume specific heat:將右邊定義為定容比熱：

Case 2 — Heating at constant pressure情況 2——定壓加熱

This defines the constant-pressure specific heat:由此定義定壓比熱：

Case 3 — Ideal gas情況 3——理想氣體

Answer to the question. At constant volume $\delta q = c_v\,dT$, and at constant pressure $\delta q = c_p\,dT$. For air, $c_p = 1.005$ and $c_v = 0.718$ kJ/kg·K, so heating at constant pressure needs about 40% more heat for the same temperature rise. The extra $R = 0.287$ kJ/kg·K is the boundary work done pushing the piston. Why $u$ depends on $T$ alone is explained in Equations of State.問題的答案。定容時 $\delta q = c_v\,dT$；定壓時 $\delta q = c_p\,dT$。以空氣為例，$c_p = 1.005$、$c_v = 0.718$ kJ/kg·K——相同溫升下，定壓加熱約需多 40% 的熱；多出的 $R = 0.287$ kJ/kg·K 即為推動活塞所做的邊界功。$u$ 為何僅與 $T$ 有關，詳見狀態方程式。

Applying the closed-system energy balance封閉系統能量平衡的應用

Practice problems練習題

Work each one yourself first, then reveal the solution step by step.先自行嘗試，再逐步顯示解答。