114 年 國立臺灣大學生物機電系碩士班乙組《熱力學》

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第 1 題

  1. For a micro-environment where dry-bulb temperature=25°C, and dew point = 15 °C, please determine:
    (a) (5%) The specific humidity in g/kg dry air of the air in the space
    (b) (5%) The relative humidity in % of the air in the space
    (c) (5%) When the direct evaporative cooling is applied in the space, how many degrees of the maximum temperature drop can be achieved?

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Simplified psychrometric chart at 1 atm total pressure

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這一題的完整詳解

本題主要考驗利用濕度圖（Psychrometric Chart）判讀空氣性質的能力，以及對蒸發冷卻原理的理解。

核心觀念：

  1. 比濕（Specific Humidity, ww）：單位質量乾空氣中所含的水蒸氣質量。
  2. 相對濕度（Relative Humidity, RH）：空氣中水蒸氣的分壓與相同溫度下飽和水蒸氣分壓的比值。
  3. 露點溫度（Dew Point Temperature）：空氣中所含水蒸氣達到飽和狀態時的溫度。
  4. 濕度圖（Psychrometric Chart）：一種圖表，用以顯示在特定總壓下，空氣的各種熱力學性質（如乾球溫度、濕球溫度、露點溫度、比濕、焓、比容等）之間的關係。
  5. 蒸發冷卻（Evaporative Cooling）：水蒸發時會吸收熱量，使周圍空氣溫度降低的過程。在理想情況下，蒸發冷卻過程近似於絕熱飽和過程（Adiabatic Saturation Process），即焓值保持不變。

解題步驟：

(a) 求比濕 (ww)

  • 已知條件：

    • 乾球溫度（Dry-bulb temperature, TdbT_{db}）= 25°C
    • 露點溫度（Dew point temperature, TdpT_{dp}）= 15°C
    • 總壓（Total pressure）= 1 atm（由濕度圖標示可知）

  • 解題方法：

    1. 在濕度圖上找到乾球溫度 25°C 的縱線。
    2. 找到露點溫度 15°C 的橫線（這條線同時也標示了飽和水蒸氣壓，以及對應的飽和比濕）。
    3. 這兩條線的交點代表了該狀態點。
    4. 從此點沿著平行於 X 軸（乾球溫度軸）的方向，讀取右側縱軸（Specific humidity, ww）的數值。
    5. 或者，更精確地，露點溫度 Tdp=15∘CT_{dp} = 15^\circ\text{C} 直接對應到飽和曲線上的某一點。在這個點，空氣是飽和的，其水蒸氣分壓 pvp_v 等於 15∘C15^\circ\text{C} 時的飽和水蒸氣壓 psat@15∘Cp_{sat@15^\circ\text{C}}。比濕 ww 可以由下式計算：
      w=0.622pvP−pvw = 0.622 \frac{p_v}{P - p_v}
      其中 PP 是總壓 (1 atm = 101.325 kPa)。
      查表或從濕度圖上讀取 15∘C15^\circ\text{C} 時的飽和水蒸氣壓 psat@15∘C≈1.7056p_{sat@15^\circ\text{C}} \approx 1.7056 kPa。
      w=0.6221.7056 kPa101.325 kPa−1.7056 kPa≈0.622×1.705699.6194≈0.01062 kg/kg dry airw = 0.622 \frac{1.7056 \text{ kPa}}{101.325 \text{ kPa} - 1.7056 \text{ kPa}} \approx 0.622 \times \frac{1.7056}{99.6194} \approx 0.01062 \text{ kg/kg dry air}
      題目要求單位為 g/kg dry air，所以 w≈10.62w \approx 10.62 g/kg dry air。

  • 從濕度圖讀取： 在濕度圖上，乾球溫度 25°C 的垂直線與露點溫度 15°C 的水平線（或相應的飽和曲線上的點）的交點，對應到右側的 Specific humidity 軸，大約是 10.5 到 11 g/kg 之間。

  • 【答案】 約 10.6 g/kg dry air。

(b) 求相對濕度 (RH)

  • 已知條件：

    • 乾球溫度 TdbT_{db} = 25°C
    • 露點溫度 TdpT_{dp} = 15°C
    • 總壓 PP = 1 atm

  • 解題方法：

    1. 相對濕度定義為：
      RH=pvpsat@TdbRH = \frac{p_v}{p_{sat@T_{db}}}
      其中 pvp_v 是空氣中的水蒸氣分壓，而 psat@Tdbp_{sat@T_{db}} 是在乾球溫度 TdbT_{db} 時的飽和水蒸氣壓。
    2. 由於已知露點溫度 Tdp=15∘CT_{dp} = 15^\circ\text{C}，這表示空氣中的水蒸氣分壓 pvp_v 等於 15∘C15^\circ\text{C} 時的飽和水蒸氣壓 psat@15∘Cp_{sat@15^\circ\text{C}}。
    3. 查表或從濕度圖上讀取 15∘C15^\circ\text{C} 時的飽和水蒸氣壓 psat@15∘C≈1.7056p_{sat@15^\circ\text{C}} \approx 1.7056 kPa。
    4. 查表或從濕度圖上讀取 25∘C25^\circ\text{C} 時的飽和水蒸氣壓 psat@25∘C≈3.1698p_{sat@25^\circ\text{C}} \approx 3.1698 kPa。
    5. 計算相對濕度：
🔒

後續完整解題步驟與【答案】

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第 2 題

  1. A refrigerator uses refrigerant-134a as the working fluid, and operates on an ideal vapor-compression refrigeration cycle between 4 °C and 42 °C. The refrigerant enters the compressor as a saturated vapor and leaves at 1200 kPa. The outlet of the condenser is at the saturated liquid line. The outlet of the evaporator is at the saturated vapor line.
    (a) (5%) the compressor outlet temperature if the compression is isentropic.
    (b) (9%) the change in enthalpy during the compression process. (kJ/kg).
    (c) (9%) the enthalpy at the inlet of the evaporator if the throttling process is isentropic.
    (d) (7%) the actual COP using the result from (b) and (c).
    (e) (5%) the theoretical maximum COP.

TABLE A-11: Saturated refrigerant-134a—Temperature table

Temp., °CPsat kPaSpecific volume, m³/kgInternal energy, kJ/kgEnthalpy, kJ/kgEntropy, kJ/kg-K
VfVgUfUfgUg
0293.010.00077220.06933551.61178.58
2314.840.00077610.06469054.28177.01
4337.900.00078020.06041256.97175.44
..................
18537.520.00081120.03831776.07163.92
20572.070.00081600.03601278.85162.19
22608.270.00082090.03386781.64160.45
24646.180.00082600.03186984.44158.68
26685.840.00083120.03000887.26156.89
28727.310.00083660.02827190.09155.08
30770.640.00084210.02664892.93153.24
32815.890.00084770.02513195.79151.37
34863.110.00085350.02371298.67149.48
36912.350.00085950.022383101.56147.55
38963.680.00086570.021137104.47145.60
401017.10.00087200.019968107.39143.61
421072.80.00087860.018870110.34141.59
441130.70.00088540.017837113.30139.53

Superheated refrigerant-134a—Temperature table
P = 0.80 MPa (Tsat = 31.31°C) | P = 0.90 MPa (Tsat = 35.51°C) | P = 1.20 MPa (Tsat = 46.29°C) | P = 1.40 MPa (Tsat = 52.40°C)
T | v | u | h | s | v | u | h | s | v | u | h | s | v | u | h | s
°C | m³/kg | kJ/kg | kJ/kg | kJ/kg-K | m³/kg | kJ/kg | kJ/kg | kJ/kg-K | m³/kg | kJ/kg | kJ/kg | kJ/kg-K | m³/kg | kJ/kg | kJ/kg | kJ/kg-K
... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ...
Sat. | 0.025645 | 246.82 | 267.34 | 0.9185 | 0.022686 | 248.82 | 269.25 | 0.9169 | 0.016728 | 253.84 | 273.92 | 0.9132 | 0.014119 | 256.40 | 276.17 | 0.9107
40 | 0.027035 | 254.84 | 276.46 | 0.9481 | 0.023375 | 253.15 | 274.19 | 0.9328 | 0.017201 | 257.64 | 278.28 | 0.9268 | 0.015005 | 264.46 | 285.47 | 0.9389
... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ...
1200 | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ...

  1. A refrigerator uses refrigerant-134a as the working fluid, and operates on an ideal vapor-compression refrigeration cycle between 4 °C and 42 °C. The refrigerant enters the compressor as a saturated vapor and leaves at 1200 kPa. The outlet of the condenser is at the saturated liquid line. The outlet of the evaporator is at the saturated vapor line.
    (a) (5%) the compressor outlet temperature if the compression is isentropic.
    (b) (9%) the change in enthalpy during the compression process. (kJ/kg).
    (c) (9%) the enthalpy at the inlet of the evaporator if the throttling process is isentropic.
    (d) (7%) the actual COP using the result from (b) and (c).
    (e) (5%) the theoretical maximum COP.

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TABLE A-11: Saturated refrigerant-134a—Temperature table

Temp., °CPsat kPaSpecific volume, m³/kgInternal energy, kJ/kgEnthalpy, kJ/kgEntropy, kJ/kg-K
VfVgUfUfgUg
0293.010.00077220.06933551.61178.58
2314.840.00077610.06469054.28177.01
4337.900.00078020.06041256.97175.44
..................
18537.520.00081120.03831776.07163.92
20572.070.00081600.03601278.85162.19
22608.270.00082090.03386781.64160.45
24646.180.00082600.03186984.44158.68
26685.840.00083120.03000887.26156.89
28727.310.00083660.02827190.09155.08
30770.640.00084210.02664892.93153.24
32815.890.00084770.02513195.79151.37
34863.110.00085350.02371298.67149.48
36912.350.00085950.022383101.56147.55
38963.680.00086570.021137104.47145.60
401017.10.00087200.019968107.39143.61
421072.80.00087860.018870110.34141.59
441130.70.00088540.017837113.30139.53

Superheated refrigerant-134a—Temperature table
P = 0.80 MPa (Tsat = 31.31°C) | P = 0.90 MPa (Tsat = 35.51°C) | P = 1.20 MPa (Tsat = 46.29°C) | P = 1.40 MPa (Tsat = 52.40°C)
T | v | u | h | s | v | u | h | s | v | u | h | s | v | u | h | s
°C | m³/kg | kJ/kg | kJ/kg | kJ/kg-K | m³/kg | kJ/kg | kJ/kg | kJ/kg-K | m³/kg | kJ/kg | kJ/kg | kJ/kg-K | m³/kg | kJ/kg | kJ/kg | kJ/kg-K
... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ...
Sat. | 0.025645 | 246.82 | 267.34 | 0.9185 | 0.022686 | 248.82 | 269.25 | 0.9169 | 0.016728 | 253.84 | 273.92 | 0.9132 | 0.014119 | 256.40 | 276.17 | 0.9107
40 | 0.027035 | 254.84 | 276.46 | 0.9481 | 0.023375 | 253.15 | 274.19 | 0.9328 | 0.017201 | 257.64 | 278.28 | 0.9268 | 0.015005 | 264.46 | 285.47 | 0.9389
... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ...
1200 | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ... | ...

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這一題的完整詳解

核心觀念

本題考查理想蒸氣壓縮冷凍循環的四個狀態：

  • 狀態 1：蒸發器出口、壓縮機入口，為 4∘C4^\circ\text{C} 飽和蒸氣。
  • 狀態 2：壓縮機出口，壓力為 1200 kPa1200\ \text{kPa}，壓縮為等熵。
  • 狀態 3：冷凝器出口，為 42∘C42^\circ\text{C} 飽和液體。
  • 狀態 4：蒸發器入口，經節流閥降壓。

主要公式為：

s2=s1s_2=s_1 wc=h2−h1w_c=h_2-h_1 qL=h1−h4q_L=h_1-h_4 COPR=qLwc\mathrm{COP}_R=\frac{q_L}{w_c}

解題方法與查表

由 4∘C4^\circ\text{C} 飽和液體表得狀態 1：

h1=hg(4∘C)=252.82 kJ/kgh_1=h_g(4^\circ\text{C})=252.82\ \text{kJ/kg} s1=sg(4∘C)=0.92946 kJ/(kg\cdotK)s_1=s_g(4^\circ\text{C})=0.92946\ \text{kJ/(kg\cdot K)}

由 42∘C42^\circ\text{C} 飽和液體表得狀態 3：

h3=hf(42∘C)=111.28 kJ/kgh_3=h_f(42^\circ\text{C})=111.28\ \text{kJ/kg}

(a) 壓縮機出口溫度

壓縮過程為等熵，因此：

s2=s1=0.92946 kJ/(kg\cdotK)s_2=s_1=0.92946\ \text{kJ/(kg\cdot K)}

在 P2=1.2 MPaP_2=1.2\ \text{MPa} 的過熱 R-134a 表中：

TTsshh
40∘C40^\circ\text{C}0.92680.9268278.28 kJ/kg278.28\ \text{kJ/kg}
50∘C50^\circ\text{C}0.96150.9615289.66 kJ/kg289.66\ \text{kJ/kg}

以熵線性內插：

T2=40+0.92946−0.92680.9615−0.9268(50−40)T_2 =40+\frac{0.92946-0.9268}{0.9615-0.9268}(50-40) T2≈40.8∘CT_2\approx 40.8^\circ\text{C}

因此：

T2≈40.8∘C\boxed{T_2\approx 40.8^\circ\text{C}}

(b) 壓縮過程的焓變

先內插狀態 2 的焓：

h2=278.28+0.92946−0.92680.9615−0.9268(289.66−278.28)h_2 =278.28+ \frac{0.92946-0.9268}{0.9615-0.9268} (289.66-278.28) h2≈279.15 kJ/kgh_2\approx279.15\ \text{kJ/kg}

所以壓縮過程焓變為：

Δhcomp=h2−h1\Delta h_{\text{comp}}=h_2-h_1 Δhcomp=279.15−252.82=26.33 kJ/kg\Delta h_{\text{comp}} =279.15-252.82 =26.33\ \text{kJ/kg}

因此：

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後續完整解題步驟與【答案】

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第 3 題

  1. A lead-acid battery is an energy storage device that is commonly used in cars and motorcycles for the start-up purpose. A single lead-acid cell when discharged runs on the following reaction:
    Pb(s) + PbO₂(s) + 4H⁺(aq) + 2SO₄²⁻(aq) → 2PbSO₄(s) + 2H₂O(l)
    Please answer the following questions.
    (a) Use data in Table 1 to determine the values of ΔrH\Delta_r H, ΔrS\Delta_r S and ΔrG\Delta_r G for this discharge reaction occurred in a lead-acid cell (for one mole of Pb). Assume that the reaction occurs at ambient temperature (298 K) and pressure (1 bar). (9%)
    (b) Based on (a), show that the thermodynamic identity ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S is valid. (6%)
    (c) In a lead-acid cell, the sulfate ions will first combine with protons through the reaction 2SO₄²⁻(aq) + 2H⁺(aq) → 2HSO₄⁻(aq). Then, half reactions for a lead-acid cell are
    at (-) electrode: Pb(s) + HSO₄⁻(aq) → PbSO₄(s) + H⁺(aq) + 2e⁻
    at (+) electrode: PbO₂(s) + 3H⁺(aq) + HSO₄⁻(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
    Use the identity ΔG=−nFV\Delta G = -nFV, where nn is the charge number, FF the Faraday constant, to calculate the voltage VV of a lead-acid cell. (5%)
    (d) Following (c), the work generated from a lead-acid cell (for each mole of Pb used) is ΔG\Delta G. Explain why it is not ΔH\Delta H. (5%)
    (e) The electrolyte of a lead-acid battery is usually composed of 6 M sulfuric acid instead of 1 M. Is the actual cell voltage higher or lower than the result in (c)? Explain how you reach your conclusion. (5%)
    (f) A lead-acid battery is a secondary battery, which means it can be charged; i.e., the aforementioned full-cell reaction will be running backward during charging. But this means cell entropy decreases during charging. How could this be possible? (5%)

Table 1. Thermodynamics Properties of Selected Substances

Substance (form)ΔfH\Delta_f H (kJ)ΔfG\Delta_f G (kJ)S (J/K)Cₚ (J/K)V (cm³)
H₂(g)00130.6828.82
H(g)217.97203.25114.7120.78
H⁺(aq)0000
H₂O(l)-285.83-237.1369.9175.2918.068
H₂O(g)-241.82-228.57188.8333.58
O₂(g)00205.1429.38
O₂(aq)-11.716.4110.9
OH⁻(aq)-229.99-157.24-10.75-148.5
Pb(s)0064.8126.4418.3
PbO₂(s)-277.4-217.3368.664.64
PbSO₄(s)-920.0-813.0148.5103.2
SO₄²⁻(aq)-909.27-744.5320.1-293
HSO₄⁻(aq)-887.34-755.91131.8-84

All of the values in this table are for one mole of material at 298 K and 1 bar. The form of the substance, either solid (s), liquid (l), gas (g) or aqueous solution (aq), is denoted after the chemical formula. Data for aqueous solutions are at a standard concentration of 1 mole per kilogram water. The enthalpy and Gibbs free energy of formation, ΔfH\Delta_f H and ΔfG\Delta_f G, represent the changes in H and G upon forming one more of the material starting with elements in their most stable pure states (e.g., O₂(g), etc.).

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這一題的完整詳解

核心觀念

本題考查：

  • 反應焓、反應熵與反應 Gibbs 自由能的計算
  • 熱力學基本關係式
    ΔrG=ΔrH−TΔrS\Delta_rG=\Delta_rH-T\Delta_rS
  • 電池電動勢與 Gibbs 自由能的關係
    ΔrG=−nFV\Delta_rG=-nFV
  • 濃度（更嚴格地說是活度）對電池電壓的影響
  • 第二定律與充電過程的熵變

所有熱力學量皆依「生成物總和減反應物總和」計算。

(a) 計算 ΔrH\Delta_rH、ΔrS\Delta_rS 與 ΔrG\Delta_rG

反應為

Pb(s)+PbO2(s)+4H+(aq)+2SO42−(aq)→2PbSO4(s)+2H2O(l)\mathrm{Pb(s)+PbO_2(s)+4H^+(aq)+2SO_4^{2-}(aq)} \rightarrow \mathrm{2PbSO_4(s)+2H_2O(l)}

反應焓 ΔrH\Delta_rH

ΔrH=[2ΔfH(PbSO4)+2ΔfH(H2O)]−[ΔfH(PbO2)+2ΔfH(SO42−)]\begin{aligned} \Delta_rH &=\left[2\Delta_fH(\mathrm{PbSO_4})+ 2\Delta_fH(\mathrm{H_2O})\right]\\ &\quad-\left[\Delta_fH(\mathrm{PbO_2})+ 2\Delta_fH(\mathrm{SO_4^{2-}})\right] \end{aligned}

代入數值：

ΔrH=[2(−920.0)+2(−285.83)]−[(−277.4)+2(−909.27)]=−315.72 kJ/mol\begin{aligned} \Delta_rH &=\left[2(-920.0)+2(-285.83)\right]\\ &\quad-\left[(-277.4)+2(-909.27)\right]\\ &=-315.72\ \mathrm{kJ/mol} \end{aligned}

因此，

ΔrH=−315.72 kJ/mol\boxed{\Delta_rH=-315.72\ \mathrm{kJ/mol}}

反應熵 ΔrS\Delta_rS

ΔrS=[2S(PbSO4)+2S(H2O)]−[S(Pb)+S(PbO2)+2S(SO42−)]\begin{aligned} \Delta_rS &=\left[2S(\mathrm{PbSO_4})+ 2S(\mathrm{H_2O})\right]\\ &\quad-\left[S(\mathrm{Pb})+ S(\mathrm{PbO_2})+ 2S(\mathrm{SO_4^{2-}})\right] \end{aligned}

代入數值：

ΔrS=[2(148.5)+2(69.91)]−[64.81+68.6+2(20.1)]=263.21 J/(K⋅mol)\begin{aligned} \Delta_rS &=\left[2(148.5)+2(69.91)\right]\\ &\quad-\left[64.81+68.6+2(20.1)\right]\\ &=263.21\ \mathrm{J/(K\cdot mol)} \end{aligned}

因此，

ΔrS=263.21 J/(K⋅mol)\boxed{\Delta_rS=263.21\ \mathrm{J/(K\cdot mol)}}

亦可寫成

ΔrS=0.26321 kJ/(K⋅mol)\Delta_rS=0.26321\ \mathrm{kJ/(K\cdot mol)}

反應 Gibbs 自由能 ΔrG\Delta_rG

ΔrG=[2ΔfG(PbSO4)+2ΔfG(H2O)]−[ΔfG(PbO2)+2ΔfG(SO42−)]\begin{aligned} \Delta_rG &=\left[2\Delta_fG(\mathrm{PbSO_4})+ 2\Delta_fG(\mathrm{H_2O})\right]\\ &\quad-\left[\Delta_fG(\mathrm{PbO_2})+ 2\Delta_fG(\mathrm{SO_4^{2-}})\right] \end{aligned}

代入數值：

ΔrG=[2(−813.0)+2(−237.13)]−[(−217.33)+2(−744.53)]=−393.87 kJ/mol\begin{aligned} \Delta_rG &=\left[2(-813.0)+2(-237.13)\right]\\ &\quad-\left[(-217.33)+2(-744.53)\right]\\ &=-393.87\ \mathrm{kJ/mol} \end{aligned}

因此，

ΔrG=−393.87 kJ/mol\boxed{\Delta_rG=-393.87\ \mathrm{kJ/mol}}

負值表示在標準狀態下，放電反應具有自發性。

(b) 驗證 ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta S

在 T=298 KT=298\ \mathrm{K} 時，

TΔrS=298(0.26321)=78.44 kJ/molT\Delta_rS =298(0.26321) =78.44\ \mathrm{kJ/mol}

因此，

ΔrH−TΔrS=−315.72−78.44=−394.16 kJ/mol\begin{aligned} \Delta_rH-T\Delta_rS &=-315.72-78.44\\ &=-394.16\ \mathrm{kJ/mol} \end{aligned}

由表中 ΔfH\Delta_fH、ΔfG\Delta_fG 與 SS 的四捨五入誤差，直接計算得到的結果為 −394.16 kJ/mol-394.16\ \mathrm{kJ/mol}；由生成 Gibbs 自由能計算得到 −393.87 kJ/mol-393.87\ \mathrm{kJ/mol}，差異僅約 0.29 kJ/mol0.29\ \mathrm{kJ/mol}，相對誤差小於 0.1%0.1\%。

故可視為

ΔrG≈ΔrH−TΔrS\boxed{\Delta_rG\approx\Delta_rH-T\Delta_rS}

熱力學恆等式成立。

(c) 計算鉛酸電池電壓

兩個半反應相加為：

負極：

Pb+HSO4−→PbSO4+H++2e−\mathrm{Pb+HSO_4^-\rightarrow PbSO_4+H^++2e^-}

正極：

PbO2+3H++HSO4−+2e−→PbSO4+2H2O\mathrm{PbO_2+3H^++HSO_4^-+2e^- \rightarrow PbSO_4+2H_2O}

相加後：

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第 4 題

  1. Multiple choice (Please select the most appropriate answer, 5% each)
    (a) Which of the following statement regarding heat and work is NOT true?
    (1) heat and work are path dependent
    (2) work can be completely converted to heat
    (3) heat can be completely converted to work
    (4) entropy always flows along with heat
    (b) What is the most important reason that makes a fuel cell usually much more efficient than a heat engine?
    (1) use different fuels
    (2) operate at different temperatures
    (3) produce different amount of waste heat
    (4) built with different materials
    (c) Which of the following statement regarding dilute solutions is false?
    (1) solute interacts with solvent
    (2) solute interacts with solute
    (3) solvent interacts with solvent
    (4) solvent and solute homogeneously mixed
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這一題的完整詳解

核心觀念

本題考查三個熱力學概念：

  1. 熱與功是否為狀態函數，以及熵與熱傳的關係。
  2. 燃料電池與熱機的能量轉換方式。
  3. 稀溶液中溶質與溶劑的分子作用。

(a) 熱與功

解題方法

熱 QQ 與功 WW 皆為「路徑函數」，其數值取決於系統如何由初態到終態，而非只由初、終狀態決定。

根據熱力學第二定律，熱機不可能將吸收的熱量全部轉換為功；必須有部分熱量排放至低溫熱庫。

選項分析

(1) heat and work are path dependent

正確。熱與功不是狀態函數，微分分別寫作 δQ\delta Q 與 δW\delta W，表示其值依過程路徑而定。

(2) work can be completely converted to heat

正確。功可藉由摩擦、電阻加熱等方式完全轉換為熱，並不違反熱力學第二定律。

(3) heat can be completely converted to work

錯誤。對循環熱機而言，若將熱量完全轉為功而不向低溫熱庫排熱，將違反 Kelvin–Planck 敘述的熱力學第二定律。

(4) entropy always flows along with heat

正確。在可逆熱傳中，熵傳遞量為

dS=δQrevTdS=\frac{\delta Q_{\mathrm{rev}}}{T}

熱傳通常伴隨熵的傳遞；此外，系統內部也可能因不可逆過程產生熵。

解題技巧

看到「熱量能否全部轉成功」時，應立即聯想到熱力學第二定律：

  • 功可以完全轉成熱。
  • 熱不能在循環過程中完全轉成功。

【答案】(a) (3)

(b) 燃料電池的效率

核心觀念

熱機先將燃料的化學能轉成熱，再利用高、低溫熱庫間的循環產生功，因此受到 Carnot 效率限制：

ηCarnot=1−TLTH\eta_{\mathrm{Carnot}}=1-\frac{T_L}{T_H}

燃料電池則透過電化學反應，直接將燃料的化學能轉換為電能，不必先經過燃燒產熱，因此能量轉換效率通常較高。

選項分析

(1) use different fuels

錯誤。燃料電池與熱機可能使用相同燃料，例如氫氣；燃料種類不是效率較高的主要原因。

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其他考古題

其他學校的生物生化系考古題

成功大學204中正大學10中山大學26臺灣海洋大學1