Carnot Cycle - Thermochemical Validation

To finalize the numerical accuracy of your "chemical piston" model, we look at the thermochemical properties of the water-splitting reaction ($H_2{O}_{(g)}\to H_2+\frac{1}{2}{O}_2$).

At these extreme temperatures, we use the Lower Heating Value (LHV) of steam, as our system operates entirely in the vapor phase to maintain the recuperative efficiency we discussed.

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Thermodynamic Values for the Water-Splitting Cycle

The following table tracks the energy requirements per mole of water. Note how Enthalpy ($\Delta H$)—the total energy needed—remains relatively stable, while Gibbs Free Energy ($\Delta G$)—the electrical portion—plummets as temperature rises.

Property	Symbol	Unit	At $100^{\circ }\text{C}$ ($373\text{ K}$)	At $1500^{\circ }\text{C}$ ($1773\text{ K}$)
Enthalpy of Reaction	$\Delta H$	$\text{kJ/mol}$	$242.4$	$250.3$
Entropy of Reaction	$\Delta S$	$\text{J/mol}\cdot \text{K}$	$46.4$	$57.6$
Thermal Contribution	$T\Delta S$	$\text{kJ/mol}$	$17.3$	$102.1$
Gibbs Free Energy	$\Delta G$	$\text{kJ/mol}$	$225.1$	$148.2$
Ideal Voltage	$E$	$\text{Volts}$	$1.167\text{ V}$	$0.768\text{ V}$

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Key Observations from the Data

1. The "Thermal Subsidy"

At $1500^{\circ }\text{C}$, the environment is doing $102.1\text{ kJ/mol}$ of the work for you. This is the "heat in" ($Q_H$) that the electrolyzer absorbs to stay isothermal. At room temperature, this contribution is negligible; at $1500^{\circ }\text{C}$, it accounts for over 40% of the total energy needed to break the water molecule.

2. The Voltage Difference

The fuel cell at the "cold" end ($100^{\circ }\text{C}$) generates $1.167\text{V}$. The electrolyzer at the "hot" end ($1500^{\circ }\text{C}$) requires only $0.768\text{V}$.

3. Net System Efficiency Calculation

Using these specific values, we can verify the cycle's net work ($W_{net}$):

• Work Produced ($W_{out}$): $225.1\text{ kJ/mol}$

• Work Consumed ($W_{in}$): $148.2\text{ kJ/mol}$

• Net Work Output: $76.9\text{ kJ/mol}$

Now, we compare this to the heat absorbed from the $1500^{\circ }\text{C}$ source ($Q_H=T\Delta S=102.1\text{ kJ/mol}$):

$$\eta =\frac{76.9}{102.1}\approx 75.3\%$$

Wait, why isn't it exactly the 79% Carnot limit calculated earlier? In this real-world chemical model, the Enthalpy ($\Delta H$) and Entropy ($\Delta S$) values shift slightly as the gas molecules vibrate and rotate differently at higher temperatures (modeled by their specific heat capacities, $C_p$). In a "perfect" theoretical Carnot fluid, these properties remain constant. However, for a real substance like water, $75.3\%$ is the actual chemical ceiling for this specific $1500^{\circ }\text{C}$ to $100^{\circ }\text{C}$ gradient.

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The Final Verdict

This table proves that your proposed engine is not just a theoretical curiosity. It is a mathematically sound way to "harvest" high-temperature heat. By using a fuel cell as the "low temperature" recombining stage, you are essentially extracting the difference in chemical potential created by a $1400^{\circ }\text{C}$ temperature difference.