Law Thermodynamics

Overview

The laws of thermodynamics are fundamental principles governing energy, heat, and work in physical systems.

Key Formulas

Zeroth Law

$T_A = T_B \text{ and } T_B = T_C \Rightarrow T_A = T_C$

If two systems are each in thermal equilibrium with a third, they are in equilibrium with each other.

First Law

$\Delta U = Q - W$

Change in internal energy = heat added − work done by the system.

Second Law (Clausius)

$\Delta S \geq \frac{Q}{T}$

Entropy of an isolated system never decreases.

Third Law

$\lim_{T \to 0} S = 0$

Entropy approaches zero as temperature approaches absolute zero (perfect crystal).

Variables

Entropy and the Second Law

Entropy quantifies the unavailable energy in a system and its relative ability to perform work. Key properties derived from the Second Law include:

• Entropy change conditions:

  • $\Delta S > 0$: irreversible process
  • $\Delta S = 0$: reversible process
  • $\Delta S < 0$: impossible process

• Entropy is not conserved like energy; it is produced by all real processes and associated with energy degradation.

• The entropy of the universe tends to increase as energy flows towards lower availability.

For a thermodynamic process, the entropy change is given by:

$dS = \frac{dH}{T_a}$

where $dS$ is entropy change (kJ/kg·K), $dH$ is change in enthalpy or internal energy (kJ/kg), and $T_a$ is average temperature (K).

Carnot Heat Cycle Analysis

The Carnot cycle models an ideal heat engine with four reversible stages:

1. Isothermal expansion at $T_h$: $\Delta E_1 = 0$, $w_1 = -H_1$
2. Adiabatic expansion from $T_h$ to $T_c$: $H_2 = 0$, $\Delta E_2 = w_2$
3. Isothermal compression at $T_c$: $\Delta E_3 = 0$, $w_3 = -H_3$
4. Adiabatic compression back to $T_h$: $H_4 = 0$, $\Delta E_4 = w_4$

Net work and heat relationships: $-w = H = H_1 + H_2 + H_3 + H_4$ $\oint \frac{dH}{T} = 0$

The cycle's thermodynamic efficiency is: $\eta = \frac{T_h - T_c}{T_h} = 1 - \frac{T_c}{T_h}$

Entropy Change Examples

Heating Water

• Mass: 1 kg, from 0°C (273 K) to 100°C (373 K)
• Specific enthalpy: $h_1 = 0$ kJ/kg, $h_2 = 419$ kJ/kg
• Average temperature: $T_a = (273 + 373)/2 = 323$ K
• Entropy change: $\Delta S = \frac{419 - 0}{323} = 1.297 \, \text{kJ/kg·K}$

Evaporation of Water

• Phase change at 100°C (373 K)
• Specific enthalpy: liquid $h_1 = 418$ kJ/kg, vapor $h_2 = 2675$ kJ/kg
• Temperature: $T_a = 373$ K
• Entropy change: $\Delta S = \frac{2675 - 418}{373} = 6.054 \, \text{kJ/kg·K}$

These examples illustrate entropy increase during heating and phase change, highlighting energy dispersal at constant temperature.

References

• Original Source