Thermodynamics Basics

Thermodynamics is the branch of physics that studies heat, energy, and their transformations, governing processes in systems ranging from engines to stars. It explores how energy flows as heat and work, and how systems reach equilibrium through laws that apply universally. The field is crucial for understanding natural phenomena like weather patterns, as well as engineering applications such as power generation and refrigeration. This guide provides a detailed overview with all basic formulas (first law, heat equation, ideal gas law, work, efficiency, entropy, heat transfer), numerous examples, and practical applications to illustrate thermodynamic principles.

First Law (with Ideal Gas Law, Work)

The First Law of Thermodynamics, also known as the law of energy conservation, states that the total energy of an isolated system is constant. Energy can be transferred as heat (\( Q \)) or work (\( W \)), changing the system’s internal energy (\( \Delta U \)):

\[ \Delta U = Q - W \]

Where:

\( \Delta U \): Change in internal energy (in Joules, J)
\( Q \): Heat added to the system (in J)
\( W \): Work done by the system (in J)

Related Formulas:

Ideal Gas Law: \( PV = nRT \) (where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R = 8.314 \, \text{J/mol·K} \), \( T \) is temperature in Kelvin)
Work (Isobaric Process): \( W = P \Delta V \)

Example 1: First Law Application

A system absorbs \( Q = 500 \, \text{J} \) and does \( W = 200 \, \text{J} \). Find \( \Delta U \):

\[ \Delta U = Q - W \] \[ = 500 - 200 \] \[ = 300 \, \text{J} \]

Example 2: Ideal Gas Law

1 mole of gas at \( T = 300 \, \text{K} \), \( V = 0.0248 \, \text{m}^3 \). Find pressure:

\[ PV = nRT \] \[ P \cdot 0.0248 = 1 \cdot 8.314 \cdot 300 \] \[ P \cdot 0.0248 = 2494.2 \] \[ P = \frac{2494.2}{0.0248} \] \[ \approx 100572.58 \, \text{Pa} \]

Example 3: Work in Expansion

A gas expands at constant \( P = 100000 \, \text{Pa} \), from \( V_1 = 0.01 \, \text{m}^3 \) to \( V_2 = 0.03 \, \text{m}^3 \). Find work done:

\[ W = P \Delta V \] \[ \Delta V = V_2 - V_1 \] \[ = 0.03 - 0.01 \] \[ = 0.02 \, \text{m}^3 \] \[ W = 100000 \cdot 0.02 \] \[ = 2000 \, \text{J} \]

Heat Equation (with Efficiency, Entropy, Heat Transfer)

The heat required to change an object’s temperature depends on its mass, specific heat, and temperature change:

\[ Q = m c \Delta T \]

Where:

\( Q \): Heat energy (in J)
\( m \): Mass (in kg)
\( c \): Specific heat (in J/kg°C)
\( \Delta T \): Temperature change (in °C)

Related Formulas:

Efficiency (Heat Engine): \( \eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H} \) (where \( Q_H \) is heat input, \( Q_C \) is heat rejected)
Entropy Change: \( \Delta S = \frac{Q}{T} \) (for reversible process, \( T \) in Kelvin)
Heat Transfer (Conduction): \( Q = k A \frac{\Delta T}{\Delta x} t \) (where \( k \) is thermal conductivity, \( A \) is area, \( \Delta x \) is thickness, \( t \) is time)

Example 1: Heat Equation

Heat 1 kg of water (\( c = 4186 \, \text{J/kg°C} \)) by 10°C:

\[ Q = m c \Delta T \] \[ = 1 \cdot 4186 \cdot 10 \] \[ = 41860 \, \text{J} \]

Example 2: Efficiency of a Heat Engine

An engine takes \( Q_H = 1000 \, \text{J} \), rejects \( Q_C = 600 \, \text{J} \). Find efficiency:

\[ \eta = 1 - \frac{Q_C}{Q_H} \] \[ = 1 - \frac{600}{1000} \] \[ = 1 - 0.6 \] \[ = 0.4 \] \[ \eta = 40\% \]

Example 3: Entropy Change

A system absorbs 400 J at \( T = 300 \, \text{K} \). Find \( \Delta S \):

\[ \Delta S = \frac{Q}{T} \] \[ = \frac{400}{300} \] \[ \approx 1.333 \, \text{J/K} \]

Example 4: Heat Transfer by Conduction

Heat transfer through a copper wall (\( k = 385 \, \text{W/m°C} \)), \( A = 0.5 \, \text{m}^2 \), \( \Delta x = 0.02 \, \text{m} \), \( \Delta T = 50 \, \text{°C} \), over \( t = 3600 \, \text{s} \):

\[ Q = k A \frac{\Delta T}{\Delta x} t \] \[ = 385 \cdot 0.5 \cdot \frac{50}{0.02} \cdot 3600 \] \[ = 385 \cdot 0.5 \cdot 2500 \cdot 3600 \] \[ = 1732500000 \, \text{J} \]

Applications

Thermodynamics drives numerous technologies and natural systems. Here are detailed examples with calculations:

Example 1: Engine Work Output

An engine with \( Q_H = 1500 \, \text{J} \), \( Q_C = 900 \, \text{J} \). Find work done:

\[ W = Q_H - Q_C \] \[ = 1500 - 900 \] \[ = 600 \, \text{J} \]

Example 2: Refrigerator Cooling

A refrigerator removes \( Q_C = 500 \, \text{J} \), does \( W = 200 \, \text{J} \). Find heat rejected:

\[ Q_H = Q_C + W \] \[ = 500 + 200 \] \[ = 700 \, \text{J} \]

Example 3: Climate System (Heat Transfer)

Heat transfer through a wall (\( k = 0.8 \, \text{W/m°C} \)), \( A = 10 \, \text{m}^2 \), \( \Delta x = 0.1 \, \text{m} \), \( \Delta T = 20 \, \text{°C} \), over 1 hour:

\[ Q = k A \frac{\Delta T}{\Delta x} t \] \[ = 0.8 \cdot 10 \cdot \frac{20}{0.1} \cdot 3600 \] \[ = 0.8 \cdot 10 \cdot 200 \cdot 3600 \] \[ = 5760000 \, \text{J} \]

Example 4: Gas Expansion

Gas at \( P = 200000 \, \text{Pa} \), expands from \( V_1 = 0.02 \, \text{m}^3 \) to \( V_2 = 0.05 \, \text{m}^3 \):

\[ W = P \Delta V \] \[ \Delta V = 0.05 - 0.02 \] \[ = 0.03 \] \[ W = 200000 \cdot 0.03 \] \[ = 6000 \, \text{J} \]

Example 5: Heating a Gas (Ideal Gas Law)

2 moles of gas at \( P = 101325 \, \text{Pa} \), \( V = 0.0496 \, \text{m}^3 \). Find temperature:

\[ PV = nRT \] \[ 101325 \cdot 0.0496 = 2 \cdot 8.314 \cdot T \] \[ 5025.72 = 16.628 \cdot T \] \[ T = \frac{5025.72}{16.628} \] \[ \approx 302.27 \, \text{K} \]

Example 6: Entropy in Melting Ice

1 kg of ice melts at 273 K, latent heat \( L = 334000 \, \text{J/kg} \). Find \( \Delta S \):

\[ Q = m L \] \[ = 1 \cdot 334000 \] \[ = 334000 \, \text{J} \] \[ \Delta S = \frac{Q}{T} \] \[ = \frac{334000}{273} \] \[ \approx 1223.44 \, \text{J/K} \]