Thermodynamics: An Engineering Approach
Publisher: McGraw Hill
- Energy, heat and work
- Thermodynamic systems
- Properties and state
- Internal energy
- The First Law of Thermodynamics
Thermodynamics is the study of energy, heat, temperature, and work. Mechanical engineers use thermodynamics to understand how energy is transferred, transformed, and utilized in physical systems. From automobile engines and power plants to refrigerators, air conditioners, turbines, and renewable energy systems, thermodynamics provides the foundation for designing efficient machines and understanding how energy flows throughout engineering applications.
Thermodynamics is the study of energy, heat, temperature and work. Mechanical engineers use thermodynamics to understand how energy moves, changes form and affects machines and physical systems.
Energy is the ability to perform work or produce a change. Energy can exist in several different forms.
Energy cannot be created or destroyed. It can only be transferred or converted from one form into another. This principle is known as the conservation of energy.
Heat and temperature are related, but they do not mean the same thing.
Temperature describes how hot or cold a substance is. Temperature is related to the average kinetic energy of the particles inside the substance.
Common temperature units include:
Kelvin is the SI temperature unit commonly used in engineering equations.
Heat is energy transferred from one object or system to another because of a temperature difference.
Heat naturally moves from a region of higher temperature to a region of lower temperature.
When two objects at different temperatures come into contact, heat moves from the hotter object to the colder object.
Heat transfer continues until both objects reach the same temperature. This condition is called thermal equilibrium.
Before analyzing a thermodynamics problem, engineers identify the system.
A system is the object, machine or region being studied. Everything outside the system is called the surroundings.
The real or imaginary surface separating the system from the surroundings is called the boundary.
There are three main types of thermodynamic systems:
Turbines, compressors and pumps are commonly analyzed as open systems. A sealed piston-cylinder device is commonly analyzed as a closed system.
A property is a measurable characteristic that describes the condition of a thermodynamic system.
Common thermodynamic properties include:
The condition of a system at a particular moment is called its state.
Thermodynamic properties can be classified as either intensive or extensive.
Pressure is the amount of force applied over a specific area.
Formula:
P = F / A
Pressure is measured in Pascals.
One Pascal is equal to one Newton per square meter.
1 Pa = 1 N/m²
In thermodynamics, work is a transfer of energy that occurs when a force causes movement.
Examples of thermodynamic work include:
Work is represented by the symbol W and is normally measured in Joules.
Internal energy is the microscopic energy stored inside a substance.
Internal energy includes energy associated with:
Internal energy is represented by the symbol U.
The First Law of Thermodynamics applies the conservation of energy to thermodynamic systems.
Energy entering a system must either leave the system or remain stored inside it.
Formula:
ΔU = Q − W
In this sign convention, heat entering the system is positive and work performed by the system is positive.
If more heat enters the system than leaves as work, the internal energy of the system increases.
Specific heat capacity describes the amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius or one Kelvin.
Formula:
Q = mcΔT
Materials with a high specific heat capacity require more energy to increase their temperature.
Heat can be transferred in three main ways:
A metal spoon becoming hot in soup is an example of conduction. Warm air rising from a heater is an example of convection. Energy traveling from the Sun to Earth is an example of radiation.
A substance can exist as a solid, liquid or gas. A phase change occurs when a substance changes from one phase to another.
Common phase changes include:
Energy transferred during a phase change can be calculated using:
Formula:
Q = mL
Efficiency compares the useful energy output of a machine to the amount of energy supplied to it.
Formula:
η = Useful Energy Output / Energy Input
Efficiency is commonly expressed as a percentage.
Real machines are not 100 percent efficient because some energy is transferred to the surroundings, often as waste heat.
Thermodynamics problems often require engineers to use consistent units and rearrange equations to solve for different unknown values.
To convert Joules to kilojoules, divide by 1,000.
To convert kilojoules to Joules, multiply by 1,000.
To convert Pascals to kilopascals, divide by 1,000.
To convert kilopascals to Pascals, multiply by 1,000.
In the equation Q = mcΔT, the standard units are:
Temperature change can be calculated using:
ΔT = Tfinal − Tinitial
In the equation Q = mL, the standard units are:
Efficiency may be written as a decimal or a percentage.
To convert a percentage to a decimal, divide by 100.
30% = 30 / 100 = 0.30
To convert a decimal to a percentage, multiply by 100%.
0.35 × 100% = 35%
The equations from this lesson can be rearranged depending on the unknown value.
Mechanical engineers use thermodynamics to understand and improve systems that transfer or convert energy.
Examples include:
Thermodynamics provides the foundation for understanding engines, refrigeration systems, power plants, turbines and many other mechanical engineering technologies.
These sources provide additional explanations of energy, heat, temperature, thermodynamic systems, pressure, internal energy, heat transfer and the First Law of Thermodynamics.
Authors: Yunus A. Çengel and Michael A. Boles
Publisher: McGraw Hill
Authors: Moran, Shapiro, Boettner and Bailey
Publisher: Wiley
Organization: OpenStax
Access: Free Online
Organization: OpenStax
Access: Free Online
Organization: OpenStax
Access: Free Online
Organization: National Institute of Standards and Technology
Abbreviation: NIST
Important: Thermodynamics equations depend on the selected system, sign convention and assumptions. The equation ΔU = Q − W on this page uses the convention that heat entering the system and work performed by the system are positive.
These problems use only the thermodynamics concepts and formulas introduced on the lesson page.
A gas pushes against a piston with a force of 600 N. The piston has an area of 0.003 m². Find the pressure exerted on the piston.
Step 1: Use the pressure formula
P = F / A
Step 2: Substitute the known values
P = 600 / 0.003
Step 3: Calculate
P = 200,000 Pa
Step 4: Convert to kilopascals
200,000 Pa ÷ 1,000 = 200 kPa
Answer: The pressure exerted on the piston is 200,000 Pa, or 200 kPa.
A pressure of 150,000 Pa acts on a piston with an area of 0.02 m². Find the force acting on the piston.
Step 1: Start with the pressure formula
P = F / A
Step 2: Rearrange the formula to solve for force
F = PA
Step 3: Substitute the known values
F = 150,000 × 0.02
Step 4: Calculate
F = 3,000 N
Answer: The force acting on the piston is 3,000 N.
A piston experiences a force of 2,500 N while the pressure acting on it is 500,000 Pa. Find the area of the piston.
Step 1: Start with the pressure formula
P = F / A
Step 2: Rearrange the formula to solve for area
A = F / P
Step 3: Substitute the known values
A = 2,500 / 500,000
Step 4: Calculate
A = 0.005 m²
Answer: The area of the piston is 0.005 m².
A closed thermodynamic system receives 900 J of heat and performs 350 J of work on its surroundings. Find the change in the system's internal energy.
Step 1: Use the First Law of Thermodynamics
ΔU = Q − W
Step 2: Identify the sign convention
Heat entering the system is positive, and work performed by the system is positive.
Step 3: Substitute the known values
ΔU = 900 − 350
Step 4: Calculate
ΔU = 550 J
Answer: The internal energy of the system increases by 550 J.
A gas receives 400 J of heat and performs 650 J of work. Find the change in its internal energy.
Step 1: Use the First Law of Thermodynamics
ΔU = Q − W
Step 2: Substitute the known values
ΔU = 400 − 650
Step 3: Calculate
ΔU = −250 J
Answer: The change in internal energy is −250 J. The negative sign means the system's internal energy decreases by 250 J.
A thermodynamic system has an internal energy increase of 700 J while performing 200 J of work. Find the amount of heat added to the system.
Step 1: Start with the First Law
ΔU = Q − W
Step 2: Rearrange the formula to solve for heat
Q = ΔU + W
Step 3: Substitute the known values
Q = 700 + 200
Step 4: Calculate
Q = 900 J
Answer: A total of 900 J of heat was added to the system.
A system receives 1,200 J of heat, and its internal energy increases by 750 J. Find the work performed by the system.
Step 1: Start with the First Law
ΔU = Q − W
Step 2: Rearrange the formula to solve for work
W = Q − ΔU
Step 3: Substitute the known values
W = 1,200 − 750
Step 4: Calculate
W = 450 J
Answer: The system performs 450 J of work.
A 2 kg metal object has a specific heat capacity of 500 J/(kg·°C). Its temperature increases by 15°C. Find the amount of heat added to the object.
Step 1: Use the specific heat formula
Q = mcΔT
Step 2: Substitute the known values
Q = 2 × 500 × 15
Step 3: Calculate
Q = 15,000 J
Step 4: Convert to kilojoules
15,000 J ÷ 1,000 = 15 kJ
Answer: The object receives 15,000 J, or 15 kJ, of heat.
A 4 kg substance receives 24,000 J of heat. Its specific heat capacity is 1,000 J/(kg·°C). Find its temperature change.
Step 1: Start with the specific heat formula
Q = mcΔT
Step 2: Rearrange the formula to solve for temperature change
ΔT = Q / (mc)
Step 3: Substitute the known values
ΔT = 24,000 / (4 × 1,000)
Step 4: Calculate the denominator
4 × 1,000 = 4,000
Step 5: Calculate the temperature change
ΔT = 24,000 / 4,000
ΔT = 6°C
Answer: The temperature of the substance increases by 6°C.
A 3 kg object receives 18,000 J of heat and increases in temperature by 20°C. Find its specific heat capacity.
Step 1: Start with the specific heat formula
Q = mcΔT
Step 2: Rearrange the formula to solve for specific heat capacity
c = Q / (mΔT)
Step 3: Substitute the known values
c = 18,000 / (3 × 20)
Step 4: Calculate the denominator
3 × 20 = 60
Step 5: Calculate
c = 18,000 / 60
c = 300 J/(kg·°C)
Answer: The specific heat capacity is 300 J/(kg·°C).
A 2 kg substance changes phase. Its latent heat is 225,000 J/kg. Find the energy required for the phase change.
Step 1: Use the latent heat formula
Q = mL
Step 2: Substitute the known values
Q = 2 × 225,000
Step 3: Calculate
Q = 450,000 J
Step 4: Convert to kilojoules
450,000 J ÷ 1,000 = 450 kJ
Answer: The phase change requires 450,000 J, or 450 kJ, of energy.
A 0.5 kg substance requires 100,000 J of energy to complete a phase change. Find its latent heat.
Step 1: Start with the latent heat formula
Q = mL
Step 2: Rearrange the formula to solve for latent heat
L = Q / m
Step 3: Substitute the known values
L = 100,000 / 0.5
Step 4: Calculate
L = 200,000 J/kg
Answer: The latent heat of the substance is 200,000 J/kg.
A machine receives 5,000 J of energy and produces 1,750 J of useful energy output. Find the machine's efficiency.
Step 1: Use the efficiency formula
η = Useful Energy Output / Energy Input
Step 2: Substitute the known values
η = 1,750 / 5,000
Step 3: Calculate the decimal efficiency
η = 0.35
Step 4: Convert the decimal to a percentage
Efficiency = 0.35 × 100%
Efficiency = 35%
Answer: The machine is 35% efficient.
A heat engine receives 8,000 J of energy and operates at an efficiency of 30%. Find its useful energy output.
Step 1: Convert the percentage to a decimal
30% = 0.30
Step 2: Start with the efficiency formula
η = Useful Energy Output / Energy Input
Step 3: Rearrange the formula to solve for useful energy output
Useful Energy Output = η × Energy Input
Step 4: Substitute the known values
Useful Energy Output = 0.30 × 8,000
Step 5: Calculate
Useful Energy Output = 2,400 J
Answer: The heat engine produces 2,400 J of useful energy.
A machine produces 3,000 J of useful energy and has an efficiency of 60%. Find the total energy supplied to the machine.
Step 1: Convert the percentage to a decimal
60% = 0.60
Step 2: Start with the efficiency formula
η = Useful Energy Output / Energy Input
Step 3: Rearrange the formula to solve for energy input
Energy Input = Useful Energy Output / η
Step 4: Substitute the known values
Energy Input = 3,000 / 0.60
Step 5: Calculate
Energy Input = 5,000 J
Answer: The machine receives 5,000 J of energy.
Pressure:
P = F / A
P = Pressure (Pa or N/m²)
F = Force acting on the surface (N)
A = Area over which the force acts (m²)
Units: N / m² = Pa
Force from Pressure:
F = PA
F = Force (N)
P = Pressure (Pa or N/m²)
A = Area (m²)
Units: N/m² × m² = N
Area from Pressure:
A = F / P
A = Area (m²)
F = Force (N)
P = Pressure (Pa or N/m²)
Units: N ÷ N/m² = m²
First Law of Thermodynamics:
ΔU = Q − W
ΔU = Change in Internal Energy (J)
Q = Heat added to the system (J)
W = Work performed by the system (J)
Units: J − J = J
Heat from the First Law:
Q = ΔU + W
Q = Heat added to the system (J)
ΔU = Change in Internal Energy (J)
W = Work performed by the system (J)
Units: J + J = J
Work from the First Law:
W = Q − ΔU
W = Work performed by the system (J)
Q = Heat added to the system (J)
ΔU = Change in Internal Energy (J)
Units: J − J = J
Specific Heat:
Q = mcΔT
Q = Heat transferred (J)
m = Mass of the substance (kg)
c = Specific heat capacity (J/(kg·°C) or J/(kg·K))
ΔT = Temperature change (°C or K)
Units:
kg × J/(kg·°C) × °C = J
Temperature Change:
ΔT = Q / (mc)
ΔT = Temperature change (°C or K)
Q = Heat transferred (J)
m = Mass (kg)
c = Specific heat capacity (J/(kg·°C) or J/(kg·K))
Units:
J ÷ [kg × J/(kg·°C)] = °C
Specific Heat Capacity:
c = Q / (mΔT)
c = Specific heat capacity (J/(kg·°C) or J/(kg·K))
Q = Heat transferred (J)
m = Mass (kg)
ΔT = Temperature change (°C or K)
Units:
J ÷ (kg × °C) = J/(kg·°C)
Mass from Specific Heat:
m = Q / (cΔT)
m = Mass (kg)
Q = Heat transferred (J)
c = Specific heat capacity (J/(kg·°C) or J/(kg·K))
ΔT = Temperature change (°C or K)
Units:
J ÷ [J/(kg·°C) × °C] = kg
Phase Change Energy:
Q = mL
Q = Heat transferred during the phase change (J)
m = Mass of the substance (kg)
L = Latent heat of the substance (J/kg)
Units: kg × J/kg = J
Latent Heat:
L = Q / m
L = Latent heat (J/kg)
Q = Heat transferred during the phase change (J)
m = Mass of the substance (kg)
Units: J ÷ kg = J/kg
Mass During a Phase Change:
m = Q / L
m = Mass of the substance (kg)
Q = Heat transferred during the phase change (J)
L = Latent heat (J/kg)
Units: J ÷ J/kg = kg
Efficiency as a Decimal:
η = Euseful / Einput
η = Efficiency as a decimal (no unit)
Euseful = Useful energy output (J)
Einput = Total energy input (J)
Units: J / J = no unit
Efficiency as a Percentage:
Efficiency (%) =
(Euseful / Einput) × 100%
Euseful = Useful energy output (J)
Einput = Total energy input (J)
Units: J / J × 100% = %
Useful Energy Output:
Euseful = ηEinput
Euseful = Useful energy output (J)
η = Efficiency written as a decimal (no unit)
Einput = Total energy input (J)
Units: no unit × J = J
Energy Input:
Einput = Euseful / η
Einput = Total energy input (J)
Euseful = Useful energy output (J)
η = Efficiency written as a decimal (no unit)
Units: J ÷ no unit = J
Temperature Difference:
ΔT = Tfinal − Tinitial
ΔT = Temperature change (°C or K)
Tfinal = Final temperature (°C or K)
Tinitial = Initial temperature (°C or K)
Units: °C − °C = °C, or K − K = K
Common Energy Conversion:
1 kJ = 1,000 J
kJ = Kilojoule
J = Joule
Common Pressure Conversion:
1 kPa = 1,000 Pa
kPa = Kilopascal
Pa = Pascal
Important Sign Convention:
For ΔU = Q − W:
Positive Q means heat enters the system.
Negative Q means heat leaves the system.
Positive W means the system performs work.
Negative W means work is performed on the system.
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Thermodynamics is used whenever engineers study how energy, heat, temperature, pressure, and work move through or affect a system. It helps engineers design engines, refrigeration systems, turbines, power plants, heating systems, and many other machines that transfer or convert energy.
Automobile engines convert chemical energy from fuel into thermal energy and then into mechanical work. Engineers study combustion, pressure, temperature, and efficiency to improve engine performance.
Concepts used:
Refrigerators remove heat from the inside compartment and transfer it to the surrounding room. A refrigerant circulates through the system and changes pressure, temperature, and phase.
Concepts used:
Air-conditioning systems cool buildings by removing heat from indoor air and releasing it outdoors. Engineers design these systems to control temperature while using as little energy as possible.
Concepts used:
Many power plants convert thermal energy into mechanical work and then into electrical energy. Heat may be used to produce steam that rotates a turbine connected to a generator.
Concepts used:
Turbines use high-energy fluids such as steam or hot gas to rotate blades and produce shaft work. Engineers analyze the energy entering and leaving the turbine to improve performance.
Concepts used:
Heat pumps transfer heat from one location to another. They can move heat into a building during cold weather and remove heat during warm weather.
Concepts used:
Boilers and water heaters transfer energy from fuel or electricity into water. The added heat raises the water temperature or changes liquid water into steam.
Concepts used:
Aircraft and rocket engines release energy through combustion and use expanding gases to produce thrust. Engineers study high temperatures, pressures, heat transfer, and energy efficiency.
Concepts used:
Solar thermal systems collect radiation from the Sun and use it to heat water, air, or another working fluid. Engineers design collectors and storage systems to capture and use this energy efficiently.
Concepts used:
Manufacturing processes often involve heating, cooling, melting, freezing, drying, or compressing materials. Engineers use thermodynamics to control temperatures and energy use during production.
Concepts used:
Computers and electronic devices produce heat while operating. Engineers use fans, heat sinks, cooling liquids, and thermal materials to prevent components from overheating.
Concepts used:
Insulation slows heat transfer between a building and the outdoor environment. Engineers select materials that help buildings remain warmer in winter and cooler in summer.
Concepts used:
Piston-cylinder devices are used in engines, compressors, and laboratory equipment. A gas inside the cylinder can expand or compress while heat and work are transferred.
Concepts used:
Cooking and food storage both involve thermodynamics. Ovens transfer heat into food, while refrigerators and freezers remove heat to slow spoilage.
Concepts used:
This section can later include your own demonstrations, such as comparing how quickly different materials heat up, observing conduction through metal, measuring cooling over time, or testing insulation materials.
Concepts used:
This video demonstrates thermodynamics by showing how heat naturally transfers from a hotter object to a colder object until both reach the same temperature. Using only two cups of water and a thermometer, this simple experiment illustrates heat transfer and thermal equilibrium.
Fill one cup with warm water at approximately 65°C and another cup with cold water at approximately 15°C. Measure and show both temperatures on camera. Pour the two cups together, stir the mixture, and measure the final temperature. The water should settle at approximately 40°C, demonstrating that heat naturally transfers from the warmer water to the cooler water until thermal equilibrium is reached.