02Introduction to Thermodynamics
Apply the zeroth, first, and second laws of thermodynamics to analyze heat engines, entropy, and energy efficiency in thermal systems.
Thermodynamics governs every engine, refrigerator, and power plant. Its laws set fundamental limits on what technology can achieve.
Essential Question: What fundamental limits do the laws of thermodynamics place on the conversion of heat into useful work, and why can no real engine ever be perfectly efficient?
Lesson Overview
Thermodynamics is the branch of physics that studies the relationships between heat, work, and internal energy. Every engine, refrigerator, and living cell operates under its laws. In this lesson you will classify thermodynamic systems, apply the first law (energy conservation) to calculate internal energy changes, analyze four idealized processes (isothermal, adiabatic, isochoric, isobaric), and use the second law to understand why heat flows spontaneously from hot to cold and why entropy always increases. You will also calculate the maximum possible efficiency of a heat engine using the Carnot formula, explore refrigerators and heat pumps through their coefficients of performance, and connect entropy to the third law of thermodynamics.
Four Thermodynamic Processes
Isothermal
Constant: Temperature (T = const)
W = nRT ln(V₂/V₁)
ΔU = 0 → Q = W
Adiabatic
Constant: No heat exchange (Q = 0)
W = −ΔU
ΔU = −W (gas cools when expanding)
Isochoric
Constant: Volume (V = const)
W = 0 (no expansion)
ΔU = Q
Isobaric
Constant: Pressure (P = const)
W = PΔV
ΔU = Q − PΔV
Laws of Thermodynamics
If system A is in thermal equilibrium with system B, and B with C, then A is in equilibrium with C. This defines temperature.
Energy is conserved: ΔU = Q − W. The change in internal energy equals heat added to the system minus work done by the system.
The total entropy of an isolated system never decreases. Heat flows spontaneously from hot to cold; no engine converts heat entirely into work.
As temperature approaches absolute zero (0 K), the entropy of a perfect crystal approaches zero. Absolute zero is unattainable in a finite number of steps.
Key Equations
ΔU = Q − W
First law of thermodynamics
W = PΔV
Work done by gas at constant pressure
η = W/Q_h = 1 − Q_c/Q_h
Thermal efficiency of a heat engine
η_Carnot = 1 − T_c/T_h
Maximum (Carnot) efficiency
ΔS = Q/T
Entropy change (reversible process)
Worked Examples
A gas absorbs Q = 500 J of heat and does W = 200 J of work on its surroundings. What is the change in internal energy ΔU?
Identify: Q = +500 J (heat added), W = +200 J (work done by gas)
Apply the first law: ΔU = Q − W
ΔU = 500 J − 200 J
ΔU = 300 J
A gas expands isobarically (constant pressure) at P = 2 × 10⁵ Pa. The volume increases by ΔV = 0.003 m³. How much work does the gas do?
Identify: P = 2 × 10⁵ Pa, ΔV = 0.003 m³
Apply: W = PΔV
W = (2 × 10⁵ Pa)(0.003 m³)
W = 600 J
A heat engine absorbs Q_h = 1000 J from a hot reservoir and exhausts Q_c = 600 J to a cold reservoir. Find the work output W and the thermal efficiency η.
Work output: W = Q_h − Q_c = 1000 J − 600 J = 400 J
Efficiency: η = W / Q_h = 400 J / 1000 J = 0.40
Convert to percent: η = 40%
A Carnot engine operates between a hot reservoir at T_h = 500 K and a cold reservoir at T_c = 300 K. What is the maximum (Carnot) efficiency?
Apply Carnot efficiency formula: η_Carnot = 1 − T_c / T_h
η_Carnot = 1 − 300 K / 500 K
η_Carnot = 1 − 0.60 = 0.40
Convert to percent: η_Carnot = 40%
800 J of heat flows reversibly into a thermal reservoir at a constant temperature of T = 400 K. What is the entropy change ΔS of the reservoir?
Identify: Q = +800 J (heat added), T = 400 K
Apply: ΔS = Q / T
ΔS = 800 J / 400 K
ΔS = 2 J/K
Guided Problems
A gas releases Q = 300 J of heat to its surroundings and has W = 100 J of work done on it (compression). Find ΔU.
Hint: When heat is released, Q is negative (Q = −300 J). When work is done ON the gas, W is negative (W = −100 J). Apply ΔU = Q − W carefully with signs.
A piston compresses a gas at constant pressure P = 1.5 × 10⁵ Pa, reducing the volume by ΔV = 0.002 m³. How much work is done on the gas?
Hint: The gas is compressed, so ΔV is negative: ΔV = −0.002 m³. Use W = PΔV to find the work done by the gas, then note that work done on the gas is the negative of that.
A heat engine operates between T_h = 800 K and T_c = 400 K. What is the Carnot efficiency? If Q_h = 2000 J, what is the maximum work output?
Hint: First find η_Carnot = 1 − T_c/T_h. Then use W_max = η_Carnot × Q_h.
A refrigerator removes Q_c = 500 J from a cold reservoir and requires W = 200 J of work input. What is the coefficient of performance (COP)?
Hint: For a refrigerator, COP = Q_c / W. Substitute the given values directly.
600 J of heat flows out of a hot reservoir at T = 300 K. What is the entropy change of the reservoir?
Hint: Heat leaves the reservoir, so Q is negative: Q = −600 J. Apply ΔS = Q/T.
Key Vocabulary
Internal Energy (U)
The total microscopic kinetic and potential energy of all particles in a system. It depends on temperature and the state of the system.
Example: ΔU = Q − W; if Q = 500 J and W = 200 J, then ΔU = 300 J
First Law of Thermodynamics
Energy is conserved: the change in internal energy equals heat added to the system minus work done by the system (ΔU = Q − W).
Example: Gas absorbs 400 J and does 150 J of work → ΔU = 250 J
Second Law of Thermodynamics
The total entropy of an isolated system always increases or stays the same. Heat flows spontaneously only from hot to cold objects.
Example: A hot coffee cup cools in a room — entropy of the universe increases
Entropy (S)
A measure of the disorder or number of possible microscopic arrangements of a system. For a reversible process: ΔS = Q/T.
Example: 800 J absorbed at 400 K → ΔS = 2 J/K
Heat Engine
A device that converts heat from a hot reservoir into work while exhausting some heat to a cold reservoir. Efficiency η = W/Q_h = 1 − Q_c/Q_h.
Example: Q_h = 1000 J, Q_c = 600 J → W = 400 J, η = 40%
Carnot Efficiency
The maximum theoretical efficiency of a heat engine operating between two temperatures: η_Carnot = 1 − T_c/T_h (temperatures in Kelvin).
Example: T_h = 500 K, T_c = 300 K → η_Carnot = 40%
Isothermal Process
A thermodynamic process that occurs at constant temperature. For an ideal gas, ΔU = 0, so all heat absorbed equals work done: Q = W.
Example: Slow expansion of gas in a heat bath at constant T
Adiabatic Process
A thermodynamic process with no heat exchange between the system and surroundings (Q = 0). The first law gives ΔU = −W.
Example: Rapid compression of gas in an insulated cylinder
Workbook Check — Interactive Quiz
Interactive Practice — 5 Questions
A gas absorbs Q = 800 J of heat and does W = 300 J of work. What is ΔU?
A gas expands at constant pressure P = 3 × 10⁵ Pa with ΔV = 0.002 m³. What work does the gas do?
A Carnot engine operates between T_h = 600 K and T_c = 300 K. What is its efficiency?
Which thermodynamic process has no heat exchange with the surroundings?
600 J of heat flows reversibly into a reservoir at T = 200 K. What is the entropy change?
Independent Practice
A gas absorbs 800 J of heat and does 300 J of work on its surroundings. What is the change in internal energy of the gas?
A heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. Calculate the maximum (Carnot) efficiency.
A Carnot engine has an efficiency of 40% and absorbs 1,000 J of heat per cycle. Find (a) the work done per cycle and (b) the heat rejected to the cold reservoir.
An ideal gas undergoes an isothermal expansion at 400 K, absorbing 500 J of heat. Calculate the change in entropy of the gas.
★ A heat engine operates between 800 K and 400 K. In each cycle it absorbs 2,400 J from the hot reservoir. (a) Calculate the Carnot efficiency. (b) Find the maximum work output per cycle. (c) Find the heat rejected per cycle. (d) If the actual efficiency is 30%, how much extra heat is rejected compared to the Carnot engine? (e) Explain why the actual efficiency is always less than the Carnot efficiency.
ChallengeCommon Mistakes
Thinking the 1st law ΔU = Q − W means work always decreases internal energy
W is positive when the gas does work ON the surroundings (expansion). If the surroundings do work ON the gas (compression), W is negative and ΔU increases
Calculating Carnot efficiency using Celsius temperatures
Carnot efficiency: η = 1 − T_C/T_H requires ABSOLUTE temperatures in Kelvin. Using Celsius gives a wrong answer
Believing a heat engine can be 100% efficient if well-engineered
The 2nd law forbids 100% efficiency for any heat engine operating between two reservoirs. Maximum efficiency is the Carnot efficiency
Confusing isothermal (ΔT = 0, so ΔU = 0) with adiabatic (Q = 0) processes
Isothermal: temperature constant, Q = W. Adiabatic: no heat exchange, ΔU = −W. These are different processes with different constraints
Math Tips
First law: ΔU = Q − W. Sign convention: Q > 0 when heat flows INTO the system; W > 0 when the system does work (expands). Both increase/decrease U accordingly
Carnot efficiency: η_C = 1 − T_C/T_H (Kelvin!). This is the MAXIMUM possible efficiency. Real engines are always less efficient due to irreversibilities
COP for refrigerators: COP = Q_C/W = T_C/(T_H − T_C). For heat pumps: COP = Q_H/W = T_H/(T_H − T_C). Both use Kelvin temperatures
Entropy change: ΔS = Q_rev/T (reversible process). For the universe: ΔS_universe ≥ 0 always. Spontaneous processes increase total entropy