Thermodynamics · Chapter 12

The Laws of Thermodynamics

Thermodynamics governs the direction of natural processes and the ultimate limits of heat engines — including every engine ever built or conceivable.

PrerequisitesHeat & Temperature (Ch. 10) \cdot Ideal Gas Law (Ch. 11) \cdot Familiarity with calculus notation helps for entropy; no integrals required beyond recognition
Learning Goals
  • State all four laws of thermodynamics and explain the physical principle each encodes.
  • Applythefirstlaw\DeltaU=QWtoisothermal,isochoric,andisobaricprocessesApply the first law \DeltaU = Q - W to isothermal, isochoric, and isobaric processes.
  • Calculate Carnot efficiency from reservoir temperatures and identify it as the upper bound.
  • ComputeentropychangesforreversibleheattransfersusingdS=dQrev/TCompute entropy changes for reversible heat transfers using dS = dQ_rev/T.
  • Analyze refrigerator and heat pump performance using coefficients of performance.

12.1 The Zeroth and First Laws

The laws of thermodynamics are numbered oddly because the first two were well established before physicists recognized that an even more fundamental principle — the zeroth law — was being assumed without statement. The zeroth law defines temperature itself.

Definition 12.1Zeroth Law of Thermodynamics
If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other. This law is what makes temperature a well-defined, transitive quantity — it licenses the use of thermometers.

The first law is the conservation of energy, stated for thermodynamic systems. It formalizes the equivalence of heat and work established experimentally by Joule in 1843.

Definition 12.2First Law of Thermodynamics
Thechangeininternalenergy\DeltaUofasystemequalstheheatQaddedtothesystemminustThe change in internal energy \DeltaU of a system equals the heat Q added to the system minus the work W done by the system:
ΔU=QW\Delta U=Q-W(12.1)

The sign convention matters: Q > 0 means heat flows into the system; W > 0 means the system does work on its surroundings. For a gas expanding against pressure, W = ∫P dV.

For a process at constant pressure (isobaric), W = PΔV, so ΔU = Q − PΔV. For a constant-volume (isochoric) process, W = 0, so all heat goes directly into internal energy. For an isothermal process, ΔU = 0 (for an ideal gas), so Q = W — all heat input is converted to work.

12.2 The Second Law and Entropy

The first law says energy is conserved, but it says nothing about direction. A hot coffee cup cools to room temperature; a cold cup never spontaneously heats itself by drawing heat from the room, even though energy conservation permits it. The second law captures this directional asymmetry.

Definition 12.3Second Law of Thermodynamics (Clausius Statement)
Heat never flows spontaneously from a cold body to a hot body. Equivalently: no process is possible whose sole result is the transfer of heat from a cooler body to a warmer one.
Definition 12.4Second Law (Kelvin–Planck Statement)
No heat engine operating in a cycle can convert heat entirely into work. There must always be some heat rejected to a cold reservoir.

These statements are equivalent. The quantitative measure of the second law is entropy, introduced by Clausius in 1865. For a reversible process:

dS=dQrevTdS=\frac{dQ_\mathrm{rev}}{T}(12.2)
Theorem 12.1Entropy and the Second Law
For any process in an isolated system, the total entropy either increases or remains constant — it never decreases:\DeltaStotal0\DeltaS_total \ge 0Equality holds for reversible (quasi-static) processes. All real processes are irreversible and increase the total entropy of the universe.

Entropy has a statistical interpretation, given by Boltzmann: S = k_B ln Ω, where Ω is the number of microscopic configurations (microstates) consistent with the macroscopic state, and k_B = 1.38 × 10⁻²³ J/K. Systems evolve toward states with overwhelmingly more microstates — not because of a mysterious force, but because probability demands it.

12.3 Heat Engines and the Carnot Cycle

A heat engine is any device that converts heat into work by operating in a thermodynamic cycle. It absorbs heat Q_h from a hot reservoir at temperature T_h, converts some of it into net work W_net, and rejects the remainder Q_c to a cold reservoir at T_c. By energy conservation: W_net = Q_h − Q_c.

The thermal efficiency η of a heat engine is the ratio of net work output to heat absorbed:

η=WnetQh=1QcQh\eta=\frac{W_\mathrm{net}}{Q_h}=1-\frac{Q_c}{Q_h}(12.3)

The question of how efficient an engine can theoretically be was answered by Sadi Carnot in 1824 — decades before the first and second laws were formally stated. The Carnot cycle is the most efficient possible cycle operating between two temperature reservoirs. It consists of four reversible steps: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

Figure 12.1. The Carnot cycle on a P–V diagram. The shaded area equalsthenetworkoutputpercycle.Adjustthereservoirtemperaturestoseehowefficiencyηs the net work output per cycle. Adjust the reservoir temperatures to see how efficiency \etaime.
Theorem 12.2Carnot Efficiency
The maximum possible efficiency of any heat engine operating between reservoirs at temperatures T_c (in kelvin) is the Carnot efficiency:ηCarnot=1Tc/Th\eta_Carnot = 1 - T_{c} / T_{h}No engine can exceed this limit. A real engine always falls short because real processes are irreversible — they generate entropy.

This is a profound result. The maximum efficiency depends only on the temperatures of the two reservoirs — not on the working substance, the pressure, the volume, or any other detail of the engine. A steam turbine with T_h = 800 K and T_c = 300 K cannot exceed η = 1 − 300/800 = 62.5%, no matter how well engineered.

Example 12.1Carnot Engine Efficiency

A coal-fired power plant operates with steam at 580°C and rejects heat to a river at 20°C. What is the maximum possible thermal efficiency?

Convert to kelvin:Th=580+273=853KTc=20+273=293KT_{h} = 580 + 273 = 853 K \qquad T_{c} = 20 + 273 = 293 K
Carnot efficiency:η=1Tc/Th=1293/853=10.344=0.656=65.6\eta = 1 - T_{c}/T_{h} = 1 - 293/853 = 1 - 0.344 = 0.656 = 65.6%
Interpretation:Even in principle, 34.4% of the fuel's heat energy must be rejected to the river. Real plants achieve 35–45% due to additional irreversibilities.

12.4 Refrigerators and Heat Pumps

A refrigerator runs a heat engine in reverse: it uses work input to move heat from a cold reservoir to a hot one. By the second law, this requires net work input — you cannot cool your kitchen by leaving the refrigerator door open. The coefficient of performance (COP) of a refrigerator is:

COPref=QcW=QcQhQc\mathrm{COP}_\mathrm{ref}=\frac{Q_c}{W}=\frac{Q_c}{Q_h-Q_c}(12.4)

A heat pump also moves heat from cold to hot, but the desired output is Q_h delivered to the hot space (e.g., heating a building). Its COP is:

COPhp=QhW=1+COPref\mathrm{COP}_\mathrm{hp}=\frac{Q_h}{W}=1+\mathrm{COP}_\mathrm{ref}(12.5)

The Carnot COP sets the upper bound: COP_ref,max = T_c / (T_h − T_c) and COP_hp,max = T_h / (T_h − T_c). Under normal operating conditions, a heat pump can deliver more heat to a building per joule of electrical input than direct electrical resistance heating (COP = 1), which is why ground-source heat pumps often deliver 3–5 units of heat per unit of electrical energy.

12.5 The Third Law

Definition 12.5Third Law of Thermodynamics (Nernst's Theorem)
Asthetemperatureofasystemapproachesabsolutezero(T0K),itsentropyapproachesAs the temperature of a system approaches absolute zero (T \to 0 K), its entropy approaches a minimum value — typically zero for a perfect crystal in its ground state:limT0S=0lim_{T\to0} S = 0A consequence: absolute zero is unattainable by any finite sequence of processes.

The third law has practical importance: it sets the reference point for absolute entropy measurements, and it explains why liquefying helium requires progressively more effort as you approach 0 K — each stage of cooling is less effective than the last.

Example 12.2Entropy Change in Heat Transfer

100 J of heat flows from a reservoir at 400 K to a reservoir at 200 K. What is the total entropy change of the universe?

Hot reservoir loses heat:\DeltaSh=Q/Th=100/400=0.250J/K\DeltaS_h = -Q/T_{h} = -100/400 = -0.250 J/K
Cold reservoir gains heat:\DeltaSc=+Q/Tc=+100/200=+0.500J/K\DeltaS_c = +Q/T_{c} = +100/200 = +0.500 J/K
Total:\DeltaStotal=0.250+0.500=+0.250J/K\DeltaS_total = -0.250 + 0.500 = +0.250 J/K > 0 ✓
Lesson:This irreversible process increased the entropy of the universe by 0.25 J/K. No process can reverse this — the entropy increase is permanent.
Definition 12.6Common Traps
  • Sign convention matters: hereWisworkdonebythesystem,soexpansionworkreducesinternalenergyifQ=0here W is work done by the system, so expansion work reduces internal energy if Q = 0
  • Carnot temperatures must be kelvin: Celsius ratios give physically meaningless efficiencies.
  • Entropy of a subsystem can decrease: the second law constrains the total entropy of an isolated system.
  • COP can exceed 1: refrigerators and heat pumps move heat; they do not convert work directly into heat output one-for-one.
Exercises — 12.1–12.5 Laws of Thermodynamics
1.
A heat engine operates between 580°C and 20°C. What is its maximum possible efficiency?
%
Straightforward
2.
ACarnotenginerejectsheattoa300Kreservoirandhasefficiency25A Carnot engine rejects heat to a 300 K reservoir and has efficiency 25%. What is T_{h}?
K
Straightforward
3.
100 J of heat flows irreversibly from a 400 K reservoir to a 200 K reservoir. What is the total entropy change of the universe?
J/K
Intermediate
4.Aheatpumpheatsahouseat20°C(293K)bydrawingfromoutdoorairat10°C(263K).CaA heat pump heats a house at 20°C (293 K) by drawing from outdoor air at -10°C (263 K). Calculate the maximum COP and compare to resistance heating.
Intermediate
5.AheatengineabsorbsQh=500Jfroma600KreservoirandrejectsQc=300Jtoa300A heat engine absorbs Q_{h} = 500 J from a 600 K reservoir and rejects Q_{c} = 300 J to a 300 K reservoir per cycle. Find: (a) actual efficiency, (b) Carnot efficiency, (c) entropy generated per cycle.
Challenging
Key Takeaways
  • Zeroth law: thermal equilibrium is transitive — this defines temperature.
  • Firstlaw:\DeltaU=QW.Energyisconserved;heatandworkareequivalentFirst law: \DeltaU = Q - W. Energy is conserved; heat and work are equivalent.
  • Second law: entropy of an isolated system never decreases. Heat flows spontaneously only from hot to cold.
  • EntropyS=kBlnΩconnectsmacroscopicthermodynamicstomicroscopicprobabilityEntropy S = k_{B} ln \Omega connects macroscopic thermodynamics to microscopic probability.
  • Carnotefficiencyη=1Tc/ThisthemaximumpossibleforanyheatenginesetbythesCarnot efficiency \eta = 1 - Tc/Th is the maximum possible for any heat engine — set by the second law alone.
  • Thirdlaw:S0asT0K,makingabsolutezerounattainablebyanyfiniteprocessThird law: S \to 0 as T \to 0 K, making absolute zero unattainable by any finite process.