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Class XI · Chapter 11

Thermodynamics

Why no engine — not even a hypothetical perfect one — can turn heat completely into work, and the one honest upper limit every real engine has to live under.

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1Thermal Equilibrium and the Zeroth Law

Two objects are in thermal equilibrium when no net heat flows between them — meaning they're at the same temperature. This sounds almost too obvious to state as a "law," but it's exactly what makes the entire concept of temperature logically sound.

Zeroth Law of Thermodynamics If object A is in thermal equilibrium with object B, and object B is in thermal equilibrium with object C, then A is also in thermal equilibrium with C.

This is precisely why a thermometer works at all: it reaches equilibrium with whatever it's measuring, and by the zeroth law, its reading tells you something objectively true about that object's temperature — not just about the thermometer itself.

2Heat, Internal Energy and Work

Internal energy (U) is the total energy contained within a system — the kinetic and potential energy of all its particles, on the microscopic scale. Unlike heat or work, internal energy is a genuine property of the system's current state, regardless of how it got there. Heat and work are simply two different routes by which energy can be added to or removed from that internal energy store.

3The First Law of Thermodynamics

The first law is nothing more than energy conservation, applied specifically to heat and work.

First Law of Thermodynamics
ΔU = Q − W
Q = heat added to the system, W = work done by the system on its surroundings. Add heat, and internal energy rises; have the system do work, and internal energy falls — accounted for exactly, with nothing created or lost.
Work Done by an Expanding Gas
W = ∫ P dV
For a constant-pressure process specifically, this simplifies to W = PΔV.

4Specific Heats of a Gas

Gases need two different specific heats, depending on whether they're heated at constant volume or constant pressure — because at constant pressure, some of the added heat goes into doing expansion work rather than raising temperature.

Mayer's Relation
Cp − Cv = R
Cp (constant pressure) is always larger than Cv (constant volume) for exactly this reason — heating at constant pressure has to supply extra energy for the gas to expand against its surroundings.

5Thermodynamic Processes

A thermodynamic process describes how a system moves from one state to another. Several idealised types come up repeatedly:

  • Isothermal: temperature constant throughout. For an ideal gas, ΔU = 0, so Q = W entirely.
  • Adiabatic: no heat exchange with surroundings (Q = 0), so ΔU = −W — any work done comes entirely at the expense of internal energy, changing temperature.
  • Isochoric: constant volume, so no expansion work is done (W = 0), meaning ΔU = Q entirely.
  • Isobaric: constant pressure, with W = PΔV.
  • Cyclic: the system eventually returns to its exact starting state, so ΔU = 0 over the full cycle, meaning net heat absorbed equals net work done — the foundation of every heat engine.

6Heat Engines

A heat engine converts heat into useful work by operating in a repeating cycle, absorbing heat from a hot reservoir and rejecting some (never all) of it to a cold reservoir.

Efficiency
η = W / Qh = 1 − Qc/Qh
Qh = heat absorbed from the hot reservoir, Qc = heat rejected to the cold reservoir. No real engine reaches η = 1 (100%) — some heat rejection is always unavoidable, as the second law (Section 8) makes precise.

7Refrigerators and Heat Pumps

A refrigerator is essentially a heat engine run in reverse: instead of producing work from a heat flow, it consumes work to force heat to move from a cold space to a warmer one — something that would never happen spontaneously.

Coefficient of Performance
COP = Qc / W = Qc / (Qh − Qc)
Qc = heat extracted from the cold space, W = work input. Unlike engine efficiency, COP can exceed 1 — a refrigerator can move more heat energy than the work it consumes, which is exactly why refrigerators and heat pumps are more energy-efficient than direct electrical heating.

8The Second Law of Thermodynamics

The first law only forbids creating or destroying energy — it says nothing about which energy conversions are actually possible in practice. The second law fills that gap, and can be stated in two equivalent ways:

Kelvin-Planck Statement No engine operating in a cycle can convert heat completely into work, with 100% efficiency — some heat must always be rejected to a cold reservoir.
Clausius Statement Heat cannot spontaneously flow from a colder body to a hotter one without external work being done — a refrigerator always needs power input; it can never run "for free."

9Reversible and Irreversible Processes

A reversible process could, in principle, be run backward through exactly the same sequence of states, leaving no net change anywhere in the universe — an idealisation, since it requires infinitely slow, perfectly balanced (quasi-static) changes. Every real process is, to some degree, irreversible — friction, turbulence, and rapid unbalanced changes all generate some permanent, one-directional loss, which is exactly why real engines can never quite reach the theoretical Carnot efficiency below.

10The Carnot Engine

The Carnot engine is a theoretical, perfectly reversible heat engine, operating between two fixed temperatures — and it sets the absolute upper bound on efficiency that any real engine, however cleverly designed, can never exceed.

Carnot Efficiency
η = 1 − Tc/Th
Temperatures in kelvin. Efficiency depends only on the two reservoir temperatures — never on the working substance or engine design. A larger temperature difference always allows for higher possible efficiency, which is exactly why power plants aim for the hottest practical operating temperature.

Formula Summary

First Law
ΔU = Q − W
Mayer's Relation
Cp − Cv = R
Isothermal (Ideal Gas)
ΔU = 0, Q = W
Adiabatic
Q = 0, ΔU = −W
Engine Efficiency
η = 1 − Qc/Qh
Refrigerator COP
COP = Qc/W
Carnot Efficiency
η = 1 − Tc/Th

Solved Examples

Example 1 · First Law of Thermodynamics

A gas absorbs 500 J of heat and does 200 J of work on its surroundings. Find the change in its internal energy.

Solution: ΔU = Q − W = 500 − 200 = 300 J.

Example 2 · Carnot Engine Efficiency

A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. Find its efficiency.

Solution: η = 1 − Tc/Th = 1 − 300/500 = 1 − 0.6 = 0.4 = 40%.

No real engine operating between these same two temperatures could ever exceed 40% efficiency — this is the absolute theoretical ceiling.

Quick Check

1. In an adiabatic process:
Temperature is constant
Volume is constant
✓ No heat is exchanged with the surroundings (correct)
Pressure is constant
2. According to the second law of thermodynamics:
Energy can be created in a cyclic process
✓ No engine can convert heat completely into work (correct)
Heat always flows from cold to hot
Internal energy is always conserved
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Thermal Properties of Matter

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