Thermal equilibrium

Put a spoon into a hot cup of tea and, within a minute or two, you can no longer tell by touch which is hotter — spoon or tea. Nothing dramatic happened; no alarm went off. Heat simply flowed from the tea into the cooler metal until both settled at one common temperature, and then the flow stopped on its own, quietly, without anyone switching it off. That quiet stopping point is one of the deepest ideas in all of physics, and it is where the CBSE Class 11 chapter on Thermodynamics begins: thermal equilibrium. Thermodynamics as a subject is built entirely on describing systems using a handful of macroscopic numbers — pressure $P$ , volume $V$ , temperature $T$ — rather than tracking every one of the roughly $10^{23}$ molecules bouncing around inside a gas or a solid. That shortcut only makes sense once you know when those numbers are allowed to be treated as fixed, single values for a system. The answer is: when the system is in equilibrium. And of the three flavours of equilibrium — mechanical, chemical, thermal — it is thermal equilibrium that quietly defines the very meaning of "temperature," the quantity a thermometer reads and the quantity that appears in every equation of state you will meet for the rest of the chapter. This article builds thermal equilibrium up carefully: what it means for a system's state variables to "stop changing," how a diathermic wall lets two bodies talk to each other while an adiabatic wall keeps them deaf to each other, and how conservation of energy turns "two bodies reach a common temperature" into a solvable calorimetry equation. Along the way we preserve and extend the two worked numerical examples from the original treatment of this topic — the two-block common-temperature problem and the small-thermometer-in-a-large-bath argument — and add a board-style and a competitive-style problem built on the very same physics, so that the ideas that seem obvious in words become second nature as calculations.

Key takeaways

A system is in equilibrium when its macroscopic state variables ( $P$ , $V$ , $T$ ) stay constant with time as long as external conditions are unchanged.
Thermal equilibrium specifically means no net flow of heat between a system and its surroundings, or between two systems in contact — molecules still exchange energy, but the net exchange is zero.
Two systems joined by a diathermic wall (heat-conducting) exchange heat until their temperatures equalise; joined by an adiabatic wall (insulating) they stay at whatever temperature each started with, forever.
Mechanical equilibrium (no unbalanced force/pressure) and chemical equilibrium (no net reaction or diffusion) must also hold, together with thermal equilibrium, for a system to be in full thermodynamic equilibrium and describable by one clean set of $(P, V, T)$ values.
The common temperature two bodies in an insulated system settle at follows directly from conservation of energy: heat lost by the hotter body equals heat gained by the cooler one, $m_A c_A(T_A - T_f) = m_B c_B(T_f - T_B)$ .
Thermal equilibrium is not just a qualitative idea — it is the physical basis the zeroth law of thermodynamics (the very next topic in this chapter) will formalise, and it is what makes a thermometer reading trustworthy in the first place.
Key idea: thermal equilibrium is the state of zero net heat flow that two systems in contact through a diathermic wall settle into at a common temperature, and it is this settling-down that gives the word "temperature" its physical meaning.

The core insight

Thermal equilibrium is the condition in which the macroscopic state variables of two or more systems in contact through a heat-conducting wall stop changing because the net flow of heat between them has fallen to zero, defining a single shared temperature.

Where it fits in the syllabus

This topic is the opening idea of the CBSE Class 11 Physics unit on Thermodynamics, and everything that follows in the chapter — the zeroth law, the first law $\Delta Q = \Delta U + \Delta W$ , and the qualitative statements of the second law — leans on it. The syllabus expects you to describe a thermodynamic system using state variables $P$ , $V$ , $T$ , to say precisely what it means for such a system to be in mechanical, chemical, and thermal equilibrium, and to connect thermal equilibrium to the definition of temperature that the zeroth law makes formal. Quantitative gas processes (isothermal, adiabatic, $PV=\text{const}$ , $PV^\gamma=\text{const}$ ) and the first law build on top of this idea but are not this article's job — they belong to later nodes in the same chapter.

The physical picture

A thermodynamic system is any part of the universe we choose to study — a gas in a cylinder, a metal block, a cup of tea — separated from its surroundings by a boundary. The system's macroscopic condition at any instant is captured by its state variables: pressure $P$ , volume $V$ , and temperature $T$ . A system is said to be in equilibrium when these state variables do not change with time, provided nothing external is changed to disturb it. There are three independent conditions bundled into that one word "equilibrium," and a system needs all three simultaneously to be in full thermodynamic equilibrium:

\text{Mechanical} + \text{Chemical} + \text{Thermal} \;\Longrightarrow\; \text{Thermodynamic equilibrium}

where: mechanical equilibrium means no unbalanced force or pressure difference exists between the system and its surroundings (otherwise volume would keep changing); chemical equilibrium means no net chemical reaction or diffusion of matter is occurring between parts of the system; and thermal equilibrium — the star of this article — means no net flow of heat is occurring between the system and its surroundings, or between two systems placed in contact. Only when all three hold simultaneously can the system's state be written down as one definite triple $(P, V, T)$ that stays put. Focus now on thermal equilibrium alone. Imagine two systems, A and B, at different temperatures $T_A > T_B$ , brought into contact through a wall that allows heat to pass — a diathermic wall, such as a thin metal sheet. Heat begins to flow spontaneously from the hotter body A to the colder body B. As it does, $T_A$ falls and $T_B$ rises. This is not a symmetric tug of war: heat only ever flows net from hot to cold, never the reverse, however long you wait. Eventually the two temperatures become equal, macroscopic heat flow stops, and the system is said to have reached thermal equilibrium. The molecules on both sides of the wall never actually stop exchanging energy with each other — collisions and vibrations continue at the microscopic level forever — but once temperatures are equal, energy crosses the wall in both directions at exactly matched rates, so the net transfer is zero. That common, unchanging value of temperature is, by definition, what it means for A and B to be "at the same temperature." If instead the wall between A and B is an adiabatic wall — a perfect thermal insulator, such as an idealised vacuum-jacketed wall — then no heat passes between them regardless of how different their temperatures are. Each system can then be held at its own independent temperature indefinitely, changed only by doing work on it (compressing or expanding it), never by heat leaking through the wall.

Figure

Two blocks joined by a diathermic wall showing heat flowing from the hotter block to the colder block until both reach the same temperature, next to two blocks joined by an adiabatic wall showing no heat flow and each block keeping its own separate temperature

A useful shorthand for the equilibrium condition, applied to an isolated pair of bodies exchanging heat only with each other, is conservation of energy in the form

Q_{\text{lost by hotter body}} = Q_{\text{gained by colder body}}

where: $Q$ is heat transferred, measured in joules, and the equality holds exactly when no heat escapes to any third body (an insulated container, for instance). This is the calorimetry principle you will use repeatedly, and it is nothing more than the statement that energy is conserved while the two bodies march toward a shared final temperature.

Derivation ACommon temperature reached by two bodies exchanging heat

The goal is to derive, from nothing but conservation of energy and the definition of specific heat capacity, a formula for the single common temperature $T_f$ that two bodies of different mass, specific heat, and starting temperature settle at once thermal equilibrium is reached inside an insulated enclosure — this is the calculation that turns the qualitative idea of thermal equilibrium into a number you can compute and check.

Step 1. Set up the physical situation. Two bodies, A and B, with masses $m_A$ , $m_B$ , specific heat capacities $c_A$ , $c_B$ , and initial temperatures $T_A > T_B$ , are placed in contact inside a perfectly insulated enclosure so that no heat escapes to the surroundings and only A and B exchange heat with each other. Left alone, heat flows from A to B until both reach a common final temperature $T_f$ , with $T_B < T_f < T_A$ .

Step 2. Write the heat lost by the hotter body. As A cools from $T_A$ to $T_f$ , it releases heat $Q_A = m_A c_A (T_A - T_f)$ , a positive quantity since $T_A > T_f$ . This follows directly from the definition of specific heat capacity, $Q = mc\,\Delta T$ , applied to a temperature drop.

Step 3. Write the heat gained by the colder body. As B warms from $T_B$ to $T_f$ , it absorbs heat $Q_B = m_B c_B (T_f - T_B)$ , also positive since $T_f > T_B$ . The same definition of specific heat capacity applies, this time to a temperature rise.

Step 4. Apply conservation of energy. Because the enclosure is insulated, every joule that leaves A must enter B — none escapes and none is created. So $Q_A = Q_B$ , which gives the working equation of calorimetry:

m_A c_A (T_A - T_f) = m_B c_B (T_f - T_B)

Step 5. Solve for the common temperature. Expanding both sides and collecting all terms containing $T_f$ on one side gives $m_A c_A T_A + m_B c_B T_B = T_f (m_A c_A + m_B c_B)$ , so

T_f = \dfrac{m_A c_A T_A + m_B c_B T_B}{m_A c_A + m_B c_B}

This is a weighted average of the two starting temperatures, weighted by each body's heat capacity $mc$ — the body with the larger heat capacity pulls the final temperature closer to its own starting value, exactly as a large bath of water barely changes temperature when a small hot object is dropped into it. Verified: setting $m_A=2$ kg, $c_A=450$ , $T_A=80$ , $m_B=3$ kg, $c_B=450$ , $T_B=20$ into the boxed formula gives $T_f = (2\times450\times80 + 3\times450\times20)/(2\times450+3\times450) = (72000+27000)/2250 = 44^\circ\text{C}$ , matching the worked example below by an independent route.

Worked examples

Board level - two blocks reaching a common temperature

Two blocks of the same metal are placed in contact inside an insulated box (so no heat escapes to the surroundings). Block A has mass 2 kg at 80°C and Block B has mass 3 kg at 20°C. Specific heat capacity of the metal is $c = 450\ \text{J kg}^{-1}\text{K}^{-1}$ for both blocks. Find the common temperature at thermal equilibrium.

Step 1. Since heat lost by A equals heat gained by B (conservation of energy inside an insulated box), write $m_A c (T_A - T_f) = m_B c (T_f - T_B)$ . The specific heat $c$ is the same for both blocks, so it cancels, leaving $2(80 - T_f) = 3(T_f - 20)$ .

Step 2. Expand both sides: $160 - 2T_f = 3T_f - 60$ .

Step 3. Collect the $T_f$ terms: $160 + 60 = 3T_f + 2T_f$ , so $220 = 5T_f$ , giving $T_f = 44^\circ\text{C}$ .

Step 4. Check by computing heat lost and gained separately. A cools from $80^\circ\text{C}$ to $44^\circ\text{C}$ , a drop of $36^\circ\text{C}$ , so heat lost $= 2 \times 450 \times 36 = 32{,}400$ J. B warms from $20^\circ\text{C}$ to $44^\circ\text{C}$ , a rise of $24^\circ\text{C}$ , so heat gained $= 3 \times 450 \times 24 = 32{,}400$ J. The two match exactly, confirming energy conservation and the equilibrium temperature.

The common temperature at thermal equilibrium is $T_f = 44^\circ\text{C}$ .

Competitive level - three bodies including the container

The same two metal blocks — A: 2 kg at 80°C, and B: 3 kg at 20°C, both with $c = 450\ \text{J kg}^{-1}\text{K}^{-1}$ — are now placed inside a calorimeter (container) of mass 0.5 kg and specific heat $c_{\text{cal}} = 900\ \text{J kg}^{-1}\text{K}^{-1}$ , initially also at 20°C. The whole assembly is insulated from the outside. Find the final common temperature, and state how much lower it is than the two-body answer.

Step 1. Recognise that the calorimeter is a third body absorbing heat, and set up conservation of energy with all three bodies reaching the same final temperature $T_f$ : heat lost by A equals heat gained by B plus heat gained by the calorimeter. $m_A c (T_A - T_f) = m_B c (T_f - T_B) + m_{\text{cal}} c_{\text{cal}} (T_f - T_{\text{cal}})$ .

Step 2. Substitute numbers. Left side: $2 \times 450 \times (80 - T_f) = 900(80 - T_f)$ . Right side: $3 \times 450 \times (T_f - 20) + 0.5 \times 900 \times (T_f - 20) = 1350(T_f-20) + 450(T_f-20) = 1800(T_f - 20)$ .

Step 3. Solve. $900(80 - T_f) = 1800(T_f - 20)$ , so $72{,}000 - 900T_f = 1800T_f - 36{,}000$ , giving $108{,}000 = 2700 T_f$ , so $T_f = 40^\circ\text{C}$ .

Step 4. Compare with the two-body case. Without the calorimeter the answer was $44^\circ\text{C}$ ; including the calorimeter's own heat capacity as an extra sink pulls the equilibrium temperature down to $40^\circ\text{C}$ , a difference of $4^\circ\text{C}$ — exactly the mark-losing trap of forgetting the container, made numerical.

The final common temperature is $T_f = 40^\circ\text{C}$ , which is $4^\circ\text{C}$ lower than the container-free estimate. Timing note: this is a two-line-setup, three-line-algebra problem — budget about 3 minutes in a competitive paper: write the conservation equation with all three terms first (10 seconds to avoid missing the calorimeter), substitute numbers in one pass, and solve the linear equation for $T_f$ without re-deriving from scratch.

Common traps

Confusing "equal heat exchanged" with "equal temperature." Thermal equilibrium is about temperature becoming equal, not about the heat exchanged being some fixed absolute amount — the amounts of heat lost and gained only match each other because the system is isolated, which is a separate fact.
Thinking heat flow stops instantly on contact. Equilibrium is the final steady state reached only after heat has flowed for some time (which could be seconds or hours depending on the materials); it is never the starting condition the moment two bodies touch.
Ignoring the adiabatic-wall case. Not every two objects in contact equilibrate — only if the boundary between them is diathermic (heat-conducting). A perfectly adiabatic (insulating) wall keeps two systems at independent temperatures indefinitely, however long you wait.
Forgetting a third body that also absorbs heat. In calorimetry problems, missing a container, thermometer, or stirrer that shares in the heat exchange gives a systematically wrong equilibrium temperature — as the competitive example above shows numerically, a 4°C error from a single overlooked calorimeter.
Treating thermal equilibrium as sufficient for full thermodynamic equilibrium. A system can be at uniform temperature (thermal equilibrium) while still undergoing a chemical reaction or having an unbalanced pressure — thermodynamic equilibrium additionally requires mechanical and chemical equilibrium.

You should be able to…

State, in your own words, the conditions for mechanical, chemical, and thermal equilibrium, and explain why all three together define thermodynamic equilibrium.
Explain physically why net heat flows from a hotter body to a colder body and why that flow falls to zero once temperatures equalise.
Distinguish a diathermic wall from an adiabatic wall and predict, for a given pair of systems in contact, whether they will equilibrate.
Derive the common final temperature formula $T_f = \dfrac{m_Ac_AT_A + m_Bc_BT_B}{m_Ac_A+m_Bc_B}$ from conservation of energy, and apply it to two- and three-body calorimetry problems.
Recognise, in a numerical, when a container or instrument must be included as an additional heat-absorbing body.
Connect thermal equilibrium to the definition of temperature as a physical, measurable state variable, in preparation for the zeroth law.

Ask the tutor: Still unsure why a system needs mechanical and chemical and thermal equilibrium to count as fully equilibrated, or want to practise more three-body calorimetry problems before the zeroth law builds on this idea? Ask the AI tutor to walk through additional thermal-equilibrium examples.

If molecules never actually stop moving, how can we say heat flow "stops" at thermal equilibrium?

Good catch — heat flow does not literally stop in the sense of molecules freezing in place. At equilibrium, molecules on both sides of the wall keep colliding and exchanging kinetic energy constantly. What becomes zero is the net transfer: energy crosses from the hotter side to the colder side at exactly the same rate as it crosses back, so the two rates cancel and the macroscopic temperatures stop changing. It is a dynamic balance, not a frozen stillness — similar to how a crowded room can have people constantly walking through a doorway in both directions while the number of people in each room stays constant.

Why does a good calorimetry problem always mention that the container is insulated?

Because the entire heat-lost-equals-heat-gained equation depends on it. If the box were not insulated, some heat would leak out to the surroundings while A and B are equilibrating, and then $Q_{\text{lost by A}}$ would be larger than $Q_{\text{gained by B}}$ — the missing energy having escaped rather than crossed to B. The insulation assumption is what lets you write a clean conservation equation with only the bodies you have listed; forgetting it, or forgetting that the container itself absorbs some heat, is exactly the kind of mark-losing slip examiners test for.

Is thermal equilibrium the same thing as thermodynamic equilibrium?

No, and this distinction is worth exam marks. Thermal equilibrium is specifically about temperature becoming uniform — no net heat flow. Thermodynamic equilibrium is the broader condition requiring mechanical equilibrium (no unbalanced forces or pressures), chemical equilibrium (no net reaction or diffusion), and thermal equilibrium all at once. A gas could be at a perfectly uniform temperature throughout (thermally equilibrated) while still being compressed by an unbalanced piston (not mechanically equilibrated) — so thermal equilibrium alone is necessary but not sufficient for full thermodynamic equilibrium.

Why does this topic matter if there is no formula to memorise for the exam?

Thermal equilibrium is conceptually foundational rather than formula-heavy, but it is exactly the kind of topic CBSE tests with reasoning-based and assertion-reason questions, and it is the physical basis of the zeroth law that comes right after it in the syllabus — you cannot properly justify the zeroth law, or explain why a thermometer works at all, without this idea. It also shows up disguised inside every calorimetry numerical, where the "formula" you do use, $m_A c_A(T_A-T_f) = m_B c_B(T_f-T_B)$ , is nothing but thermal equilibrium written as an energy-conservation equation.