Thermal Energy Transfer and Equilibrium: Study Guide (Physics 2 9.3)

Systems in thermal contact

Start by choosing your systems. A system is the object or group of objects you decide to study, for example a cup of tea, the air in a room, or a sample of gas in a sealed box.

Two systems are in thermal contact when energy can move between them by thermal processes. A spoon standing in hot soup is in thermal contact with the soup. Two boxes of gas that share a thin metal wall are in thermal contact. Two boxes separated by thick foam are, to a good approximation, not.

The words for the direction of the transfer matter:

• Heating: energy enters a system through a thermal process (conduction, convection or radiation).
• Cooling: energy leaves a system through a thermal process.

So when hot soup and a cold spoon touch, the spoon is heated and the soup is cooled. It is the same energy transfer, described from each system’s point of view. If the two systems are isolated from everything else, the energy the soup loses equals the energy the spoon gains.

Keep three ideas separate:

Idea	What it is	Example of correct use
Temperature T	A property of a system, set by the average kinetic energy of its atoms	“The gas is at 400 K.”
Internal energy	The total energy of the particles inside a system	“The large tank holds more internal energy.”
Heating / cooling	A transfer of energy caused by a temperature difference	“The gas was heated by 37 J.”

A system has a temperature and internal energy. It does not “contain heat”.

Three thermal processes

Energy moves between systems at different temperatures in three ways.

Process	How energy moves	Needs matter?	Everyday example
Conduction	Atoms collide with their neighbours and pass kinetic energy along. The material itself does not flow.	Yes	A metal pan handle gets hot from the pan
Convection	A fluid (liquid or gas) moves and carries its energy with it. Warmer fluid is usually less dense and rises; cooler fluid sinks.	Yes, a fluid	Warm air rising above a heater and circulating round a room
Radiation	Electromagnetic waves (mostly infrared at everyday temperatures) carry energy from one surface to another.	No	Energy from the Sun crossing empty space to reach Earth

Points to notice:

• Every object emits radiation, not just hot ones. A hotter object emits more. Between two objects at different temperatures, the net radiation transfer is from the hotter to the colder.
• Convection cannot happen inside a solid, because the material cannot flow.
• In real situations, two or three processes usually act together. A cup of tea cools by conduction into the table, convection in the air above it, and radiation from its surface. (Evaporation also carries energy away, but you only need the three processes above.)
• A vacuum gap stops conduction and convection, because both need matter. Only radiation can cross it. This is why a vacuum flask has a vacuum between two walls, and shiny surfaces to reduce radiation.

You will meet the rate of conduction quantitatively in Topic 9.5. In this topic the processes are described qualitatively.

Why energy flows from hot to cold

Topic 9.1 linked temperature to atomic motion: the average kinetic energy of an atom is K_avg = (3/2)k_B T. So a hotter system has atoms with a larger average kinetic energy. Not every atom in the hot system is faster than every atom in the cold one, because each system has a spread of speeds (Figure 2). But on average, the hot system’s atoms carry more energy.

When two systems touch, atoms at the boundary collide with atoms of the other system. In a collision, kinetic energy can pass from one atom to the other. The key rule is about probability:

• In a single collision, energy is most likely to pass from the higher-energy atom to the lower-energy atom. It is not certain. Worked example 1 shows a collision where the lower-energy atom gives energy to the higher-energy one.
• Over an enormous number of collisions, the most probable outcome is that the hotter system loses energy and the colder system gains it.
• After many collisions, the most likely end state is one where the two systems share the same temperature.

So energy moves spontaneously (without any outside help) from the higher-temperature system to the lower-temperature system. The reverse, energy flowing by itself from cold to hot, is not forbidden by energy conservation. It is just so improbable for a large system that it is never observed. Topic 9.6 builds on this with entropy and the second law.

Notice what decides the direction: temperature, not internal energy. A swimming pool at 25 °C has far more internal energy than a cup of coffee at 80 °C. Drop the coffee cup into the pool and energy still flows from the coffee into the pool.

Thermal equilibrium

Two systems in thermal contact are in thermal equilibrium when there is no net energy transfer between them by thermal processes. This happens when they are at the same temperature.

“No net transfer” does not mean “nothing happens”. Atoms keep moving and colliding at the boundary, and individual collisions still pass energy both ways. At equilibrium the flow from A to B and the flow from B to A balance on average, so neither temperature changes.

This is also how a thermometer works. You put it in thermal contact with a system and wait until there is no net energy transfer. Then the thermometer is at the same temperature as the system, and its reading tells you that temperature. If systems A and B are each in thermal equilibrium with a thermometer showing the same reading, then A and B would be in equilibrium with each other. (Some books call this the zeroth law of thermodynamics.)

Figure 1 shows how two systems approach equilibrium. The temperature difference is largest at the start, so the net transfer is fastest at the start. As the difference shrinks, the curves flatten.

The distribution of atomic speeds

Topic 9.1 introduced the Maxwell–Boltzmann distribution: a graph showing how many atoms in a gas have each speed. You do not need its formula. You do need to read its features:

• A hotter gas has a distribution that is wider and shifted to higher speeds, with a lower peak (the area under each curve represents the same number of atoms).
• Even in a cold gas, some atoms are fast. Even in a hot gas, some are slow. That overlap is why a single collision can transfer energy “the wrong way”.

When the two samples reach equilibrium, both end up with the same distribution, in between the two starting ones.

Worked example 1: energy transfer in one collision

Question. An atom A from a hot gas (mass 4.0 × 10⁻²⁶ kg) collides head-on and elastically with an atom B from a cold gas (mass 6.0 × 10⁻²⁶ kg), along a line. Take the positive direction to the right.

(a) Before: v_A = +900 m/s and v_B = −300 m/s. After: v_A = −540 m/s and v_B = +660 m/s. Check that momentum and kinetic energy are conserved, and find the energy transferred and its direction.

(b) Repeat with v_B = −700 m/s before. After: v_A = −1020 m/s and v_B = +580 m/s.

(a)

1. Momentum before: (4.0 × 10⁻²⁶)(900) + (6.0 × 10⁻²⁶)(−300) = 3.6 × 10⁻²³ − 1.8 × 10⁻²³ = 1.8 × 10⁻²³ kg·m/s.
2. Momentum after: (4.0 × 10⁻²⁶)(−540) + (6.0 × 10⁻²⁶)(660) = −2.16 × 10⁻²³ + 3.96 × 10⁻²³ = 1.8 × 10⁻²³ kg·m/s. Conserved.
3. Kinetic energies before: K_A = ½(4.0 × 10⁻²⁶)(900)² = 1.62 × 10⁻²⁰ J; K_B = ½(6.0 × 10⁻²⁶)(300)² = 2.7 × 10⁻²¹ J. Total 1.89 × 10⁻²⁰ J.
4. After: K_A = ½(4.0 × 10⁻²⁶)(540)² = 5.832 × 10⁻²¹ J; K_B = ½(6.0 × 10⁻²⁶)(660)² = 1.3068 × 10⁻²⁰ J. Total 1.89 × 10⁻²⁰ J. Conserved, so the collision is elastic.
5. Atom A lost 1.62 × 10⁻²⁰ − 5.832 × 10⁻²¹ = 1.04 × 10⁻²⁰ J, and atom B gained the same amount.

Answer (a). About 1.04 × 10⁻²⁰ J passes from A (higher energy) to B (lower energy). That is 64% of A’s starting kinetic energy. This is the usual direction.

(b)

1. Momentum before: 3.6 × 10⁻²³ + (6.0 × 10⁻²⁶)(−700) = 3.6 × 10⁻²³ − 4.2 × 10⁻²³ = −6.0 × 10⁻²⁴ kg·m/s. After: (4.0 × 10⁻²⁶)(−1020) + (6.0 × 10⁻²⁶)(580) = −4.08 × 10⁻²³ + 3.48 × 10⁻²³ = −6.0 × 10⁻²⁴ kg·m/s. Conserved.
2. Before: K_A = 1.62 × 10⁻²⁰ J; K_B = ½(6.0 × 10⁻²⁶)(700)² = 1.47 × 10⁻²⁰ J. Total 3.09 × 10⁻²⁰ J.
3. After: K_A = ½(4.0 × 10⁻²⁶)(1020)² = 2.0808 × 10⁻²⁰ J; K_B = ½(6.0 × 10⁻²⁶)(580)² = 1.0092 × 10⁻²⁰ J. Total 3.09 × 10⁻²⁰ J. Conserved.
4. Atom B lost 1.47 × 10⁻²⁰ − 1.0092 × 10⁻²⁰ = 4.61 × 10⁻²¹ J, and atom A gained it.

Answer (b). About 4.6 × 10⁻²¹ J passes from B (lower energy) to A (higher energy).

Interpretation. Both collisions obey the same conservation laws. Part (b) shows that energy can pass from a lower-energy atom to a higher-energy one. So the hot-to-cold rule is not a rule about each collision. It is about what happens most often. Collisions that move energy from the higher-energy atom to the lower-energy one, like (a), happen more often than collisions like (b). With an enormous number of collisions every second, the hot system loses energy overall.

Worked example 2: two gas samples reach equilibrium

Question. Two rigid sealed boxes share a thin metal wall and are insulated from everything else. Box 1 holds 2.0 × 10²² atoms of a monatomic ideal gas at 400 K. Box 2 holds 3.0 × 10²² atoms of the same gas at 250 K. Treat the total kinetic energy of the atoms as the internal energy of each sample (an ideal monatomic gas has no other internal energy). Find (a) the direction of energy transfer, (b) the final temperature and (c) the energy transferred.

1. Average kinetic energy at the start: box 1: (3/2)(1.38 × 10⁻²³ J/K)(400 K) = 8.28 × 10⁻²¹ J. Box 2: (3/2)(1.38 × 10⁻²³)(250) = 5.175 × 10⁻²¹ J.
2. Internal energy at the start: U₁ = N₁K_avg = (2.0 × 10²²)(8.28 × 10⁻²¹ J) = 165.6 J. U₂ = (3.0 × 10²²)(5.175 × 10⁻²¹ J) = 155.25 J. Total 320.85 J.
3. (a) Box 1 is at the higher temperature, so energy moves from box 1 to box 2. Box 1 is cooled; box 2 is heated.
4. (b) The boxes are rigid and insulated from outside, so the total energy stays at 320.85 J. At equilibrium both have the same T_f: (3/2)k_B(N₁ + N₂)T_f = (3/2)k_B(N₁T₁ + N₂T₂), so T_f = (N₁T₁ + N₂T₂)/(N₁ + N₂) = (2.0 × 400 + 3.0 × 250)/5.0 = 1550/5.0 = 310 K.
5. (c) Energy lost by box 1: (2.0 × 10²²)(3/2)(1.38 × 10⁻²³)(400 − 310) = 37.26 J. Energy gained by box 2: (3.0 × 10²²)(3/2)(1.38 × 10⁻²³)(310 − 250) = 37.26 J.

Answer. Energy moves from box 1 to box 2; the final temperature is 310 K; about 37 J is transferred.

Check and interpretation. The energy lost equals the energy gained, as conservation requires. Final energies: U₁ = 128.34 J and U₂ = 192.51 J, which still add up to 320.85 J. T_f lies between 250 K and 400 K, and nearer to 250 K, because box 2 has more atoms. The simple average, 325 K, would be wrong here. Notice that at equilibrium box 2 has more internal energy than box 1 but the same temperature. No net energy flows, because the average energy per atom is the same: 6.417 × 10⁻²¹ J.

Worked example 3: explaining a claim

Question. A hot metal block is placed in a cool, sealed, insulated container of air. A student says: “After an hour the block and the air are at the same temperature, so the atoms of the block have stopped exchanging energy with the air.” Evaluate the claim.

1. Equal temperatures mean the systems are in thermal equilibrium: there is no net energy transfer.
2. The atoms of the block still vibrate, and the air molecules still move and hit the block. Individual collisions still pass energy one way or the other.
3. At equilibrium the energy passing from block to air balances, on average, the energy passing from air to block. So neither temperature changes.

Answer. The first half is correct; the second half is not. Energy exchange continues, but the net transfer is zero. Equilibrium is a balance, not a stop.

Common misconceptions

• “Cold flows into the warm object.” There is no such thing as a flow of cold. Energy flows out of the warm object (cooling) and into the cold one (heating).
• “Energy flows from the object with more internal energy.” Temperature decides the direction. A large, cool object can have more internal energy than a small, hot one and still be heated by it.
• “Every collision moves energy from hot to cold.” Single collisions can go either way (Worked example 1b). The rule is about the most probable result of very many collisions.
• “At equilibrium the atoms stop moving or stop colliding.” Atoms keep moving; only the net transfer is zero.
• “The metal feels colder, so it is at a lower temperature.” Objects left in the same room reach the same temperature. Metal feels colder because it conducts energy away from your hand faster.
• “Only hot objects radiate.” All objects emit radiation. The net transfer by radiation goes from hotter to colder.
• “Convection happens in solids.” Convection needs a moving fluid. Solids transfer energy by conduction (and their surfaces by radiation).
• Taking the simple average for the final temperature. For two monatomic ideal gas samples, the simple average works only when they hold equal numbers of atoms. Otherwise, weight each starting temperature by its number of atoms.

Where this leads

Topic 9.4 turns this into an energy equation. Heating becomes the term Q in the first law, alongside work done on a gas: see the Topic 9.4 study guide. Topic 9.5 then adds specific heat and the rate of conduction. Back in Topic 9.2 you modelled the gas in each box with the ideal gas law.

Test yourself now with the practice questions, then use the revision notes and the topic checklist to consolidate.