AP Physics 2 · Topic 9.3

Topic 9.3: Thermal Energy Transfer and Equilibrium

Unit 9: Thermodynamics15-18% of the multiple-choice section

Two systems are in thermal contact if energy can pass between them by conduction, convection, or radiation. Energy moves spontaneously from the hotter system to the cooler one. Thermal equilibrium is reached when no net energy crosses between them, which happens once their temperatures are equal.

AP Physics: Unit 9 (topics 9.3 Thermal Energy Transfer and Equilibrium). AP Physics 2 Unit 9, Topic 9.3. One learning objective. 9.3.A asks students to describe the transfer of energy between two systems in thermal contact due to temperature differences of those two systems. It is supported by eight essential knowledge statements: 9.3.A.1 (two systems are in thermal contact if the systems may transfer energy by thermal processes), 9.3.A.1.i (heating is the transfer of energy into a system by thermal processes), 9.3.A.1.ii (cooling is the transfer of energy out of a system by thermal processes), 9.3.A.2 (the thermal processes by which energy may be transferred between systems at different temperatures are conduction, convection, and radiation), 9.3.A.3 (energy is transferred through thermal processes spontaneously from a higher-temperature system to a lower-temperature system), 9.3.A.3.i (in collisions between atoms from different systems, energy is most likely to be transferred from higher-energy atoms to lower-energy atoms), 9.3.A.3.ii (after many collisions of atoms from different systems, the most probable state is one in which both systems have the same temperature), and 9.3.A.4 (thermal equilibrium results when no net energy is transferred by thermal processes between two systems in thermal contact with each other). The topic prints no boundary statement and no relevant equation; the conduction rate equation Q/deltat = k A deltaT / L is filed under 9.5.B.1 and Q = m c deltaT under 9.5.A.1. The CED's suggested skills here are 1.A, 2.C, 3.B, and 3.C, with no calculation skill listed. Unit 9 carries 15 to 18 percent of the multiple-choice section and a suggested 10 to 16 class periods.

What Topic 9.3 requires

Topic 9.3 carries one learning objective and eight essential knowledge statements beneath it, four at the top level and four sub-statements. Unit 9 is 15 to 18 percent of the multiple-choice section and a suggested 10 to 16 class periods.

9.3.A, describe the transfer of energy between two systems in thermal contact due to temperature differences of those two systems.

9.3.A.1 states that two systems are in thermal contact if the systems may transfer energy by thermal processes.
9.3.A.1.i states that heating is the transfer of energy into a system by thermal processes.
9.3.A.1.ii states that cooling is the transfer of energy out of a system by thermal processes.
9.3.A.2 states that the thermal processes by which energy may be transferred between systems at different temperatures are conduction, convection, and radiation.
9.3.A.3 states that energy is transferred through thermal processes spontaneously from a higher-temperature system to a lower-temperature system.
9.3.A.3.i states that in collisions between atoms from different systems, energy is most likely to be transferred from higher-energy atoms to lower-energy atoms.
9.3.A.3.ii states that after many collisions of atoms from different systems, the most probable state is one in which both systems have the same temperature.
9.3.A.4 states that thermal equilibrium results when no net energy is transferred by thermal processes between two systems in thermal contact with each other.

Three features of that list shape everything below.

First, Topic 9.3 prints no boundary statement and no relevant equation. Three of Unit 9's six topics do print a boundary statement (9.1, 9.5, and 9.6); 9.2, 9.3, and 9.4 do not. Nothing here is explicitly fenced off, and nothing here is handed to you as an equation to plug into either.

Second, the objective verb is "describe", and it appears once. Compare Topic 9.4, which has two objectives, or Topic 9.5, which has two and prints an equation under each. Topic 9.3 is the definitional spine of the unit: it fixes what thermal contact, heating, cooling, and thermal equilibrium mean, and the rest of Unit 9 uses those words as defined here.

Third, the suggested skills the CED lists for this topic are 1.A (create diagrams, tables, charts, or schematics to represent physical situations), 2.C (compare physical quantities between two or more scenarios or at different times and locations in a single scenario), 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim), and 3.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws).

No calculation skill appears on that list. Skill 2.B, calculate or estimate an unknown quantity with units, sits on Topic 9.5's list instead, alongside 1.B, 2.D, 3.A, and 3.B. That split tells you where the arithmetic lives: Topic 9.3 asks which way energy goes and why, Topic 9.5 asks how much and how fast.

Thermal contact means energy can move, not that surfaces are touching (9.3.A.1)

Read 9.3.A.1 slowly, because the modal verb is doing the work: two systems are in thermal contact if the systems may transfer energy by thermal processes. The definition is about possibility, not about physical contact.

That matters because one of the three thermal processes in 9.3.A.2 is radiation, and radiation needs no medium and no touching. A hot filament inside an evacuated bulb is in thermal contact with the glass. The Earth is in thermal contact with the Sun across 150 million kilometers of near-vacuum. Neither pair touches.

The reverse case is the one exam questions are built on. Two systems separated by an ideal insulator are not in thermal contact, because no thermal process can move energy between them, and so no amount of temperature difference will produce a transfer. A vacuum flask is an attempt to break all three pathways at once: a gap to stop conduction, an evacuated gap to stop convection, and a silvered surface to reduce radiation.

When you draw the situation for skill 1.A, draw the system boundary first, then draw an arrow for each thermal process that crosses it. A boundary with no arrows crossing it is an insulated boundary. A boundary with arrows crossing it in both directions that cancel is thermal equilibrium, which is not the same thing at all, and is the subject of a later section on this page.

Heating and cooling are transfers, not stuff a system holds (9.3.A.1.i and 9.3.A.1.ii)

The CED defines heating as the transfer of energy into a system by thermal processes and cooling as the transfer of energy out of a system by thermal processes. Both definitions are about a transfer across a boundary. Neither describes a property that the system has before or after.

This is why the sentence "the hot block has more heat in it" does not survive the CED's own definitions. "Heat" is not a quantity a system stores. What a system stores is internal energy, which is Topic 9.4's subject. Heating is one of the ways that stored quantity gets changed.

The symbol $Q$ is the amount of energy transferred by heating, so its sign carries the direction:

$Q > 0$ means the system was heated, and energy entered it.
$Q < 0$ means the system was cooled, and energy left it.
$Q = 0$ means no energy crossed the boundary by a thermal process, which is what "adiabatic" names in 9.4.B.3.

Those signs are not a house convention. They are forced by the first law as the AP Physics 2 sheet prints it, $\Delta U = Q + W$ : for $Q$ to add to internal energy when the system is heated, $Q$ must be positive on the way in. Keep that reading to the end of Unit 9, because the sheet's companion equation $W = -P\Delta V$ is signed the same way, with $W$ the work done on the system. The PV diagrams guide works that convention through in full.

The three thermal processes, and the one rate equation on the sheet (9.3.A.2)

9.3.A.2 names conduction, convection, and radiation as the thermal processes by which energy may be transferred between systems at different temperatures. Three, no more, and the CED does not rank them.

Process	What carries the energy	Needs matter?	Bulk motion of matter?
Conduction	collisions between neighboring atoms and free electrons	yes	no
Convection	a moving mass of fluid that carries its internal energy with it	yes, a fluid	yes
Radiation	electromagnetic waves emitted by the surface	no	no

Conduction is the one 9.3.A.3.i describes microscopically: energy moves atom to atom along the material while the material itself stays put. Convection moves the atoms themselves, since warmer fluid is usually less dense, rises, and carries its energy with it. Radiation is the odd one out because it crosses a vacuum and because every surface above absolute zero emits it.

A useful habit for exam scenarios is to go pathway by pathway. Two objects that touch have a conduction pathway; each object touching air or water has a convection pathway; every pair of objects that can see each other has a radiation pathway, whether or not they touch. A hot can and a cold can sitting apart on the same table are connected to each other only by radiation, and connected to the room by all three.

The AP Physics 2 equation sheet does print a rate equation for conduction, in the Thermal Physics group:

\frac{Q}{\Delta t} = \frac{k A \Delta T}{L}

The CED attaches that equation to essential knowledge 9.5.B.1, not to 9.3, so the calculations belong on Topic 9.5. Read it here as a description of what governs conduction. The rate rises with the thermal conductivity $k$ of the material, rises with the cross-sectional area $A$ the energy flows through, rises with the temperature difference $\Delta T$ across the material, and falls as the thickness $L$ grows.

Two cautions about that $k$ . It is the thermal conductivity, in watts per meter per kelvin. It is not the Coulomb constant, printed as $k = 9.0 \times 10^9\ \mathrm{N \cdot m^2/C^2}$ on the same Physics 2 sheet, and it is not the spring constant in $\vec{F}_s = -k\Delta \vec{x}$ in the sheet's Mechanics and Fluids group. Three quantities share the letter on one page.

The Unit 9 opener poses the essential question of why the tile floor in the bathroom feels so much colder than the bathroom mat. The answer sits in that rate equation, not in a temperature difference. Tile and mat have both been in the same room and are at the same temperature. Tile has a far larger thermal conductivity, so it carries energy away from your foot much faster. "Feels cold" is your skin reporting a rate of energy loss.

Why energy goes from hot to cold, and why the CED says most likely (9.3.A.3)

9.3.A.3 states the direction: energy is transferred through thermal processes spontaneously from a higher-temperature system to a lower-temperature system. The two statements under it are the reason, and their wording is careful in a way that is easy to flatten.

9.3.A.3.i says that in collisions between atoms from different systems, energy is most likely to be transferred from higher-energy atoms to lower-energy atoms. Not "always". A single collision between a fast atom and a slow one can leave the fast one faster still, and nothing in mechanics forbids it. The statement claims a tendency over a population.

9.3.A.3.ii then says that after many collisions of atoms from different systems, the most probable state is one in which both systems have the same temperature. Again a probability claim, and again about many collisions rather than one.

So the CED grounds the hot-to-cold direction in statistics, not in a rule banning the reverse. That framing is the seed of Topic 9.6, where the second law and entropy make the same argument formally. When 9.6 calls a process irreversible, it means the reverse is overwhelmingly improbable for a system of $10^{23}$ particles, not that it is mechanically impossible.

The microscopic picture comes from Topic 9.1, which ties temperature to the average kinetic energy of the atoms through the sheet equation

K_{\text{avg}} = \frac{3}{2} k_B T = \frac{1}{2} m v_{\text{rms}}^2

Put the atoms of a hotter gas in collisional contact with those of a cooler one and the average energy flows from the energetic population toward the less energetic one until the two averages match. Topic 9.1's boundary statement fixes how far you must take this: "AP Physics 2 only expects students to perform qualitative and quantitative analysis of collisions in one and two dimensions. Students are not expected to know the functional form of the Maxwell-Boltzmann distribution but are expected to be familiar with how features of the distribution are related to the temperature of the gas."

Note what 9.3.A.3 does not say. It says nothing about the size of the systems, their masses, their materials, or their internal energies. Temperature alone sets the direction, so a small hot object transfers energy to a large cold one even though the large one holds far more internal energy in total.

Thermal equilibrium means no net transfer, not no transfer (9.3.A.4)

9.3.A.4 states that thermal equilibrium results when no net energy is transferred by thermal processes between two systems in thermal contact with each other. The word "net" is the whole statement, and dropping it produces the most common wrong answer on this topic.

At thermal equilibrium the atoms of the two systems are still colliding across the boundary. Individual collisions still hand energy across in both directions. What has stopped is the imbalance: as much energy crosses one way as the other, so neither system's internal energy changes. Equilibrium here is dynamic, in the same sense that a saturated solution is still dissolving and precipitating at equal rates.

Two corollaries worth having ready:

Equilibrium is not the absence of thermal contact. Two systems separated by a perfect insulator also exchange no net energy, but they are not in thermal equilibrium; they are not in thermal contact, so 9.3.A.4 does not apply. If one sits at 800 K and the other at 200 K, opening the boundary produces an immediate transfer, which is not something an equilibrium state does.
Equilibrium is reached asymptotically. 9.3.A.3.ii calls equal temperature the "most probable state" after many collisions. The approach slows as the temperature difference shrinks, because the difference is what drives the transfer. Two temperatures still 5 kelvin apart and still moving have not reached equilibrium, however flat the curves look.

Combine 9.3.A.3 with 9.3.A.4 and the condition for equilibrium falls out: while a temperature difference exists there is a spontaneous transfer, so there is a net flow, so the systems are not yet in equilibrium. Equal temperature is the condition.

Same temperature does not mean same energy (skill 2.C)

Skill 2.C, comparing physical quantities between two or more scenarios, is on this topic's list, and the comparison the CED sets up is between temperature and internal energy. They behave differently and students routinely swap them.

Temperature is intensive. Cut a system in half and each half has the same temperature. The sheet's $K_{\text{avg}} = \frac{3}{2} k_B T$ says why: temperature tracks the average kinetic energy per atom, and an average does not care how many atoms there are.

Internal energy is extensive. Cut a system in half and each half has half the internal energy. The sheet writes it for an ideal monatomic gas as

U = \frac{3}{2} n R T = \frac{3}{2} N k_B T

with the atom count $N$ or the mole count $n$ sitting right there in the expression.

So two systems that have reached thermal equilibrium have equal temperatures and equal average atomic kinetic energies, and can still have wildly different internal energies. A bathtub of water at 310 K holds far more internal energy than a cup of water at 310 K, and putting them in thermal contact transfers nothing net, because the direction of transfer is set by temperature alone.

Turn that around and you get the other half of the comparison. Energy transferred is not temperature change. How much a system's temperature moves for a given amount of energy transferred depends on its mass and its specific heat, through $Q = mc\Delta T$ from essential knowledge 9.5.A.1. That routine belongs to Topic 9.5, and the worked examples below deliberately avoid it so the two pages do not teach the same procedure twice.

Writing the answer: what skills 1.A, 3.B, and 3.C are asking for

Three of this topic's four suggested skills are about representing and arguing rather than computing, so the free-response version of Topic 9.3 is usually a paragraph, not a number.

For skill 1.A, draw a closed curve for each system boundary, label each system with its temperature, and draw one labeled arrow per thermal process crossing a boundary, pointing the way energy actually moves. Name the process on the arrow, not just $Q$ , and mark an insulated boundary as insulated rather than leaving it bare.

For skills 3.B and 3.C, the claim goes first and names the relationship behind it, then the evidence points at something specific: a number in the stem, a row of a data table, a feature of a graph. "The temperatures are 340 K and 295 K, so X is hotter" is evidence. "It seems hotter" is not.

A structure that fits almost every 9.3 justification:

Name the systems, say whether they are in thermal contact, and by which process or processes.
Compare the temperatures using the actual values.
State the direction of net transfer, citing that spontaneous transfer runs from higher to lower temperature.
State the end condition: net transfer stops when the temperatures are equal, which is thermal equilibrium.
If the question asks about rate rather than direction, switch to the conduction relationship and name which of $k$ , $A$ , $\Delta T$ , or $L$ changed.

Step 5 is where partial credit goes missing, because the question asked "which cools faster" and the answer explained "which is hotter". Direction and rate are separate questions with separate evidence.

How Topic 9.3 is tested, and where it leads

The recurring task types are narrow.

Given two systems at stated temperatures, state the direction of net energy transfer and justify it (skills 3.B and 3.C).
Identify which of conduction, convection, and radiation operates along each pathway in a described setup, and say which pathway an insulator or a vacuum blocks (skill 1.A).
Decide from a data table or a pair of cooling curves whether thermal equilibrium has been reached, and estimate the final common temperature (skills 2.C and 3.C).
Explain why a surface feels colder than another at the same temperature, or why one object cools faster than another, using the conduction relationship rather than a temperature argument.
Compare temperature with internal energy for two systems of different size, and say which is equal at equilibrium and which is not (skill 2.C).
Explain, at the level of atomic collisions, why the transfer runs from hot to cold, using the "most likely" and "most probable" language of 9.3.A.3.

Forwards, Topic 9.4 turns the $Q$ of this page into one term of the first law, $\Delta U = Q + W$ , and adds the other way to change a system's internal energy: doing work on it. Topic 9.5 makes both the amount and the rate quantitative, with $Q = mc\Delta T$ and $\frac{Q}{\Delta t} = \frac{k A \Delta T}{L}$ . Topic 9.6 takes the statistical hedge in 9.3.A.3 seriously and builds the second law and entropy on it.

Backwards, Topic 9.1 supplies the atomic picture of temperature that 9.3.A.3.i assumes, and Topic 9.2 supplies the ideal gas law. If "average kinetic energy per atom" is not yet automatic, start there. The PV diagrams guide is the fastest route into the calculation side of the unit, and /formulas/ap-physics-2 has the full sheet, including all eight equations in its Thermal Physics group.

Two gas samples reach thermal equilibrium: direction, final temperature, and what is actually equal

Sample A is $3.0 \times 10^{22}$ atoms of a monatomic ideal gas at $500\ \mathrm{K}$ . Sample B is $9.0 \times 10^{22}$ atoms of the same gas at $300\ \mathrm{K}$ . The two samples are placed in thermal contact inside a rigid, perfectly insulated container, so no energy leaves the pair and no work is done on either sample. (a) Which way does energy transfer, and why? (b) Find the common final temperature. (c) Find how much energy is transferred. (d) At equilibrium, are the average atomic kinetic energies equal? Are the internal energies equal? Use $k_B = 1.38 \times 10^{-23}\ \mathrm{J/K}$ from the sheet.

(a) Compare temperatures, because temperature alone sets the direction (9.3.A.3). $T_A = 500\ \mathrm{K} > T_B = 300\ \mathrm{K}$ , so energy transfers spontaneously from A to B. Sample A is cooled and sample B is heated.
Check that against the atomic picture in 9.3.A.3.i using $K_{\text{avg}} = \frac{3}{2} k_B T$ . For A, $1.5 \times (1.38 \times 10^{-23})(500) = 1.04 \times 10^{-20}\ \mathrm{J}$ . For B, $1.5 \times (1.38 \times 10^{-23})(300) = 6.21 \times 10^{-21}\ \mathrm{J}$ . A's atoms are the higher-energy population by a factor of $5/3$ , so collisions across the boundary most likely move energy from A to B.
(b) The container is rigid and insulated, so the total internal energy of the pair is fixed. Using $U = \frac{3}{2} N k_B T$ for each sample, $\frac{3}{2} k_B (N_A T_A + N_B T_B) = \frac{3}{2} k_B (N_A + N_B) T_f$ . The factor $\frac{3}{2} k_B$ cancels, leaving a count-weighted average of the two temperatures.
$T_f = \dfrac{N_A T_A + N_B T_B}{N_A + N_B} = \dfrac{(3.0)(500) + (9.0)(300)}{3.0 + 9.0} = \dfrac{1500 + 2700}{12} = 350\ \mathrm{K}$ , working in units of $10^{22}$ atoms so the powers of ten cancel. It lands between 300 K and 500 K and nearer B, which is right, because B has three times as many atoms and so moves three times less.
(c) Track sample A. Before: $U_A = \frac{3}{2}(3.0 \times 10^{22})(1.38 \times 10^{-23})(500) = 310.5\ \mathrm{J}$ . After: $U_A' = \frac{3}{2}(3.0 \times 10^{22})(1.38 \times 10^{-23})(350) = 217.35\ \mathrm{J}$ . A's internal energy falls by $93.15\ \mathrm{J}$ , or $93\ \mathrm{J}$ to the two significant figures the atom counts carry.
Confirm with sample B: $U_B = 558.9\ \mathrm{J}$ before and $U_B' = 652.05\ \mathrm{J}$ after, a rise of $93.15\ \mathrm{J}$ . The two changes are equal and opposite, as they must be inside a rigid insulated container.
In first-law language, using the sheet's $\Delta U = Q + W$ with $W$ the work done on the gas: the walls are rigid, so $\Delta V = 0$ and $W = -P\Delta V = 0$ for both samples. That leaves $\Delta U_A = Q_A = -93\ \mathrm{J}$ and $\Delta U_B = Q_B = +93\ \mathrm{J}$ . Negative $Q$ is cooling and positive $Q$ is heating, exactly as 9.3.A.1.i and 9.3.A.1.ii define them.
(d) At 350 K, $K_{\text{avg}} = 1.5 \times (1.38 \times 10^{-23})(350) = 7.245 \times 10^{-21}\ \mathrm{J}$ for both samples, because both are at the same temperature. Equal. But the internal energies are $217.35\ \mathrm{J}$ for A and $652.05\ \mathrm{J}$ for B, a ratio of exactly 3 because B has exactly three times as many atoms. Not equal, and not required to be.

(a) From A to B, because A is at the higher temperature. (b) $T_f = 350\ \mathrm{K}$ . (c) $93\ \mathrm{J}$ transferred, so $Q_A = -93\ \mathrm{J}$ and $Q_B = +93\ \mathrm{J}$ . (d) The average atomic kinetic energies are equal at $7.245 \times 10^{-21}\ \mathrm{J}$ ; the internal energies are not, standing at $217.35\ \mathrm{J}$ and $652.05\ \mathrm{J}$ . Thermal equilibrium equalizes temperature, never total energy.

Reading a cooling data table: has equilibrium been reached?

Two blocks, P and Q, are placed in thermal contact inside an insulated box. Their temperatures in degrees Celsius are recorded every two minutes.

Time (min)	0	2	4	6	8	10
Block P	80.0	68.0	60.8	56.5	53.9	52.3
Block Q	20.0	32.0	39.2	43.5	46.1	47.7

(a) State the direction of net energy transfer and justify it from the data. (b) Has thermal equilibrium been reached by $t = 10$ minutes? (c) Estimate the common temperature the blocks are heading for. (d) What does the data say about how the transfer rate changes as the process runs?

(a) Block P falls from 80.0 to 52.3 degrees Celsius while block Q rises from 20.0 to 47.7, so energy is leaving P and entering Q. That matches 9.3.A.3: P starts hotter, and spontaneous transfer runs from the higher-temperature system to the lower-temperature one. The box is insulated, so the energy P loses is the energy Q gains.
(b) Take the difference at each reading: $60.0$ , $36.0$ , $21.6$ , $13.0$ , $7.8$ , $4.6$ degrees. At $t = 10$ minutes the blocks still differ by 4.6 degrees and both readings are still moving. Any temperature difference guarantees a net transfer by 9.3.A.3, so thermal equilibrium has not been reached. The curves are flattening, not flat.
(c) Average the two readings in each column: $(80.0 + 20.0)/2 = 50.0$ , then $50.0$ , $50.0$ , $50.0$ , $50.0$ , $50.0$ again for every later column. The mean holds at 50.0 degrees Celsius throughout and the two curves close on it from opposite sides, so the common final temperature is about 50 degrees Celsius.
Name why the mean holds still rather than assuming it: P falls by exactly as much as Q rises at every step, which happens only when the two blocks need the same energy per degree. That is a statement about mass and specific heat, from $Q = mc\Delta T$ , and belongs to Topic 9.5. With different masses or specific heats the final temperature would sit nearer the start of the block with the larger $mc$ , and the midpoint shortcut would fail.
(d) The difference falls 24.0 degrees over the first two minutes and only 3.2 degrees over the last two. The transfer rate drops as the temperature difference shrinks, which is what $\frac{Q}{\Delta t} = \frac{k A \Delta T}{L}$ predicts, since the rate is proportional to $\Delta T$ . This is why equilibrium is approached asymptotically and never quite arrives on a data table.

(a) From P to Q, because P starts at the higher temperature and its readings fall while Q's rise. (b) No. The blocks still differ by 4.6 degrees at $t = 10$ minutes and both temperatures are still changing. (c) About 50 degrees Celsius, which is where both curves are heading and is the constant mean of the two columns. (d) The rate falls as the temperature difference falls: 24.0 degrees of closing in the first interval against 3.2 degrees in the last.

Frequently asked questions

What is thermal equilibrium in AP Physics 2?

Thermal equilibrium results when no net energy is transferred by thermal processes between two systems in thermal contact with each other. That is essential knowledge 9.3.A.4, in AP Physics 2 Topic 9.3. The condition for it is equal temperature: while a temperature difference exists, energy transfers spontaneously from the hotter system to the cooler one, so there is a net flow and equilibrium has not been reached. The word net matters. At equilibrium, atoms are still colliding across the boundary and exchanging energy both ways; the two rates have simply become equal.

Does thermal equilibrium mean no energy is transferred at all?

No. The AP Physics 2 definition is that no NET energy is transferred, which is not the same as no energy being transferred. Two objects at the same temperature and in thermal contact are still exchanging energy: their atoms keep colliding across the boundary, and both surfaces keep emitting and absorbing radiation. What has changed is that as much energy crosses each way, so neither object's internal energy changes. Thermal equilibrium is a dynamic balance, not a stop.

Do two objects have to be touching to be in thermal contact?

No. AP Physics 2 essential knowledge 9.3.A.1 defines thermal contact as two systems that MAY transfer energy by thermal processes, and one of the three thermal processes is radiation, which needs no medium and no contact. The Earth and the Sun are in thermal contact across empty space. Conversely, two objects that touch through a perfect insulator are not in thermal contact, because no thermal process can move energy between them.

What are the three ways thermal energy is transferred?

Conduction, convection, and radiation. AP Physics 2 essential knowledge 9.3.A.2 names exactly these three as the thermal processes by which energy may be transferred between systems at different temperatures. Conduction moves energy through a material by collisions between neighboring atoms while the material itself stays put. Convection moves energy by the bulk motion of a fluid that carries its internal energy along with it. Radiation moves energy as electromagnetic waves emitted by any surface above absolute zero, and it is the only one of the three that crosses a vacuum.

Why does energy always flow from hot to cold?

AP Physics 2 grounds this in statistics rather than in a prohibition. Essential knowledge 9.3.A.3.i says that in collisions between atoms from different systems, energy is MOST LIKELY to be transferred from higher-energy atoms to lower-energy atoms, and 9.3.A.3.ii says that after many collisions the MOST PROBABLE state is one in which both systems have the same temperature. A single collision can move energy the other way. Across the number of atoms in a real system the tendency is overwhelming, which is why 9.3.A.3 states flatly that spontaneous transfer runs from the higher-temperature system to the lower-temperature one. Topic 9.6 turns the same argument into the second law of thermodynamics.

Why does a tile floor feel colder than a bath mat at the same temperature?

Because what your skin reports is a rate of energy loss, not a temperature. Tile and mat have both been sitting in the same room, so they are at the same temperature. Tile has a much higher thermal conductivity, so it conducts energy away from your foot far faster than the mat does. The AP Physics 2 equation sheet prints the conduction rate as Q divided by delta-t equals k A delta-T over L, so a larger thermal conductivity k gives a larger rate for the same temperature difference. This is one of the essential questions the College Board lists in the Unit 9 opener.

Do two systems at the same temperature have the same energy?

No, and this is a standard trap. Temperature is an intensive property: it tracks the average kinetic energy per atom, through K_avg = (3/2) k_B T on the AP Physics 2 sheet, and an average does not depend on how many atoms there are. Internal energy is extensive: for an ideal monatomic gas the sheet gives U = (3/2) n R T = (3/2) N k_B T, with the amount of gas sitting in the expression. A bathtub and a cup of water at the same temperature are in thermal equilibrium and transfer no net energy to each other, yet the bathtub holds far more internal energy.