Conduction vs Convection: The Difference

Conduction moves energy through a material by collisions between neighboring particles, with the material itself staying put. Convection moves energy by carrying the fluid bodily from one place to another. AP Physics 2 gives conduction a rate equation and treats convection only by name.

AP Physics: Unit 9 (topics 9.3 Thermal Energy Transfer and Equilibrium, 9.5 Specific Heat and Thermal Conductivity). Both processes are named in AP Physics 2 essential knowledge 9.3.A.2, which states that the thermal processes by which energy may be transferred between systems at different temperatures are conduction, convection, and radiation. That clause is the whole of the CED's treatment of convection apart from one further mention in the appendix's Heat vs. Heating vs. Cooling section: the word appears exactly twice in the AP Physics 2 CED, there is no learning objective or essential knowledge statement devoted to it, no equation, no constant and no boundary statement, and it appears nowhere in the AP Physics 1 CED. Conduction is treated quantitatively. Learning objective 9.5.B, in Topic 9.5, is to describe the rate at which energy is transferred by conduction through a given material; 9.5.B.1 states that the rate is related to the thermal conductivity, the physical dimensions of the material, and the temperature difference across the material, with the relevant equation Q over delta t equals k A delta T over L; and 9.5.B.2 states that the thermal conductivity of a material is an intrinsic property of that material that depends on the arrangement and interactions of the atoms that make up the material. That equation is printed in the Thermal Physics panel of the AP Physics 2 Table of Information, whose symbol list defines k as thermal conductivity, A as area, L as length and Q as energy transferred to a system by heating. No values of k are printed anywhere on the sheet. The parallel first objective of Topic 9.5 is 9.5.A, with 9.5.A.1 giving Q equals m c delta T and 9.5.A.2 giving specific heat as an intrinsic property. Radiation becomes quantitative only in Unit 15: essential knowledge 15.4.A.3.iii gives the Stefan-Boltzmann law as P equals A sigma T to the fourth, printed in the Modern Physics panel, with sigma in the constants box as 5.67 times 10 to the minus 8 watts per square meter per kelvin to the fourth. The rest of Topic 9.3 supplies the shared framing: 9.3.A.1 on thermal contact, 9.3.A.1.i and 9.3.A.1.ii defining heating and cooling, 9.3.A.3 on spontaneous transfer from higher to lower temperature with 9.3.A.3.i and 9.3.A.3.ii giving the collision-level account, and 9.3.A.4 defining thermal equilibrium. Unit 9 prints three boundary statements, under Topics 9.1, 9.5 and 9.6; Topics 9.2, 9.3 and 9.4 carry none. The Topic 9.5 statement reads that AP Physics 2 will model specific heat as independent of temperature, which constrains 9.5.A rather than the conduction objective 9.5.B. The Topic 9.1 statement reads that AP Physics 2 only expects students to perform qualitative and quantitative analysis of collisions in one and two dimensions, and that 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. The Topic 9.6 statement reads that only qualitative treatment of the second law of thermodynamics is within the scope of AP Physics 2. None of the three mentions conduction or convection. Unit 9 is weighted at 15 to 18 percent of the multiple-choice section across a suggested 10 to 16 class periods; the suggested skills listed for Topic 9.3 are 1.A, 2.C, 3.B and 3.C, and for Topic 9.5 they are 1.B, 2.B, 2.D, 3.A and 3.B. One of the unit's printed essential questions asks why the tile floor in the bathroom feels so much colder than the bathroom mat.

The distinction, stated once

Both are ways energy crosses from a hotter place to a cooler one, and essential knowledge 9.3.A.2 lists them together: the thermal processes by which energy may be transferred between systems at different temperatures are conduction, convection, and radiation. Three processes, one list, and the difference between the first two is whether the matter travels.

Conduction passes energy along without moving the material. Particles at the hot end are moving faster, they collide with their neighbors, the neighbors speed up and collide with theirs, and the disturbance propagates. The metal spoon in a hot drink is in the same place after the handle warms as it was before. Because nothing has to flow, conduction is the process that works in solids.

Convection moves the material and lets it carry its energy along. Warm fluid moves bodily from one region to another and arrives with the internal energy it already had. Nothing is passed from particle to particle; the particles themselves relocate. Because the matter has to travel, convection needs a fluid and cannot happen in a solid.

So the test is a question about the material, not about the energy: did the stuff move, or only the energy?

That is the physics. Before going further it is worth being straight about the exam, because the two processes are not treated equally by the course at all, and the next section is about that.

Check the scope first: one of these has an equation and the other does not

AP Physics 2 gives conduction its own learning objective and a printed rate equation. It gives convection a single mention.

Conduction is quantitative. Learning objective 9.5.B, in Topic 9.5, Specific Heat and Thermal Conductivity, is to describe the rate at which energy is transferred by conduction through a given material. Essential knowledge 9.5.B.1 says that rate is related to the thermal conductivity, the physical dimensions of the material, and the temperature difference across the material, with the relevant equation

QΔt=kAΔTL\frac{Q}{\Delta t} = \frac{kA\Delta T}{L}

That equation is printed on the AP Physics 2 equation sheet, in the Thermal Physics panel, and the panel's symbol list defines kk as thermal conductivity, AA as area and LL as length. Essential knowledge 9.5.B.2 adds that the thermal conductivity of a material is an intrinsic property of that material that depends on the arrangement and interactions of the atoms that make up the material.

Convection is not quantitative, and it is not really qualitative either. The word convection appears in the AP Physics 2 CED exactly twice. Once in 9.3.A.2, as one of the three named thermal processes. Once in the appendix's vocabulary section on heat, again inside a list of thermal processes. That is the whole of it. There is no learning objective about convection, no essential knowledge statement of its own, no equation, no constant in the constants box, and no boundary statement limiting how it is treated, because there is nothing to limit. Do not learn a convection formula for this exam. Whatever coefficient of convective heat transfer you may have met elsewhere, the AP course does not use it and could not ask you to.

Neither word appears anywhere in the AP Physics 1 CED. Thermal energy transfer is an AP Physics 2 topic in this course sequence.

Radiation, the third item on the list, sits in between and it is worth knowing where. In Unit 9 it is named alongside the other two in 9.3.A.2 and left there. It becomes quantitative in a different unit: 15.4.A.3.iii, in Unit 15, Modern Physics, says the rate at which energy is emitted by a blackbody is proportional to the surface area of the body and to the temperature of the body raised to the fourth power, as described by the Stefan-Boltzmann law, with the relevant equation P=AσT4P = A\sigma T^4. That equation is printed in the Modern Physics panel of the sheet, and the constants box supplies σ=5.67×108 W/(m2K4)\sigma = 5.67 \times 10^{-8}\ \text{W/(m}^2 \cdot \text{K}^4).

So of the three processes named in one sentence of 9.3.A.2, two acquire equations in different units of the course and one never does.

Conduction against convection, row by row

PropertyConductionConvection
Does the material movenoyes, in bulk
Mechanismcollisions between neighboring particlesthe fluid carries its internal energy with it
Works in a solidyesno
Works in a fluidyesyes
Named in 9.3.A.2yesyes
Learning objective of its ownyes, 9.5.Bno
Printed equationQ/Δt=kAΔT/LQ/\Delta t = kA\Delta T/Lnone
Material property involvedthermal conductivity kk, intrinsic per 9.5.B.2none named in the CED
Appearances in the AP Physics 2 CEDfourtwo
What the exam can requirea rate, a comparison, an experimental designidentifying it as the process at work
Direction of transferhigh to low temperature, per 9.3.A.3high to low temperature, per 9.3.A.3

Rows six through nine are the practical content of this page, and row seven is the one that decides how to revise. There is a number to compute on one side and nothing to compute on the other.

Row eleven is where the two are genuinely alike, and it is worth saying because it is the part that is examinable for both. 9.3.A.3 states that energy is transferred through thermal processes spontaneously from a higher-temperature system to a lower-temperature system, and the statement covers all three processes without distinguishing them. It is supported by 9.3.A.3.i, which 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, which says 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 then defines the endpoint: thermal equilibrium results when no net energy is transferred by thermal processes between two systems in thermal contact with each other.

So the direction and the destination are shared. Only the mechanism, and the exam treatment, differ.

The conduction rate equation, factor by factor

Since this is the half of the pair you can be asked to compute, it is worth reading the equation slowly. 9.5.B.1 gives

QΔt=kAΔTL\frac{Q}{\Delta t} = \frac{kA\Delta T}{L}

Four things on the right, and each one behaves differently.

  • kk, the thermal conductivity. 9.5.B.2 calls it an intrinsic property of the material, depending on the arrangement and interactions of its atoms. Intrinsic means it belongs to the substance, not to the slab: cutting a copper bar in half does not change kk. Its unit follows from the equation, watts per meter per kelvin. No values of kk are printed on the equation sheet, so a question that needs one has to supply it.
  • AA, the area across which the energy flows, perpendicular to the flow. Doubling the area doubles the rate, which is why a large window loses more than a small one made of the same glass.
  • ΔT\Delta T, the temperature difference across the material, not the temperature of either side. A difference in kelvin and a difference in degrees Celsius are the same number, so this is one of the few places you do not need to convert. Celsius vs Kelvin sets out when you do.
  • LL, the thickness the energy has to cross. It is in the denominator, so a thicker layer conducts more slowly. Doubling the thickness halves the rate.

Notice what the left side is. Q/ΔtQ/\Delta t is a rate, energy per unit time, measured in watts. It is not an amount of energy, and a question asking how much energy crosses in a given interval wants you to multiply by that interval. This is the commonest arithmetic slip on the topic.

Notice also what is missing. There is no time dependence, no mass, and no specific heat. The equation gives the steady rate at which energy crosses a slab, and it says nothing about what happens to the energy once it arrives. That is the other half of Topic 9.5: 9.5.A.1 gives Q=mcΔTQ = mc\Delta T for the energy required to change a material's temperature, and 9.5.A.2 says the specific heat of a material is an intrinsic property that depends on the arrangement and interactions of the atoms making it up, in wording deliberately parallel to 9.5.B.2. Two intrinsic properties, one governing how fast energy moves through and the other how much energy a temperature change costs.

The case that separates them: why the tile floor feels colder than the mat

This is Unit 9's own opening question. The CED prints it as one of the unit's essential questions: why does the tile floor in the bathroom feel so much colder than the bathroom mat?

The two are at the same temperature. They have both been sitting in the same room all night. A thermometer laid on each reads the same number. So whatever your foot is detecting, it is not temperature.

What your foot detects is the rate at which energy leaves it, and that is a conduction question with kk as the only thing that differs.

Take a contact patch of 0.010 m20.010\ \text{m}^2, a surface layer 0.005 m0.005\ \text{m} thick, and skin at 33 C33\ \text{C} against a room at 20 C20\ \text{C}, so ΔT=13 K\Delta T = 13\ \text{K}. Suppose a question supplies k=1.1 W/(mK)k = 1.1\ \text{W/(m} \cdot \text{K)} for the tile and k=0.050 W/(mK)k = 0.050\ \text{W/(m} \cdot \text{K)} for the mat.

tile: QΔt=(1.1)(0.010)(13)0.005=28.6 W\text{tile:}\ \frac{Q}{\Delta t} = \frac{(1.1)(0.010)(13)}{0.005} = 28.6\ \text{W}
mat: QΔt=(0.050)(0.010)(13)0.005=1.3 W\text{mat:}\ \frac{Q}{\Delta t} = \frac{(0.050)(0.010)(13)}{0.005} = 1.3\ \text{W}

A factor of 2222. Over ten seconds the tile draws 286 J286\ \text{J} out of your foot and the mat draws 13 J13\ \text{J}. Same temperature, same area, same thickness, same temperature difference, and one of them is pulling energy out twenty-two times faster.

This is the cleanest demonstration on the page of what conduction is, and it makes two points at once.

First, conduction is a rate, not a state. Cold is not a substance the tile has more of. The tile is not colder, it conducts faster, and your nerve endings report a rate of energy loss rather than a temperature. Heat vs temperature is the page for that confusion in general.

Second, convection is not in this story at all. Nothing about the tile or the mat is flowing anywhere. Both are solids, and a solid cannot convect. If the same question were asked about standing in a draft rather than on a floor, convection would be the answer and there would be no equation to reach for. That asymmetry is the exam-relevant one: the situations the course can make you compute are conduction situations, by construction.

What an exam can actually ask about convection

Not a calculation. With no equation and no material constant in the course, no question can require a convective number. What is left is identification and reasoning, and both are supported by the skill list attached to Topic 9.3: 1.A, create diagrams, tables, charts, or schematics to represent physical situations; 2.C, compare physical quantities between two or more scenarios; 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.

So the reasonable expectations are these.

  • Name the process operating in a described situation, given that 9.3.A.2 supplies the list of three.
  • State the direction of the transfer, which 9.3.A.3 fixes as spontaneously from higher temperature to lower temperature, whichever of the three processes is at work.
  • Say when the transfer stops, which 9.3.A.4 defines as thermal equilibrium, when no net energy is transferred between two systems in thermal contact.
  • Recognize that all three deliver energy to the same first law. Whichever mechanism carried it, the energy arrives as QQ in ΔU=Q+W\Delta U = Q + W, and the mechanism leaves no trace in the result.

The distinguishing feature to have ready, if you are asked to justify a choice, is the one from the first section: convection requires the material itself to move in bulk, conduction does not. A sealed solid wall cannot convect. A pot of water on a stove does both at once, conducting through the metal base and convecting within the water.

What you should not write is a convective analogue of Q/Δt=kAΔT/LQ/\Delta t = kA\Delta T/L. The AP Physics 2 course does not define one, and inventing a coefficient in a free-response answer spends time on something no rubric can credit.

When it costs a mark

  • Reporting a rate as an energy. Q/ΔtQ/\Delta t is in watts. A question asking how much energy crossed a wall in an hour needs the rate multiplied by 3600 s3600\ \text{s}, and 5400 W5400\ \text{W} becomes 1.944×107 J1.944 \times 10^7\ \text{J}.
  • Putting LL in the numerator. Thickness is in the denominator: thicker insulates better. Checking the direction of one factor before substituting catches this instantly.
  • Using one side's temperature instead of the difference. The equation contains ΔT\Delta T across the material. A wall between 21 C21\ \text{C} and 3 C3\ \text{C} has ΔT=18 K\Delta T = 18\ \text{K}, not 294 K294\ \text{K}.
  • Converting a temperature difference to kelvin. A difference is already the same in both scales, so 18 C18\ \text{C} of difference is 18 K18\ \text{K} of difference. Adding 273273 to a difference is a real and costly error.
  • Treating kk as a property of the slab. 9.5.B.2 calls thermal conductivity an intrinsic property of the material. The slab's area and thickness are separate factors in the equation, and confusing the two turns a material comparison into a geometry comparison.
  • Confusing thermal conductivity with specific heat. Both are intrinsic, both are described in parallel wording by 9.5.B.2 and 9.5.A.2, and both live in Topic 9.5. kk sets how fast energy crosses a material, cc sets how much energy a temperature change costs, and they appear in different equations.
  • Answering a convection question with a formula. There is none in this course, and writing one signals that the scope was not checked.
  • Calling any transfer through a fluid convection. A fluid conducts too. Convection specifically requires the fluid to move in bulk and carry energy with it, so a still layer of trapped air between two panes is conducting, not convecting.

When they run together, and why that lulls you

In almost every real situation both processes are working at once, which is why they are hard to keep separate in the first place.

A saucepan on a stove conducts through the metal base and then convects inside the water. A house loses energy by conducting through the walls and by convecting through the draughts. A cup of coffee conducts into the ceramic, convects within the liquid, and radiates from the surface, all simultaneously and all in the same direction, from hot to cold. Nothing in the outcome labels which fraction went by which route.

And that is the deeper reason the distinction is easy to lose: the first law does not care. All three processes deliver their energy as QQ, the same QQ that appears in ΔU=Q+W\Delta U = Q + W, and the equation sheet's symbol list defines QQ simply as the energy transferred to a system by heating, with no subdivision by mechanism. Once the energy has arrived, the route it took has left no trace in the internal energy or the temperature.

So the three mechanisms matter for exactly one thing: predicting how fast, or in the case of an insulated container, whether at all. They never matter for the bookkeeping afterwards.

One more overlap worth naming, because it is the standard exam trap. Insulation usually works by trapping a fluid so it cannot convect. Double glazing, foam, a duvet, and the fur on an animal all rely on a layer of still air, and still air conducts poorly. The situation looks like it is about convection, since preventing convection is the design goal, but the physics that survives is conduction, and conduction is the half with the equation. If a question about an insulating layer expects a number, the number comes from Q/Δt=kAΔT/LQ/\Delta t = kA\Delta T/L applied to whatever is left after the bulk motion has been stopped.

What the CED asks of Topics 9.3 and 9.5

Both processes belong to AP Physics 2 Unit 9, Thermodynamics, weighted at 1515 to 18 %18\ \% of the multiple-choice section across a suggested 1010 to 1616 class periods.

Topic 9.3, Thermal Energy Transfer and Equilibrium, has one learning objective, 9.3.A, to describe the transfer of energy between two systems in thermal contact due to temperature differences of those two systems. Its essential knowledge is:

  • 9.3.A.1, two systems are in thermal contact if the systems may transfer energy by thermal processes, with 9.3.A.1.i defining heating as the transfer of energy into a system by thermal processes and 9.3.A.1.ii defining cooling as 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, with 9.3.A.3.i and 9.3.A.3.ii giving the collision-level account.
  • 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.

That is the entirety of what the framework says about convection: one clause of one sentence in 9.3.A.2.

Topic 9.5, Specific Heat and Thermal Conductivity, has two learning objectives, and conduction gets the second one to itself. 9.5.A concerns the energy required to change a temperature, with Q=mcΔTQ = mc\Delta T at 9.5.A.1 and specific heat as an intrinsic property at 9.5.A.2. 9.5.B is to describe the rate at which energy is transferred by conduction through a given material, with 9.5.B.1 giving Q/Δt=kAΔT/LQ/\Delta t = kA\Delta T/L and 9.5.B.2 giving thermal conductivity as an intrinsic property.

Topic 9.5 carries a boundary statement, and it is not about conduction: it reads that AP Physics 2 will model specific heat as independent of temperature, which constrains the topic's other half. Unit 9 prints three boundary statements altogether, under Topics 9.1, 9.5 and 9.6, the other two being that AP Physics 2 only expects qualitative and quantitative analysis of collisions in one and two dimensions with no functional form of the Maxwell-Boltzmann distribution required, and that only qualitative treatment of the second law of thermodynamics is within the scope of the course. Topic 9.3 carries none. So there is no boundary statement anywhere in the unit about conduction, and none about convection.

The suggested skills listed for Topic 9.3 are 1.A, 2.C, 3.B and 3.C. For Topic 9.5 they are 1.B, 2.B, 2.D, 3.A and 3.B, and the presence of 3.A, create experimental procedures that are appropriate for a given scientific question, is a hint that conduction is a plausible subject for the experimental design question, which it can be precisely because it has a measurable rate.

Topic 9.3 and Topic 9.5 carry the framing in full, and heat vs internal energy covers why all three mechanisms deliver the same kind of quantity.

A single-glazed window, and the three factors you can change

A window pane of area 1.5 m21.5\ \text{m}^2 and thickness 4.0×103 m4.0 \times 10^{-3}\ \text{m} separates a room at 21 C21\ \text{C} from outside air at 3 C3\ \text{C}. The glass has thermal conductivity k=0.80 W/(mK)k = 0.80\ \text{W/(m} \cdot \text{K)}. (a) Find the rate at which energy is conducted through the pane. (b) Find the energy conducted in one hour. (c) Find the new rate if the pane is replaced by one twice as thick. (d) Find the new rate if instead the area is doubled at the original thickness.

  1. (a) Use 9.5.B.1's printed relation, Q/Δt=kAΔT/LQ/\Delta t = kA\Delta T/L. First the temperature difference: ΔT=213=18\Delta T = 21 - 3 = 18. A difference of 1818 degrees Celsius is a difference of 1818 kelvin, so no conversion is needed and none should be attempted.

  2. Numerator: kAΔT=(0.80)(1.5)(18)kA\Delta T = (0.80)(1.5)(18). Take it in two steps, (0.80)(1.5)=1.2(0.80)(1.5) = 1.2, then (1.2)(18)=21.6(1.2)(18) = 21.6.

  3. Divide by the thickness: Q/Δt=21.6/(4.0×103)=5400 WQ/\Delta t = 21.6 / (4.0 \times 10^{-3}) = 5400\ \text{W}, that is 5.4 kW5.4\ \text{kW}.

  4. Sanity-check the units: W/(mK)×m2×K/m\text{W/(m} \cdot \text{K)} \times \text{m}^2 \times \text{K} / \text{m} leaves watts. The answer is a rate, not an energy.

  5. (b) One hour is 3600 s3600\ \text{s}, so the energy conducted is (5400)(3600)=1.944×107 J(5400)(3600) = 1.944 \times 10^7\ \text{J}, or about 19.4 MJ19.4\ \text{MJ}.

  6. This is the step most often skipped. The equation returns watts and the question asked for joules, and the two differ by a factor of the time interval.

  7. (c) Thickness is in the denominator, so doubling LL to 8.0×103 m8.0 \times 10^{-3}\ \text{m} halves the rate: Q/Δt=21.6/(8.0×103)=2700 WQ/\Delta t = 21.6 / (8.0 \times 10^{-3}) = 2700\ \text{W}.

  8. (d) Area is in the numerator, so doubling AA to 3.0 m23.0\ \text{m}^2 doubles the rate: (0.80)(3.0)(18)/(4.0×103)=43.2/(4.0×103)=10,800 W(0.80)(3.0)(18)/(4.0 \times 10^{-3}) = 43.2/(4.0 \times 10^{-3}) = 10{,}800\ \text{W}.

  9. The pattern is worth carrying because skill 2.D asks for exactly this kind of factor-of-change reasoning. Rate is proportional to kk, to AA and to ΔT\Delta T, and inversely proportional to LL. Nothing in the equation is squared, so every factor of change passes through unchanged or inverted.

  10. One thing this calculation does not include: convection. Air moving across either face carries energy too, and AP Physics 2 provides no way to quantify it. The number above is the conduction through the glass alone, which is the only part the course can ask you to compute.

(a) 5400 W5400\ \text{W}. (b) 1.944×107 J1.944 \times 10^7\ \text{J} in one hour. (c) 2700 W2700\ \text{W}, halved by doubling the thickness. (d) 10,800 W10{,}800\ \text{W}, doubled by doubling the area.

Two materials, and how much glass it takes to match the insulation

Two slabs each have area 2.0 m22.0\ \text{m}^2 and thickness 0.10 m0.10\ \text{m}, with a temperature difference of 25 K25\ \text{K} maintained across each. Slab A is a foam insulator with k=0.040 W/(mK)k = 0.040\ \text{W/(m} \cdot \text{K)} and slab B is glass with k=0.80 W/(mK)k = 0.80\ \text{W/(m} \cdot \text{K)}. (a) Find the conduction rate through each. (b) Find the ratio. (c) Find the thickness of glass that would conduct at the same rate as the foam. (d) State what changed and what did not, in the language of 9.5.B.2.

  1. (a) Slab A: Q/Δt=(0.040)(2.0)(25)/0.10Q/\Delta t = (0.040)(2.0)(25)/0.10. The numerator is (0.040)(2.0)=0.080(0.040)(2.0) = 0.080, then (0.080)(25)=2.0(0.080)(25) = 2.0. Dividing, 2.0/0.10=20 W2.0/0.10 = 20\ \text{W}.

  2. Slab B: (0.80)(2.0)(25)/0.10(0.80)(2.0)(25)/0.10. Numerator (0.80)(2.0)=1.6(0.80)(2.0) = 1.6, then (1.6)(25)=40(1.6)(25) = 40. Dividing, 40/0.10=400 W40/0.10 = 400\ \text{W}.

  3. (b) The ratio is 400/20=20400/20 = 20. That is exactly the ratio of the thermal conductivities, 0.80/0.040=200.80/0.040 = 20, which it has to be, since every other factor in the equation was identical.

  4. (c) Set the glass rate equal to 20 W20\ \text{W} and solve for the thickness: L=kAΔT/(Q/Δt)=(0.80)(2.0)(25)/20=40/20=2.0 mL = kA\Delta T / (Q/\Delta t) = (0.80)(2.0)(25)/20 = 40/20 = 2.0\ \text{m}.

  5. So it would take a 2.0 m2.0\ \text{m} slab of glass to conduct as slowly as a 0.10 m0.10\ \text{m} slab of the foam, a thickness ratio of 2020 that mirrors the conductivity ratio exactly.

  6. Check that against the equation's structure. Holding the rate, the area and the temperature difference fixed makes LL proportional to kk, so a conductivity twenty times larger needs a thickness twenty times larger. 0.10×20=2.0 m0.10 \times 20 = 2.0\ \text{m}. It agrees.

  7. (d) What changed is the material. 9.5.B.2 says the thermal conductivity of a material is an intrinsic property of that material that depends on the arrangement and interactions of the atoms that make up the material, so kk travels with the substance and not with the slab.

  8. What did not change is the geometry in part (a): both slabs had the same AA, the same LL and the same ΔT\Delta T. That is what made part (b) a clean comparison of materials rather than a comparison of slabs.

  9. And part (c) shows the trade the other way. Geometry can compensate for a bad material, but only in proportion, and here the price of using glass instead of foam is a wall two meters thick.

  10. Neither slab is convecting. Both are solids, and 9.3.A.2's second process needs a fluid that can move in bulk. The foam works precisely because it holds air still enough that only conduction remains.

(a) 20 W20\ \text{W} through the foam and 400 W400\ \text{W} through the glass. (b) A ratio of 2020, equal to the ratio of the conductivities. (c) 2.0 m2.0\ \text{m} of glass. (d) Only the material changed, and 9.5.B.2 makes thermal conductivity an intrinsic property of the material rather than of the slab.

The bathroom floor question, answered with numbers

A tile floor and a fabric mat have been in the same room all night and are at the same temperature. A bare foot at 33 C33\ \text{C} makes contact over 0.010 m20.010\ \text{m}^2 with a surface layer 0.005 m0.005\ \text{m} thick, backed by material at the room temperature of 20 C20\ \text{C}. Take k=1.1 W/(mK)k = 1.1\ \text{W/(m} \cdot \text{K)} for the tile and k=0.050 W/(mK)k = 0.050\ \text{W/(m} \cdot \text{K)} for the mat. (a) Find the rate at which energy leaves the foot in each case. (b) Find the energy lost in ten seconds on each. (c) Find the ratio and explain the sensation. (d) State whether convection plays any part.

  1. The temperature difference is the same for both: ΔT=3320=13\Delta T = 33 - 20 = 13 degrees Celsius, which is 13 K13\ \text{K} as a difference.

  2. (a) Tile: Q/Δt=(1.1)(0.010)(13)/0.005Q/\Delta t = (1.1)(0.010)(13)/0.005. Numerator, (1.1)(0.010)=0.011(1.1)(0.010) = 0.011, then (0.011)(13)=0.143(0.011)(13) = 0.143. Dividing, 0.143/0.005=28.6 W0.143/0.005 = 28.6\ \text{W}.

  3. Mat: (0.050)(0.010)(13)/0.005(0.050)(0.010)(13)/0.005. Numerator, (0.050)(0.010)=5.0×104(0.050)(0.010) = 5.0 \times 10^{-4}, then (5.0×104)(13)=6.5×103(5.0 \times 10^{-4})(13) = 6.5 \times 10^{-3}. Dividing, 6.5×103/0.005=1.3 W6.5 \times 10^{-3}/0.005 = 1.3\ \text{W}.

  4. (b) Over ten seconds, the tile removes (28.6)(10)=286 J(28.6)(10) = 286\ \text{J} and the mat removes (1.3)(10)=13 J(1.3)(10) = 13\ \text{J}.

  5. (c) The ratio is 28.6/1.3=22.028.6/1.3 = 22.0, and it equals the ratio of the conductivities, 1.1/0.050=221.1/0.050 = 22, because every other factor was held identical by the setup.

  6. The sensation follows from the rate, not from a temperature. Both surfaces are at 20 C20\ \text{C}. What differs is that one draws energy from your skin twenty-two times faster than the other, and the nerve endings that report cold are responding to that rate.

  7. This is worth stating as a general principle, because it answers the unit's own essential question about the tile floor and the bathroom mat: cold is not a quantity a floor holds. The floor conducts, and how fast it conducts is a property of what it is made of.

  8. (d) No. Both the tile and the mat are solids, so nothing is moving in bulk and convection is not available. 9.3.A.2 lists it as one of three thermal processes but it requires a fluid that can carry its internal energy from place to place.

  9. There is a small convective contribution from the air around the foot, and AP Physics 2 supplies no way to quantify it, which is one more reason the question is posed about two solids in contact.

  10. One check on scope before finishing. Every number here came from the printed Q/Δt=kAΔT/LQ/\Delta t = kA\Delta T/L, and the two conductivities were supplied by the problem. The equation sheet prints no table of kk values, so any question of this kind has to hand them to you.

(a) 28.6 W28.6\ \text{W} into the tile and 1.3 W1.3\ \text{W} into the mat. (b) 286 J286\ \text{J} and 13 J13\ \text{J} over ten seconds. (c) A ratio of 2222, equal to the ratio of thermal conductivities. The tile feels colder because it removes energy faster, not because it is colder. (d) No, both are solids and convection needs a fluid that moves in bulk.

Frequently asked questions

What is the difference between conduction and convection?

Whether the material itself moves. Conduction passes energy along through a material by collisions between neighboring particles while the material stays where it is, which is why it works in solids. Convection moves energy by carrying the fluid bodily from one place to another, so the particles themselves relocate and take their internal energy with them, which is why it needs a fluid. AP Physics 2 essential knowledge 9.3.A.2 lists both, saying the thermal processes by which energy may be transferred between systems at different temperatures are conduction, convection, and radiation. Both move energy spontaneously from a higher-temperature system to a lower-temperature one, per 9.3.A.3.

Is convection on the AP Physics 2 exam?

Only by name. The word convection appears in the AP Physics 2 CED exactly twice: once in essential knowledge 9.3.A.2, as one of the three named thermal processes, and once in the appendix's vocabulary section on heat, again inside a list. There is no learning objective about convection, no essential knowledge statement of its own, no equation on the sheet, no constant in the constants box, and no boundary statement. So a question can expect you to identify convection as the process at work in a described situation and to know that energy moves from hot to cold, but it cannot ask for a convective calculation. Do not learn a convection formula for this exam.

What is the equation for the rate of heat conduction?

The rate is Q divided by delta t, equal to k times A times delta T, all over L. It is printed in the Thermal Physics panel of the AP Physics 2 equation sheet and it comes from essential knowledge 9.5.B.1, which says the rate at which energy is transferred by conduction through a given material is related to the thermal conductivity, the physical dimensions of the material, and the temperature difference across the material. In the symbol list, k is the thermal conductivity, A is the area, L is the length or thickness, and Q is the energy transferred by heating. The left side is a rate in watts, so an answer in joules needs it multiplied by the time interval.

Is thermal conductivity on the AP equation sheet?

The symbol is, but no values are. The Thermal Physics panel prints the conduction rate equation and its symbol list defines k as thermal conductivity, so you will always have the relationship. The sheet contains no table of conductivities for glass, copper or anything else, which means a question that requires a number has to supply it in the stem. Essential knowledge 9.5.B.2 says the thermal conductivity of a material is an intrinsic property of that material that depends on the arrangement and interactions of the atoms that make up the material, phrased in deliberate parallel with 9.5.A.2 on specific heat. Intrinsic means it belongs to the substance: cutting a bar in half does not change k.

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

Because your skin reports the rate at which energy leaves it, not the temperature of what it is touching. Both surfaces have been in the same room and are at the same temperature. Put a foot at 33 degrees Celsius on each over a contact patch of 0.010 square meters and a surface layer 0.005 meters thick, with the room at 20 degrees Celsius. If the tile has a thermal conductivity of 1.1 and the mat 0.050 watts per meter per kelvin, the conduction rates are 28.6 watts and 1.3 watts, a factor of 22. Over ten seconds the tile draws 286 joules out of your foot and the mat draws 13 joules. The tile is not colder, it conducts faster.

How does radiation compare with conduction and convection in the AP course?

All three are named together in essential knowledge 9.3.A.2 as the thermal processes by which energy may be transferred between systems at different temperatures, but they are not treated equally. Conduction gets its own learning objective, 9.5.B, and a printed rate equation in Unit 9. Radiation becomes quantitative in a different unit: essential knowledge 15.4.A.3.iii, in Unit 15 on modern physics, gives the Stefan-Boltzmann law as power equals A sigma T to the fourth, printed in the Modern Physics panel of the sheet, with sigma supplied in the constants box as 5.67 times 10 to the minus 8 watts per square meter per kelvin to the fourth. Convection never becomes quantitative anywhere in the course.

Does insulation work by stopping conduction or convection?

Usually by stopping convection so that only conduction is left, and then relying on the fact that still air conducts poorly. Foam, double glazing, a duvet and animal fur all trap air in small pockets where it cannot circulate. The design goal is convective, but the physics that remains is conduction, and conduction is the half of the pair with an equation. So if a question about an insulating layer expects a number, the number comes from the rate equation Q over delta t equals k A delta T over L applied to whatever material is left after the bulk motion has been stopped. Note also that a still fluid conducts: transfer through a fluid is not automatically convection.