4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A 0.200 kg aluminum block is heated from 20.0°C to 60.0°C. The specific heat of aluminum is 900 J/(kg·K). How much energy is transferred to the block?
Answer and reasoning
A1.1 × 10⁴ J A student who puts the final temperature, 60.0°C, in place of the temperature change picks this: (0.200 kg)(900 J/(kg·K))(60.0) = 1.1 × 10⁴ J. Q = mcΔT needs the change, 40.0 K.
B7.2 × 10³ JCorrect The temperature change is ΔT = 60.0°C − 20.0°C = 40.0 K; a change of one Celsius degree is a change of one kelvin. Q = mcΔT = (0.200 kg)(900 J/(kg·K))(40.0 K) = 7.2 × 10³ J.
C5.6 × 10⁴ J A student who adds 273 to the temperature change picks this, using ΔT = 313 K. Adding 273 converts a temperature, not a difference: a rise of 40.0°C is a rise of 40.0 K.
D3.6 × 10⁴ J A student who treats the specific heat as the energy needed to warm the whole block by 1 K picks this, leaving out the mass: (900 J/(kg·K))(40.0 K). Specific heat is energy per kilogram per kelvin, so it must be multiplied by the mass, 0.200 kg.
Working ΔT = 60.0°C − 20.0°C = 40.0 K. Q = mcΔT = (0.200 kg)(900 J/(kg·K))(40.0 K) = 7200 J = 7.2 × 10³ J.
The specific heat of liquid water is about twice the specific heat of ice. Which statement best accounts for this difference?
Answer and reasoning
ALiquid water is denser than ice, so each kilogram of it contains more molecules. A student who thinks a denser material packs more molecules into each kilogram picks this. Density is mass per unit volume; a kilogram of ice and a kilogram of liquid water contain the same number of molecules, because every water molecule has the same mass.
BFreezing changes how the molecules interact, so the energy per kilogram per kelvin changes.Correct Specific heat is an intrinsic property that depends on how a material's atoms or molecules are arranged and interact. Ice and liquid water are the same substance, but freezing changes the arrangement of the molecules and how they interact, so each phase has its own specific heat.
CLiquid water is warmer than ice, so it contains more heat to begin with. A student who thinks of heat as something an object contains picks this. Specific heat is the energy needed per kilogram per kelvin of temperature change, not energy stored in a sample, and ice and liquid water can both be at 0°C.
DLiquid water conducts energy better than ice, so it takes in more energy. A student who treats thermal conductivity and specific heat as one property picks this. Conductivity describes how fast energy passes through a material; specific heat describes how much energy a kilogram needs per kelvin. In fact ice conducts better than liquid water.
A copper rod, insulated along its sides, connects two reservoirs held at different temperatures, and energy is conducted along it at a steady rate. The rod is replaced by a copper rod of the same length but twice the diameter, and the temperature difference between the reservoirs is halved. By what factor does the rate of energy transfer along the rod change?
Answer and reasoning
A×1 A student who doubles the area when the diameter doubles picks this: 2 × ½ = 1. The area of a circle is πd²/4, so doubling the diameter makes it four times as great.
B×½ A student who thinks the rate does not depend on the rod's cross-sectional area picks this, counting only the halved temperature difference. A wider rod provides more paths for conduction: the rate is proportional to A.
C×2Correct Q/Δt = kAΔT/L. The cross-sectional area is proportional to the square of the diameter, so doubling the diameter makes the area 4 times as great. Halving ΔT halves the rate: 4 × ½ = 2.
D×⅛ A student who puts the area underneath and the length on top picks this: (1/4) × (1/2) = 1/8. In Q/Δt = kAΔT/L the area is in the numerator: a wider rod conducts energy faster.
Working Q/Δt = kAΔT/L. A = πd²/4, so d → 2d gives A → 4A; ΔT → ΔT/2; k and L unchanged. Factor = 4 × ½ = 2. (Area ∝ d: 2 × ½ = 1. Area ignored: ½. L/A instead of A/L: ¼ × ½ = ⅛.)
A metal block and a wooden block have been on the same table in a room at 18°C for several hours. When a student touches them, the metal block feels colder. Which statement correctly explains this?
Answer and reasoning
AThe metal has the greater thermal conductivity, so it conducts energy from the hand faster.Correct After hours in the room both blocks are at 18°C. Thermal conductivity is a property of each material, set by how its atoms are arranged and interact, and metals conduct far better than wood. Energy leaves the hand faster into the metal, so the skin cools faster and the metal feels colder.
BThe metal block is at a lower temperature than the wooden block, which is at room temperature. A student who judges temperature by touch picks this. Objects left together in a room reach the same temperature; what differs is how fast they conduct energy away from the skin.
CThe wooden block releases thermal energy of its own, which keeps the hand warm. A student who thinks some materials are warm in themselves picks this. Wood has no internal source of energy; it feels warmer only because it conducts energy away from the hand slowly.
DThe metal has the greater specific heat, so it takes in more energy from the hand. A student who treats specific heat as the property that governs conduction picks this. How fast the hand loses energy depends on thermal conductivity, and in fact the specific heat of wood is greater than that of most metals.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
9.5.A.1 Specific heat, c Fix
Specific heat, c
The energy needed to raise the temperature of 1 kg of a material by 1 K. A material with a large specific heat, such as water, changes temperature little for a given energy per kilogram. SI unit: J/(kg·K).
Energy to change temperature, Q = mcΔT
The energy Q transferred to a sample of mass m and specific heat c when its temperature changes by ΔT. Q is positive when the temperature rises (energy transferred into the sample) and negative when it falls. SI unit: joule (J).
Temperature change, ΔT
ΔT = Tf − Ti. A change of 1°C is the same size as a change of 1 K, so a temperature change has the same value on both scales; 273 is added only to convert a temperature, not a change. SI unit: kelvin (K).
Reaching a common temperature (calorimetry)
When objects at different temperatures reach a common temperature in an isolated system, the energy transferred from the warmer objects equals the energy transferred to the cooler ones: the sum of mcΔT over all the objects is zero. The common temperature lies closer to the starting temperature of the object with the larger mc.
Students often think The temperature of a sample (for example its final temperature) can be used in Q = mcΔT in place of its change in temperature. In fact No. ΔT is the change in temperature, Tf − Ti. A sample heated from 20°C to 60°C has ΔT = 40 K, whatever its final temperature.
Students often think A temperature change measured in degrees Celsius must be converted to kelvins by adding 273. In fact No. A change of 1°C is a change of 1 K, so a temperature change has the same value on both scales. Adding 273 converts a temperature, as in PV = nRT, not a difference between two temperatures.
9.5.A.2 Specific heat as an intrinsic property Fix
Specific heat as an intrinsic property
Specific heat is the same for every sample of a given material, whatever its size or shape, because it depends on how the material's atoms or molecules are arranged and interact. Different phases of one substance, such as ice and liquid water, have different specific heats. AP Physics 2 models specific heat as independent of temperature.
Energy needed per kelvin by a whole object, mc
The energy needed to raise the temperature of a particular object by 1 K is the product mc. It depends on the object's mass, unlike the specific heat c, which is a property of the material. SI unit: J/K.
Students often think Specific heat is the energy needed to raise the temperature of the whole object by 1 K, so a larger object of the same material has a larger specific heat and the mass need not be included. In fact No. Specific heat is the energy needed per kilogram per kelvin, a property of the material. The energy needed to warm a whole object by 1 K is mc, which is larger for a larger object of the same material.
Students often think A denser material has more particles in each kilogram, so it has a greater specific heat. In fact No. Density is mass per unit volume. The number of molecules in a kilogram of a substance depends on the mass of each molecule, so a kilogram of ice and a kilogram of liquid water contain the same number of molecules.
9.5.B.1 Thermal conduction Fix
Thermal conduction
Transfer of energy through a material from a region of higher temperature to a region of lower temperature by interactions between neighboring particles, without any bulk movement of the material.
Rate of conduction, Q/Δt = kAΔT/L
The energy conducted per unit time through a slab or rod of thermal conductivity k, cross-sectional area A and thickness (length) L, with a temperature difference ΔT between its faces. The rate is proportional to A and ΔT and inversely proportional to L. SI unit: watt (W = J/s).
Students often think The rate of conduction depends on the temperature of the hotter side, not on the temperature difference across the material. In fact No. It depends on the temperature difference across the material, ΔT. Two slabs with faces at 40°C and 20°C and at 100°C and 80°C conduct at the same rate if they are otherwise identical.
Students often think The rate of conduction depends only on the material, its thickness and the temperature difference, not on the area. In fact Yes. Q/Δt = kAΔT/L: doubling the area through which energy flows doubles the rate, as a larger window loses energy faster than a smaller one of the same glass.
9.5.B.2 Thermal conductivity, k Fix
Thermal conductivity, k
An intrinsic property of a material that describes how readily it conducts energy; it depends on how the material's atoms are arranged and interact, not on the size or shape of the sample. Metals have large thermal conductivities; wood, plastics and air have small ones. SI unit: W/(m·K).
Students often think A material that conducts energy well also has a large specific heat: conducting energy and taking in energy are the same property. In fact No. Thermal conductivity describes how fast energy passes through a material; specific heat describes how much energy a kilogram needs per kelvin of temperature change. Metals conduct well yet have small specific heats.
Students often think Objects that feel colder are at a lower temperature: in the same room, metal is colder than wood. In fact No. Objects left in the same room reach the room's temperature. Metal feels colder because it conducts energy away from the skin faster, so the skin cools faster.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
A 0.300 kg iron block at 80.0°C is placed in 0.500 kg of water at 20.0°C in an insulated container, and the block and the water reach a common temperature. The specific heat of iron is 450 J/(kg·K), and the specific heat of water is 4180 J/(kg·K). Energy transfer to the container and the surroundings is negligible. What is the common temperature?
Answer and reasoning
A50.0°C A student who takes the common temperature to be halfway between the starting temperatures picks this. That holds only when the two objects have equal values of mc; here the water's mc is about 15 times the iron's, so the common temperature is much closer to 20.0°C.
B42.5°C A student who weights the two temperatures by mass alone picks this, as if a kilogram of iron and a kilogram of water needed the same energy per kelvin. Their specific heats differ by a factor of more than 9, and the energy balance must include them.
C70.9°C A student who thinks the material with the greater specific heat changes temperature more picks this: in effect the two specific heats are swapped, so the water warms by about 51 K while the iron cools by only about 9 K. The reverse is true: water's large specific heat keeps its temperature change small.
D23.6°CCorrect The energy transferred from the iron equals the energy transferred to the water: (0.300 kg)(450 J/(kg·K))(80.0°C − T) = (0.500 kg)(4180 J/(kg·K))(T − 20.0°C), which gives T = 23.6°C. The water, with much the larger mc, changes temperature much less than the iron.
Working Energy from iron = energy to water: (0.300)(450)(80.0 − T) = (0.500)(4180)(T − 20.0). 135(80.0 − T) = 2090(T − 20.0), so 10 800 + 41 800 = 2225T and T = 52 600/2225 = 23.6°C.
Blocks X and Y are made of different materials. The graph shows the temperature T of each block as energy Q is transferred to it; the mass of each block is marked beside its line. What is the ratio of the specific heat of Y to the specific heat of X?
Answer and reasoning
AcY/cX = 4.0 A student who compares the energy each whole block needs per kelvin picks this: Y needs 400 J/K and X needs 100 J/K. That comparison leaves out the masses; Y has twice the mass of X, so per kilogram its specific heat is only twice as great.
BcY/cX = 0.5 A student who thinks a greater specific heat goes with a greater temperature rise picks this, taking c to be proportional to mΔT/Q. A greater specific heat means a smaller temperature change for the same energy per kilogram: c = Q/(mΔT).
CcY/cX = 2.0Correct Specific heat is energy per kilogram per kelvin, c = Q/(mΔT). For the same 4.0 kJ, X (0.20 kg) warms by 40 K and Y (0.40 kg) by 10 K, so cX = 500 J/(kg·K) and cY = 1000 J/(kg·K): Y's specific heat is twice X's.
DcY/cX = 1.0 A student who uses the final temperatures, 60°C and 30°C, instead of the temperature changes picks this, since 0.20 kg × 60 and 0.40 kg × 30 are equal. Both blocks start at 20°C, so the changes are 40 K and 10 K.
Working From the graph, Q = 4.0 kJ raises X (0.20 kg) from 20°C to 60°C (ΔT = 40 K) and Y (0.40 kg) from 20°C to 30°C (ΔT = 10 K). c = Q/(mΔT): cX = 4000 J/(0.20 kg × 40 K) = 500 J/(kg·K); cY = 4000 J/(0.40 kg × 10 K) = 1000 J/(kg·K). cY/cX = 2.0.
The diagram shows three slabs, X, Y and Z, made of the same material. Each slab's thickness, the area of its faces and the temperatures of its two faces are marked. Energy is conducted through each slab from its hotter face to its cooler face. Which ranking of the rates at which energy is conducted through the slabs is correct?
Answer and reasoning
AZ > X > YCorrect Q/Δt = kAΔT/L, and k is the same for all three. In units of kA/L: X gives (1)(20 K)/1 = 20, Y gives (1)(20 K)/2 = 10, and Z gives (2)(30 K)/2 = 30. So Z > X > Y.
BZ > Y > X A student who ranks the slabs by the temperature of the hotter face, 70°C, 60°C and 20°C, picks this. The rate depends on the temperature difference across a slab: X and Y both have ΔT = 20 K, and Y is twice as thick, so Y conducts more slowly than X.
CX > Z > Y A student who leaves out the area picks this, ranking by ΔT/L: 20, 15 and 10 for X, Z and Y. Z's faces have twice the area, which doubles its rate to 30 in these units, the largest of the three.
DY > Z > X A student who puts the thickness on top and the area underneath picks this: ΔT·L/A gives 40 for Y, 30 for Z and 20 for X. A thicker slab conducts energy more slowly and a larger area more quickly: Q/Δt = kAΔT/L.
Working Q/Δt = kAΔT/L, same k. X: A, L, ΔT = 20 − 0 = 20 K → 20 kA/L. Y: A, 2L, ΔT = 60 − 40 = 20 K → 10 kA/L. Z: 2A, 2L, ΔT = 70 − 40 = 30 K → 30 kA/L. Ranking Z > X > Y.
A section of brick wall has an area of 4.0 m² and a thickness of 0.10 m. Its inner surface is at 19.0°C and its outer surface is at 3.0°C. The thermal conductivity of the brick is 0.70 W/(m·K). At what rate is energy conducted through this section of wall?
Answer and reasoning
A5.3 × 10² W A student who uses the temperature of the warmer surface, 19.0°C, instead of the temperature difference picks this. Conduction depends on the difference between the two surfaces, 16.0 K.
B1.1 × 10² W A student who leaves out the area picks this: (0.70 W/(m·K))(16.0 K)/(0.10 m) = 112 W is the rate through each square meter of wall. The section's area, 4.0 m², multiplies it.
C8.1 × 10³ W A student who adds 273 to the temperature difference picks this. A difference of 16.0°C is a difference of 16.0 K; 273 is added only when converting a temperature.
D4.5 × 10² WCorrect Q/Δt = kAΔT/L with ΔT = 19.0°C − 3.0°C = 16.0 K: (0.70 W/(m·K))(4.0 m²)(16.0 K)/(0.10 m) = 448 W, or 4.5 × 10² W to two significant figures.
Working ΔT = 19.0°C − 3.0°C = 16.0 K. Q/Δt = kAΔT/L = (0.70 W/(m·K))(4.0 m²)(16.0 K)/(0.10 m) = 448 W ≈ 4.5 × 10² W.
A wall is made of two layers, X and Y, of equal area, pressed together with good contact. Layer X has thickness L and thermal conductivity k. Layer Y has thickness 2L and thermal conductivity 4k. The outer surfaces of the wall are held at fixed temperatures that differ by ΔT, and energy is conducted through the wall at a steady rate, so that the temperature at every point in the wall is constant. Which expression gives the temperature difference across layer X?
Answer and reasoning
A(1/2)ΔT A student who thinks the temperature difference is shared equally between the layers picks this. The same rate passes through both, so the layer that conducts less readily for its size needs the larger temperature difference: X has k/L against 2k/L for Y, so ΔTX = 2ΔTY = (2/3)ΔT.
B(4/5)ΔT A student who leaves out the thicknesses picks this: kΔTX = 4kΔTY gives ΔTX = 4ΔTY = (4/5)ΔT. Y is twice as thick as X, which halves the rate through it for a given temperature difference, so ΔTX = 2ΔTY = (2/3)ΔT.
C(2/3)ΔTCorrect At a steady rate the energy conducted through X each second equals that through Y; otherwise the boundary between them would warm or cool. The rate through X is kAΔTX/L, and through Y it is (4k)AΔTY/(2L) = 2kAΔTY/L. So ΔTX = 2ΔTY, and since ΔTX + ΔTY = ΔT, ΔTX = (2/3)ΔT.
D(8/9)ΔT A student who puts the thickness in the numerator and the area in the denominator picks this: kLΔTX = (4k)(2L)ΔTY gives ΔTX = 8ΔTY = (8/9)ΔT. The rate is kAΔT/L, so the thicker layer conducts less for a given temperature difference: ΔTX = 2ΔTY = (2/3)ΔT.
Working Steady rate: the rate through X equals the rate through Y (no energy builds up at the boundary). X: kAΔTX/L. Y: (4k)AΔTY/(2L) = 2kAΔTY/L. Equal rates: ΔTX = 2ΔTY. ΔTX + ΔTY = ΔT, so ΔTX = (2/3)ΔT. Errors: equal shares → (1/2)ΔT; thickness left out, kΔTX = 4kΔTY → (4/5)ΔT; thickness in the numerator (rate ∝ kLΔT/A), kLΔTX = (4k)(2L)ΔTY → (8/9)ΔT.
A pot contains water of mass m and specific heat c. The pot has mass m/2, and the specific heat of the material it is made of is c/5. An electric heater transfers energy to the pot and the water at a constant rate P, and the pot and the water are always at the same temperature. No energy is transferred to the surroundings. Which expression gives the time needed to raise the temperature of the pot and the water by ΔT?
Answer and reasoning
A1.00mcΔT/P A student who thinks only the water needs energy picks this. The pot warms by ΔT too and takes in (m/2)(c/5)ΔT = 0.10mcΔT of its own, so the total is 1.10mcΔT and the time is 1.10mcΔT/P.
B1.50mcΔT/P A student who thinks the energy needed depends only on mass, not on the material, picks this: the pot is treated as another m/2 of water, (m + m/2)cΔT. The pot's material has one-fifth the specific heat of water, so the pot needs only (m/2)(c/5)ΔT = 0.10mcΔT, and the time is 1.10mcΔT/P.
C0.90mcΔT/P A student who gives the pot and water together the average specific heat, (c + c/5)/2 = 0.6c, for their total mass 1.5m picks this: 0.9mcΔT. The water is twice as massive as the pot, so the specific heats cannot simply be averaged; adding the parts, mcΔT + (m/2)(c/5)ΔT, gives 1.10mcΔT and a time of 1.10mcΔT/P.
D1.10mcΔT/PCorrect Both the water and the pot warm by ΔT. The water needs mcΔT, and the pot needs (m/2)(c/5)ΔT = 0.10mcΔT, so the total is 1.10mcΔT. The heater supplies energy at the rate P, so the time is Δt = 1.10mcΔT/P.
Working Energy needed: Q = (m/2)(c/5)ΔT + mcΔT = (1/10 + 1)mcΔT = 1.10mcΔT. Power P = Q/Δt, so Δt = Q/P = 1.10mcΔT/P. Errors: pot ignored → 1.00mcΔT/P; pot treated as if it were water, only the mass counting (m06) → (m + m/2)cΔT/P = 1.50mcΔT/P; average specific heat (c + c/5)/2 = 0.6c for the total mass 1.5m → 0.90mcΔT/P.
Compiled from the AP Physics 2 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
