EduNinja
IB Physics HL/Notes/B.1 Thermal energy transfers

IB Physics HLB.1 Thermal energy transfersNotes

Anchor Molecular states

Use the particle model to connect macroscopic properties to molecular behaviour. Fixed particle positions explain why solids keep shape. Mobile but close particles explain why liquids keep volume but flow. Large spacing and random motion explain why gases expand to fill a container and are compressible. Melting, freezing, boiling, evaporating, and condensing involve changes in molecular separation and potential energy.

In a solid, particles are closely packed in a fixed arrangement and vibrate about fixed positions.
In a liquid, particles remain close together but are disordered and can move past one another, so liquids flow.
In a gas, particles are far apart, move rapidly and randomly, and have negligible intermolecular forces except during collisions.
During heating within one phase, average random kinetic energy increases, so temperature rises.
During a phase change, temperature stays constant because energy changes intermolecular potential energy rather than average random kinetic energy.

Sort each particle-model statement by state of matter.

Sort
Unsorted
6
solid
0
liquid
0
gas
0

Compare solids, liquids, and gases in terms of particle arrangement, spacing, and motion. Then explain why temperature remains constant during a phase change.

Common mark losses are listing macroscopic properties only, or saying added energy always raises temperature during a phase change.

Compare solids, liquids, and gases in terms of particle arrangement, spacing, and motion. Then explain why temperature remains constant during a phase change.

Choose

Density

Density connects a macroscopic measurement to particle spacing. For the same mass, a smaller volume means a larger density. Gases have low density and are compressible because there is much empty space between particles. In calculations, most mistakes come from unit conversion rather than the formula itself.

Density is mass per unit volume: ρ = m/V.
The SI unit of density is kg m^-3 when mass is in kg and volume is in m^3.
Density is a scalar quantity; for a non-uniform object, m/V gives average density.
Solids and liquids are usually much denser than gases because their particles are much closer together.
Convert volume units carefully before substituting, especially cm^3 to m^3 or g cm^-3 to kg m^-3.

Build the density relation and unit check.

Formula
Target formula ρ = m/V
ρ
density
kg m^-3
m
mass of the sample
kg
V
volume of the sample
m^3
1Use mass and volume for the same material sample.same sample
2Convert mass to kg if using SI density.m in kg
3Convert volume to m^3.V in m^3
4Divide mass by volume.ρ = m/V
5If the sample is non-uniform, interpret the result as average density.average ρ

Define density and state the SI unit. Explain why gases are usually much less dense than solids or liquids using the particle model.

Common mark losses are giving only the equation without “mass per unit volume”, or using inconsistent mass and volume units.

Define density and state the SI unit. Explain why gases are usually much less dense than solids or liquids using the particle model.

Choose

Model Temperature scales

Celsius is convenient for everyday reference points, but Kelvin is the thermodynamic temperature scale used in physics equations that depend on absolute temperature. Because the scales have the same interval size, ΔT in kelvin has the same numerical value as Δt in degrees Celsius. The common trap is adding 273 to a temperature change or using Celsius in equations that need absolute temperature.

The Kelvin scale starts at absolute zero, the lowest possible temperature: 0 K = -273 °C approximately.
Kelvin and Celsius have the same size interval, so a change of 1 K equals a change of 1 °C.
Convert Celsius temperature to Kelvin using T/K = t/°C + 273 for IB calculations.
Kelvin is written K, not °K.
Use Kelvin for absolute temperature in gas, kinetic theory, and radiation equations; Celsius may be used for temperature differences when the size of the interval is what matters.

Repair the temperature-scale mistakes.

Spot Errors

Explain the relationship between the Celsius and Kelvin scales, including absolute zero and the numerical size of a temperature change.

Common mark losses are using Celsius as absolute temperature, adding 273 to a temperature difference, or writing kelvin with a degree sign.

Explain the relationship between the Celsius and Kelvin scales, including absolute zero and the numerical size of a temperature change.

Choose

Model Temperature scale changes

There are two different tasks: converting a temperature reading and converting a temperature change. A temperature reading needs the 273 offset because Kelvin and Celsius have different zero points. A temperature change does not need the offset because both scales step by the same amount. Always look for Δ before deciding what to do.

For an absolute temperature, convert Celsius to Kelvin using T(K) = t(°C) + 273.
Convert Kelvin to Celsius using t(°C) = T(K) - 273.
A temperature change has the same numerical size in kelvin and degrees Celsius: ΔT(K) = Δt(°C).
Use kelvin for ratios, gas laws, kinetic theory, Wien’s law, and Stefan-Boltzmann calculations.
In heating equations such as Q = mcΔT, the temperature change can be written in K or °C because the interval size is the same.

Match each temperature statement to the correct conversion move.

Match
Reasons
0/5

A sample warms from 20 °C to 55 °C. State the initial temperature in kelvin and the temperature change in kelvin.

Common mark losses are adding 273 to the temperature change or forgetting to use kelvin for the absolute initial temperature.

A sample warms from 20 °C to 55 °C. State the initial temperature in kelvin and the temperature change in kelvin.

Choose

Model Kelvin temperature and kinetic energy

The kinetic theory link is about microscopic random motion. A hotter ideal gas has faster molecules on average, but individual molecules still have a range of speeds. Do not use Celsius in this relationship, and do not confuse average energy per molecule with total energy, which also depends on the number of molecules.

For an ideal gas, Kelvin temperature is directly proportional to the average random translational kinetic energy of the molecules.
For one molecule, the average translational kinetic energy is <E_k> = (3/2)k_B T.
T must be in kelvin; doubling the Kelvin temperature doubles the average molecular kinetic energy.
Temperature describes average kinetic energy, not the total internal energy of the whole sample.
At a constant-temperature phase change, average kinetic energy stays constant even though internal energy can change.

Build the ideal-gas temperature and kinetic-energy relation.

Formula
Target formula <E_k> = (3/2)k_B T
<E_k>
average translational kinetic energy per molecule
J
k_B
Boltzmann constant
J K^-1
T
absolute temperature
K
1Check that the model is an ideal gas molecular model.ideal gas
2Convert temperature to kelvin.T in K
3Use average kinetic energy per molecule, not total energy.<E_k>
4Apply the Boltzmann relation.<E_k> = (3/2)k_B T
5Interpret proportional changes.<E_k> ∝ T

State how the average random translational kinetic energy of ideal gas molecules depends on temperature, and explain why the temperature must be in kelvin.

Common mark losses are using Celsius, discussing total internal energy instead of average kinetic energy, or omitting that the relation is for an ideal gas model.

State how the average random translational kinetic energy of ideal gas molecules depends on temperature, and explain why the temperature must be in kelvin.

Choose

Internal energy

Internal energy is a microscopic energy store of the whole sample. Heating within a phase usually increases the random kinetic component, so temperature rises. During a phase change, energy changes the potential component as particles separate or bond, so temperature can remain constant while internal energy changes.

Internal energy is the total random kinetic energy plus total intermolecular potential energy of the particles in a substance.
The random kinetic part is associated with particle motion and is linked to temperature.
The intermolecular potential part is associated with particle separation and bonding, especially during phase changes.
Temperature is related to average molecular kinetic energy, not total internal energy.
Thermal energy or heat refers to energy transferred because of a temperature difference; it is not a substance stored inside an object.

Sort each statement into the internal-energy model.

Sort
Unsorted
5
random kinetic component
0
intermolecular potential component
0
related but not the total definition
0

Define internal energy and explain why internal energy can change during a phase change even when temperature is constant.

Common mark losses are defining internal energy as temperature, omitting intermolecular potential energy, or saying an object contains heat.

Define internal energy and explain why internal energy can change during a phase change even when temperature is constant.

Choose

Model Thermal transfer direction

Temperature tells the direction of net energy transfer, not the amount of energy stored. A small hot object can have less internal energy than a large cooler object, but net thermal transfer still begins from hot to cold. Mechanisms such as conduction, convection, and radiation explain how the transfer occurs; this card focuses on the direction.

Net thermal energy transfer is from the higher-temperature body or region to the lower-temperature body or region.
A temperature difference drives thermal transfer; a larger temperature difference usually gives a greater rate, all else equal.
Thermal transfer continues until thermal equilibrium is reached.
At thermal equilibrium, bodies in contact have the same temperature and there is no net thermal energy transfer.
Energy is transferred, not “coldness”; the colder body gains internal energy while the hotter body loses internal energy.

Sort each situation by the net thermal transfer direction.

Sort
Unsorted
5
from hotter to colder
0
thermal equilibrium / no net transfer
0
incorrect model
0

A small metal ball at 90 °C is placed in a large water bath at 30 °C. State the direction of net thermal energy transfer and describe when the transfer stops.

Common mark losses are saying cold flows, or using total mass/internal energy rather than temperature to decide the net transfer direction.

A small metal ball at 90 °C is placed in a large water bath at 30 °C. State the direction of net thermal energy transfer and describe when the transfer stops.

Choose

Model Phase change

Phase change is not just “particles get hotter”. While a pure substance is melting or boiling, added energy goes into changing the molecular arrangement and separation rather than increasing average kinetic energy. After the phase change is complete, further energy transfer within one phase changes temperature again.

During a phase change at constant pressure, temperature remains constant while the change of state occurs.
Average random kinetic energy stays constant during the phase change because temperature is constant.
Energy transferred changes intermolecular potential energy as particles separate or come closer together.
Melting, boiling, and evaporation require energy input; freezing and condensation release energy.
On a heating or cooling curve, a phase change appears as a temperature plateau.

Sort each process by energy direction or temperature effect.

Sort
Unsorted
6
absorbs energy at constant temperature
0
releases energy at constant temperature
0
changes temperature within one phase
0

Explain why the temperature of a pure substance remains constant while it is boiling, even though energy is being supplied.

Common mark losses are saying no energy is transferred during the plateau, or saying the average kinetic energy increases during boiling.

Explain why the temperature of a pure substance remains constant while it is boiling, even though energy is being supplied.

Choose

Model Specific heat and latent heat

The decision is made from the physical process, not from the numbers given. A sloped section of a heating curve uses Q = mcΔT because temperature changes. A plateau uses Q = mL because the state changes while temperature stays constant. Use latent heat of fusion for solid-liquid changes and latent heat of vaporisation for liquid-gas changes.

Use Q = mcΔT when energy transfer changes temperature without a change of state.
Specific heat capacity c is the energy needed to raise 1 kg of a substance by 1 K, with unit J kg^-1 K^-1.
Use Q = mL when energy transfer changes state at constant temperature.
Specific latent heat L is the energy needed to change the state of 1 kg of substance, with unit J kg^-1.
For a process with both warming/cooling and phase change, calculate each stage separately and add the energy transfers.

Build the correct thermal energy equation for the process.

Formula
Target formula temperature change: Q = mcΔT; phase change: Q = mL; multi-stage total: Q_total = ΣQ_stage
Q
thermal energy transferred
J
m
mass
kg
c
specific heat capacity
J kg^-1 K^-1
ΔT
temperature change
K
L
specific latent heat
J kg^-1
1Decide whether the stage changes temperature or state.slope or plateau
2For temperature change in one phase, use specific heat capacity.Q = mcΔT
3For phase change at constant temperature, use latent heat.Q = mL
4Choose fusion for solid-liquid or vaporisation for liquid-gas.L_f or L_v
5Add energy transfers for multi-stage processes.Q_total = ΣQ_stage

A sample is heated from below its melting point to above it. Explain how to decide which parts of the calculation use Q=mcΔT and which use Q=mL.

Common mark losses are using one formula for the whole process, forgetting the latent-heat plateau, or using ΔT during a phase change.

A sample is heated from below its melting point to above it. Explain how to decide which parts of the calculation use Q=mcΔT and which use Q=mL.

Choose

Model Thermal transfer mechanisms

The mechanism depends on what is carrying the energy. In conduction, neighbouring particles or free electrons pass energy through a material. In convection, warmer and cooler regions of a fluid move because of density differences. In radiation, energy is carried by electromagnetic waves, so no particles are needed between source and absorber.

Conduction transfers energy through particle vibrations, collisions, and in metals by free electrons; it requires a material medium.
Convection transfers energy by bulk motion of a fluid, so it occurs in liquids and gases, not in solids or vacuum.
Thermal radiation transfers energy by electromagnetic waves, mainly infrared, and does not require a medium.
Radiation can transfer energy through a vacuum, which is how energy from the Sun reaches Earth.
In real situations, more than one mechanism can occur, but each mechanism has a different physical cause.

Sort each transfer cue into the correct mechanism.

Sort
Unsorted
6
conduction
0
convection
0
radiation
0

Distinguish conduction, convection, and radiation in terms of how energy is transferred and whether a material medium is required.

Common mark losses are saying radiation needs air, or describing convection without bulk fluid motion.

Distinguish conduction, convection, and radiation in terms of how energy is transferred and whether a material medium is required.

Choose

Model Conduction

At the hot end of a solid, particles vibrate more strongly. These vibrations pass energy to neighbouring particles, so energy moves through the solid from hot to cold. Metals conduct especially well because mobile free electrons can move through the lattice and transfer kinetic energy rapidly. The particles of the solid do not travel from one end to the other as a bulk flow.

Conduction is thermal energy transfer through a material without bulk movement of the material.
Particles in a hotter region have greater average random kinetic energy and transfer energy to neighbouring particles by collisions and vibrations.
In metals, free electrons carry energy quickly from hotter regions to cooler regions.
Conduction requires a material medium; it cannot occur through a vacuum.
Good conductors transfer energy quickly, while insulators reduce the rate of conduction.

Repair the conduction misconceptions.

Spot Errors

Explain how thermal energy is transferred by conduction along a metal rod heated at one end.

Common mark losses are omitting free electrons in metals, or describing a bulk movement of the solid instead of energy transfer through particles.

Explain how thermal energy is transferred by conduction along a metal rod heated at one end.

Choose

Model Conduction rate

The equation applies when the temperature difference is maintained and the slab has uniform material and thickness. The temperature difference can be in kelvin or degrees Celsius because it is a change. Use the magnitude of ΔT for the rate; direction is from the hotter side to the colder side.

For steady-state conduction through a uniform slab, the rate of thermal energy transfer is P = ΔQ/Δt = kAΔT/L.
k is thermal conductivity, with unit W m^-1 K^-1.
A larger area A, larger temperature difference ΔT, or larger k increases the conduction rate.
A larger thickness L decreases the conduction rate.
Insulation reduces conduction by using low-k materials, trapping air, or increasing thickness.

Build the conduction-rate equation for a uniform slab.

Formula
Target formula P = ΔQ/Δt = kAΔT/L
P
rate of thermal energy transfer
W
ΔQ
thermal energy transferred
J
Δt
time interval
s
k
thermal conductivity
W m^-1 K^-1
A
area normal to energy transfer
m^2
ΔT
temperature difference
K
L
slab thickness
m
1Check steady-state conduction through a uniform slab.constant ΔT, uniform k
2Use area perpendicular to the heat flow.A
3Use temperature difference, not absolute temperature.ΔT
4Put thickness in the denominator.L
5Assemble the conduction-rate equation.P = kAΔT/L

A wall has area A, thickness L, thermal conductivity k, and temperature difference ΔT across it. State the conduction-rate equation and explain one way to reduce the rate of energy transfer.

Common mark losses are using absolute temperature instead of ΔT, putting L in the numerator, or omitting the steady-state slab condition.

A wall has area A, thickness L, thermal conductivity k, and temperature difference ΔT across it. State the conduction-rate equation and explain one way to reduce the rate of energy transfer.

Choose

Model Convection

Convection begins with a temperature difference inside a fluid. Heating reduces density in one region, so buoyancy makes that warmer fluid rise while cooler fluid sinks. This circulation carries internal energy through the fluid. The key distinction from conduction is that the material itself moves in convection.

Convection is thermal energy transfer by bulk movement of a fluid.
A heated region of fluid expands, becomes less dense, and rises in a gravitational field.
Cooler, denser fluid sinks to replace it, forming a convection current.
Convection occurs in liquids and gases, not in solids or vacuum.
Forced convection uses a fan or pump to move the fluid, but energy is still carried by bulk fluid motion.

Repair the convection misconceptions.

Spot Errors

Explain how a convection current forms in water heated from below.

Common mark losses are omitting density differences, or describing particle collisions without bulk motion.

Explain how a convection current forms in water heated from below.

Choose

Model Black-body radiation

Black-body radiation is thermal electromagnetic radiation from an ideal absorber and emitter. The model removes material details: the spectrum depends on temperature, not on paint colour or surface chemistry. For a fixed area, the total power emitted rises with the fourth power of absolute temperature, so T must be in kelvin.

A black body is an ideal object that absorbs all incident electromagnetic radiation.
It is also an ideal emitter: the emitted radiation forms a continuous spectrum determined only by temperature.
As Kelvin temperature increases, the emitted intensity increases and the peak wavelength shifts to shorter wavelengths.
For an ideal black body, total emitted power is P = σAT^4, where A is surface area and T is absolute temperature in kelvin.
Stars are often modelled approximately as black bodies when relating temperature, spectrum, and luminosity.

Repair the black-body radiation misconceptions.

Spot Errors

Define a black body and state how the total power it emits depends on surface area and absolute temperature.

Common mark losses are saying only that the object is black, forgetting perfect emission, or using Celsius in T^4.

Define a black body and state how the total power it emits depends on surface area and absolute temperature.

Choose

Apparent brightness

Apparent brightness describes what reaches the detector, not how much power the source produces in total. If a source radiates equally in all directions, its luminosity is spread over a sphere whose area grows as d^2. That geometric spreading gives the inverse-square relation.

Apparent brightness b is the power received per unit area at the observer, with unit W m^-2.
Luminosity L is the total power emitted by the source, with unit W.
For isotropic emission with negligible absorption, b = L/(4πd^2).
The factor 4πd^2 is the surface area of a sphere centred on the source.
Doubling the distance reduces apparent brightness to one quarter if luminosity is unchanged.

Build the apparent-brightness equation from luminosity and distance.

Formula
Target formula b = L/(4πd^2)
b
apparent brightness, received power per unit area
W m^-2
L
luminosity, total power emitted
W
d
distance from source to observer
m
1Start with total power emitted by the source.L
2Spread that power over a sphere of radius d.area = 4πd^2
3Divide power by area to get received power per unit area.b = L/(4πd^2)
4Check isotropic emission and negligible absorption between source and observer.inverse-square model

A star has luminosity L and is distance d from Earth. Define apparent brightness and state the equation linking b, L, and d.

Common mark losses are confusing luminosity with apparent brightness, omitting the sphere area 4πd^2, or using inconsistent distance units.

A star has luminosity L and is distance d from Earth. Define apparent brightness and state the equation linking b, L, and d.

Choose

Model Luminosity and brightness

The model has two different areas. The star surface area A controls how much power the star emits, through the Stefan-Boltzmann law. The sphere of radius d around the star controls how that emitted power is spread out before it reaches the observer. Keeping R and d separate prevents most calculation errors.

Luminosity L is the total power emitted by a source, measured in watts.
For an ideal black-body surface, L = σAT^4; for a spherical star, A = 4πR^2 so L = 4πR^2σT^4.
Apparent brightness b is the power received per unit area by an observer, measured in W m^-2.
For isotropic spreading over distance d, b = L/(4πd^2).
Radius R affects luminosity; distance d affects apparent brightness after the light has left the star.

Repair the luminosity and brightness setup errors.

Spot Errors

A star may be modelled as a black body of radius R and surface temperature T. State an expression for its luminosity and then for its apparent brightness at distance d.

Common mark losses are mixing up R and d, forgetting that luminosity is total power, or omitting the inverse-square spreading factor.

A star may be modelled as a black body of radius R and surface temperature T. State an expression for its luminosity and then for its apparent brightness at distance d.

Choose

Model Wien’s displacement law

On a black-body spectrum, λ_max is the wavelength at which the emitted intensity is greatest. Wien’s law says this peak wavelength is inversely proportional to absolute temperature. The law is useful for estimating the surface temperature of stars from the wavelength at the peak of their spectrum.

Wien’s displacement law links the peak wavelength of a black-body spectrum to absolute temperature.
The equation is λ_max T = 2.90×10^-3 m K.
λ_max must be in metres and T must be in kelvin for SI use.
A hotter black body has a shorter peak wavelength, so its spectrum shifts toward the blue/ultraviolet side.
Wien’s law locates the peak of the spectrum; Stefan-Boltzmann describes total emitted power.

Build Wien’s displacement law from the spectrum peak and temperature.

Formula
Target formula λ_max T = 2.90×10^-3 m K
λ_max
wavelength at maximum intensity
m
T
absolute temperature of the black body
K
2.90×10^-3 m K
Wien displacement constant
m K
1Identify the wavelength at the maximum intensity of the spectrum.λ_max
2Use absolute temperature for the black body.T in K
3Multiply peak wavelength by absolute temperature.λ_max T
4Set the product equal to Wien’s displacement constant.λ_max T = 2.90×10^-3 m K

A star approximates a black body and its spectrum peaks at wavelength λ_max. State Wien’s displacement law and explain how the peak changes if the star is hotter.

Common mark losses are using Celsius, forgetting the unit m K for the constant, or describing total power instead of the peak wavelength.

A star approximates a black body and its spectrum peaks at wavelength λ_max. State Wien’s displacement law and explain how the peak changes if the star is hotter.

Choose

Retrieve the B.1 Thermal energy transfers Model

Review

B.1 is strongest when each calculation is tied to its physical model. Start with the particle description of matter, decide whether energy changes temperature or phase, identify the transfer mechanism, then use the radiation and luminosity equations only under their black-body or inverse-square assumptions.

Describe solids, liquids, and gases using particle spacing, motion, and intermolecular forces.
Use ρ = m/V and temperature conversion T/K = t/°C + 273; temperature changes have the same size in K and °C.
Connect Kelvin temperature to average ideal-gas molecular kinetic energy and define internal energy as total random KE plus intermolecular PE.
Choose Q = mcΔT for temperature change and Q = mL for phase change at constant temperature.
Distinguish conduction, convection, and radiation by the physical mechanism and whether a medium is required.
Use conduction rate P = kAΔT/L for a steady uniform slab.
Use black-body laws P = σAT^4 and λ_max T = 2.90×10^-3 m K with absolute temperature.
Relate luminosity and apparent brightness using b = L/(4πd^2).

Match each B.1 retrieval cue to the physics move it should trigger.

Match
Reasons
0/9