B.1 Thermal energy transfers

Particle view

Matter is made of particles in continuous random motion. The state depends mainly on how closely particles are packed, how freely they move and how strongly intermolecular forces hold them together.

Compare the three states

Use temperature carefully

At the same temperature, particles have the same average kinetic energy in the kinetic-theory model. The different states are then distinguished by separation and intermolecular forces, not by claiming that one state automatically has hotter particles.

Common trap

Do not describe a solid as having motionless particles. “Fixed position” means the particles oscillate about equilibrium positions; it does not mean their kinetic energy is zero.

Density

Density is mass per unit volume:

$$\rho=\frac{m}{V}$$

It describes how much mass is concentrated in a given volume.

Calculation method

1. Identify the mass of the object or sample.
2. Use the volume occupied by that same sample.
3. Convert units before substituting.
4. Report density with units such as kg m⁻³ or g cm⁻³.

Useful conversion: 1 g cm⁻³ = 1000 kg m⁻³.

Interpret the result

For equal volumes, the denser sample has the greater mass. For equal masses, the denser sample occupies the smaller volume. A non-uniform object requires its total mass divided by its total external volume unless the question specifies a particular material region.

Worked example from local Question Bank row 36355

A spherical hydrogen nebula has radius $9.0\times10^{15}\,\mathrm{m}$ and number density $1.0\times10^{10}\,\mathrm{atoms\,m^{-3}}$. With $m_H=1.67\times10^{-27}\,\mathrm{kg}$, its mass density is $\rho=(1.0\times10^{10})(1.67\times10^{-27})=1.67\times10^{-17}\,\mathrm{kg\,m^{-3}}$.

$$V=\frac43\pi r^3=3.05\times10^{48}\,\mathrm{m^3}$$
$$m=\rho V=(1.67\times10^{-17})(3.05\times10^{48})=5.1\times10^{31}\,\mathrm{kg}$$

Check the boundary

Do not mix the volume of displaced fluid with the object’s mass, and do not use a material’s density formula with inconsistent units. Density is a scalar, so it has no direction.

Two temperature scales

Celsius is convenient for everyday temperature differences. Kelvin is the absolute thermodynamic scale used when temperature is linked to particle energy or radiation.

Convert between them

$$T_{\mathrm K}=\theta_{\circ\mathrm C}+273.15$$

So 0 °C = 273.15 K and 100 °C = 373.15 K. Kelvin is written without a degree symbol.

Choose the scale

Use Celsius when a question asks for a familiar temperature or a change described on the Celsius scale. Use Kelvin in equations such as $$E_k=\frac32k_BT$$, $$L=\sigma AT^4$$ and $$\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}$$.

Common trap

Never substitute a Celsius value directly into a formula that uses absolute temperature. Convert the temperature first.

Same size of change

Because the Celsius and Kelvin scales have the same interval size, a temperature change has the same numerical value in both scales:

$$\Delta T(\mathrm K)=\Delta\theta(^{\circ}\mathrm C)$$

Read a change, not an absolute value

If a sample falls from +10 °C to −10 °C, then

$$\Delta\theta=-10-10=-20^{\circ}\mathrm C$$

The same change is −20 K. The zero point shifts, but the spacing between adjacent temperatures does not.

Common trap

Do not add 273.15 when converting a temperature difference. Add 273.15 only when converting an absolute Celsius temperature to Kelvin.

Absolute temperature and motion

For particles in an ideal gas, Kelvin temperature is proportional to their average random translational kinetic energy:

$$\overline{E_k}=\frac{3}{2}k_BT$$

Here $$k_B$$ is the Boltzmann constant and T must be in kelvin.

What the equation says

If the Kelvin temperature doubles, the average translational kinetic energy doubles. A higher temperature means greater average random kinetic energy, not that every particle has exactly the same kinetic energy.

Scope of the model

The relation describes average random translational motion. It does not include the whole internal energy of a substance, which also contains intermolecular potential energy.

Worked example from local Question Bank row 31728

For helium atoms at $T=320\,\mathrm{K}$ with $m=6.6\times10^{-27}\,\mathrm{kg}$, equate mean translational kinetic energy to $\tfrac12mv^2$:

$$\frac12mv^2=\frac32k_BT\Rightarrow v=\sqrt{\frac{3k_BT}{m}}$$
$$v=\sqrt{\frac{3(1.38\times10^{-23})(320)}{6.6\times10^{-27}}}=1.4\times10^3\,\mathrm{m\,s^{-1}}$$

This is a characteristic speed derived from the average energy, not a claim that every atom has that speed.

Common trap

A Celsius temperature cannot be used in this equation. Convert first; 0 °C corresponds to about 273 K, not zero particle kinetic energy.

Internal energy

The internal energy of a system is the sum of:

• random molecular kinetic energy; and
• intermolecular potential energy associated with forces between particles.

Temperature is only one part

For a fixed phase and amount of substance, raising temperature usually increases the particles’ average random kinetic energy. During a phase change, temperature can stay constant while intermolecular potential energy changes.

Do not equate heat with internal energy

Internal energy is a state property of the system. Thermal energy transfer is energy crossing the system boundary because of a temperature difference.

Temperature difference drives net transfer

When two bodies at different temperatures can exchange energy, the net thermal energy transfer is from the higher-temperature body to the lower-temperature body.

What equilibrium means

Transfer can occur in both directions microscopically, but at thermal equilibrium the opposing transfers balance and there is no net transfer. Equal temperature is the condition for zero net thermal transfer, not necessarily equal internal energy.

Apply the direction rule

First compare temperatures, then draw the net energy arrow. The arrow is independent of which object is heavier or contains more total internal energy.

Common trap

A larger object can contain more internal energy while still receiving energy from a smaller, hotter object. “Hotter” means higher temperature, not “more total energy”.

What changes in a phase change

Melting, freezing, boiling, condensing and other phase changes alter how particles are arranged and how freely they move. Energy transfer changes the balance of intermolecular potential energy.

Why temperature stays constant

During a phase change of a pure substance at constant pressure, the supplied or removed energy changes particle interactions rather than increasing the average random kinetic energy. Therefore the temperature remains constant until the phase change is complete.

Read a heating curve

A sloped section represents temperature changing within one phase. A flat section represents energy transfer during a phase change. The flat section can be long even though the thermometer reading does not change.

Common trap

“Constant temperature” does not mean “no energy transfer”. It means the transfer is not increasing average particle kinetic energy at that stage.

Temperature change within a phase

Use

$$Q=mc\Delta T$$

where c is the specific heat capacity. For a given mass, a larger c means more energy is required for the same temperature rise.

Energy during a phase change

Use

$$Q=mL$$

where L is the specific latent heat of fusion or vaporization. This energy changes particle interactions while the temperature remains constant.

Choose the equation

• temperature changes, no phase change: $$Q=mc\Delta T$$
• phase changes at constant temperature: $$Q=mL$$

If a process contains both stages, calculate the energy for each stage and add the signed or positive magnitudes consistently.

Worked example from local Question Bank row 22716

A cable receives $30\,\mathrm{W}$ and initially warms at $35\,\mathrm{mK\,s^{-1}}=3.5\times10^{-2}\,\mathrm{K\,s^{-1}}$. For copper, $c=390\,\mathrm{J\,kg^{-1}\,K^{-1}}$. Using $P=mc(\Delta T/\Delta t)$,

$$m=\frac{30}{390(3.5\times10^{-2})}=2.2\,\mathrm{kg}$$

The rate form is valid during the initial interval when losses are negligible.

Common trap

Do not use a temperature difference in $$Q=mL$$, and do not use $$Q=mc\Delta T$$ across a phase-change plateau.

Three mechanisms

Thermal energy can be transferred by conduction, convection or thermal radiation. The mechanism depends on what connects the hot and cold regions and on whether bulk matter moves.

Choose the mechanism

Real situations can combine them

A saucepan may conduct energy through its metal, transfer energy through moving water by convection and radiate energy from its surfaces. Identify the dominant mechanism being asked about rather than insisting that only one process exists.

Common trap

Radiation does not require air, and convection is not the same as “hot molecules vibrating faster through a solid”.

Conduction

In conduction, particles in a hotter region have greater average kinetic energy. Through collisions and intermolecular forces, they transfer energy to neighbouring particles in the cooler region.

What moves and what does not

Energy propagates through the material, but the material does not need to undergo bulk flow. In a solid, particles usually vibrate about fixed positions while transferring energy to neighbours.

Compare with other mechanisms

Conduction needs matter and microscopic contact. Convection transfers energy through bulk motion of a fluid. Radiation transfers energy by electromagnetic waves and can cross a vacuum.

Common trap

Conduction is not the same as particles travelling from the hot end to the cold end. The net transfer is through local interactions.

Conduction rate

The rate of thermal energy transfer through a uniform slab is

$$\frac{\Delta Q}{\Delta t}=\frac{kA\Delta T}{\Delta x}$$

where k is the material’s thermal conductivity, A is cross-sectional area, ΔT is the temperature difference and Δx is the transfer distance.

Read the proportionalities

The rate increases with larger k, larger area and larger temperature difference. It decreases when the material is thicker, because Δx is in the denominator.

Calculation checks

Use consistent SI units: area in m², distance in m, temperature difference in K or °C, and k in W m⁻¹ K⁻¹. The rate is measured in watts, because 1 W = 1 J s⁻¹.

Worked example from local Question Bank row 127628

Ice has $k=2.3\,\mathrm{W\,m^{-1}\,K^{-1}}$, thickness $0.019\,\mathrm{m}$ and temperature difference $6\,\mathrm{K}$. Per unit area,

$$\frac{1}{A}\frac{\Delta Q}{\Delta t}=\frac{k\Delta T}{\Delta x}=\frac{(2.3)(6)}{0.019}=7.3\times10^2\,\mathrm{W\,m^{-2}}$$

The result is a heat flux; multiply by area to obtain total power.

Common trap

Use the temperature difference across the slab, not an absolute temperature. A temperature gradient is a change per distance, so do not omit Δx.

Density difference drives convection

When part of a liquid or gas is heated, it generally expands and becomes less dense. The warmer region experiences greater buoyancy and rises while cooler, denser fluid sinks.

A convection current

The rising warm fluid and sinking cool fluid form a circulation. The fluid’s bulk motion carries internal energy from the warmer region to other parts of the fluid.

What the syllabus asks

This objective is qualitative: identify the density change, the direction of motion and how that motion transfers energy. It does not require a detailed fluid-dynamics calculation.

Common trap

Convection occurs in fluids, not in a rigid solid. A solid can conduct energy even though it does not circulate as a bulk fluid.

Black-body emission

A black body is an ideal surface that emits electromagnetic radiation according to its absolute temperature. Its total emitted power, or luminosity, is modelled by

$$L=\sigma AT^4$$

Read the variables

$$A$$ is the emitting surface area, $$T$$ is absolute temperature in kelvin and $$\sigma$$ is the Stefan–Boltzmann constant. The equation gives total power emitted, not the brightness received by a particular observer.

Use proportional reasoning

At fixed area, doubling T multiplies L by $$2^4=16$$. At fixed temperature, doubling the emitting area doubles L. The fourth-power dependence makes temperature especially important.

Worked comparison from local Question Bank row 29005

Treat Mars at $200\,\mathrm{K}$ and Earth at $300\,\mathrm{K}$ as black bodies. For equal emitting area,

$$\frac{L_{Mars}}{L_{Earth}}=\left(\frac{200}{300}\right)^4=0.198\approx0.20$$

Mars emits about one fifth as much power per unit area in this ideal model.

Common trap

Do not use Celsius in the fourth-power term, and do not confuse luminosity with apparent brightness, which also depends on distance.

Apparent brightness

Apparent brightness, $$b$$, describes how much power from a distant source is received per unit area at the observer. It is an observation-dependent quantity.

Why distance matters

Radiation from an approximately point-like source spreads over larger spherical areas as it travels outward. The same emitted power is distributed over more area, so the received power per unit area decreases.

Do not confuse the quantities

Luminosity is the source’s total emitted power. Apparent brightness is what reaches a specified observer per unit area. A source can be intrinsically luminous but appear faint when it is far away.

Common trap

Apparent brightness is not simply the source’s total power. Always ask whether the question concerns emission by the source or reception at a distance.

Brightness–luminosity relation

For isotropic emission without absorption,

$$b=\frac{L}{4\pi d^2}$$

where $$L$$ is total luminosity and $$d$$ is the source–observer distance.

Use the inverse-square pattern

At fixed luminosity, doubling distance makes apparent brightness one quarter as large. At fixed distance, doubling luminosity doubles apparent brightness.

Rearrange before calculating

$$L=4\pi d^2b$$

so

$$d=\sqrt{\frac{L}{4\pi b}}$$

Keep luminosity in watts, distance in metres and brightness in W m⁻².

Worked example from local Question Bank row 30016

Mars is about $1.5$ times farther from the Sun than Earth. If solar intensity at Earth is $1.36\times10^3\,\mathrm{W\,m^{-2}}$,

$$b_{Mars}=b_{Earth}\left(\frac{d_E}{d_M}\right)^2=(1.36\times10^3)\frac{1}{1.5^2}=6.04\times10^2\,\mathrm{W\,m^{-2}}$$

The same solar luminosity is spread over a sphere with larger radius.

Common trap

The factor is $$d^2$$, not $$d$$. Also distinguish a source’s total emitted power from the power received per square metre.

Black-body spectrum

A black body emits a continuous spectrum of wavelengths. The wavelength at which the emitted intensity is greatest is $$\lambda_{\max}$$.

Wien’s law

The peak wavelength and absolute temperature obey

$$\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}$$

Therefore

$$T=\frac{2.9\times10^{-3}}{\lambda_{\max}}$$

Interpret the shift

A hotter black body has a smaller peak wavelength, so its spectrum shifts toward shorter wavelengths. A cooler black body peaks at a longer wavelength.

Worked example from local Question Bank row 31596

A star's spectrum peaks at $740\,\mathrm{nm}=740\times10^{-9}\,\mathrm{m}$.

$$T=\frac{2.9\times10^{-3}}{740\times10^{-9}}=3.9\times10^3\,\mathrm{K}\approx4000\,\mathrm{K}$$

The wavelength conversion is essential because Wien's constant is in metres kelvin.

Calculation checks

Use $$\lambda_{\max}$$ in metres and T in kelvin. The law identifies the peak of the spectrum; it does not say that the object emits only that one wavelength.