B.1 Thermal energy transfers

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.

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 $\rho=(1.0\times10^{10})(1.67\times10^{-27})=1.67\times10^{-17}\,\mathrm{kg\,m^{-3}}$.

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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.