HomeLearning HubIB DP PhysicsB.1 Thermal energy transfers
B.1

Thermal energy transfers

Theme B · The particulate nature of matter · SL and HL

🎯What you need to be able to do

  • Describe solids, liquids and gases in terms of molecular spacing, motion and forces.
  • Distinguish temperature, internal energy and thermal energy, and use the Kelvin scale.
  • Use \( Q = mc\Delta T \) and \( Q = mL \) for heating and for phase change.
  • Explain why temperature stays constant during a change of phase.
  • Describe conduction, convection and radiation.
  • Use the Stefan–Boltzmann law and Wien’s displacement law for black bodies.

📚The physics

Three words that are not synonyms. Temperature is proportional to the average random kinetic energy of the molecules. Internal energy is the total of all the molecular kinetic energies plus all the intermolecular potential energies. Thermal energy (heat) is energy in transit from hotter to colder. A bathtub of warm water has far more internal energy than a spark, while the spark has the far higher temperature. Examiners test this distinction constantly.

Absolute temperature. \( T(\text{K}) = \theta(^\circ\text{C}) + 273 \). Kelvin is not an alternative unit of convenience: average kinetic energy is proportional to the absolute temperature, so doubling from 20 °C to 40 °C does not double anything. Doubling from 293 K to 586 K does. Every proportionality in this theme requires kelvin.

Specific heat capacity \(c\) is the energy needed to raise 1 kg by 1 K. Specific latent heat \(L\) is the energy needed to change the phase of 1 kg with no temperature change.

\( Q = mc\Delta T \)
\( Q = mL \)

Water’s specific heat capacity is unusually large, about 4200 J kg\(^{-1}\) K\(^{-1}\), which is why coastal climates are mild and why water is the coolant of choice. Fusion is solid–liquid, vaporisation is liquid–gas, and for water the latter is roughly seven times larger — separating molecules entirely costs far more than merely loosening them.

Why the temperature stops during a phase change is the question most worth being able to answer in words. The energy supplied goes into potential energy, breaking the intermolecular bonds, not into kinetic energy. Since temperature tracks average kinetic energy, and that is not changing, the thermometer sits still even though energy is pouring in. On a temperature–time graph this is the flat plateau, and a longer plateau simply means a larger latent heat.

The three transfer mechanisms. Conduction passes energy through a material by molecular collision and, in metals, by free electrons — which is why metals conduct so much better. It needs a medium. Convection moves energy by bulk motion of a fluid: warmed fluid expands, becomes less dense, and rises. It needs a fluid. Radiation transfers energy as electromagnetic waves and is the only one that works through a vacuum, which is how the Sun’s energy reaches us at all.

Black-body radiation. A black body is a perfect absorber and a perfect emitter. Two laws describe what it emits.

Stefan–Boltzmann\( P = \sigma A T^{4} \)
Wien\( \lambda_{\max} T = 2.9 \times 10^{-3}\ \text{m K} \)

That fourth power does a great deal of work: a modest rise in temperature produces a dramatic rise in power. Wien’s displacement law says hotter bodies peak at shorter wavelengths — the reason a heated metal glows dull red, then orange, then white, and the reason blue stars are hotter than red ones.

✏️Worked example

How much energy is needed to turn 0.30 kg of ice at −15 °C into water at 25 °C? Take \( c_{\text{ice}} = 2100 \), \( c_{\text{water}} = 4200 \) J kg\(^{-1}\) K\(^{-1}\), \( L_f = 3.34 \times 10^{5} \) J kg\(^{-1}\).

Break it into stages and never skip the plateau.

  1. Warm the ice from −15 °C to 0 °C: \( Q = 0.30 \times 2100 \times 15 = 9450 \) J.
  2. Melt it at 0 °C: \( Q = 0.30 \times 3.34 \times 10^{5} = 100200 \) J.
  3. Warm the water from 0 °C to 25 °C: \( Q = 0.30 \times 4200 \times 25 = 31500 \) J.
\[ Q_{\text{total}} = 141150\ \text{J} \approx 1.4 \times 10^{5}\ \text{J} \]
Notice the proportions. Melting alone accounts for about 71% of the total, even though the temperature does not budge during it. Students who treat the problem as one continuous heating from −15 to 25 get roughly 40 kJ — less than a third of the right answer. The plateau is not a detail.

🔭See it happen

PhET, States of Matter: Basics. Heat the solid and watch the temperature readout stall while the molecules visibly break out of their lattice. It is the graph plateau and the molecular explanation shown as the same event, which is exactly the link the exam question is testing.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. How much energy is needed to raise the temperature of 2.0 kg of water from 20 °C to 80 °C? Take \( c = 4200 \) J kg\(^{-1}\) K\(^{-1}\).
\( Q = mc\Delta T = 2.0 \times 4200 \times 60 = 5.0 \times 10^{5} \) J. Note that a temperature difference is the same number in °C and K, so no conversion is needed here — unlike in any proportionality.
2. 0.50 kg of water at 80 °C is mixed with 0.50 kg of water at 20 °C in an insulated container. Find the final temperature.
Equal masses of the same substance, so the energy lost by the hot water equals the energy gained by the cold at the midpoint: 50 °C. Formally, \( mc(80 - T) = mc(T - 20) \), and \(m\) and \(c\) cancel.
3. How much energy is needed to boil away 0.25 kg of water already at 100 °C? Take \( L_v = 2.26 \times 10^{6} \) J kg\(^{-1}\).
\( Q = mL = 0.25 \times 2.26 \times 10^{6} = 5.7 \times 10^{5} \) J. No \( \Delta T \) term appears, because the temperature does not change during the phase change.
4. Explain why the temperature of a substance stays constant while it melts, even though energy is still being supplied.
The supplied energy goes into potential energy, breaking the intermolecular bonds that hold the solid lattice together, rather than into kinetic energy. Temperature is a measure of the average random kinetic energy of the molecules, and that is not changing, so the thermometer reading holds steady. On a temperature–time graph this is the flat plateau.
5. A lamp filament of surface area \( 5.0 \times 10^{-5} \) m\(^{2}\) reaches 2500 K. Treating it as a black body with \( \sigma = 5.67 \times 10^{-8} \) W m\(^{-2}\) K\(^{-4}\), find the power it radiates.
\( P = \sigma A T^{4} = 5.67 \times 10^{-8} \times 5.0 \times 10^{-5} \times 2500^{4} \). With \( 2500^{4} = 3.91 \times 10^{13} \), \( P = 1.1 \times 10^{2} \) W.
6. A star’s spectrum peaks at 480 nm. Find its surface temperature, using \( \lambda_{\max}T = 2.9 \times 10^{-3} \) m K.
\( T = \dfrac{2.9 \times 10^{-3}}{480 \times 10^{-9}} = 6.0 \times 10^{3} \) K — close to the Sun, which is why the Sun peaks in the visible.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Hypertextbook — thermal properties of matter
  • HyperPhysics — heat transfer and blackbody radiation