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Thermal Energy Transfer

Thermal Energy Transfer

Info: Board Coverage AQA Paper 2 | Edexcel CP9/CB6 | OCR (A) Paper 2 | CIE P3

Energy moves from hot to cold by three mechanisms: conduction, convection and radiation. This page consolidates the transfer mechanisms, U-values, and the latent-heat calculations that sit alongside the specific heat capacity work in Thermal Properties.

1. Conduction

Conduction is the transfer of energy through a material without any bulk movement of the material itself. In metals, the dominant mechanism is free (delocalised) electrons colliding with the lattice; in non-metals, energy passes through lattice vibrations (phonons). This is why metals conduct heat far better than insulators: the same free electrons that carry current carry thermal energy.

The Conduction Equation (Fourier’s Law)

The rate of energy flow through a slab of material is

dQdt=kA Δθd\frac{\mathrm{d}Q}{\mathrm{d}t} = \frac{kA\,\Delta\theta}{d}

where kk is the thermal conductivity (W m−1 K−1\mathrm{W\,m^{-1}\,K^{-1}}), AA the cross-sectional area, dd the thickness, and Δθ\Delta\theta the temperature difference across the slab.

Worked example. A window pane has area 2.0 m22.0\ \mathrm{m^2}, thickness 4.0 mm4.0\ \mathrm{mm} and k=0.80 W m−1K−1k = 0.80\ \mathrm{W\,m^{-1}K^{-1}}. The inside is at 20 ∘C20\ ^\circ\mathrm{C}, the outside at 5.0 ∘C5.0\ ^\circ\mathrm{C}.

dQdt=0.80×2.0×154.0×10−3=6.0×103 W\frac{\mathrm{d}Q}{\mathrm{d}t} = \frac{0.80 \times 2.0 \times 15}{4.0\times10^{-3}} = 6.0\times10^{3}\ \mathrm{W}

In reality the glass surfaces are not at room temperature; the temperature drops mostly across boundary air layers. That is exactly why double glazing works: a trapped argon gap with low kk adds a large effective thickness.

2. Convection

Convection requires a fluid. The heated fluid expands, becomes less dense, and rises; cooler fluid replaces it, forming a convection current. Energy is then carried by the bulk motion of the fluid itself.

  • Natural convection: driven by density differences alone (radiator warming a room).
  • Forced convection: driven by an external pump or fan (a fan heater, or coolant pumped through a nuclear reactor — see Nuclear Power and Safety).

The rate of transfer is commonly modelled as

dQdt=hA (θsurface−θfluid)\frac{\mathrm{d}Q}{\mathrm{d}t} = hA\,(\theta_\text{surface} - \theta_\text{fluid})

with hh a heat transfer coefficient. You are rarely asked to calculate with hh at A-level, but you are expected to explain designs that reduce convection: trapping air in foam or fibreglass works because air is a poor conductor and the fibres prevent the air from circulating.

3. Radiation

All bodies emit electromagnetic radiation. The power radiated by a body at Kelvin temperature TT is given by the Stefan—Boltzmann law in the form used at A-level:

dQdt=e σAT4\frac{\mathrm{d}Q}{\mathrm{d}t} = e\,\sigma A T^4

where ee is the emissivity (0≤e≤10 \le e \le 1; e=1e = 1 for a perfect black body) and σ=5.67×10−8 W m−2K−4\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}} is the Stefan—Boltzmann constant.

For an object in surroundings at temperature TsT_s, the net exchange is

dQdt=e σA (T4−Ts4)\frac{\mathrm{d}Q}{\mathrm{d}t} = e\,\sigma A\,(T^4 - T_s^4)

Key qualitative points examiners award marks for:

  • Emission depends on T4T^4: doubling the Kelvin temperature multiplies the radiated power by 16. Using the Celsius temperature here is one of the most common errors in the paper.
  • Matte black surfaces are the best absorbers and the best emitters; shiny silver surfaces are the worst at both.
  • The Earth’s radiative equilibrium temperature follows from balancing absorbed solar flux against eσT4e\sigma T^4 — a standard application question.

4. U-Values

For a building element, the U-value bundles conduction through all layers into one figure:

dQdt=UA Δθ\frac{\mathrm{d}Q}{\mathrm{d}t} = U A\, \Delta\theta

A double-glazed unit might have U≈1.5 W m−2K−1U \approx 1.5\ \mathrm{W\,m^{-2}K^{-1}} against U≈5.0U \approx 5.0 for single glazing. Lower U-value means better insulation. Exam questions typically give several layers in series and ask you to compare energy loss before and after an upgrade, then comment on cost savings from a stated energy price.

5. Specific Latent Heat and Phase Changes

During a phase change the temperature of a pure substance is constant even though energy is still being supplied: the energy goes into breaking intermolecular bonds, not into kinetic energy. The energy required is

Q=mLQ = m L

with LfL_f the specific latent heat of fusion (solid ↔\leftrightarrow liquid) and LvL_v the specific latent heat of vaporisation (liquid ↔\leftrightarrow gas). For water, Lv≈2.26×106 J kg−1L_v \approx 2.26\times10^{6}\ \mathrm{J\,kg^{-1}} — roughly seven times LfL_f, which is why scalding steam burns are so severe: the steam releases mLvmL_v plus the cooling water releases mcΔθm c \Delta\theta as it condenses and cools.

Worked example. An electric kettle delivers 2.5 kW2.5\ \mathrm{kW} to 0.30 kg0.30\ \mathrm{kg} of boiling water. How long to boil it dry? (Lv=2.26×106 J kg−1L_v = 2.26\times10^{6}\ \mathrm{J\,kg^{-1}})

t=mLvP=0.30×2.26×1062500≈270 s (≈4.6 minutes)t = \frac{m L_v}{P} = \frac{0.30 \times 2.26\times10^{6}}{2500} \approx 270\ \mathrm{s}\ (\approx 4.6\ \text{minutes})

Mixing Problems (Calorimetry)

Multi-stage problems chain latent heat with specific heat capacity. A standard ice-to-steam calculation has five stages: warm the ice to 0 ∘C0\ ^\circ\mathrm{C}, melt it, heat the water to 100 ∘C100\ ^\circ\mathrm{C}, boil it, superheat the steam. Set total energy supplied equal to the sum over stages and solve for the unknown. Always compute and compare stage-by-stage when the question asks “does the ice all melt?” — answer that sub-question explicitly first.

6. Common Pitfalls

  • Celsius vs Kelvin: T4T^4 radiation laws and the ideal gas law need Kelvin. Specific heat questions only need differences, where Celsius is fine.
  • Area vs radius: radiation questions give a diameter. A=4πr2A = 4\pi r^2 needs r=d/2r = d/2.
  • Latent heat has no Δθ\Delta\theta: writing Q=mLΔθQ = mL\Delta\theta in a phase-change stage is an automatic mark loss.
  • Direction of energy flow: in net-radiation questions, quote the sign convention you are using (net loss from the object vs net gain by the surroundings).
  • Conduction in a steady state: the rate is the same through every layer in series; the temperature differences distribute in proportion to the thermal resistances.

7. Summary

MechanismRequires a medium?Key equationMain source of error
ConductionSolid/fluid, no bulk flowkAΔθd\frac{kA\Delta\theta}{d}Unit conversion of thickness
ConvectionFluid + bulk motionhAΔθhA\Delta\thetaVague descriptions (say density change)
RadiationNothing (vacuum fine)eσA(T4−Ts4)e\sigma A(T^4 - T_s^4)Using Celsius in T4T^4

Phase changes: Q=mLQ = mL at constant temperature. Mixing problems: sum the stages.

Cross-References