Thermal Properties of Matter · TOPIC LESSON
Connect conduction networks, cooling transients and radiative exchange through one energy balance.
BY THE END, YOU CAN
BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
Heat transfer moves energy because of a temperature difference. Conduction transfers energy through microscopic interactions without requiring bulk transport of material. Convection includes energy carried by moving fluid. Thermal radiation transfers electromagnetic energy and can cross vacuum. Several mechanisms can operate simultaneously, so identifying one visible mechanism does not automatically exclude the others.
Total heat Q is measured in joules, while heat-transfer rate H = dQ/dt is measured in watts. A constant rate gives Q = HΔt, but a cooling body usually has a changing rate because its temperature difference changes. The same distinction appears between energy stored in a hot object and the rate at which it loses that energy.
The temperature difference drives net transfer, but geometry and material properties control its magnitude. Two objects at the same temperature can feel different to touch because they transfer energy at different rates. A material with low thermal conductivity is not necessarily at a lower temperature; it may simply exchange energy slowly.
Steady state means the temperature at each location no longer changes with time. It does not mean all locations have the same temperature or that heat flow has stopped. A rod held between hot and cold reservoirs can have a permanent gradient and constant nonzero heat current. Thermal equilibrium is the special no-net-transfer state under the given contacts.
02 / UNDERSTAND & APPLY
For one-dimensional conduction through area A, Fourier’s law is H_x = −kA dT/dx, where thermal conductivity k has units W m⁻¹ K⁻¹. The minus sign sends heat toward lower temperature. If the temperature decreases along positive x, its derivative is negative and positive-x heat current is positive.
For a uniform rod of length L, constant k and cross section A, insulated laterally and in steady state without internal heat generation, the temperature profile is linear. The magnitude of heat current is H = kA(T_hot − T_cold)/L. A greater area provides more parallel routes, while greater length increases resistance to transfer.
Temperature differences can be expressed in kelvin or Celsius degrees in this linear conduction formula because the increments are equal. The absolute temperatures remain important for radiation and temperature-dependent material properties. Do not insert Celsius values into a fourth-power radiation relation just because a conduction difference allowed Celsius.
If the rod area varies, steady heat current is still constant without side losses or generation, but the gradient varies as 1/[k(x)A(x)]. Integrating gives thermal resistance R_th = ∫dx/[k(x)A(x)]. A uniform linear temperature graph is therefore a consequence of constant kA, not a universal property of steady conduction.
ORIGINAL WORKED EXAMPLE
A slab has k = 0.50 W m⁻¹ K⁻¹, area 2.0 m² and thickness 0.10 m. Its surfaces are held at 40°C and 10°C. Find steady heat current and energy transferred in 5.0 min, ignoring side losses.
Answer 300 W; 90 kJ in 5.0 min.
03 / UNDERSTAND & APPLY
Define R_th = L/(kA) for a uniform slab, so H = ΔT/R_th. Thermal resistance has units K W⁻¹. This algebra resembles electrical resistance, with temperature difference analogous to voltage and heat current analogous to electric current. The analogy helps organize a steady energy balance, but it does not turn temperature into a flow quantity.
For layers in series, the same steady heat current crosses each layer when there is no energy storage, side loss or internal generation. Resistances add, and each temperature drop is H R_i. The less conducting or thicker layer generally takes a larger drop for the same area. Interface temperature follows from the drop across either side.
For parallel conduction paths linking the same hot and cold reservoirs, each sees the same temperature difference and currents add. Conductances 1/R_i therefore add. A thermal bridge is a relatively conducting parallel route that can carry a disproportionate share of heat despite occupying a small area.
Perfect contact assumes no temperature discontinuity at an interface. Real thermal contact resistance can produce an additional temperature drop and belongs as another series element. Convective surface resistances can also be included if their heat-transfer coefficients are supplied. Omitting them while treating ambient air temperatures as exact solid-surface temperatures is an extra approximation.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
Two slabs of equal area 1.0 m² and thickness 0.10 m are in series. Their conductivities are 1.0 and 2.0 W m⁻¹ K⁻¹. The first touches a 100°C reservoir and the second a 10°C reservoir. Find steady heat current and junction temperature with ideal contact.
Answer 600 W; junction temperature 40°C.
04 / UNDERSTAND & APPLY
Convection involves fluid motion near a surface. Natural convection is driven by buoyancy associated with density differences, while forced convection uses a fan, pump or imposed flow. The detailed flow determines heat transfer, so it cannot generally be captured by thermal conductivity of the solid alone.
A common surface model writes outward heat rate H = hA(T_body − T_surroundings), where h is a heat-transfer coefficient in W m⁻² K⁻¹. Unlike a simple material conductivity, h depends on geometry, fluid properties, flow speed and other conditions. Treating it as constant is a modeling assumption over the stated operating range.
To describe a whole body with one temperature T(t), its internal conduction must be fast enough to keep temperature nearly uniform compared with its surface cooling. This is the lumped-capacitance approximation. A small Biot number Bi = hL_c/k_body, with characteristic length often volume/surface area, is a useful criterion; values much less than one support the approximation.
An insulated interior point in a large poorly conducting object may cool much more slowly than the surface. In that case, one exponential for the entire object can be inaccurate even if a surface convection law remains locally useful. Separate the model of energy crossing the boundary from the model of temperature distribution inside the body.
05 / UNDERSTAND & APPLY
If a body of total heat capacity C is internally uniform, surroundings stay at T_s and the effective surface coefficient hA is constant, energy balance gives C dT/dt = −hA(T − T_s). Define excess temperature θ = T − T_s and cooling constant λ = hA/C > 0. The solution is θ(t) = θ₀e^(−λt).
The body approaches the surroundings asymptotically in this model. It does not lose the same number of degrees in each equal interval; it loses the same fraction of its remaining temperature excess. The cooling rate becomes smaller as the temperature difference shrinks. If initially colder than the surroundings, the same equation predicts warming toward T_s.
For two measured temperatures at times separated by Δt, λ = ln[(T₁ − T_s)/(T₂ − T_s)]/Δt when both excesses have the same nonzero sign. A half-time refers to the excess above ambient, not to half the Celsius temperature. The time constant τ = 1/λ gives a surviving excess of e⁻¹, not one half.
For a narrow temperature range, an average-rate approximation uses the mean excess over that interval. But the exact exponential is usually as easy when a constant λ model is given. If room temperature changes, wind speed changes or the body evaporates appreciably, the parameters or energy balance may change and one fitted exponential need not extrapolate correctly.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
A lumped body cools in a room at 20°C from 80°C to 50°C in 10 min under a constant Newton-cooling model. Find λ and its temperature after another 10 min.
Answer λ ≈ 0.0693 min⁻¹; later temperature 35°C.
06 / UNDERSTAND & APPLY
A blackbody is an ideal absorber of all incident radiation and an ideal thermal emitter. Its total emitted power is P_emit = σAT⁴, where σ is the Stefan–Boltzmann constant and T is kelvin. A gray-surface approximation uses P_emit = εσAT⁴ with emissivity 0 ≤ ε ≤ 1 treated as approximately wavelength independent over the relevant spectrum.
Emission is not the same as net cooling. For a small gray body exchanging radiation with a large isothermal surrounding enclosure under the usual view-factor assumptions, net outward rate is εσA(T⁴ − T_s⁴). At equal temperatures, the body still emits and absorbs radiation, but the net exchange vanishes. The formula does not describe every pair of finite surfaces without geometric factors.
At thermal equilibrium, spectral emissivity and absorptivity match for corresponding wavelength, direction and conditions, a form of Kirchhoff’s radiation law. A good absorber is also a good emitter in the same spectral conditions. A surface’s appearance in visible light does not by itself specify its emissivity at all thermal infrared wavelengths.
Radiation travels through vacuum, unlike conduction or convection through a material medium. A polished surface can reduce radiative exchange when its emissivity is low in the relevant band, while an evacuated gap suppresses gas conduction and convection. A practical insulating design can need several mechanisms addressed independently.
ORIGINAL WORKED EXAMPLE
A gray surface of area 0.010 m² and emissivity 0.80 is at 400 K inside a large enclosure at 300 K. Use σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. Find emitted power and net outward radiative power.
Answer Approximately 12 W emitted; 7.9 W net outward.
07 / UNDERSTAND & APPLY
A thermal spectrum has a broad range of wavelengths. For an ideal blackbody, the wavelength at the peak of spectral power per unit wavelength obeys λ_max T = b, with b approximately 2.90 × 10⁻³ m K. Raising temperature shifts this wavelength peak shorter and increases the total area under the spectrum according to T⁴.
The peak depends on the spectral variable. A curve plotted per unit frequency has a peak that does not convert to the wavelength-domain peak by merely setting λ = c/f at the maximum, because the density changes under variable transformation. In elementary Wien problems, specify that λ_max refers to the wavelength spectrum.
For temperatures close to ambient, write T = T_s + θ with |θ| much smaller than T_s in kelvin. Expanding gives T⁴ − T_s⁴ ≈ 4T_s³θ. Thus net radiative loss is approximately 4εσAT_s³(T − T_s), a Newton-like linear cooling term. This explains a limited connection between the fourth-power law and exponential cooling.
For large temperature differences, the neglected higher powers of θ matter and pure radiative cooling of a lumped body follows C dT/dt = −εσA(T⁴ − T_s⁴), not generally a single exponential. A constant effective cooling coefficient over a wide range should therefore be justified empirically or by the dominant mechanism, not assumed from the word “cooling.”
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
An ideal blackbody has its wavelength-spectrum peak at 2.0 μm. Estimate temperature using b = 2.90 × 10⁻³ m K. If temperature doubles, find new peak wavelength and emitted power per area ratio.
Answer Approximately 1.5 × 10³ K; new peak 1.0 μm; total emitted power per area increases sixteenfold.
08 / UNDERSTAND & APPLY
When a body loses heat simultaneously by convection and radiation, a lumped model gives C dT/dt = −hA(T − T_s) − εσA(T⁴ − T_s⁴), with any other inputs added separately. Near ambient, linearizing radiation gives an effective conductance hA + 4εσAT_s³. Then the local cooling time constant is C divided by this total conductance.
If a constant heater supplies power P_in, steady temperature satisfies P_in = total outward loss. Steady state can therefore have T > T_s and nonzero continuous heat flow. A change in surface area or emissivity changes the equilibrium through the loss law; thermal mass mainly changes how quickly the body approaches it in a lumped model.
For a body heated through a thermal resistance R_th and storing energy in heat capacity C, the simplest linear time constant is R_thC. This is analogous to an electrical RC transient only under the corresponding single-temperature and constant-resistance assumptions. A multi-layer body with internal gradients can require several time scales rather than one.
ORIGINAL WORKED EXAMPLE
A body has heat capacity 600 J K⁻¹ and effective linear heat-loss conductance 3.0 W K⁻¹ to a 20°C room. A constant 90 W heater is switched on. Find steady temperature and time constant under this linear model.
Answer Steady temperature 50°C; time constant 200 s.
ORIGINAL PRACTICE · 1
It halves.
H = kAΔT/L.
ORIGINAL PRACTICE · 2
600 J.
Q = Ht when the rate is constant.
ORIGINAL PRACTICE · 3
5 K W⁻¹.
The same heat current crosses both and their temperature drops add.
ORIGINAL PRACTICE · 4
1.2 K W⁻¹.
Conductances add: 1/R = 1/2 + 1/3 = 5/6.
ORIGINAL PRACTICE · 5
No.
A time-independent spatial gradient can support constant nonzero heat current.
ORIGINAL PRACTICE · 6
45°C.
Initial excess is 50 K; halving gives 25 K, then add ambient 20°C.
ORIGINAL PRACTICE · 7
No; it leaves e⁻¹, about 36.8%.
The solution is θ/θ₀ = e^(−t/τ).
ORIGINAL PRACTICE · 8
Yes; it warms toward the room.
Negative θ gives positive dT/dt in dT/dt = −λθ.
ORIGINAL PRACTICE · 9
It increases sixteenfold.
P ∝ T⁴ with T in kelvin.
ORIGINAL PRACTICE · 10
No.
Radiation laws require absolute temperatures in kelvin.
ORIGINAL PRACTICE · 11
No.
Emission and absorption continue but balance in net energy exchange.
ORIGINAL PRACTICE · 12
Temperature doubles.
Wien’s wavelength law gives λ_maxT constant.
ORIGINAL PRACTICE · 13
Internal temperature differences must remain small.
Internal conduction must be fast compared with boundary cooling; a small Biot number supports the approximation.
ORIGINAL PRACTICE · 14
Ambient air temperature need not equal the wall surface temperature.
The surface exchange can cause an additional temperature drop before heat crosses the solid.
09 / UNDERSTAND & APPLY
Identify which temperatures are reservoir values, surface values or body averages. Choose a conduction geometry, a convection coefficient model or a radiative-exchange geometry accordingly. Separate a steady spatial profile from a changing lumped temperature. If mechanisms act simultaneously, add their energy-transfer rates instead of choosing only one by habit.
Check units before interpreting the result: conductivity is W m⁻¹ K⁻¹, surface coefficient is W m⁻² K⁻¹, resistance is K W⁻¹ and heat capacity is J K⁻¹. Use kelvin for fourth powers and Wien’s law. Finally test limiting cases: zero temperature difference gives zero net exchange, a larger resistance reduces steady current, and a larger heat capacity slows a given linear cooling process.
KEEP THE BIG IDEAS
READ WITH CONTEXT
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Factual cross-check only. Explanations, numerical examples and practice are independently authored.
Analytics use a separately reviewed, incomplete historical corpus. Opening a topic does not change its counts or denominator.