Thermal Properties of Matter · Lesson 5 of 8
Conduction and Composite Materials
“Develop steady conduction as a conserved heat current, derive series-bar results and apply them to insulation, ice melting and a steam boiler.”
• Distinguish transient heating from steady conduction without bulk matter transport. • Use conductivity, geometry and temperature difference to calculate heat current. • Derive junction temperature and equivalent conductivity for series bars. • Combine conduction power with latent heat in practical energy balances.
Three transport mechanisms, one energy account
Energy can travel from a hotter region to a colder one by conduction, convection or radiation. Conduction transfers energy through microscopic interactions without bulk transport of material. Convection transports energy with moving fluid. Radiation transfers electromagnetic energy and can cross a vacuum. These mechanisms often act together: a metal pan conducts energy through its base, moving water carries it around inside, and exposed surfaces exchange radiation with the surroundings.
Conduction through a laterally insulated bar
Touch one end of a metal bar to a hot source. Initially nearby regions warm first, and the temperature at each position changes with time. This is the transient stage. If the ends are maintained at fixed temperatures and the sides are well insulated, a steady state can eventually develop. At steady state a small element receives as much energy per second as it passes onward; otherwise its temperature would still be changing. The same heat current therefore crosses every section.
A larger area provides more parallel microscopic paths, so current is proportional to A. A greater temperature difference drives more transfer, whereas a longer path reduces the gradient and the current. Temperature differences can be in kelvins or Celsius degrees because their interval sizes agree. The unit check is (W m⁻¹ K⁻¹)(m²)(K)/m = W. To find energy over a steady time interval, multiply current by that time.
Conduction occurs in solids, liquids and gases through microscopic interactions. Mobile electrons help make metals especially good conductors; energy is not carried by a bulk migration of the rod from hot to cold. A steady temperature profile is not necessarily uniform: for a uniform bar with constant conductivity it is linear between the maintained end temperatures. Uniform temperature would give zero current in this model, not the nonzero steady current between unequal-temperature reservoirs.
| Material | K / W m⁻¹ K⁻¹ |
|---|---|
| Silver | 406 |
| Copper | 385 |
| Aluminium | 205 |
| Brass | 109 |
| Steel | 50.2 |
| Lead | 34.7 |
| Mercury | 8.3 |
| Insulating brick | 0.15 |
| Concrete | 0.8 |
| Body fat | 0.20 |
| Felt | 0.04 |
| Glass | 0.8 |
| Ice | 1.6 |
| Glass wool | 0.04 |
| Wood | 0.12 |
| Water | 0.8 |
| Air | 0.024 |
| Argon | 0.016 |
| Hydrogen | 0.14 |
The table compares representative material properties, not whole-object heat currents. A thick short wood layer and a very thin long metal part cannot be compared by conductivity alone. Copper spreads energy readily across cookware bases. Wood and plastic handles limit the current toward a hand. Foams and glass wool contain small air pockets: air conducts poorly, and confinement suppresses large convection currents. A wide free air gap can permit convection, so trapped air and freely circulating air are not equivalent insulation.
Concrete conducts far less strongly than copper but can still pass substantial energy through a large roof. Adding a suitable low-conductivity layer reduces that transfer. In other settings the goal is efficient removal: a reactor core or engine needs adequate heat transport to avoid accumulating internal energy and overheating. Insulation is useful only when reducing the current serves the physical purpose.
Bars in series share current, not temperature drop
Write G1 = K1A1/L1 and G2 = K2A2/L2. Expanding the current equation gives G1Th − G1Tj = G2Tj − G2Tc. Collecting the junction terms gives Tj = (G1Th + G2Tc)/(G1 + G2). The simple average is valid only for equal conductances. The section that resists heat flow more strongly has the larger temperature drop.
Problem
A steel rod of length 0.150 m and area 2A joins a copper rod of length 0.100 m and area A. The outer steel end is 300 °C and the outer copper end 0 °C. Use Ksteel = 50.2 and Kcopper = 385 W m⁻¹ K⁻¹.
- 1.Set 50.2(2A)(300 − Tj)/0.150 = 385A(Tj − 0)/0.100.
- 2.Cancel A, which is common to both currents. Expanding gives 669.333(300 − Tj) = 3850Tj.
- 3.Thus Tj ≈ 44.43 °C, about 44.4 °C. The steel drop is about 255.6 °C and copper drop about 44.4 °C.
- 4.The junction is not 150 °C because the two segments do not have equal thermal conductance.
For equal-area bars, define an equivalent conductivity as that of a single bar of area A and total length L1 + L2 with the same current and end temperatures. Equating its resistance to the sum gives the following result. If areas differ, adding thermal resistances still works, but an equivalent conductivity requires an explicitly chosen reference geometry.
Problem
An iron bar and brass bar each have length 0.100 m and area 0.0200 m², joined end to end with outer temperatures 373 K and 273 K. Conductivities are 79 and 109 W m⁻¹ K⁻¹. Find the junction temperature, equivalent conductivity and current.
- 1.Equal lengths and areas reduce current matching to 79(373 − Tj) = 109(Tj − 273).
- 2.Therefore Tj = (79 × 373 + 109 × 273)/(79 + 109) = 315.02 K ≈ 315 K.
- 3.For the total 0.200 m length, Keq = 2 × 79 × 109/(79 + 109) = 91.606 W m⁻¹ K⁻¹.
- 4.The total current is Keq × 0.0200 × (373 − 273)/0.200 = 916.06 W.
- 5.Checking either individual bar gives the same current. That agreement is the steady-state conservation test, not an additional assumption.
Insulation plus melting: turn current into mass
Problem
A cubical thermocol icebox is modelled with six 0.300 m square faces, wall thickness 0.0500 m and K = 0.0100 W m⁻¹ K⁻¹. It contains 4.00 kg ice at 0 °C, with outside at 45 °C. Estimate ice remaining after 6 h using Lf = 3.35 × 10⁵ J kg⁻¹.
- 1.Use the chapter’s plane-wall approximation: total area A = 6 × (0.300)² = 0.540 m². Neglect corners, seams and changing inner/outer area.
- 2.The current is 0.0100 × 0.540 × 45/0.0500 = 4.86 W.
- 3.Over 21600 s, entering energy is 104976 J and melted mass is 104976/335000 = 0.31336 kg.
- 4.Remaining ice is about 3.69 kg, or 3.7 kg. Because ice remains, the inner ice–water mixture can stay near 0 °C, supporting the assumed temperature difference.
Conduction can supply a phase-change power
Problem
Water boils at 100 °C above a brass base of area 0.150 m² and thickness 0.0100 m. Steam forms at 6.00 kg min⁻¹. Use K = 109 W m⁻¹ K⁻¹ and Lv = 2.256 × 10⁶ J kg⁻¹. Find the flame-side base temperature in the steady ideal model.
- 1.Mass rate is 6/60 = 0.100 kg s⁻¹, so the required latent-energy power is H = 0.100 × 2.256 × 10⁶ = 225600 W.
- 2.Conduction supplies this current: H = KA(Thot − 100)/L.
- 3.Thus Thot − 100 = 225600 × 0.0100/(109 × 0.150) = 137.98 K, giving Thot ≈ 238 °C.
- 4.This estimates the hot face of the brass base, idealized as the flame-contact temperature. Actual flame gas can be hotter because interface transfer, heating of feedwater and other losses have been omitted.
Quick Quiz
Quiz
A uniform bar has different fixed end temperatures and a nonzero steady heat current. Which statement is correct?
Steady temperature means no local energy accumulation, not zero current.
For a fixed material, area and temperature difference, doubling a bar’s length changes the steady current to
The conduction law makes current inversely proportional to length.
Two rods in lossless series at steady state necessarily have the same
Junction energy conservation requires equal current, while resistance determines each drop.
Why can glass wool insulate effectively?
The structure combines low gas conduction with restricted convection.
An icebox admits a steady 10 W for 100 s while ice remains at 0 °C. With Lf = 3.34 × 10⁵ J kg⁻¹, melted mass is approximately
Entering energy is 1000 J, so m = 1000/334000 kg ≈ 0.0030 kg.
Practice Problems
- A uniform rod has K = 200 W m⁻¹ K⁻¹, area 2.0 cm² and length 0.400 m. Its ends are held at 100 °C and 20 °C with insulated sides. Find the steady heat current and energy transferred in 5.0 min.
- Two equal-area bars of equal length are joined in series. Their conductivities are 40 and 120 W m⁻¹ K⁻¹; the first outer end is 100 °C and the second 0 °C. Derive and find the junction temperature and equivalent conductivity.
- A plane insulating wall of area 2.0 m² and thickness 0.050 m has K = 0.040 W m⁻¹ K⁻¹. Find the steady current for a 15 K temperature difference. Predict the new current if a second identical layer is added with negligible contact resistance.
- Ice at 0 °C receives a constant heat current of 8.0 W for 2.0 h. Find the melted mass using Lf = 3.34 × 10⁵ J kg⁻¹. State what must be checked before assuming its temperature remains 0 °C throughout.
- A metal boiler base has A = 0.100 m², L = 0.0080 m and K = 100 W m⁻¹ K⁻¹. Its hot face is 200 °C and boiling-water face 100 °C. If all conducted energy vaporizes water, find the steam mass rate using Lv = 2.256 × 10⁶ J kg⁻¹.
Key Takeaways
• Steady conduction means equal current through each section, with no local energy accumulation. • Conductivity is a material property; whole-body current also depends on geometry and temperature difference. • Series bars share current, and their thermal resistances and temperature drops add. • Combine H with elapsed time or latent-energy rate only after checking the maintained temperature and phase assumptions.