Thermal Properties of Matter · Lesson 8 of 8
Chapter Summary and Practice
“Connect thermometry, expansion, calorimetry, phase conversion and heat transport through model selection, units, assumptions and mixed worked reasoning.”
• Choose an appropriate thermal model from the given physical conditions. • Use a connected formula and unit reference without confusing rates, energies or temperature scales. • Combine phase change and heat transport while checking energy and mass inventories. • Recognize common conceptual and numerical errors across the chapter.
Begin with the physical question
Thermal problems ask several different kinds of question. A thermometer problem converts a reproducible property into a temperature. An expansion problem asks how dimensions respond to a temperature change. Calorimetry balances energy between bodies. A phase-change problem tracks both state and temperature. Heat-transfer laws ask how quickly energy crosses a boundary, while a cooling law predicts how the temperature evolves as the driving difference changes. Choose the question before choosing a formula.
| Question | Appropriate starting model | Essential condition |
|---|---|---|
| What temperature does a sensor indicate? | Linear calibration or specified gas/resistance relation | Use the given property law and reference conditions |
| How does a free dimension change? | Δl = αl l0ΔT; isotropic area/volume factors | Small change, suitable coefficient, free expansion |
| How much energy warms a body? | Q = mcΔT | One phase, approximately constant c |
| How much energy converts a phase? | Q = mL | Specified transition pressure and temperature |
| What steady current crosses a layer? | H = KAΔT/L | Maintained ends, insulated sides, steady flow |
| What radiative power is exchanged? | eσAT⁴ or eσA(T⁴ − Ts⁴) | Kelvin temperatures and appropriate surface/enclosure model |
| How long does passive cooling take? | Exponential decay of T − Ts | Stable κ and surroundings, small-difference regime |
A formula reference with units and limits
| Quantity | SI unit | What to distinguish |
|---|---|---|
| Temperature T | K | A reading versus a temperature difference |
| Energy transferred Q | J | A total energy versus current in W |
| Heat capacity S | J K⁻¹ | Whole body versus specific and molar quantities |
| Specific heat c | J kg⁻¹ K⁻¹ | Single-phase warming versus latent conversion |
| Molar heat capacity C | J mol⁻¹ K⁻¹ | Constant-volume versus constant-pressure process |
| Latent heat Lf or Lv | J kg⁻¹ | Mass-specific energy, not a heat capacity |
| Thermal conductivity K | W m⁻¹ K⁻¹ | Material property versus whole-layer resistance |
| Expansion coefficient α | K⁻¹ | Linear, area and volume definitions |
| Emissivity e | Dimensionless | Effective surface response, not temperature |
Connections that prevent common errors
Temperature is not a stock of heat. Energy transferred because of a temperature difference is heat; drilling can increase internal energy through work. Equal-temperature brass and wood can feel different because they conduct energy from skin at different rates. Equilibrium means no net temperature-driven energy transfer, not that all microscopic motion or opposing radiation ceases.
Coexisting liquid and vapour in an elementary equilibrium model share a temperature and pressure while having different densities. Very tall fluid columns can have hydrostatic pressure differences, so the uniform-pressure picture is the usual small-system idealization. During boiling at maintained pressure, adding energy changes phase proportions rather than requiring a temperature rise. The triple point is not ordinary atmospheric boiling or freezing, and thermometry’s historical water-reference convention must be distinguished from today’s fixed-Boltzmann-constant Kelvin definition.
Geometry and constraints matter as much as material labels. A hole expands with its sheet. A freely expanding rod need not develop axial stress, whereas a fixed rod can have zero total extension and nonzero stress. A poor conductor in a thin, large-area layer can still pass substantial energy. A vacuum gap suppresses matter-based transport across the gap but still allows radiation. Running water can involve both local conduction and bulk convection.
Check units before arithmetic. Convert centimetres to metres for conduction lengths, square centimetres to square metres for area, minutes to seconds for watts, and grams to kilograms when specific heat is in J kg⁻¹ K⁻¹. Use absolute temperatures in gas and radiation laws, but use numerical temperature differences directly in Celsius degrees or kelvins. For cooling, work with excess above the surroundings, not the body temperature alone.
Worked synthesis: insulation admits energy while ice melts
Problem
An ideal plane insulating layer has K = 0.040 W m⁻¹ K⁻¹, area 0.50 m² and thickness 0.050 m. The outside is 25 °C and the inside contains 0.200 kg of ice at 0 °C. Find the conducted energy and melted mass after 1.0 h, taking Lf = 3.34 × 10⁵ J kg⁻¹ and neglecting all other paths.
- 1.As long as ice remains, the inside can stay at 0 °C. The steady current is 0.040 × 0.50 × 25/0.050 = 10.0 W.
- 2.Over 3600 s, energy admitted is 36000 J. The melted mass is 36000/334000 = 0.1078 kg.
- 3.About 0.0922 kg ice remains, confirming the assumed phase plateau and constant 25 K difference for this interval.
- 4.After all ice melts, the same constant-current calculation cannot automatically continue: the liquid can warm, changing the wall’s temperature difference. A new energy balance is needed.
Worked synthesis: compare a surface law with a cooling law
Problem
Two otherwise comparable surfaces exchange radiation with a large 300 K enclosure. One is at 310 K and the other at 320 K. Why is net radiation not exactly doubled, even though a Newton approximation might suggest a factor of two?
- 1.The exact radiation-model ratio is (320⁴ − 300⁴)/(310⁴ − 300⁴), approximately 2.10 for equal area and emissivity.
- 2.Newton’s linear approximation retains only the first-order term in excess above 300 K, predicting the ratio 20/10 = 2.
- 3.The difference comes from higher-order terms. A linear law can be useful without being the exact fourth-power law.
- 4.A complete cooling-time prediction also needs the body heat capacity and stable transfer conditions; a surface power alone is not a temperature history.
Quick Quiz
Quiz
Which observation by itself implies no net thermal exchange at equilibrium?
Equal equilibrium temperature removes the driving temperature difference; mass and colour alone do not.
An insulated ideal mixture still contains ice and liquid water at 1 atm. Further small energy input initially goes mainly into
The phase inventory includes unconverted ice, so energy can change phase proportions at the melting temperature.
Which calculation specifically requires absolute rather than Celsius temperatures?
Fourth powers must use Kelvin temperatures; temperature intervals have equal Celsius and Kelvin magnitudes.
A fixed-end rod is uniformly heated within its elastic range. The simple axial model predicts
Support compression cancels the free thermal strain so the constrained total strain is zero.
With constant Newton relaxation rate, equal cooling times produce equal
Excess temperature changes exponentially, so successive equal-time ratios are the same.
Practice Problems
- A resistance thermometer reads 100 Ω at 273.16 K and 140 Ω at 373.16 K, with a linear resistance–temperature relation. Find the temperature at 118 Ω in K and °C. Explain why its calibration law, rather than touch, defines the reading.
- A copper sheet with a 20.0 mm circular hole is freely heated by 100 K. Use αl = 1.7 × 10⁻⁵ K⁻¹ to find the hole’s diameter change and the sheet’s approximate fractional area increase. Explain why no axial thermal stress need arise.
- An insulated vessel of negligible heat capacity contains 0.300 kg of water at 40 °C. Add 0.100 kg of ice at 0 °C. Use cwater = 4186 J kg⁻¹ K⁻¹ and Lf = 3.34 × 10⁵ J kg⁻¹. Test complete melting, then find the final equilibrium state and temperature.
- A laterally insulated wall has area 1.0 m² and two series layers, each 0.020 m thick, with conductivities 0.040 and 0.080 W m⁻¹ K⁻¹. Its outer temperatures are 30 °C and 0 °C. Find the steady current and the temperature at the junction with the poorer conductor on the hot side.
- A blackbody thermal spectrum peaks at 10 μm. Estimate its temperature using b = 2.9 × 10⁻³ m K. Separately calculate the net radiative loss from an e = 0.80, A = 0.010 m² surface at 400 K in a large 300 K enclosure, using σ = 5.67 × 10⁻⁸ in SI units.
- A body in a 20 °C room cools from 60 °C to 40 °C in 5 min. Predict the later time from 30 °C to 25 °C with the constant-κ exponential model, and state three conditions that should be checked experimentally.
Key Takeaways
• Select the physical model first, then check phase, geometry, constraints, units and energy inventory. • Temperature, energy and power answer different questions; their units prevent many common mistakes. • The chapter’s laws work together, but each retains its own conditions and approximations. • Verify results against equilibrium limits, conserved energy and plausible physical trends.
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Newton’s Law of Cooling
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