熱輸送、熱量計法および膨張
導入済み| English | 意味 |
|---|---|
| thermal resistance/ˈθɜːml rɪˈzɪstəns/ | 熱抵抗 |
| latent heat/ˈleɪtənt hiːt/ | 潜熱 |
A decision before an answer
- Two wall layers in series carry the same steady heat rate, but their temperature drops need not be equal.
- Your goal: Combine conductive thermal resistances and distinguish heat rate from heat flux.
Add thermal resistances
- For one-dimensional steady conduction through a uniform slab, Fourier’s law gives heat flux q_x=−κ dT/dx. With constant conductivity κ, area A, length L and hot-to-cold temperature difference ΔT, heat rate is Qdot=κAΔT/L. Define thermal resistance R_th=L/(κA), measured in K/W; temperature drop is Qdot R_th.
- Series layers without internal sources carry the same Qdot and their resistances add. Parallel paths at the same endpoint temperatures have heat rates that add, so inverse resistances add. Heat flux is rate per area, W/m², and need not match between layers of different area. Interface contact resistance and heat leakage must be included if specified.
Two slabs in series without sources have thermal resistances R and 3R. The steady temperature-drop ratio is:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Their heat rate is the same; ΔT_i=Qdot R_i gives 1:3.
Balance sensible and latent energy
- Heat capacity C=dQ/dT depends on the thermodynamic path and is not the same as specific heat c per mass. For a material interval without a phase transition, Q=∫mc(T)dT, or mcΔT when c is constant. In an isolated calorimeter, sum the energy changes of all objects, including the container when relevant, and set the sum to zero.
- At a phase transition at its equilibrium temperature, added energy can change phase fraction rather than temperature: Q=mℓ uses latent heat ℓ in J/kg. First supply the sensible heat to reach the transition, then budget latent heat. Do not let a simple weighted-temperature average predict an impossible temperature when melting or freezing is part of the process.
An isotropic unconstrained plate with a circular hole is warmed under small uniform expansion. Its hole radius:
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
All in-plane lengths, including the hole boundary, scale by 1+αΔT; volume scaling is a different quantity.
Check expansion constraints
- For a freely expanding rod and small temperature change, ΔL≈αL₀ΔT with linear expansion coefficient α. For an isotropic solid with small strain, each dimension acquires factor 1+αΔT, so ΔA/A≈2αΔT and ΔV/V≈3αΔT. A hole expands with the surrounding material as if its missing region had expanded too.
- These relations assume a nearly constant coefficient and no mechanical constraint; anisotropic crystals need directional coefficients. If an elastic rod is prevented from changing length, its mechanical strain cancels thermal strain. With tensile stress positive, σ≈−YαΔT during heating, where Y is Young’s modulus. This small-strain estimate requires elastic response without yielding or buckling.
Two slabs of area 0.5 m² have L₁=0.02 m, κ₁=0.5 W/(m·K), L₂=0.03 m and κ₂=0.25 W/(m·K). R₁=0.08 and R₂=0.24 K/W; for boundaries 60°C and 20°C, Qdot=40/0.32=125 W. Drops are 10 K and 30 K, so the interface is 50°C and flux 250 W/m². Separately equal specific heats with masses 2 kg at 80°C and 1 kg at 20°C give isolated final 60°C without phase change. A free 2 m rod with α=10^−5 K^−1 heated 50 K lengthens 1 mm; if constrained and Y=100 GPa, σ=−50 MPa within the elastic model.
A body has C=200 J/K and hA=10 W/K under lumped Newton cooling. Its time constant is ____ s.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
τ=C/(hA)=200/10=20 s.
Choose the transport model
- Conduction transfers energy through a temperature gradient, convection transports it with moving matter, and thermal radiation can cross a vacuum. For a surface at T facing large surroundings at T_env, a simple grey-body model gives net radiative rate εσ_SB A(T⁴−T_env⁴). Use absolute kelvin in these fourth powers, rather than Celsius values.
- A lumped body whose internal temperature remains nearly uniform can obey C dT/dt=−hA(T−T_env) under Newton cooling with constant h. Then the temperature excess decays with τ=C/(hA), rather than the entire Celsius or kelvin temperature decaying toward zero. This approximation requires sufficiently small internal gradients; a conduction-limited body needs a spatial temperature model.
Keep W separate from W/m², include the calorimeter or latent energy when needed, and do not use free-expansion length together with constrained-stress assumptions.
このケースに当てはまる答えはどれか？
熱抵抗（コンダクティブ）を合成し、熱流束と熱伝達率を区別する
A fully constrained elastic rod heated with positive α develops compressive stress if buckling and yield are excluded.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Mechanical strain must cancel positive thermal strain; tensile-signed stress is −YαΔT.
Keep the distinctions
- thermal resistance 热阻 — Temperature difference per steady heat-transfer rate, measured in K/W.
- latent heat 潜热 — Energy transferred during a phase change at its transition temperature; specific latent heat is per mass.
- Combine conductive thermal resistances and distinguish heat rate from heat flux.
- Balance sensible and latent heat with stated isolation and heat-capacity conditions.
- Calculate free thermal expansion and constrained thermal stress with their limits.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.