TS.3 · Thermal transport, calorimetry and expansion

직렬로 연결된 두 벽층은 동일한 정상 상태 열유속을 전달하지만, 그들 사이의 온도 강하는 반드시 동일할 필요는 없다.

선행 학습 과목: 4.

• 전도성 열저항을 합산하고 열유속(heat rate)과 열플럭스(heat flux)를 구분한다
• 주어진 단열 조건과 열용량 조건 하에서 현열과 잠열의 균형을 맞춘다
• 자유 열팽창과 구속된 열응력을 계산하고 그 한계를 파악한다

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.

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.

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.

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.

직렬로 연결된 두 판은 동일한 정적 열유속을 전달합니다. 저항을 각각 0.08 및 0.24 K/W라고 합시다. 경계 온도는 각기 60 및 20 °C입니다.

W와 W/m²를 분리하고, 필요할 때는 계기나 잠열을 포함하며, 자유 팽창 길이를 제약된 응력 가정과 함께 사용하지 마십시오.

열저항(thermal resistance): 정상 상태 열전달 율당 온도 차이로, 단위는 K/W이다.

잠열(latent heat): 상변화 온도에서 상변화 시 전달되는 에너지; 비잠열은 질량당 값이다.