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GRE Subject Test · GRE Physics 주제 34 6:28 영어 내레이션 · 영어 + 중국어 자막 burned-in
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Two wall layers in series carry the same steady heat rate, yet the temperature can drop much more across one than the other.
串联的两层墙传递相同的稳定热流，但温度在其中一层上的下降可能远大于另一层。
This topic adds thermal resistances, balances sensible and latent heat in a calorimeter, and compares free expansion with the stress in a clamped rod.
本主题把热阻相加，在量热器中平衡显热和潜热，并比较自由膨胀与被夹紧杆中的应力。
For steady conduction through a uniform slab, the heat rate is kappa A delta T over L.
对于通过均匀平板的稳定热传导，热流率等于 κ A ΔT 除以 L。
Define the thermal resistance L over kappa A, in kelvin per watt, so each temperature drop is the heat rate times that resistance.
定义热阻为 L 除以 κA，单位为开尔文每瓦，这样每段的温度降等于热流率乘以该热阻。
Layers in series carry the same heat rate and their resistances add.
串联各层的热流率相同，热阻相加。
Parallel paths share the end temperatures, so their heat rates add.
并联路径两端温度相同，热流率相加。
Heat rate is in watts; heat flux divides it by the area.
热流率的单位是瓦；热流密度要再除以面积。
Heat capacity C equals d Q by d T depends on the thermodynamic path and is not the same as specific heat c per mass.
热容 C = dQ/dT 取决于热力学路径，与比热容 c 不同。
For a material interval without a phase transition, Q equals the integral of m times c of T d T, or m times c times delta T when c is constant.
对于无相变的材料区间，Q = ∫mc(T)dT；若 c 为常数，则 Q = mcΔT。
In an isolated calorimeter, sum the energy changes of all objects, including the container when relevant, and set the sum to zero.
在孤立量热计中，将所有物体（包括相关容器）的能量变化求和并设为零。
At a phase transition at its equilibrium temperature, added energy can change phase fraction rather than temperature: Q equals m times l uses latent heat l in J per kg.
在平衡温度下的相变过程中，加入的能量可能改变相分数而非温度：Q = mℓ，其中 ℓ 为潜热，单位为 J/kg。
First supply the sensible heat to reach the transition, then budget latent heat.
首先提供显热以达到转变温度，再计算潜热预算。
Do not let a simple weighted-temperature average predict an impossible temperature when melting or freezing is part of the process.
当熔化或凝固是过程的一部分时，切勿用简单的加权平均温度预测不可能的温度。
For a freely expanding rod and a small temperature change, delta L is about alpha L nought delta T, with linear expansion coefficient alpha.
对于自由膨胀的杆和小的温度变化，ΔL 约等于 α 乘以 L0 乘以 ΔT，其中 α 为线膨胀系数。
For an isotropic solid with small strain, each dimension acquires a factor of 1 plus alpha delta T, so delta A over A is about 2 alpha delta T, and delta V over V is about 3 alpha delta T.
对于小应变的各向同性固体，每个尺寸都乘以 1 加 α ΔT 的因子，因此面积相对变化约为 2 α ΔT，体积相对变化约为 3 α ΔT。
A hole expands with the surrounding material as if its missing region had expanded too.
孔洞随周围材料一起膨胀，就像缺失部分也膨胀了一样。
These relations assume a nearly constant coefficient and no mechanical constraint; anisotropic crystals need directional coefficients.
这些关系假设系数近似恒定且没有机械约束；各向异性晶体需要按方向给出系数。
If an elastic rod is prevented from changing length, its mechanical strain cancels the thermal strain.
若弹性杆被阻止改变长度，其机械应变抵消热应变。
With tensile stress positive, the stress is about minus Y alpha delta T during heating, where Y is Young's modulus.
以拉应力为正，加热时应力约等于负 Y α ΔT，其中 Y 为杨氏模量。
This small-strain estimate requires elastic response without yielding or buckling.
这一小应变估计要求材料处于弹性范围，不发生屈服或屈曲。
Conduction transfers energy through a temperature gradient. Convection transports it with moving matter.
热传导通过温度梯度传递能量，对流随流动物质传输热量，而热辐射可穿过真空。
And thermal radiation can cross a vacuum. For a surface at T facing large surroundings at T sub env, a simple grey-body model gives net radiative rate epsilon sigma SB A times T to the fourth minus T sub env to the fourth.
对于温度为 T、面对温度为 T_env 的大环境表面，简单的灰体模型给出的净辐射速率为 εσ_SB A(T⁴−T_env⁴)。
Use absolute kelvin in these fourth powers, rather than Celsius values.
这些四次方项需使用绝对开尔文温标，而非摄氏度数值。
A lumped body whose internal temperature remains nearly uniform can obey C dT by d t equals negative h A times T minus T sub env under Newton cooling with constant h.
若一个集总参数体内温度基本均匀，则在牛顿冷却定律（h 为常数）下满足 C dT/dt = −hA(T−T_env)。
Then the temperature excess decays with tau equals C divided by h A, rather than the entire Celsius or kelvin temperature decaying toward zero.
此时温度过剩量按 τ = C/(hA) 衰减，而非整个摄氏或开尔文温度趋向零。
This approximation requires sufficiently small internal gradients. A conduction-limited body needs a spatial temperature model.
该近似要求内部温差足够小；若受热传导限制，则需采用空间温度分布模型。
Two slabs of 0.08 and 0.24 kelvin per watt sit in series between 60 and 20 degrees Celsius.
热阻分别为 0.08 和 0.24 开尔文每瓦的两块平板串联在 60 摄氏度和 20 摄氏度之间。
The same heat rate passes through both: 40 kelvin over 0.32 kelvin per watt, which is 125 watts.
通过两者的热流率相同：40 开尔文除以 0.32 开尔文每瓦，等于 125 瓦。
The first slab drops 125 watts times 0.08: 10 kelvin, so the interface is at 50 degrees.
第一块平板的温度降等于 125 瓦乘 0.08：10 开尔文，所以界面处为 50 摄氏度。
The second slab takes the remaining 30 kelvin.
第二块平板承担剩下的 30 开尔文。
The same heat rate, but three times the drop across three times the resistance.
热流率相同，但三倍的热阻上有三倍的温度降。
Mix 0.20 kg of water at 40 degrees Celsius with 0.050 kg ice at zero degrees Celsius in an insulated container of negligible heat capacity.
将 0.20 kg、温度为 40 摄氏度的水与 0.050 kg、温度为 0 摄氏度的冰混合在一个热容量可忽略的绝热容器中。
Use c equals 4200 J per kg K and latent heat equals 334000 J per kg.
取比热容 c 为 4200 J/(kg·K)，潜热为 334000 J/kg。
Determine final temperature and whether ice remains.
求最终温度，并判断是否有冰剩余。
Cooling water to zero releases m c delta T equals 0.20, times 4200, times 40, which is 33600 joules.
将水冷却至零度释放的热量 mcΔT 等于 0.20 乘以 4200 乘以 40，即 33600 焦耳。
Melting all ice needs m ell equals 0.050, times 334000, which is 16700 joules, so all ice melts.
融化所有冰所需热量 mℓ 等于 0.050 乘以 334000，即 16700 焦耳，因此所有冰都会融化。
Remaining energy heats 0.25 kg water: final temperature T sub f equals 33600 minus 16700, all over 0.25, times 4200, which is 16.10 degrees Celsius.
剩余能量用于加热 0.25 千克的水：最终温度 Tf 等于 33600 减 16700，除以 0.25 乘 4200，结果为 16.10 摄氏度。
Award cooling budget, melt budget, phase decision, final heat capacity and temperature.
评分要点包括冷却预算、熔化预算、相变判断、最终热容及温度。
If cooling energy were insufficient, final temperature would be zero with some ice left.
若冷却能量不足，最终温度将为零度且仍有部分冰残留。
Keep watts separate from watts per square metre, include the calorimeter or latent energy when needed, and do not use free-expansion length together with constrained-stress assumptions.
区分瓦特与瓦特每平方米。 必要时计入量热器或潜热。 不要把自由膨胀的长度与受约束的应力假设混用。
thermal resistance: Temperature difference per steady heat-transfer rate, measured in K/W.
热阻：单位稳态传热速率下的温差，单位为K/W。
latent heat: Energy transferred during a phase change at its transition temperature; specific latent heat is per mass.
潜热：相变过程中在相变温度下传递的能量；比潜热是单位质量对应的热量。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆相关含义，然后改变原始传递过程中的一个参数，并检验极限情况。