Transferts d'énergie thermique
Baccalauréat international · IB Diploma · Physics · SL Sujet 4 4:08 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
Two equal masses receive the same energy but show different temperature rises.
两个等质量的物体吸收相同能量，但温度升高不同。
Material properties determine how energy transfer changes temperature.
材料性质决定了能量传递如何引起温度变化。
Specific heat capacity is the energy needed to raise the temperature of unit mass by one degree.
比热容是使单位质量的物质温度升高一度所需的能量。
Specific latent heat relates energy to a change of state without temperature change, for the idealized process.
比潜热把能量与物态变化联系起来，在理想过程中温度不变。
Specific heat capacity is energy per mass per temperature rise.
比热容是单位质量、单位温升所需的能量。
Latent heat is energy associated with a change of state.
潜热是与物态变化相关的能量。
Temperature relates to particle motion in a model. Internal energy includes kinetic and potential contributions.
温度与模型中的粒子运动相关；内能包括动能和势能的贡献。
During a change of state, energy can change particle arrangements rather than temperature.
在物态变化过程中，能量可能改变粒子的排列方式，而非温度。
In the lab: Measure mass, electrical input and temperature change for an insulated block.
在实验室中，测量绝热块的质量、电输入量和温度变化。
Ensure the temperature sensor has good contact, allow time for equilibration, and consider energy transferred to the surroundings.
确保温度传感器接触良好，给予足够时间使其达到平衡，并考虑向环境散失的能量。
Known: a 0.50 kg block gains 2,000 J and rises 10 degrees Celsius.
已知：一个质量为 0.50 kg 的物块吸收 2,000 J 的热量，温度升高 10 °C。
Use E equals m c delta T.
使用公式 E = mcΔT。
Rearranging gives c equals E divided by the product of m and delta T.
整理得 c 等于 E 除以 m 与 ΔT 的乘积。
So c equals 2,000 divided by the product of 0.50 and 10, which equals 400 joules per kilogram per degree.
所以 c 等于 2,000 除以 0.50 与 10 的乘积，即 400 J/(kg·°C)。
Heat loss would make the value inferred from electrical input too large.
若存在热损失，则根据电输入计算出的比热容值会偏大。
Your turn.
轮到你了。
A 2 kilogram sample with c equal to 500 joules per kilogram per degree warms by 3 degrees Celsius.
一个质量为 2 千克的样品，比热容 c 为每千克每摄氏度 500 焦耳，温度升高了 3 摄氏度。
Find the energy gained.
求其吸收的能量。
A flat section of a heating curve can show a phase change, not absence of energy transfer.
加热曲线的平坦段可能表示相变，而非能量传递的缺失。
Do not substitute temperature for a temperature difference in E equals m c delta T.
在公式 E = mcΔT 中，请勿用温度替代温差。
Not true: A temperature plateau always means zero energy transfer.
这种说法不正确。 温度平台并不总是意味着能量转移为零。
A plateau on a heating curve marks a phase change, where energy goes into changing particle arrangement as latent heat.
加热曲线上的平台标志着相变，此时能量以潜热形式用于改变粒子排列。
Energy transfer continues; only the temperature pauses.
能量转移仍在继续；只是温度暂时保持不变。
Specific heat capacity is energy needed to raise the temperature of unit mass by one degree.
比热容是指使单位质量的物质升高一度所需的热量。
Temperature relates to particle motion in a model; internal energy includes kinetic and potential contributions.
在模型中，温度与粒子的运动有关；内能包括动能和势能的贡献。
Watch for: A flat section of a heating curve can show a phase change, not absence of energy transfer.
请注意：加热曲线中的平坦段可能表示相变，而非没有能量传递。
A 0.40 kg sample with specific heat capacity 2000 J per kilogram per degree warms by 5 degrees.
一个质量为 0.40 kg、比热容为 2000 J/(kg·°C) 的样品温度升高了 5°C。
Find the energy gain.
求其吸收的能量。
A temperature plateau can still transfer energy.
即使处于温度平台期，系统仍可传递能量。
The result is an energy gain of 4000 joules.
结果是能量增加4000焦耳。
The energy needed rises in proportion to the temperature rise: 1,000 joules for every 5 degrees.
所需能量与温度升高成正比：每升高5度需要1000焦耳。
The gradient, 200 joules per degree, is the mass times the specific heat capacity.
斜率每度200焦耳，等于质量乘以比热容。
A fictional 0.50 kg block receives 2,000 J electrical energy and warms 8 K; 200 J heats its container and 200 J escapes.
一个虚构的零点五千克物块接收了两千焦耳的电能量，温度升高了八开尔文。 其中两百焦耳用于加热容器，另有两百焦耳散失。
Find c and compare using all input.
求比热容c，并利用全部输入能量进行比较。
The block's energy is 1,600 J, so c is 1,600 divided by the product of 0.50 and 8, which is 400 joules per kilogram per kelvin.
物块获得的能量为 1600 J，所以 c 等于 1600 除以 0.50 与 8 的乘积，即 400 J/(kg·K)。
The uncorrected 500 joules per kilogram per kelvin is too high.
未校正的 500 J/(kg·K) 偏高。
State the system boundary and loss assumptions.
要说明系统边界和热损失假设。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。