IB Physics: Three Modes of Thermal Energy Transfer and Their Calculations | IB物理：热能传递的三种方式与计算方法

📚 IB Physics: Three Modes of Thermal Energy Transfer and Their Calculations | IB物理：热能传递的三种方式与计算方法

In IB Physics, thermal energy transfer is a core topic that connects mechanics, thermodynamics, and real-world applications. Understanding how heat moves through conduction, convection, and radiation is essential not only for exams but also for explaining phenomena from boiling water to the greenhouse effect. This article provides a systematic guide to the three modes of transfer, their governing equations, and worked examples aligned with the IB syllabus.

在IB物理中，热能传递是连接力学、热力学与现实应用的核心主题。理解热量如何通过传导、对流和辐射进行传递，不仅对考试至关重要，也能帮助你解释从烧开水到温室效应等各种现象。本文系统讲解三种传递方式、它们所遵循的方程，以及紧扣IB考纲的例题。

1. Conduction: Energy Transfer Through Matter | 传导：通过物质的能量传递

Conduction is the transfer of thermal energy between particles within a substance due to a temperature gradient. In solids, this occurs through lattice vibrations and, in metals, through the movement of free electrons. In IB Physics, you must understand the microscopic mechanism and be able to apply the steady-state heat conduction equation.

传导是由于温度梯度，物质内部粒子之间发生的热能传递。在固体中，这种传递通过晶格振动实现；在金属中，则主要依靠自由电子的移动。IB物理要求你理解微观机制，并能熟练运用稳态热传导方程。

The rate of heat transfer by conduction is given by Fourier’s Law. For a uniform rod of cross-sectional area A, length L, and thermal conductivity k, with a temperature difference ΔT between its ends:

传导的热传递速率由傅里叶定律给出。对于横截面积为A、长度为L、导热系数为k的均匀棒，两端温差为ΔT时：

P = kA(ΔT)/L

Where P is the power (heat transfer rate) in watts, k is in W·m⁻¹·K⁻¹, A in m², ΔT in K, and L in m. Note that in IB, temperature differences in kelvin and degrees Celsius can be used interchangeably because the scale interval is identical.

其中P是热传递功率（热流速率），单位为瓦特；k的单位为W·m⁻¹·K⁻¹；A的单位为m²；ΔT的单位为K；L的单位为m。注意在IB中，因为温差间隔相同，开尔文和摄氏度在计算温差时可互换使用。

Example 1 | 例题1

A copper rod has length 0.50 m and cross-sectional area 2.0 × 10⁻⁴ m². The thermal conductivity of copper is 385 W·m⁻¹·K⁻¹. If one end is at 100 °C and the other at 20 °C, calculate the rate of heat transfer.

一根铜棒长0.50 m，横截面积为2.0 × 10⁻⁴ m²，铜的导热系数为385 W·m⁻¹·K⁻¹。一端温度为100 °C，另一端为20 °C，计算热传递速率。

P = (385)(2.0 × 10⁻⁴)(100 − 20) / 0.50 = 12.3 W

Always check that the units are consistent: area in m², length in m, and temperature difference in K (or °C).

做题时务必检查单位一致性：面积要用m²，长度用m，温差用K（或°C）。

2. Convection: Energy Transfer by Fluid Motion | 对流：流体运动引起的能量传递

Convection involves the transfer of thermal energy by the bulk movement of a fluid (liquid or gas). It occurs when warmer, less dense regions of the fluid rise, while cooler, denser regions sink, creating convection currents. In IB Physics, you are expected to describe natural and forced convection and understand why convection does not occur in solids.

对流是通过流体（液体或气体）的整体运动来传递热能。当流体中较热且密度较小的区域上升，而较冷且密度较大的区域下沉时，就会形成对流。IB物理要求你能够描述自然对流和强制对流，并理解为什么固体中不会发生对流。

Natural convection is driven by buoyancy due to density differences caused by thermal expansion. Forced convection occurs when an external agent, such as a pump or fan, moves the fluid. The rate of convective heat transfer is often modeled using Newton’s Law of Cooling for a surface at temperature T_s surrounded by a fluid at temperature T_∞:

自然对流由热膨胀引起的密度差异所导致的浮力驱动。强制对流则是由外部装置（如泵或风扇）推动流体运动。表面对流换热速率常用牛顿冷却定律建模，表面温度为T_s，周围流体温度为T_∞：

P = hA(T_s − T_∞)

Here, h is the convective heat transfer coefficient in W·m⁻²·K⁻¹, and A is the surface area. Note that this equation is only an approximation; in IB exams, you may be asked to describe factors affecting h, such as fluid viscosity, flow speed, and surface geometry.

其中，h是对流换热系数，单位为W·m⁻²·K⁻¹；A是表面积。请注意该方程只是近似模型；IB考试可能要求你描述影响h的因素，如流体黏度、流速和表面几何形状等。

Example 2 | 例题2

A hot water tank has a surface area of 1.5 m² and is at 60 °C. The surrounding air is at 20 °C. The convective heat transfer coefficient is 10 W·m⁻²·K⁻¹. Find the rate of heat loss by convection.

一个热水箱表面积为1.5 m²，表面温度为60 °C，周围空气温度为20 °C，对流换热系数为10 W·m⁻²·K⁻¹。求对流散热速率。

P = (10)(1.5)(60 − 20) = 600 W

This simple calculation shows how a 40 K temperature difference across a moderate area can drive substantial energy loss.

这个简单计算表明，40 K的温差在中等面积上就能引起可观的热量损失。

3. Radiation: Energy Transfer by Electromagnetic Waves | 辐射：通过电磁波传递能量

Thermal radiation is the emission of electromagnetic waves (mainly infrared) from the surface of an object due to its temperature. Unlike conduction and convection, radiation does not require a medium and can travel through a vacuum. All objects emit radiation, with the total power depending on their surface temperature and emissivity.

热辐射是物体因其温度而从表面发射电磁波（主要是红外线）的现象。与传导和对流不同，辐射不需要介质，可以在真空中传播。所有物体都会发出辐射，总功率取决于其表面温度和发射率。

The Stefan-Boltzmann Law gives the total power radiated by a black body of surface area A and absolute temperature T:

斯特藩-玻尔兹曼定律给出了表面积为A、绝对温度为T的黑体辐射总功率：

P = εσAT⁴

where ε is the emissivity (0 ≤ ε ≤ 1), and σ is the Stefan-Boltzmann constant, σ = 5.67 × 10⁻⁸ W·m⁻²·K⁻⁴. For a perfect black body, ε = 1. In IB problems, you often need to calculate the net radiation loss by considering both emission and absorption of radiation from the surroundings.

其中ε是发射率（0 ≤ ε ≤ 1），σ是斯特藩-玻尔兹曼常数，σ = 5.67 × 10⁻⁸ W·m⁻²·K⁻⁴。完美黑体时ε = 1。在IB题目中，常需要同时考虑物体向外辐射和吸收周围环境辐射，从而计算净辐射损失。

P_net = εσA(T⁴ − T_surr⁴)

Note that temperatures here must be in kelvin, because the fourth-power dependence is nonlinear.

注意，此处温度必须使用开尔文，因为四次方关系是非线性的。

Example 3 | 例题3

A sphere has radius 0.10 m and emissivity 0.80. Its surface temperature is 500 K, and the surroundings are at 300 K. Calculate the net radiative power loss.

一个球体半径为0.10 m，发射率为0.80，表面温度为500 K，周围环境温度为300 K。计算净辐射散热功率。

First, find the surface area of the sphere: A = 4πr² = 4π(0.10)² = 0.1257 m².

首先求球的表面积：A = 4πr² = 4π(0.10)² = 0.1257 m²。

P_net = (0.80)(5.67 × 10⁻⁸)(0.1257)(500⁴ − 300⁴) ≈ 61.5 W

This example highlights that radiation becomes increasingly significant at high temperatures because of the T⁴ relationship.

该例说明，由于T⁴关系，辐射在高温下会变得尤为重要。

4. Comparison of the Three Modes | 三种方式的比较

In IB Physics, you must be able to compare conduction, convection, and radiation in terms of medium requirement, mechanism, governing equation, and typical examples. The table below summarises the key points.

在IB物理中，你需要能够从介质需求、机制、控制方程和典型例子等方面比较传导、对流和辐射。下表总结了关键要点。

Property 属性	Conduction 传导	Convection 对流	Radiation 辐射
Medium required 是否需要介质	Yes, usually solid 需要，通常是固体	Yes, fluid only 需要，仅限流体	No, can travel in vacuum 不需要，可在真空中传播
Mechanism 微观机制	Particle collisions and free electrons 粒子碰撞与自由电子	Bulk fluid motion due to density differences 密度差引起的流体整体运动	Electromagnetic wave emission 电磁波发射
Key equation 主要方程	P = kAΔT/L	P = hAΔT	P = εσA(T⁴ − T_surr⁴)
Occurs in solids? 能否在固体中发生	Yes 能	No 不能	Yes 能
Examples 典型例子	Metal spoon in hot soup 热汤中的金属勺	Sea breeze, hot air rising 海陆风、热空气上升	Sun’s heat reaching Earth 太阳热量到达地球

You should remember that in many real situations, all three modes operate simultaneously. For example, a thermos flask minimises conduction and convection using a vacuum, and minimises radiation using a reflective silver coating.

你应该记住，在许多真实情境中，三种方式同时发生。例如，保温瓶利用真空减少传导和对流，利用反射银涂层减少辐射。

5. Energy Balance and Combined Heat Transfer | 能量平衡与组合热传递

In IB problems, you may be asked to determine the net rate of energy loss or gain of an object considering multiple mechanisms. The principle is simple: the net rate of internal energy change equals the sum of all heat transfer rates into the object minus all rates out of the object.

在IB题目中，你可能需要综合考虑多种机制来确定物体的净能量损失或增益速率。基本原则很简单：内能变化速率等于进入物体的所有热传递速率之和减去流出物体的所有速率之和。

ΔU/Δt = P_in − P_out

For example, an electric heater inside a room provides thermal energy to the air; this energy is then transferred to the walls by convection and radiation, and through the walls by conduction. In steady state, the heater power equals the total heat loss through the walls.

例如，室内电加热器向空气提供热能；这些能量通过对流和辐射传给墙壁，再通过墙壁传导出去。在稳态下，加热器功率等于墙壁的总热损失功率。

Example 4 | 例题4

A metal plate at 400 K, with emissivity 0.90 and surface area 0.020 m², loses energy by radiation to surroundings at 300 K. At the same time, it receives energy from a 50 W electrical heater. What is the net rate of internal energy change of the plate?

一块金属板温度为400 K，发射率为0.90，表面积为0.020 m²，向300 K的环境辐射散热。同时，它从功率为50 W的电加热器接收能量。求金属板内能的净变化速率。

First, find the radiative loss rate:

先求辐射散热速率：

P_rad = (0.90)(5.67 × 10⁻⁸)(0.020)(400⁴ − 300⁴) ≈ 4.46 W

Then apply energy balance:

然后应用能量平衡：

ΔU/Δt = 50 − 4.46 ≈ 45.5 W

The plate’s internal energy increases at 45.5 W, causing its temperature to rise until equilibrium is reached.

金属板内能以45.5 W的速度增加，导致其温度不断上升，直到达到平衡。

6. Applications and Real-World Contexts | 应用与真实情境

IB Physics exams often use real-life contexts to assess thermal transfer. You should be familiar with the following applications:

IB物理考试常用生活情境来考查热传递。你需要熟悉以下应用：

Thermos flask (Dewar flask): Vacuum prevents conduction and convection; silvered surfaces reduce radiation; low-conductivity stopper minimises heat loss. 保温瓶：真空防止传导和对流；镀银表面减少辐射；低导热塞子减小热损失。
Greenhouse effect: Short-wavelength solar radiation passes through glass, warms the interior, and re-emitted long-wavelength infrared radiation is partially trapped. 温室效应：短波太阳辐射穿过玻璃加热内部，重新发射的长波红外辐射被部分困住。
Cooling fins: Large surface area increases convective and radiative heat loss. 散热片：增大表面积以增加对流和辐射散热。
Animal adaptations: Blubber has low thermal conductivity, trapping heat; large ears increase surface area for heat dissipation. 动物适应：鲸脂导热系数低，能保温；大耳朵增大表面积以散热。

These contexts require you to identify which mode dominates and to justify your reasoning using physics principles, not just memorised facts.

这些情境要求你判断哪种方式起主导作用，并用物理原理说明理由，而不仅仅是记忆事实。

7. Common Mistakes and Exam Tips | 常见错误与考试技巧

Many IB students lose marks on thermal transfer questions due to easily avoidable mistakes. Here are the most common pitfalls and how to avoid them:

许多IB学生在热传递题目中因为一些可以避免的错误而丢分。以下是最常见的陷阱及避免方法：

Using °C in radiation equations: The Stefan-Boltzmann law requires kelvin. Always convert to kelvin first. 在辐射方程中直接使用°C：斯特藩-玻尔兹曼定律要求使用开尔文，务必先转换。
Forgetting emissivity: For a non-black body, multiply by ε. 忘记发射率：对于非黑体，需要乘以ε。
Ignoring the surrounding radiation: For net radiation loss, use T⁴ − T_surr⁴, not just T⁴. 忽略环境辐射：计算净辐射损失时，使用T⁴ − T_surr⁴，而不是单独的T⁴。
Unit conversion errors: Ensure area is in m², length in m, and pressure (if included in thermal problems) in Pa. 单位换算错误：确保面积用m²，长度用m。
Confusing conduction and convection: Conduction occurs in solids with no bulk movement; convection requires fluid flow. 混淆传导和对流：传导发生在固体中，没有整体运动；对流需要流体流动。

Remember to always define symbols and show your substitution step clearly. In calculation questions, write the equation first, then substitute values, then give the final answer with units.

记住：在计算题中，先写出方程，再代入数值，最后给出带单位的答案。要明确写出符号的含义。

8. Worked Exam-Style Question | 典型考试风格例题

Let’s combine everything into a single IB-style question.

让我们把所有内容整合到一道IB风格的题目中。

A spherical water tank of radius 0.40 m has a surface temperature of 50 °C. The surrounding air temperature is 10 °C. The tank surface has an emissivity of 0.70. The convective heat transfer coefficient is 8.0 W·m⁻²·K⁻¹.

一个球形水箱半径为0.40 m，表面温度为50 °C，周围空气温度为10 °C，表面发射率为0.70，对流换热系数为8.0 W·m⁻²·K⁻¹。

(a) Calculate the rate of heat loss by convection.

（a）计算对流散热速率。

The surface area of the sphere: A = 4π(0.40)² = 2.011 m². Temperature difference = 50 − 10 = 40 K.

球的表面积：A = 4π(0.40)² = 2.011 m²。温差 = 50 − 10 = 40 K。

P_conv = (8.0)(2.011)(40) ≈ 643 W

(b) Calculate the rate of heat loss by radiation. Convert temperatures to kelvin: T_s = 323 K, T_surr = 283 K.

（b）计算辐射散热速率。将温度转换为开尔文：T_s = 323 K，T_surr = 283 K。

P_rad = (0.70)(5.67 × 10⁻⁸)(2.011)(323⁴ − 283⁴) ≈ 392 W

(c) If an electric heater of power 800 W is placed inside the tank, is the tank heating up or cooling down? Find the net rate of internal energy change.

（c）若在箱内放入一个800 W的电加热器，水箱是在升温还是降温？求内能净变化速率。

Total heat loss = 643 + 392 = 1035 W. Since the heater supplies only 800 W, the tank is losing energy and cooling down. Net rate: 800 − 1035 = −235 W, i.e., energy decreases at 235 W.

总散热 = 643 + 392 = 1035 W。加热器只提供800 W，因此水箱在失热降温。净速率：800 − 1035 = −235 W，即内能以235 W的速率减少。

ΔU/Δt = −235 W

This question demonstrates how conduction, convection, and radiation concepts are combined in typical IB Paper 2 problems.

这道题展示了在IB Paper 2典型问题中如何综合运用传导、对流和辐射概念。

9. Quick Formula Summary | 公式速查表

For revision, keep these equations ready in your formula booklet. The table below lists the essential expressions.

复习时，请确保公式册中这些方程已了然于心。下表列出了核心表达式。

Mode 方式	Equation 方程	Key quantities 关键量
Conduction 传导	P = kAΔT/L	k = thermal conductivity (W·m⁻¹·K⁻¹)
Convection 对流	P = hAΔT	h = convection coefficient (W·m⁻²·K⁻¹)
Radiation 辐射	P = εσA(T⁴ − T_surr⁴)	σ = 5.67 × 10⁻⁸ W·m⁻²·K⁻⁴

When using these equations, pay close attention to the direction of heat flow: positive P means heat is leaving the object if T is higher than surroundings. In energy balance problems, assign signs consistently.

使用这些方程时，请特别注意热流方向：如果物体温度高于环境，P为正值表示热量离开物体。在能量平衡问题中，要注意符号的一致性。

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IB Physics: Three Modes of Thermal Energy Transfer and Their Calculations | IB物理：热能传递的三种方式与计算方法