Investigating Specific Heat Capacity | 探究比热容

📚 Investigating Specific Heat Capacity | 探究比热容

Specific heat capacity is a fundamental thermal property of materials, quantifying how much energy is needed to change the temperature of a substance. In IGCSE Edexcel Science (Physics), practical 5.14 requires students to investigate the specific heat capacity of substances such as water and metals using an electrical method. This article explains the theory, procedure, calculations, sources of error, and exam tips to help you master this core practical.

比热容是物质的基本热学性质，它量化了改变物质温度所需的能量。在 IGCSE Edexcel 科学（物理）中，实践 5.14 要求学生使用电热法探究水或金属等物质的比热容。本文将解释理论、步骤、计算、误差来源和考试技巧，帮助你掌握这项核心实验。

1. What is Specific Heat Capacity? | 什么是比热容？

Specific heat capacity (c) is defined as the amount of energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius (1 °C) without changing its state.

比热容 (c) 的定义是：在不改变物质状态的情况下，将 1 千克物质的温度升高 1 摄氏度 (1 °C) 所需的能量。

Materials with a high specific heat capacity, such as water (4200 J/(kg °C)), can absorb a lot of heat with only a small temperature rise. This makes water an excellent coolant and thermal store. Metals have much lower values; for example, aluminium is around 900 J/(kg °C) and copper is about 385 J/(kg °C).

比热容高的物质（如水，4200 J/(kg °C)）能吸收大量热量而温升很小，这使得水成为一种优秀的冷却剂和储热体。金属的比热容则低得多，例如铝约为 900 J/(kg °C)，铜约为 385 J/(kg °C)。

2. The Specific Heat Capacity Equation | 比热容公式

The energy transferred to or from an object as heat is given by:

传递给物体的热量或从物体传出的热量由下式给出：

E = m × c × Δθ

where E is the thermal energy transferred (J), m is the mass (kg), c is the specific heat capacity (J/(kg °C)), and Δθ is the change in temperature (°C). Rearranging gives:

其中 E 是传递的热能（焦耳），m 是质量（千克），c 是比热容 (J/(kg °C))，Δθ 是温度变化 (°C)。移项可得：

c = E / (m × Δθ)

When using an electrical heater, the energy E is calculated from the electrical power and time: E = P × t, and P = I × V, so E = I × V × t, where I is current (A), V is voltage (V), and t is time (s).

使用电加热器时，能量 E 由电功率和时间计算：E = P × t，而 P = I × V，因此 E = I × V × t，其中 I 为电流 (A)，V 为电压 (V)，t 为时间 (s)。

3. Principle of the Electrical Method | 电热法原理

In this practical, a known mass of a substance (solid or liquid) is heated by an immersion heater. Assuming all electrical energy is transferred as heat to the substance, we can equate the electrical energy supplied to the thermal energy gained:

在这个实验中，已知质量的物质（固体或液体）由浸没式加热器加热。假设所有电能都转化为热能传递给物质，我们可以将供给的电能与获得的热能等价：

I × V × t = m × c × Δθ

By measuring I, V, t, m, and Δθ, we can calculate c. This assumes no heat loss to the surroundings, which is never fully achieved in a school laboratory, so the result is only an estimate.

通过测量 I、V、t、m 和 Δθ，我们可以计算出 c。该做法的前提是没有热量散失到环境中，而学校实验室里永远无法完全避免热损失，因此得到的结果只是一个估计值。

4. Apparatus Required | 所需器材

The typical apparatus for a solid (e.g. aluminium block) is:

用于固体（如铝块）的典型器材如下：

A metal block (usually aluminium) with two holes: one for the heater, one for the thermometer.

一个金属块（通常为铝），上面有两个孔：一个放加热器，一个放温度计。
12 V immersion heater of known power, or a heater connected to an ammeter and voltmeter.

12 V 浸没式加热器（已知功率），或连接有安培计和伏特计的加热器。
Ammeter, voltmeter, and a power supply (or alternatively a joulemeter).

安培计、伏特计和电源（或直接用焦耳计代替）。
Thermometer (digital or liquid-in-glass with 0.5 °C precision).

温度计（数字式或精度为 0.5 °C 的液体温度计）。
Stopwatch or timer.

秒表或计时器。
Insulating material (polystyrene jacket or cotton wool) to reduce heat loss.

隔热材料（聚苯乙烯外壳或棉花），以减少热量散失。
Electronic balance to measure mass.

电子天平，用于测量质量。

For water, a copper or aluminium calorimeter with a lid and an insulated jacket is used, along with a stirrer to ensure even temperature distribution.

对于水，需使用带有盖子和隔热套的铜或铝量热器，并配合搅拌器以确保温度均匀分布。

5. Experimental Procedure | 实验步骤

1. Measure and record the mass of the aluminium block (or calorimeter with water) using the electronic balance.

1. 用电子天平测量并记录铝块（或盛水量热器）的质量。

2. Insert the heater and thermometer securely into the holes of the block. Ensure the heater is fully in contact with the metal.

2. 将加热器和温度计牢固地插入金属块的孔中。确保加热器与金属完全接触。

3. Wrap the block with insulating material to minimise heat loss.

3. 用隔热材料包裹金属块，以尽量减少热量散失。

4. Record the initial temperature θ₁. Wait until the reading is stable.

4. 记录初始温度 θ₁。等待读数稳定。

5. Set up the circuit: connect the heater, ammeter in series, and voltmeter in parallel across the heater. Alternatively, use a joulemeter in place of ammeter and voltmeter.

5. 连接电路：将加热器、安培计串联，伏特计并联在加热器两端。或使用焦耳计代替安培计和伏特计。

6. Switch on the power supply and simultaneously start the stopwatch. Adjust the voltage to maintain a steady current, e.g. 2.0 A.

6. 接通电源并同时启动秒表。调节电压以维持恒定电流，如 2.0 A。

7. Record the current I and voltage V at regular intervals to ensure they remain constant; take average values if they fluctuate slightly.

7. 每隔一定时间记录电流 I 和电压 V，确保它们保持恒定；若有轻微波动，取平均值。

8. Stir the water continuously if using a calorimeter; for a solid block, gently stir the thermometer (if possible) to ensure even temperature distribution.

8. 若使用量热器，需持续搅拌水；对于固体金属块，可轻转温度计（若可行）以确保温度均匀。

9. After a set time (e.g. 10 minutes), switch off the heater and record the highest temperature reached θ₂. Note the exact heating time t.

9. 达到设定时间后（如 10 分钟），关闭加热器并记录达到的最高温度 θ₂。准确记录加热时间 t。

10. Calculate Δθ = θ₂ – θ₁. Repeat the experiment at least twice to check reproducibility.

10. 计算 Δθ = θ₂ – θ₁。至少重复实验两次以检验可重复性。

6. Data Recording and Sample Calculation | 数据记录与计算示例

Suppose we obtained the following data for an aluminium block:

假设我们获得了以下铝块实验数据：

Quantity / 量	Value / 数值
Mass of block, m	1.00 kg
Initial temp., θ₁	20.0 °C
Final temp., θ₂	28.0 °C
Δθ	8.0 °C
Current, I	2.00 A
Voltage, V	10.0 V
Time, t	300 s (5 min)

Electrical energy supplied: E = I × V × t = 2.00 A × 10.0 V × 300 s = 6000 J.

供给的电能：E = I × V × t = 2.00 A × 10.0 V × 300 s = 6000 J。

Calculated specific heat capacity: c = E / (m × Δθ) = 6000 J / (1.00 kg × 8.0 °C) = 750 J/(kg °C).

计算得到的比热容：c = E / (m × Δθ) = 6000 J / (1.00 kg × 8.0 °C) = 750 J/(kg °C)。

Accepted value for aluminium is about 900 J/(kg °C). The experimental result is lower, primarily due to heat losses to the surroundings and the heater itself.

铝的公认值约为 900 J/(kg °C)。实验结果偏低，主要是由于向环境和加热器本身的热量散失。

7. Sources of Error and Improvements | 误差来源与改进

Heat loss to surroundings: Even with insulation, some energy escapes. This causes the measured temperature rise to be smaller than expected, leading to an overestimate of c. Improvement: use thicker insulation, place a lid on top, and minimise gaps.

向环境散热： 即使有隔热措施，仍会有部分能量散失。这导致测得的温升比预期小，从而高估比热容。改进： 使用更厚的隔热材料、加盖并减少缝隙。

Heater thermal capacity: The heater itself absorbs some energy, which is not transferred to the block. Improvement: use a low-heat-capacity heater or perform a correction by measuring the heater’s own temperature rise.

加热器自身热容量： 加热器本身会吸收一部分能量，并未传递给金属块。改进： 使用低热容加热器，或通过测量加热器自身温升进行修正。

Uneven temperature distribution: In a solid, poor contact between heater/thermometer and the block can cause local hot spots. Improvement: use thermal paste in the holes to improve conduction, and wait for the temperature to stabilise after switching off before recording θ₂.

温度分布不均： 固体中加热器/温度计与金属块接触不良会导致局部热点。改进： 在孔内使用导热膏以改善传导，并在断电后待温度稳定再记录 θ₂。

Inaccurate timing or electrical measurements: Small errors in t, I, V affect E directly. Improvement: use a joulemeter to measure energy directly; calibrate analogue meters and take multiple readings.

计时或电学测量不准： t、I、V 的微小误差直接影响 E。改进： 使用焦耳计直接测量能量；标定指针式仪表并多次读数。

Mass of water evaporated (for liquid): If water is heated for too long, some may evaporate, changing m. Improvement: use a lid and keep the final temperature below 60 °C to minimise evaporation.

水蒸发（液体实验）： 若加热时间过长，部分水蒸发会使质量改变。改进： 使用盖子，并将最终温度控制在 60 °C 以下以减少蒸发。

8. Exam Tips and Common Mistakes | 考试技巧与常见错误

Confusion between heat capacity and specific heat capacity: Heat capacity (C = m × c) is the energy needed to raise the temperature of an object by 1 °C, whereas specific heat capacity is per unit mass. Always check units.

混淆热容量与比热容： 热容量 (C = m × c) 是将一个物体升温 1 °C 所需的能量，而比热容是单位质量的。务必检查单位。

Units of time: In E = I × V × t, t must be in seconds, not minutes. Students often forget to convert, leading to an answer that is a factor of 60 out.

时间单位： 在 E = I × V × t 中，t 必须以秒为单位，而非分钟。考生常忘记换算，导致答案差 60 倍。

Using the same relation for cooling: The same equation E = m × c × Δθ applies when a substance cools and loses energy, but the energy transferred is negative or we consider magnitude.

冷却时用同一关系： 在物质冷却并失去能量时，同样适用 E = m × c × Δθ，但能量传递为负值，或我们仅考虑大小。

Assumption of no heat loss: In exam questions, if asked to explain why the experimental c differs from the accepted value, state that not all electrical energy goes into heating the block; some is lost to the surroundings and absorbed by the heater/apparatus.

无热损失假设： 考试中若问为什么实验 c 与公认值不同，回答并非所有电能都用于加热金属块；一部分散失到环境中或被加热器/器材吸收。

Formula rearrangement: Be able to rearrange E = m × c × Δθ to find any unknown, including mass or temperature change. Practice using correct units: energy (J), mass (kg), Δθ (°C or K).

公式变形： 要能够对 E = m × c × Δθ 进行变形，求出任何未知量，包括质量或温度变化。练习使用正确单位：能量 (J)，质量 (kg)，Δθ (°C 或 K)。

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Investigating Specific Heat Capacity | 探究比热容