AQA A Level Physics復習筆記6.4.3 Specific Heat Capacity

Category: A-level課程, 教材筆記, 福利干貨 Date: 2022年7月11日下午4:38

Specific Heat Capacity

When a substance is heated, its temperature rises causing the particles within it to gain kinetic energy
- The amount of energy required to raise the temperature of a substance is given by its?specific heat capacity
The specific heat capacity of a substance is defined as:

The amount of thermal energy required to raise the temperature of 1 kg of a substance by 1 °C (or 1 K) without a change of state

This quantity determines the amount of energy needed to change the temperature of a substance
Specific heat capacity has the symbol?c?and is measured in units of?Joules per kilogram?per?Kelvin?(J kg–1?K–1) or?Joules per kilogram per Celsius?(J kg–1?°C–1)
- Different substances have different specific heat capacities
- Specific heat capacity is mainly used for liquids and solids

From the definition of specific heat capacity, it follows that:
- The heavier the material, the more thermal energy required to raise its temperature
- The larger the change in temperature, the higher the thermal energy required to achieve this change

Calculating Specific Heat Capacity

The amount of thermal energy?Q?needed to raise the temperature by Δθ for a mass?m?with specific heat capacity?c?is equal to:

ΔQ?=?mcΔθ

Where:
- ΔQ?= change in thermal energy (J)
- m?= mass of the substance you are heating up (kg)
- c?= specific heat capacity of the substance (J kg–1?K–1?or J kg–1?°C–1)
- Δθ = change in temperature (K or °C)

14.2-Specific-heat-examples

Low v high specific heat capacity

If a substance has a?low?specific heat capacity, it heats up and cools down quickly
If a substance has a?high?specific heat capacity, it heats up and cools down slowly
The specific heat capacity of different substances determines how useful they would be for a specific purpose eg. choosing the best material for kitchen appliances

Table of values of specific heat capacity for various substances

14.2.1-Table-of-values-of-specific-heat-capacity-for-various-substances

Good electrical conductors, such as copper and lead, are?excellent conductors?of heat due to their?low?specific heat capacity

Worked Example

A kettle is rated at 1.7 kW. A mass of 650 g of a liquid at 25 °C is poured into a kettle.When the kettle is switched on, it takes 3.5 minutes to start boiling.Calculate the specific heat capacity of the liquid.

Step 1:?Calculate the Energy from the power and time

Energy = Power × Time

Power = 1.7 kW = 1.7 × 103?W

Time = 3.5 minutes = 3.5 × 60 = 210 s

Energy = 1.7 × 103?× 210 = 3.57 × 105?J

Step 2: Thermal energy equation

ΔQ?=?mcΔθ

Step 3:?Rearrange for specific heat capacity

1.-Defining-Specific-Heat-Capacity-equation-1

Step 4: Substitute in values

m?= 650 g = 650 × 10–3?kg

Δθ?= 100 – 25 = 75oC

1.-Defining-Specific-Heat-Capacity-Worked-Example-equation-answer

Exam Tip

The difference in temperature Δθ will be exactly the same whether the temperature is given in Celsius or Kelvin. Therefore, there is no need to convert between the two since the?difference?in temperature will be the same for both units.

Continuous Flow

The specific heat capacity of a fluid can be found using a continuous-flow calorimeter
A fluid flows continuously over a heating element where energy is transferred to the fluid
- It is assumed that the heat transferred from the apparatus to the surroundings is constant
For this experiment, the flow rate and the potential difference is changed, keeping the change in temperature of the fluid constant

6.4.3-Continuous-Flow-Diagram

A continuous-flow calorimeter

A fluid flows through an electrical heating wire. The rise in temperature of the fluid is measured using the electric thermometers and is calculated by:

Δθ =?T2?–?T1

To find the mass of the fluid, the flow rate is recorded and multiplied by the time taken?t?to give the mass of the fluid that flows in as?m1
The current?I?and potential difference?V?are also recorded
The flow rate is then altered to give a mass m2?and the potential difference of the power supply is changed so the temperature difference, Δθ stays the same
The specific heat capacity is found by assuming the thermal losses to the surroundings are constant for both flow rates
For the first flow rate, the electrical energy supplied to the fluid in time?t1?is:

I1V1t1?=?Q1?=?m1cΔθ +?Elost