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AQA GCSE Physics

Revision Notes
(Internal Energy & Energy Transfers)

Specific Heat Capacity

Specific Heat Capacity

Definition of Specific Heat Capacity

Specific heat capacity is the amount of energy required to raise the temperature of 1 kilogram of a substance by 16C (or 1 kelvin, since the size of the degree is the same).

It is a material property, meaning each substance has its own specific heat capacity value depending on its atomic or molecular structure.

The units of specific heat capacity are joules per kilogram per degree Celsius (J/kg6C).

Formula and Calculation

The energy transferred to or from a substance when its temperature changes can be calculated using the formula:

Q=m×c×ΔθQ = m \times c \times \Delta \theta

  • QQ is the energy transferred in joules (J)
  • mm is the mass of the substance in kilograms (kg)
  • cc is the specific heat capacity of the material in joules per kilogram per degree Celsius (J/kg6C)
  • Δθ\Delta \theta is the temperature change in degrees Celsius (6C), calculated as θfinalθinitial\theta_{final} - \theta_{initial}

This formula assumes no change of state (no melting or boiling) and that all the energy goes into changing the temperature.

For instance, if 2 kg of water is heated and its temperature rises by 106C, and the specific heat capacity of water is approximately 4200 J/kg6C, the energy transferred is:

Q=2×4200×10=84,000 JQ = 2 \times 4200 \times 10 = 84,000 \text{ J}

Factors Affecting Specific Heat Capacity

Specific heat capacity varies between different materials because it depends on the atomic or molecular structure of the substance. Some materials require more energy to raise their temperature than others.

It is important to note that specific heat capacity is:

  • Independent of the mass of the substance 6 doubling the mass doubles the energy needed, but the specific heat capacity remains the same.
  • Independent of the temperature change 6 the specific heat capacity remains constant over typical temperature ranges.

For example, metals generally have lower specific heat capacities than water, which is why they heat up and cool down faster.

Practical Investigation of Specific Heat Capacity

To find the specific heat capacity of a material experimentally, you can:

  • Measure the mass of the material sample (mm)
  • Supply a known amount of energy (QQ) to the sample, usually by electrical heating
  • Measure the temperature change (Δθ\Delta \theta) of the sample
  • Calculate the specific heat capacity using the formula c=Qm×Δθc = \frac{Q}{m \times \Delta \theta}

A typical setup might include an electric heater, a thermometer, an ammeter and voltmeter (to calculate electrical energy supplied), and an insulated container to reduce heat loss.

Sources of error include:

  • Heat loss to the surroundings, causing the measured energy to be less than the energy supplied
  • Inaccurate temperature readings due to thermometer placement or response time
  • Assuming all electrical energy goes into heating the sample, ignoring inefficiencies

Improving accuracy involves better insulation, stirring the material to ensure even temperature, and using precise instruments.

For example, if a 0.5 kg aluminium block is heated by an electric heater supplying 5000 J of energy and its temperature rises by 106C, the specific heat capacity is:

\( c = \frac{Q}{m \times \Delta \theta} = \frac{5000}{0.5 \times 10} = 1000 \text{ J/kg6C} \)

PracticeExample 2

Worked Example

Example: Calculate the energy needed to raise the temperature of 3 kg of iron by 156C. The specific heat capacity of iron is 450 J/kg6C.

PracticeExample 3

Worked Example

Example: A 1.2 kg block of copper is heated, and its temperature increases by 256C. If the specific heat capacity of copper is 390 J/kg6C, how much energy is transferred to the block?

PracticeExample 4

Worked Example

Example: In an experiment, 6000 J of energy is supplied to a 0.8 kg sample of a material, causing its temperature to rise by 206C. Calculate the specific heat capacity of the material.

  • Remember the formula Q=m×c×ΔθQ = m \times c \times \Delta \theta by thinking: Energy (Q) depends on how much mass (m) you have, what material it is (c), and how much you want to change the temperature (Δθ\Delta \theta).
  • Specific heat capacity is a fixed property for each material 6 it doesn't change with how much you have or the temperature change.
  • Use consistent units: mass in kilograms, temperature in 6C, energy in joules.

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