Specific Heat Capacity (AQA A Level Physics): Revision Note

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 $∆\theta$ for a mass $m$ with specific heat capacity $c$ is equal to:

$∆Q=mc∆\theta$

• Where:

  • $∆Q$ = 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)

  • $∆\theta$ = change in temperature (K or °C)

• 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

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

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 are changed to ensure the change in temperature of the fluid is kept constant

• 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:

$∆\theta =T_2-T_1$

• To find the mass of the fluid, the flow rate (mass leaving the right hand side per second) is recorded and multiplied by the time taken $t$ to give the mass of the fluid that flows in as $m_1$

• The current $I$ and potential difference $V$ are also recorded

• The flow rate is then altered to give a mass $m_2$ and the potential difference of the power supply is changed so the temperature difference, $∆\theta$ 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:

$I_1{V}_1{t}_1=Q_1=m_{1}c\Delta \theta +E_{lost}$

• Where $E_{lost}$ is the thermal energy lost to the surroundings and $I_1{V}_1$ represents the electrical power of the heater

• The second flow rate is:

$I_2{V}_2{t}_2=Q_2=m_{2}c\Delta \theta +E_{lost}$

• Since $E_{lost}$ is assumed to be the same, subtracting the first flow rate equation from the second gives the equation:

$I_2{V}_2{t}_2-I_1{V}_1{t}_1=Q_2-Q_1=\left(m_2-m_1\right)c\Delta \theta$

• Rearranging this for the specific heat capacity of the fluid, $c$ gives the final equation:

$c=\frac{Q_2-Q_1}{\left(m_2-m_1\right)∆\theta }$