AQA A Level Physics

Revision Notes

6.4.3 Specific Heat Capacity

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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 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)

Specific heat examples, downloadable AS & A Level Physics revision notes

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

  • 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.

 

Answer:

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

c space equals space fraction numerator increment Q over denominator m increment theta end fraction

Step 4: Substitute in values

m = 650 g = 650 × 10–3 kg

Δθ = 100 – 25 = 75oC

 

c space equals space fraction numerator 3.57 space cross times space 10 to the power of 5 over denominator open parentheses 650 space cross times space 10 to the power of negative 3 end exponent close parentheses space cross times space 75 end fraction space equals space 7323.07 space equals space 7300 J kg−1 °C−1

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

Continuous Flow Diagram, downloadable AS & A Level Physics revision notes

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 = T2T1

  • 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

  • Where Elost is the thermal energy lost to the surroundings

 
  • The second flow rate is:

I2V2t2 = Q2 = m2cΔθ + Elost

  • Since Elost is assumed to be the same, subtracting the first flow rate equation from the second gives the equation:

I2V2t2I1V1t1 = Q2Q1 = (m2 – m1)cΔθ

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

Continuous Flow Equation

Worked example

Calculate the specific heat capacity of a liquid using the following data measured in two experiments using the continuous flow method:

  • Time of each experiment, t1t2 = 40 s
  • T1 in both experiments = 15 ºC
  • T2 in both experiments = 4 ºC
  • p.d across the heater in experiment 1, V1 = 14.0 V
  • p.d across the heater in experiment 2, V2 = 9.0 V
  • Current through heater in both experiments = 3.0 A
  • Mass of water flowing in experiment 1, m1 = 136.0 g
  • Mass of water flowing in experiment 2, m2 = 73.0 g

Answer:

Step 1: Calculate the change in temperature, Δθ

Δθ = 4 – 15 = −11ºC

Step 2: Calculate Q2

Q2 = I2V2t2 = 3.0 × 9.0 × 40 = 1080 J

Step 3: Calculate Q1

Q1 = I1V1t1 = 3.0 × 14.0 × 40 = 1680 J

Step 4: Substitute values into the specific heat capacity equation

 

c space equals space fraction numerator Q subscript 2 space minus space Q subscript 1 over denominator open parentheses m subscript 2 space minus space m subscript 1 close parentheses increment theta end fraction

 

c space equals space fraction numerator 1080 space minus space 1680 over denominator open parentheses open parentheses 73 space cross times space 10 to the power of negative 3 end exponent close parentheses space minus space open parentheses 136 space cross times space 10 to the power of negative 3 end exponent close parentheses close parentheses space cross times space minus 11 end fraction space equals space 866 J kg−1 °C−1

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