Specific heat capacity

Jonny Nelson explains specific heat capacity with a GCSE Physics practical experiment

If heat is absorbed by a one kilogram block of lead, the particles gain energy. Since lead is a solid and the particles are only vibrating, they vibrate faster after being heated. As the particles are closer together in a solid, they are more likely to hit each other and pass the energy around.

This means that the energy spreads through the block quickly and the temperature of the block goes up quickly. It takes less energy to raise the temperature of a one kg block of lead by 1°C than it does to raise the temperature of one kg of water by 1°C.

From this it can be seen that a change in temperature of a system depends on:

  • the mass of the material
  • the substance of the material
  • the amount of energy put into the system
curriculum-key-fact
The specific heat capacity of a material is the energy required to raise one kilogram (kg) of the material by one degree Celsius (°C).

The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of one kg of water by 1 °C.

Some other examples of specific heat capacities are:

MaterialSpecific heat capacity (J/kg/°C)
Brick840
Copper385
Lead129

Because it has a low specific heat capacity, lead will warm up and cool down quickly as it doesn't take much energy to change its temperature.

Brick will take much longer to heat up and cool down, its specific heat capacity is higher than that of lead so more energy is needed for the same mass to change the same temperature. This is why bricks are sometimes used in storage heaters, as they stay warm for a long time. Most heaters are filled with oil (1,800 J/kg/°C). Radiators in central heating systems use water (4,200 J/kg/°C) as they need to stay warm for a long time, so must have a lot of energy to lose.

Calculating thermal energy changes

The amount of thermal energy stored or released as the temperature of a system changes can be calculated using the equation:

change in thermal energy = mass × specific heat capacity × temperature change

\Delta E_{t} = m \times c \times \Delta \theta

This is when:

  • change in thermal energy (ΔEt) is measured in joules (J)
  • mass (m) is measured in kilograms (kg)
  • specific heat capacity (c) is measured in joules per kilogram per degree Celsius (J/kg°C)
  • temperature change (∆θ) is measured in degrees Celsius (°C)

Example

How much energy is needed to raise the temperature of 3 kg of copper by 10°C?

The specific heat capacity for copper is 385 J/kg°C

\Delta E_{t} = m~c~\Delta \theta

\Delta E_{t} = 3 \times 385 \times 10

\Delta  E_{t} = 11,550~J

Question

How much energy is lost when 2 kg of water cools from 100°C to 25°C?

\Delta E_{t} = m~c~ \Delta \theta

\Delta E_{t} = 2 \times 4,200 \times (100 - 25)

\Delta E_{t} = 2 \times 4,200 \times 75

\Delta E_{t} = 630,000~J

Question

How hot does a 3.5 kg brick get if it is heated from 20°C by 400,000 J (400 kJ)?

\Delta E = m \times c \times \Delta \theta

\Delta \theta = \frac{\Delta E}{m \times c}

\Delta \theta = \frac{400,000}{3.5 \times 840}

\Delta \theta = 136 ~ {\textdegree}C

final temperature = starting temperature + change in temperature

final temperature = 20 + 136

final temperature = 156°C