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
The specific heat capacityThe amount of energy needed to raise the temperature of 1 kg of substance by 1°C. 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:
| Material | Specific heat capacity (J/kg/°C) |
| Brick | 840 |
| Copper | 385 |
| Lead | 129 |
| Material | Brick |
|---|---|
| Specific heat capacity (J/kg/°C) | 840 |
| Material | Copper |
|---|---|
| Specific heat capacity (J/kg/°C) | 385 |
| Material | Lead |
|---|---|
| Specific heat capacity (J/kg/°C) | 129 |
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 energyA more formal term for heat energy. stored or released as a substance changes state 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 (Δt) 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 (∵) 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