Specific heat capacity
If 1,000 J of heat is absorbed by a one kilogram block of lead, the particles gain energy and the temperature of the block rises. If a one kilogram block of lead absorbs 2,000 J of energy then the temperature rise will be larger.
If 1,000 J of heat is absorbed by a 2 kg block of lead then the temperature of the block doesn’t rise as much since the energy is shared between more particles. If 1,000 J of energy is absorbed by a one kilogram block of copper instead of lead then the temperature of the block doesn’t rise as much.
From this it can be seen that a change in the temperature of a system depends on:
- the mass of the material
- the substance of the material - its specific heat capacityThe amount of energy needed to raise the temperature of 1 kg of substance by 1°C.
- the amount of energy put into the system
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 1 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 faster because 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 store a large amount of energy and emit it over a long period of time. Most heaters are filled with oil (1,800 J/kg°C) and where there is central heating, radiators use water (4,200 J/kg°C).
Learn more about specific heat capacity in this podcast
Listen to the full series on BBC Sounds.
Calculating thermal energy changes
The amount of thermal energyA more formal term for heat 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
\(\Delta E_tQ = m \times c \times \Delta \theta \)
This is when:
- change in thermal energy (\(\Delta E_t Q \)) 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 (\(\Delta \theta \)) 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
\(\Delta E_t Q = m c\Delta \theta \)
\(\Delta E_t Q = 3 \times 385 \times 10\)
\(\Delta E_t Q = 11,500 \: J\)
Question
How much energy is lost when 2 kg of water cools from 100°C to 25°C?
\(\Delta E_t Q = m~c~ \Delta \theta\)
\(\Delta E_t Q = 2 \times 4,200 \times (100 - 25)\)
\(\Delta E_t Q = 2 \times 4,200 \times 75\)
\(\Delta E_t Q = 630,000~J\)
Question
How hot does a 3.5 kg brick get if it's heated from 20°C by 400,000 J (400 kJ)?
\(\Delta E_t Q = m \times c \times \Delta \theta\)
\(\Delta \theta = \frac{\Delta Q}{m \times c}\)
\(\Delta \theta = \frac{400,000}{3.5 \times 840}\)
\(\Delta \theta = 136 ~ {\textdegree}C\)
final temperature = starting temperature + change in temperature
final temperature = 20 + 136
final temperature = 156°C
