Energy, temperature and specific heat capacity
If energy is absorbed by a block of lead, the particleA general term for a small piece of matter. For example, protons, neutrons, electrons, atoms, ions or molecules. gain energy. Since lead is a solid and the particles are vibrating, they vibrate faster after being heated.
Different substances require different amounts of energy to gain temperature. It takes less energy to raise the temperature of a 1 kg block of lead by 1°C, than it does to raise the temperature of 1 kg of water by 1°C.
From this it can be seen that a change in temperature of a system depends on:
- the massThe amount of matter an object contains. Mass is measured in kilograms (kg) or grams (g). of the material
- the substance of the material 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 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 |
Lead has a low specific heat capacity and 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 because 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.
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
Or:
\(\Delta \text{E} = \text{m} \times \text{c} \times \Delta \theta\)
This is when:
- change in thermal energy (ΔE) 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.
\(\text{E} = \text{mc} \Delta \theta\)
3 × 385 × 10
\(\text{E}\) = 11,550 J
Question
How much energy is lost when 2 kg of water cools from 10°C to 25°C?
\(\text{E} = \text{m} \times \text{c} \times \Delta \theta\)
2 × 4,200 × (100 - 25)
2 × 4,200 × 75
\(\text{E}\) = 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)?
\(\Delta \text{E} = \text{m} \times \text{c} \times \Delta \theta\)
\(\Delta \theta = \frac{\Delta \text{E}}{\text{m} \times \text{c}}\)
\(\frac{400,000}{3.5 \times 840}\)
\(\Delta \theta \) = 136°C
final temperature = starting temperature + change in temperature
final temperature = 20 + 136
final temperature = 156°C
Jonny Nelson explains specific heat capacity with a GCSE Physics practical experiment
More guides on this topic
- States of matter: interactive activity - AQA Synergy
- Atomic structure - AQA Synergy
- Cells in animals and plants - AQA Synergy
- Transport into and out of cells - AQA Synergy
- Cell division - AQA Synergy
- Mitosis: interactive activity - AQA Synergy
- Waves - AQA Synergy
- Sample exam questions - building blocks - AQA Synergy