Energy, temperature and specific heat capacity

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 a different amount of energy to raise the temperature of a 1 kg block of lead by 1°C, than it does to raise 1 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 specific heat capacity
  • the amount of energy put into the system
curriculum-key-fact
The specific heat capacity of a material is the energy required to raise the temperature of 1 kg of the material by 1°C.

The specific heat capacity of water is 4,200 joules per kilogram degree Celsius (J/kg°C). This means that it takes 4,200 J of energy to raise the temperature of 1 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 faster because it doesn't take much energy to change its temperature.

Brick will take much longer to heat up and cool down, as 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) and where there is central heating, radiators use water (4,200 J/kg°C), as these need to lose a lot of energy and, therefore, stay warm for a long time.

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 ( \Delta E_t) is measured in joules (J)
  • mass (m) is measured in kilograms (kg)
  • specific heat capacity (c) is measured in joules per kilogram 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

E_t = m~c~ \Delta \Theta

E_t = 3 \times 385 \times 10

E_t = 11,550~J

Question

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

E_t = m~c~ \Delta \Theta

E_t = 2 \times 4,200 \times (100 - 25)

E_t = 2 \times 4,200 \times 75

E_t = 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 = 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°C

final temperature = starting temperature + change in temperature

final temperature = 20 + 136

final temperature = 156°C