Energy is transmitted by conduction, convection or radiation.The conductivity of materials can be compared by examining the time taken to transmit energy through them.
When materials are heated, the molecules gain kinetic energy and start moving faster. The result is that the material gets hotter.
Different materials require different amounts of energy to change temperature. The amount of energy needed depends on:
It takes less energy to raise the temperature of a block of aluminium by 1°C than it does to raise the same amount of water by 1°C. The amount of energy required to change the temperature of a material depends on the specific heat capacity of the material.
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 |
Lead will warm up and cool down fastest because it doesn’t take much energy to change its temperature. Brick will take much longer to heat up and cool down. 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) or water (4,200 J/kg°C) as these emit a lot of energy as they cool down and, therefore, stay warm for a long time.
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:
Sadie is experimenting with a model steam engine. Before the 0.25 kg of water begins to boil it needs to be heated from 20°C up to 100°C. If the specific heat capacity of water is 4,180 J/kg°C, how much thermal energy is needed to get the water up to boiling point?
\[E_{t} = m~c~\Delta \theta\]
\[E_{t} = 0.25 \times 4,180 \times (100 - 20)\]
\[E_{t} = 0.25 \times 4,180 \times 80\]
\[E_{t} = 83,600~J\]
How much thermal energy does a 2 kg steel block (c = 450 J/kg°C) lose when it cools from 300°C to 20°C?
\[E_{t} = m~c~\Delta \theta\]
\[E_{t} = 2 \times 450 \times (300 - 20)\]
\[E_{t} = 2 \times 450 \times 280\]
\[E_{t} = 252,000 J\]
How hot does a 3.5 kg brick get if it’s heated from 20°C by 20,000 J (20 kJ)?
\[\Delta E = m \times c \times \Delta \theta\]
\[\Delta E = \frac{\Delta E}{m \times c}\]
\[\Delta \theta = \frac{20,000}{3.5 \times 840}\]
\[\Delta \theta = 6.8°C\]
\[final~temperature = start~temperature + change~in~temperature\]
\[final~temperature=20+6.8\]
\[final~temperature=26.8°C\]