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Temperature changes and energy - EdexcelSpecific latent heat

Changes in a material's temperature or state of matter are caused by changes to the internal energy. The energy required by different materials depends on their 'heat capacity' and 'latent heat'.

Part of Physics (Single Science)The particle model

Specific latent heat

Changing the of a material will cause it to change temperature or change state:

  • the energy required for a particular change in temperature is given by the
  • the energy required for a particular change in state is given by the

As there are two boundaries, solid/liquid and liquid/gas, each material has two specific latent heats:

  • latent heat of fusion 鈥 the amount of energy needed to or the material at its melting point
  • latent heat of vaporisation 鈥 the amount of energy needed to or the material at its boiling point

Some typical values for specific latent heat include:

SubstanceSpecific latent heat of fusion (kJ/kg)Specific latent heat of vaporisation (kJ/kg)
Water3342,260
Lead22.4855
Oxygen13.9213
SubstanceWater
Specific latent heat of fusion (kJ/kg)334
Specific latent heat of vaporisation (kJ/kg)2,260
SubstanceLead
Specific latent heat of fusion (kJ/kg)22.4
Specific latent heat of vaporisation (kJ/kg)855
SubstanceOxygen
Specific latent heat of fusion (kJ/kg)13.9
Specific latent heat of vaporisation (kJ/kg)213

An input of 334,000 joules (J) of energy is needed to change 1 kg of ice into 1 kg of water. The same amount of energy needs to be taken out of the liquid to freeze it.

Calculating thermal energy changes

The amount of stored or released as the temperature of a system changes can be calculated using the equation:

change in thermal energy = mass 脳 specific latent heat

\(\Delta E_{t}Q = m \times l\)

This is when:

  • change in thermal energy (\(\Delta E_{t}Q\)) is measured in joules (J)
  • mass (m) is measured in kilograms (kg)
  • specific latent heat (\(l\)) is measured in joules per kilogram (J/kg)

Question

How much energy is needed to freeze 500 grams (g) of water at 0掳C?

Measuring latent heat

Latent heat can be measured from a heating or cooling curve line graph. If a heater of known power is used, such as a 60 W immersion heater that provides 60 J/s, the temperature of a known mass of ice can be monitored each second. This will generate a graph that looks like this:

Graph measuring time against temperature, looking at the temperature changes between solid, liquid and gas for ice, water and steam.

The graph is at two places. These are the places where the energy is not being used to increase the speed of the particles, increasing temperature, but is being used to break the bonds between the particles to change the state.

The longer the horizontal line, the more energy has been used to cause the change of state. The amount of energy represented by these horizontal lines is equal to the latent heat.

Example

If a horizontal line that shows boiling on a heating curve is 1 hour 3 minutes long, how much energy has a 60 watts (W) heater provided to the water?

63 minutes = 3,780 s

60 W means 60 J of energy is supplied every second

energy = power 脳 time

energy = 60 脳 3,780

energy = 226,800 J

Example 2

If this energy had been applied to 100 g of water, what is the latent heat of vaporisation of water?

226,800 J for 100 g is equivalent to 2,268,000 J for 1 kg. The latent heat of vaporisation of water is 2,268,000 J/kg.