# States of Matter

## Kinetic Theory Model

The three states of matter are solid, liquid and gas, these can be explained by looking at the arrangement and movement of the particles of the material. For modeling the states of matter, and for understanding some properties of materials, such as density, pressure and the gas laws, the atom or molecules of a substance can be thought of as solid spheres.

Solids

In solids the particles are arranged in a regular pattern with bonds between the particles, this regular arrangement can form geometric shapes called a lattice. An example of simple cube shaped lattice is shown in the diagram below.

Within this lattice the particles are bonded together and vibrate. The amount of vibration depends on the temperature of the material.

Vibration increases with temperature.

__Liquids __

Within a liquid the structure has started to break up, the particles no longer form a regular shape. They are now moving about more freely. The direction of movement of the particles is random. There are still many bonds between particles, but now they are bonded into smaller independently moving groups.

Gas

The particles in a gas have more kinetic energy and the bonds between particles have been broken, although in many gases the particles may represent a molecule of two atoms bonded together, like oxygen (O2).

The speed of the particles is much greater and they are moving in random directions.

## Changes in State

The changes in state from solid to liquids and to gasses is caused by an increase in the kinetic energy of the particles. As heat energy is absorbed by a material the particles move faster until enough energy has been absorbed to change state.

Increase in Kinetic Energy of the Particles (Increases in stored energy)

 ●Melting Solid to Liquid ●Evaporation Liquid to Gas at temperatures below boiling point, a slow process. ●Boiling Liquid to Gas at boiling point, a rapid process. ●Sublimation Solid to Gas, with no liquid phase.

__A decrease in Kinetic Energy of the Particles __(Decreases in stored energy)

 ●Condensation Gas to Liquid ●Freezing Liquid to Solid

Unlike chemical changes, all these physical changes are reversible, with no loss in properties. Ice becomes water as it melts, but once re-frozen the ice has the same properties it did before melting.

## Specific Heat Capacity

As a material absorbs heat from its surroundings the particles move faster, they are gaining kinetic energy. The temperature of a material is a measure of the kinetic energy of the particles in the material.

The amount of heat energy, (in joules) required to raise the temperature of a material varies from material to material, this property of a material is known as its Specific Heat Capacity.

Specific Heat Capacity is the energy required to raise the temperature of 1 kilogram of the material by 1 °C. Specific Heat Capacity is represented by the symbol ‘c’

Examples:

Water c = 4,186 Joules per kg per 1°C (J/kg/°C)

Air c = 1,005 J/kg/°C

Glass c = 670 J/kg/°C

Nylon c = 1600 J/kg/°C

The higher the value of, c, the more energy it takes to heat up the material, and the slower it cools. Water has a very high value of c, this is why the oceans warm up and cool down slowly. On a hot day the sea can still be cold, as air heats up 4 times faster than the water in the sea.

Calculating Energy Changes

Providing a material __does not __change state the energy change when it heats up or cools down can be calculated using the following equation.

Change in Energy = mass x specific heat capacity x change in temperature

Δ__Q = m c __ΔΘ

Where

Δ__Q is the change in energy in Joules

_ _m is the mass in kg

_ c is the specific heat capacity in J/kg/°C_

_ ΔΘ is the change in temperature in __°C._

Example:

1. How much energy is required to heat up 1 litre of water from 20°C to 100°C?

1 litre of water has a mass of 1 kg

Change in temperature = 100 - 20 = 80 °C

c of Water = 4,186 J/kg/°C

Δ__Q = m c __ΔΘ = 1 x 4186 x 80 = 334,880 Joules

1. How much energy is required to heat up 1 kg of air?

Δ__Q = m c __ΔΘ = 1 x 1005 x 80 = 80,400 Joules

1. What is the energy change when a 2 kg glass window cools from 20°C to 10°C on a summers evening?

c of Glass = 670 J/kg/°C

Change in temperature = 10 - 20 = -10°C

Δ__Q = m c __ΔΘ = 2 x 670 x -10 = -13,400 Joules

## Specific Latent Heat

For a material to change state the arrangement of the particles must change, bonds must be formed or be broken. This change in arrangement requires energy.

To melt ice, the ice must be heated to turn into water. However, during this melting phase the ice __does not __change temperature. The heat energy being absorbed by the ice is being used to change the arrangement of the particles not to increase their kinetic energy. As the kinetic energy remains unchanged so does the temperature.

The amount of energy taken in or given out during a change in state is known as the materials Specific Latent Heat, this is given the symbol ‘L’ and is measured in Joules per Kilogram (J/kg).

Materials have different Latent Heat values for melting/freezing and boiling/condensing.

The value for melting/freezing is known as the Latent Heat of Fusion

The value for boiling/ condensing is known as the Latent Heat of Vaporization

Examples

 Material L, Fusion (J/kg) L, Vaporization (J/kg) Water 334,000 2,265,705,000 Carbon Dioxide 184,000 574,000 Oxygen 13,900 213,000

The amount of energy required to change state for any mass of material can be calculated using the following equation:

Energy = mass x Specific Latent Heat

Q = m L where Q is energy in Joules, m is mass in kg and L is Specific Latent Heat in J/kg

Examples

How much energy is required to melt 2 kg of ice?

L, Fusion for Water = 334,000 J/kg

Q = m L = 2 x 334,000 = 668,000 Joules

1. How much energy is required to boil 2 kg of water to steam?

Q = m L _ = 2 x 2,265,750,000 = 4,531,500,000 _Joules _or 4,531.5 _MJ

Summary:

Specific Heat Capacity: The amount of energy required to change the temperature of a material.

Specific Latent Heat: The amount of energy of change the state of a material

## Insulation

Heat will move from hot areas to colder areas through convection, conduction and radiation. It is not possible to stop this heat transfer process. In some situations the process is useful, removing excess heat from an engine or trying to keep cool in summer. Frequently, however, the transfer of heat energy from a warm place to a cold place is a problem. Keeping warm outside on a winter’s night or keeping a house warm, both require methods to stop heat from escaping into the atmosphere. Reducing heat loss requires insulating materials that slow down the movement of the heat.

Methods of Insulation

1. Trapping layers of still air or other gases.

Air, if it is not allowed to move, is not a good conductor of heat and so if layers of air can be trapped between a hot object and the colder atmosphere, the layers will act as an insulator. This is one of the most common methods of insulation. Padded coats uses this method by trapping air between the fibres, animals keep warm this way, by trapping air between the hairs on their hinds or between feathers. Double glazing traps a layer of air, or other gases, between the pains of glass.

1. Materials with low thermal conduction.

A material with a high specific heat capacity will take longer to heat up and longer to conduct heat away from a hot object, they act as insulators.

Glass for example has a specific heat capacity of 670 J/kg/°C, whereas brick has a specific heat capacity of 840 J/kg/°C and concrete 880 J/kg/°C. Of these three building materials glass will conduct heat fastest and concrete more slowly, it is therefore a better building material in terms of its thermal properties. Houses are built of breezeblocks made of concrete for this reason.

Wood has a specific heat capacity of up to 2,400 J/kg/°C making it an excellent material for building with in cold environments. If you have been skiing you might have noticed many of the builds are wooden, now you know why.

0.5 kg of ice heats up from - 5°C to 0°C then melts, the water continues to heat up to 10 °C. How much energy has been used for this change? Ice c = 2,090 J/kg/°CWater c = 4,186 J/kg/°CL fusion ice = 334,000 J/kg
Your answer should include: 5225J / 167000J / 20930J / 5225 / 167000 / 20930
Explanation: Energy to heat the ice ΔQ(ice) = m c ΔΘ = 0.5 x 2,090 x (0℃ - -5℃ ) = 0.5 x 2,090 x 5 = 5,225 Joules Energy to melt the ice Q = m L = 0.5 x 334,000 = 167,000 Joules Energy to heat the water from 0℃ to 10 ℃ ΔQ(water) = m c ΔΘ = 0.5 x 4,186 x (10℃ -0℃ ) = 0.5 x 4,186 x 10 = 20,930 Joules Total Q = 5,225 + 167,000 + 20,930 = 193,155 Joules
A 100 Kg block of marble absorbs 909,000 J of heat energy to increase its temperature by 10℃, what is the specific heat capacity of the marble?
909
Explanation: ΔQ = m c ΔΘ, therefore c = ΔQ (m ΔΘ) c = ΔQ (m ΔΘ) = 909,000 (100 x 10) = 909,000 1,000 = 909 J/Kg/℃