THERMAL PROCESSES

Conductors

• Good conductors are used whenever heat is required to travel quickly through something.
• Bad conductors (insulators) are used to reduce the amount of heat lost tot he surroundings.

Conduction

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• Conduction is the flow of heat through matter from places of higher temperature to places of lower temperature without the movement of the matter as a whole.
• In non-metals – When heat is supplied to something, its atoms vibrate faster ans pass on their vibrations to the adjacent atoms.
• In metals – Conduction happens in the previous way and in a quicker way – electrons are free to move, they travel randomly in the metal and collide with atoms and pass on the vibrations

Convection

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• Convection is the flow of heat through a fluid from places of higher temperature to places of lower temperature by movement of the fluid itself.
• As a fluid (liquid or gas) warms up, the particles which are warmer become less dense and rise.
• They then cool and fall back to the heat source, creating a cycle called convection current.
• As particles circulate, they transfer energy to the other particles. If a cooling object is above a fluid, it will create a convection current.

Radiation

• Radiation is the flow of heat from one place to another by means of electromagnetic waves.
• Thermal radiation is mainly infra-red waves, but very hot objects also give out light waves.
• Infra-red radiation is part of the electromagnetic spectrum.

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• An emitter sends out thermal radiation.
• A reflector reflects thermal radiation, therefore is a bad absorber.
• An emitter will cool down quickly, an absorber will heat up more quickly and a reflector will not heat up quickly.

Application of Energy Transfer

• Solar Panels: The sun’s thermal radiation is absorbed by a matte black surface and warms up the pipes containing water.
• Refrigerator: The freezer compartment is located at the top of the refrigerator. It cools down the air which then sinks. Any warm air rises to the top and then is cooled. This creates a convection current which maintains a cold temperature.
• Metals are used in cooking pans because they conduct heat well.

Consequences of Energy Transfer

• Metals spoons in a hot drink will warm up because it conducts heat.
• Convection currents create sea breezes. During the day the land is warmer and acts as a heat source. During the night, the sea acts as the source.
• A black saucepan cools better than a white one, white houses stay cooler than dark ones.

THERMAL PROPERTIES

Thermal Expansion of Solids, Liquids, and Gases

• Solids, liquids, and gases expand when they are heated (as their atoms vibrate more), and this causes them to become further apart, taking up greater volume.
• Due to differences in the molecular structure of the different states of matter, expansion is greatest in gases, less so in liquids, and lowest in solids.
• Applications and consequences of thermal expansion:
  • Overhead cables have to be slack so that on cold days, when they contract, they do not snap or detach.
  • Gaps have to be left in bridges to allow for expansion.
  • Bimetal Thermostat: When temperature gets too high, bimetal strips bend, to make contacts separate until temperature falls enough, then the metal strip will become straight again and contacts touch, to maintain steady temperature.
    

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• For a fixed mass of gas at a constant pressure, the volume is directly proportional to the Kelvin temperature.

Thermometers

• A physical property that varies with temperature may be used for the measurement of temperature.
• Liquid-in-glass Thermometer:
  

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  • As temperature rises or falls, the liquid (mercury or alcohol) expands or contracts.
  • The amount of expansion can be matched to the temperature on a scale.
• Thermistor Thermometer:
  

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  • The probe contains a thermistor
  • The thermistor is a material that becomes a better electrical conductor when the temperature rises (semi-conductor)
  • So when the temperature increases, a higher current flows from a battery, causing a higher reading on the meter.
• Thermocouple Thermometer:
  

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  • The probe contains 2 different metals joined to form two junctions.
  • The temperature difference causes a tiny voltage which makes a current flow.
  • A greater temperature difference gives a greater current.
  • Thermocouple thermometers are used for high temperatures which change rapidly and have a large range (-200 °C to 1100 °C)
• Fixed Points are definite temperatures at which something happens and are used to calibrate a thermometer. For example, melting and boiling point of water.
• Calibrating a thermometer:
  • Place the thermometer in melting ice, this is 0 °C.
  • Place the thermometer in boiling water, this is 100 °C.
• Sensitivity: is the change in length or volume per degree.
• To increase sensitivity:
  • Use a thinner capillary
  • Use a less dense liquid
  • Use a bigger bulb
• Range:  is the change in the upper and lower fixed points
• Linearity: is the change in the distance between intervals
• Responsiveness: is the time it takes a thermometer to react to a change in temperature.

Melting

• This is when a solid turns into a liquid.
• Temperature increases, thus the kinetic energy is solids increase, particles vibrate more rapidly but there is no increase in the temperature of the substance when melting occurs because the thermal energy supplied is being used to break the bonds between the particles of the solid, thus making it into a liquid.
• Melting point is the temperature at which a substance melts.

Boiling

• This is when a liquid turns into a gas.
• Temperature increases thus the kinetic energy in the liquid increases and particles vibrate more rapidly but there is no increase in the temperature of the substance when boiling takes place because the thermal energy supplied is used to break the bonds between the particles of the liquid, thus making it into a gas.
• Boiling point is the temperature at which a substance boils.
• The difference between boiling and evaporation is:
  • Boiling occurs at a fixed temperature and throughout the liquid, while
  • Evaporation occurs at any temperature and only on the surface of the liquid.

Latent Heat

• The latent heat of fusion is the amount of energy needed to melt 1 kg of a substance.
• The latent heat of vaporization is the amount of energy needed to boil 1 kg of a substance.
• Specific Latent Heat of Fusion/Vaporization = ^{Energy Transferred}/_Mass
• L_f/L_v = ^E/_m

Condensation and Solidification

• Condensation is when a gas turns back into a liquid.
• When a gas is cooled, the particles lose energy. They move more and more slowly.
• When they bump into each other, they do not have enough energy to bounce away again, so they stay close together, and a liquid forms.
• When a liquid cools, the particles slow down even more. Eventually, they stop moving except for vibrations and a solid forms.

THERMAL PROPERTIES OF MATERIALS

Kinetic Model of Matter

Melting, Boiling, Evaporating

• Melting and boiling occur without a change in temperature:
  • Temperature is a measure of random K.E. of the particles.
  • At phase transition, all energy is used to break bonds.
  • No change in K.E. occurs, so the temperature does not change.
• Cooling effect of evaporation:
  • Particles which escape are those with higher velocity, so average K.E. of the remaining substance decreases.
  • Temperature = Average K.E. and overall temperature decreases

Specific Heat Capacity and Latent Heat

• Specific Heat Capacity is the energy required, per unit mass of the substance, to raise the temperature by 1 Kelvin.
• c = ^E/_mΔθ

Determining Specific Heat capacity, c

• Quantities required:
  • Accurate measurement of mass
  • Temperature at time intervals
  • Voltage and current supplied

• Measure the temperature at regular time intervals and plot the graph of temperature, θ, against time, t.
• Divide the quantity of heat equation with time:
  • ^E/_Δt = mc(^Δθ/_Δt)
  • ^E/_Δt is the power supplied, P, and P = VI
  • ^Δθ/_Δt is the gradient of the graph plotted
• Replacing the quantities in original equation:
  •  VI = mc × gradient
• Substitute values, rearrange and solve.

• Specific Latent Heat of Fusion is the energy required, per unit mass of a substance, to change the substance from solid to its liquid phase without any change in temperature.
• Specific Latent Heat of Vaporization is the energy required, per unit mass of a substance, to change the substance from liquid to its gas phase without any change in temperature.
  • L_^f/v = ^E/_m
• Specific latent heat of vaporization is always greater than that of fusion for a given substance because:
  • During vaporization, there is greater increase in volume than in fusion. thus more work done against the atmosphere.
  • In vaporization, particles need to be separated further apart than in fusion, so more work is done against the forces of attraction when vaporizing.

Examples

1. To harden a sample of pure gold, silver is mixed so that the mixture contains 5% silver by weight. The initial temperature of the silver is 27 °C. Calculate the initial temperature of gold so that the final mixture is at the melting point of gold.
   
   • As mass is not provided, we will consider a mixture of 100 g with the mass of gold, 95 g, and silver 5 g.
   • Firstly, calculate the energy required for silver to be at the melting point of gold:
     • Q = mcΔT
     • 0.005 × 235 × (1340 – (273 + 27)) = 1222 J
   • As silver is being heated past its melting point, it will also melt and change state; energy required to change it form solid to liquid is required, i.e., latent heat of fusion:
     • Q = l_f × m
     • 105 × 10³ × 0.005 = 525 J
   • The quantity of energy gold should initially have must provide the answer above, therefore:
     • Q of gold = 1222 + 525 = 1747 J
   • Gold will already be past its melting point, so there is no need to calculate the heat of fusion. Now, using mcΔT:
     • 0.095 × 129 × ΔT = 1747
     • ∴ ΔT = 142.6 K
   • The initial temperature will obviously be above the melting point, so adding the final temperature of gold, i.e., the melting point:
     • Initial temperature of gold = 1340 + 142.6 ≈ 1483 K

Internal Energy

• Internal energy is the sum of random distributions of kinetic and potential energies of molecules in a system.
• Internal Energy = Total P.E. + Total K.E.
• A rise in temperature of a body is an increase in its internal energy.

First Law of Thermodynamics

• The first law of thermodynamics states that the increase in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system.
• ΔU = Q + W
  • ΔU: increase in internal energy of the system.
  • Q: heat supplied to the system
  • W: work done on the system

Examples

1. Write down the symbol ‘+’ for increase, the symbol ‘-‘ for decrease, and the number ‘0’ for no change.
   
   1. The gas is being compressed, so work is being done on the system (W = +) and when a gas is compressed, its temperature rises. As the system is not providing heat, the gas itself is heating (Q = -). Overall, the increase in work done is balanced by the gas heating, so the net remains 0 and internal energy unchanged (U = 0)
   2. The solid is being heated so (Q = +). As the solid is not expanding (W = 0), and therefore there is an increase in internal energy (U = +)
   3. The melting of ice requires heat energy provided so (Q = +). No work is done on or by the system so (W = 0). Hence, there is a net increase so (U = +)
2. Some water in a saucepan is boiling. Explain why:
   1. external work is done by the boiling water.
      • Volume increases due to evaporation (turns into a gas), hence work is done by pushing back the atmosphere.
   2. there is a change in the internal energy as water changes to steam.
      • The E_k of atoms is constant as there is no temperature change but E_p changes because separation of the atoms increases, so internal energy increases because ΔU = E_p + E_k
   3. By reference to the first law and your answer in (a), show that thermal energy must be supplied to the water during the boiling process.
      • ΔU = Q + W
      • Changing from a liquid to a gas, there is an increase in internal energy. Work is done by the liquid so W is negative. For ΔU to be positive, Q must increase.