THERMAL PROPERTIES OF MATERIALS

A2 Physics Topic 4By Allen Tobve

KINETIC MODELGases, liquids and solids are all made up of atoms, molecules, and/or ions, but the behaviour of these particles differ in the three phases.Particles in gases are well separated with no regular arrangement.liquid are close together with no regular arrangement.solid are tightly packed, usually in a regular pattern.Particles in gases vibrate and move freely at high speeds.liquid vibrate, move about, and slide past each other.solid vibrate but generally do not move from place to place

CHANGES OF STATEParticles in a body have a mixture of kinetic energy and potential energy. The molecules have kinetic energy because they are constantly in motion and potential energy because they are held by attractive intermolecular forces and energy is needed to separate them.Kinetic energy determines the temperature of the body and potential energy determines the state of the body. Matter can be in four states like solid, liquid, gas and plasma.

CHANGE OF STATEIn order to 'change state' energy is either 'taken in' or 'given out'.The temperature is constant during a change of state, e.g. during melting or boiling, even though energy is absorbed. This energy is called the latent (or hidden) heat because it does not produce a change of temperature.

The average kinetic energy of the molecules stays constant, but the latent heat supplied increases the potential energy of the molecules, as they break free from the attraction of their neighbouring particles.

The latent heat of vaporisation is higher than the latent heat of fusion for the same substance. There is a large change in volume when a liquid changes to a vapour, so the molecular separation in the gaseous phase is several times that in the liquid phase. The energy used to increase the potential energy of the molecules is greater during vaporisation. The latent heat also supplies energy for the vapour to do external work in pushing back the atmosphere when the liquid changes to a vapour.

EVAPORATION AND BOILINGEvaporation takes place over a wide range of temperatures, whereas boiling takes place takes place at a single temperature. Unlike boiling, evaporation is affected by surface area and wind. Both are affected by pressure.

Evaporation occurs because some of the molecules in the liquid have more energy than other molecules. These fast moving molecules are able to escape from the attraction of their neighbouring molecules and leave the surface to form a vapour outside. The average kinetic of the molecules left behind decreases, so the temperature of the liquid falls. Hence evaporation has a cooling effect.

The boiling point of a liquid is defined as the temperature at which its saturated vapour pressure (SVP) becomes equal to the external atmospheric pressure. The space above a liquid is saturated with its vapour and the pressure it exerts is called the saturated vapour pressure. When a liquid is heated the SVP increases, and when it becomes equal to the external atmospheric pressure boiling occurs and bubbles of vapour form in the liquid.

SPECIFIC HEAT CAPACITY AND SPECIFIC LATENT HEATSpecific Heat CapacityThe temperature rise of an object when it is heated depends on:the amount of energy supplied (ΔQ)its mass (m)what the substance is made of, e.g. water or copper

where c is the constant of proportionality, called the specific heat capacity of the material.The specific heat capacity (c) of a substance is the heat energy required to increase the temperature of 1 kg of the substance by 1 °C or 1 K. The unit of specific heat capacity (c) is J kg-1 K-1 or J kg-1 C-1. The heat capacity (C) of an object is the heat energy required to increase the object’s temperature by 1 °C or 1 K. The unit of heat capacity (C) is J K-1 or J °C-1.

MEASURING SPECIFIC HEAT CAPACITY

The principle is to supply a known amount of energy to a known mass of material and measure the rise in temperature. The metal block of mass 1 kg is heated using an electrical heater, while keeping the current and p.d. constant. Measurements of the temperature are taken with time, as shown. If the power of the heater is 50 W and the rise in temperature is 16.4 °C in 400 s.

Heat losses to the surroundings have been neglected, so the value obtained is too high. The block is insulated to reduce heat loss. Another way is to cool the block below room temperature before heating it. Then turn the heater off when the block is at an equivalent temperature above room temperature. The block will gain energy from the surroundings when it is below room temperature and lose an equal amount when it is above room temperature. The block should be heated at a low rate so the heat spreads throughout the block.

THE CONTINUOUS-FLOW CALORIMETER

This was first developed by Callender and Barnes in 1902 for the measurement of the specific heat capacity of a liquid, and is shown in Figure 2. Its main advantage is that the thermal capacity of the apparatus itself need not be known.

Liquid flows in from a constant-head apparatus at a constant rate past a thermometer (θ1). It then flows around the heater coil and out past a second thermometer where the outlet temperature (θ2) may be measured.

When steady-state conditions have been reached (a temperature difference between inlet and outlet points of 5oC is reasonable) the temperatures and the flow rate of the liquid are measured. A vacuum jacket round the heater coil reduces heat losses.

The electrical energy supplied to the heater coil (E = V I t) may be found with an ammeter and voltmeter, or with a joulemeter.

Two sets of measurements are carried out. For a first experiment we have: Electrical energy supplied

E1 = V1I1t1 = m1 c (θ2 – θ1) + H

where c is the specific heat capacity of the liquid and the heat loss to the surroundings and to the apparatus.

The flow rate and rate of energy input are now altered to give a second set of results. However, if the inlet and outlet temperatures are the same as in the first experiment the heat loss will also be the same. Therefore:Electrical energy supplied E2 = V2I2t2 = m2 c (θ2 – θ1) + H

  Eliminating the heat loss (H) gives:Specific heat capacity of the liquid

c = (E2 – E1) / (m2 – m1) (θ2 – θ1)

FIRST LAW OF THERMODYNAMICS1. Thermodynamics is the study of heat and its transformation to

mechanical energy in a system. 2. A system is a well defined group of objects, e.g. atoms or

molecules whose energy is being measured. An example of a system is the petrol air mixture in the cylinder of a car engine.

Adding heat to the mixture or igniting it increases its internal energy, and the mixture may do work on a piston and cause it to move.

FIRST LAW OF THERMODYNAMICS3. The internal energy of a system depends on the state of

the system and is the sum of the random distribution of kinetic and potential energies associated with the molecules of a system.

4. The internal energy of a gas can be increased by either(i) heating the body. The molecules gain kinetic energy when

they collide with the hot container walls and bounce off faster (or if a fuel is burnt within it), or

(ii) doing work on the body. Gas molecules are struck by a piston moving downwards and bounce off faster, gaining kinetic energy in the same way as a ball gains kinetic energy when struck by a moving bat. This explains why a gas becomes hotter when compressed.

FIRST LAW OF THERMODYNAMICS5. The First Law of

thermodynamics in equation form applied to a gas is:

Increase in internal energy (ΔU) = Heat energy supplied (ΔQ) + Work done on the gas (ΔW)

(ii) doing work

(i) heating

FIRST LAW OF THERMODYNAMICS6. The first law of thermodynamics is the law of conservation of energy,

i.e. energy cannot be created or destroyed but it can be transferred from one form to another.

7. The first law of thermodynamics states:The internal energy of a system depends

(i) only on its state; and(ii) the increase in internal energy of a system (ΔU) equals the sum of the

energy supplied to the system by heating (ΔQ) and the work done on the system (ΔW).

8. The work done by an expanding gas is given by If the volume of the gas is decreased by ΔV, then p ΔV is the work done on

the gas.

WORKED EXAMPLESExample 1

If 100 J of energy is added to a system that does no external work, by how much does the internal energy of the system increase.

Answer ΔW = 0 ΔU = ΔQ + 0 = 100 J ∆U= 100 J

Example 2.If 100 J of energy is added to a system that does 40 J of external work, what is the increase in internal energy of the system?

WORKED EXAMPLESExample 3.

A 24 W filament bulb has been switched on for some time. Apply the first law of thermodynamics during a period of 2 s of the lamps operation, and when first switched on.

Example 4Find the increase in potential energy of a molecule of water when changing from liquid

at 100 °C to vapour at 100 °C.

( lv = 2.26 x 10 6 J kg-1, steam at 100 °C and normal atmospheric pressure has a density of

0.59 kg m-3. Normal atmospheric pressure= 1.01 x 10 5 Pa. Density of water= 1000 kg m-3.)

EXAMPLE 5(A) SUPPOSE THE GAS IN A CYLINDER EXPANDS FROM P TO Q, IN SUCH A WAY THAT THE TEMPERATURE REMAINS CONSTANT. THE GAS IS THEN COMPRESSED FROM Q TO R AT CONSTANT PRESSURE AND THEN HEATED AT CONSTANT VOLUME BACK TO P. WHAT IS THE NET WORK DONE BY THE GAS?(B) IN THE GRAPH BELOW, WHAT IS (I) AN ISOTHERMAL CHANGE, (II) AN ISOBARIC CHANGE AND (III) AN ISOVOLUMETRIC CHANGE? STATE WHAT CHANGES TAKE PLACE TO ΔU, ΔQ AND ΔW DURING THESE CHANGES?

Answer(a) Pressure-volume graphs,

called indicator diagrams, can be used to show the cycle of changes taking place in the cylinder of an engine. The work done on or by a gas can be found from the area under the graph,

Since, ΔW = p ΔV = area under the graph.From P to Q, the work done by the gas = area PQSTFrom Q to R, the work done on the gas = area RQSTAlong RP, no work is done since there is no change in volume.Net work done by the gas = shaded area enclosed PQR

(b) (i) Isothermal change This is a change from P to Q. The gas is expanding at constant temperature (or

contracting at constant temperature from Q to P.) Isothermals are constant temperature curves. Boyle’s law applies,

Using, , There is no change in temperature, ΔU = 0. Work is done by the gas as it expands, ΔW is negative. The gas takes in heat from the surroundings to keep the temperature constant, so ΔQ is positive, (equal to the work done.)

(ii) Isobaric change. This is a change from Q to R when the gas is cooled at constant pressure, (or

from R to Q.) The pressure remains constant. Charles’ law applies, (at constant pressure).

From The temperature is decreasing, ΔU is negative. The gas is contracting, work is done on it, and so ΔW is positive. Heat is removed from the gas to keep the pressure constant, so ΔQ is negative.

(iii) Iso-volumetric change. This is a change from R to P. The gas is warmed at constant volume so its pressure increases. The pressure law applies

From , There is no change in volume, so no work is done, ΔW is zero. Heat goes into the system to raise the pressure, so ΔQ is positive.(= +),The internal energy increases, ΔU is positive, (and equal to the heat going in.)

Adiabatic change This is an expansion or contraction in which no heat

enters or leaves the gas.From , ΔQ is zero as no heat enters the system,Work is done by the gas as it expands, so ΔW is

negative. ΔU is negative, the internal energy is decreasing.

Work is done by the gas at the expense of the internal energy, so the gas cools.

 

THE END