Convection is prevented by the vacuum space between the walls and by closing the Mask at the top. ‘Conduction is reduced by having the container made of glass, which is a bad conductor of heat TThe stopper is made of a bad conductor ¢.g. cork or rubber.

The vacuum is also & non ~ conducting space. The outer glass wall is supported by a pad of felt ‘or cork attached to a plasti ease Radiation is minimized by the of silvered surfaces. The silvered surface reflects any Radiant heat energy coming from the outside or inside the flank

MEASUREMENT OF THERMAL ENERGY

Heat Capacity

Heat capacity is the amount of heat required to raise the temperature of an object or substance by one degree The temperature change is the difference between the final temperature (T,) and the iitial temperature (1).

‘The Factors which Determine Heat Quality of a Substance Explain the factors which determine heat quality ofa subsance Heat is a form of energy transferred between bodies due to difference in temperature between them. The energy possessed by the body due to its temperature is called the internal thermal energy. The heat comient is due to the random motion of the particles that make up the body. The heat content is determined by its mass, temperature change and the specific heat capacity of the

suibstance,

‘The Heat Capacity

Determine the heat capacity Heat capacity is the quantity of heat required to raise the temperature of a substance by one degree Celsius. Heat capacity = mass ofthe substance X specific heat capacity

Thus H.C = MC

Also ‘Heat nergy. H = Mc (/\Q) ® AeQ-W co)

H=MC (Qf- Qi)

Example 1

Find the heat capacity of a lump of copper of mass SOkg. The specific heat capacity af copper is 420. Kg.

Data Given

Mass of copper, M= SOkg

Required: To calculate heat capacity, H.C. HC=Mc

Calculating a quantity of heat

• The quantity of heat required to change the temperature of a body with mass, M Kg by Q

degree Celsius is MCQ joules.

• Imorder to raise the temperature of a body, heat must be supplied to i
• Inorder to lower its temperature, heat must be removed from i.

The Heat Equation is therefore written Heat Gained or Heat Lost = Mass X specific heat capacity X change in temperature Change in temperature: H=McQ Where H-Heat gained / lost M= Mass of the body (= change (Rise or fall) In Temperature of the body

Example 2

Water of mass 3kg is heated from 26'c to 96%C. Find the amount of heat supplied to the water siven that the specific heat eapacity of water is 4.2.x 10° /Kg%

Data Given

Initial temperature, Qi 26°C

Required The amount of heat, H H=McQ c=HMQ H=McO

The Specific Heat Capacity

The quaiity OF heat supplied to or tke aay ftom Body depends on: 1 The mass of the body, M

• Thethermal properties of he body,

Transfer of Heat

Heat energy tends to flow ftom High tenpertures to Low temperatures feel warn, IF you pick up a cool object, eat energy transfers from hands tothe objeet and your tands fee cokd

• Cdlorimeter= the spevia instrument or vosel used for mieasement of Heat.
• Calorimeter is highly polished metal can ustlly made of copper or aluminium,
• Tes ited with an insulating cover in which there are two hoes
• Twootes alow a thermometer and sie to be inserted

+The areris made ofthe same malas that ofthe calorimeter Demonstration of the specific Heat capacity of @ solid Determining specific Heat capacity by Method of Calculation. Heat lost by solid, Hs ~ Ms x Cs (Qs ~ Qf) Heat Gained by Calorimeterand stirrer, He = Me x Ce (Qf ~ Qi) Heat Gained by Water, HW = Mw x Cw (Qf Qi) But the heat lost by the solid is equal to heat gained by the calorimeter and stirer plus the heat pained by the water in the ealorimeter.

Hys=He + Hw

But Heat gained by a calorimeter and content equal to heat lost by the solid.

Thus He+ HI= Hs

Mi Cl (QE-Qi)

Example 3

A piece of metal with a mass of 200, at a temperature of 100°C is quickly transferred into 50g of

Cs =Me Ce (Qf-Qi) +Mw Cw (Of Qi Ms(Qs-Qf) Where Cs. Is the spectfic Heat eapacity of the solids, Determining the specific heat capacity of liquid, Cl By calculation method: Heat Gained by calorimeter and stimer

He=Me Ce (Qf- Qi)

Heat Gained by liquid

Hi= M,C, (Qf Q0)

Heat lost by the solid

Hs=Ms Cs (Q6-QH)

Let (Qbe the final Temperature of the system

(Q-20)

(100-Q)

(Heat gamed by water) = (Heat lost by metal)

o= ate

Change of State

Change of state is the transformation of the condition of matter from one (state) to another caused by the change In temperature. The Behaviour of Particles of Matter by Applying Kinetic Theory Explain the behaviour of particles of matter by applying kinetic theory particles which are constantly moving or ina continual state of motion. The degree to which the particles move is determined by the amount of energy they have and their relationship to other

particles. The particles might be atoms, molecules or ions. Use of the general term \particle means the precise nature of the particles does not have to be specified Particle theory helps to explain properties and behaviour of materials by providing a model which enable us to visualise what is happening on a very small scale inside those materials. As @ model, itis useful because it appears to explain many phenomena but as with all models it does

have limitations. In solids the particles 1 liquids the particles In gases the particles

• sre hol tightly and packed © are fil close together with «ave lite atraction between them

firly close together = they are seme attraction between them are free io move in all diretions and strongly attracted o each other * —areable to move around in all collide with each oiher and withthe walls of a

• ovate in fixed positions but directions but movement i Limited by container and are widely spaced out

they do vibrate satroctions between particles . Solids, liquids and gases The model can be used to help explain: 1 the properties of matter

• what happens during physical changes such as melting, boiling and evaporating

‘The properties of matter

polis Liquids Gases

have a definite shape + donot have a definite shape + donothave a definite shope maintain that shape + flow and fil the botom of a + expand to fill any container are difficult 10 compress as container, They maintain the same volume © are easily compressed © particles are already packed unless the temperature changes because there ate only few particles osely together + are difficult to compress because ina lange volume

ace ofien dense as there are there are quite © lot of particles in a small © areoften low density as there ay panicles packed closely together volume are not many particles ina large space quite alot of patos in small volume /e-o staat / The graph of temperature versus temperature for a Heated.

‘The Melting Point of a Substance from its Cooling Curve Determine experimentally the melting point of a substance from its cooling curve Melting isthe process of change of the state of matter from solid into liquid e.g iee into water Melting point (MP): Its the temperature at which solid substance tends to change into liquid.

Freezing: Itis the process of change of the state of matte from liquid to solid e.g water into ice Freezing point: Is the temperature at which liquid change into solid. E.g water change into tee at ore Evaporation:Is the process of change liquid substance into vapour (gas) Sublimation: It is the change of state of matter from solid to gas and vice versa without passing through the liquid phase.c g. ammonium Chlonde ( NH.CL) and lodine tends to sublime.

Sublimation point is the temperature at which a solid tends to change into gas and vice versa without passing through liquid state Condensation:Is the change of state of gascous state of matter into liquid state..g steam into water.

Deposition: Is the change of the state matter from gas into solid e.g. Ammonium chloride vapour and Iodine vapour into solid (NH,CD) and (Iodine).

Heating Curve for Water

x (Oye: AH yapereatin /*~ Og m* CoAT i wd ao } /-ee mest opmce sare Change aglSement nf Timer Demonstration of cooling and melting curves for (octadecanoie acid) Melting point (mp) table

Substance Meting pint)

Conner 1083 Glass 1000 ~ 1400 tron 1450 Lead x7 Pitch 40-80 Meveuy =) Prana ins Tin 22 Tungsten 377 The Effect of Impurities on the Freezing Point and the Boiling Point of a Substance Deine i ec ef tek a fect cd ve lng pt oo Sabha The effect of dissolved substances on the boiling point and melting point (M.P) means thatthe additional of impurities will result in increased (B-P) and (M.P).

Effect of impurities on Boiling Point increased when adding the solute vapour pressure of the solution becomes lower than pure solvent, Thus Thus th boiling point gots elevated. For example boiling point of waters 100°C under normal atmospheric pressure. If we add sugar oor salt to this water its vapour pressure becomes lower and boiling point mereases.

Generally, when 1 mole of any non electrolyte is dissolved in litre of water the elevation of boiling point is 0.53° Effect of impurities on freezing point The depression in freezing point increases with the increase in concentration of the solute solvent, Since freezing point isthe temperature at which vapour pressure of liguid and solid For example the freezing point of water is O°C under normal atmospheric pressure. If we add

freezing point of water is 1.86°C. Conclusion 1 The impurities present in a Tiquid pull its two fixed points away from each other be the freezing point is lowered while the boiling point is raised.

• The depression in freezing point and the elevation in boiling point inereases with increase

in the concentration ofthe solute or impurity ie. these are the colligative properties that depends only on the no, of moles of the solute, They are independent ofthe nature of the solute.

The Effect of Pressure on the Boiling Point and Freezing Point of a Substance Demonstrate the effect of pressure on the boiling point and freezing point of a substance If a substance expands on solidifying, e.g., water, then the application of pressure lowers its melting point Ifa substance contracts on freezing, the pressure raises its melting point, e.g, paraffin wax.

The fieesing point of water is lowered by 0.007 °C per atmosphere increase in pressure, whereas that of paraffin wax increases by 0.04 °C per atmosphere merease in pressure. When a is liquid heated, its temperature rises and eventually remains constant.

Boiling is the process of forming bubbles of vapour inside the body of a Tiquid. I rises to the surface of liquid. The process usually depends onexternal pressure above the liquid.

The Phenomenon of Regelation

Explain the phenomenon of regulation Regelation is the Reffeezing process which takes place when copper wire is passed through the ce BLOCK Regelation is the Reffeezing process which takes place when the wite is observed to Cuts right through the ice block and falls on the floor.

m1 fi. Fi

The Concept of Boiling and Evaporation in Respect to the Kinetic Theory of

Matter ——E Cinhunana Pye cetera Spel vanrrmateeai eee ee SSS Eventually, even particles in the middle of the liquid form bubbles of gas in the liquid. At this Suicuninsiswsnontinn the liquid except that theyhave more energy. At norma! atmospheric pressure, all materials have Sg ee ee temperature. As with the melting point, the boiling point of materials vary widely, c.g., nitrogen – Ss Any material with a boiling temperature below 20°C is likely to be a gas at room temperature.

comhtiyimmatiahanamns diffuse through the surrounding air particles. As these particles cool down and lose energy they aera particles quickly condense as the surrounding air temperature is likely to be much less that ae seen because some of the gas particles have condensed to form small droplets of water.

a Within a liquid some particles have more energy than others. These “more energetic particles" may have sufficient energy to escape from the surface of the liquid as gas or vapour. This process is called evaporation and the result of evaporation is commonly observed when puddles or clothes dry. Evaporation takes place at room temperature which is often well below the boiling point of the liquid. Evaporation happens fiom the surface of the liquid As the

temperature increases, the rate of evaporation increases. Evaporation is also assisted by windy conditions which help to remove the vapour particles from the liquid so that more escape Evaporation is a complex idea for children for a number of reasons. The process involves the apparent disappearance of a liquid which makes the process difficult for them to understand. Itis not easy to see the water particles in the air. Also, evaporation occurs in a number of quite

differing situations – such as from a puddle or bowl of water where the amount of liquid obviously changes, to situstions where the liquid is less obvious – such as clothes drying or even those where there is no obvious liquid at all to start with « such as bread drying out. A further complication is that evaporation may be of a solvent from a solution e.g. water evaporating from salt water to leave salt, These situations are quite different yet all involve evaporation

Evaporation may also involve liquids other than water e.g. perfume, petrol, air fresheners. The particle model can be used to explain how it is possible to detect smells some distance away from the source.

Latent Heat of Fusion and Vaporisation: Demonsirate latent heat of fusion and vaporisation Latent Heat is the energy when is supplied in form of heat required to change the state of the Matter fiom one form into another Latent heat is not determined (detected) by using a thermometer. So latent heat is also called hidden heat Specific latent Heat is the energy supplied to a unit Mass and change Its state from one state oF

Matter to another state of matter. Latent heat of Vaporization is the heat required to change a liquid into a gaseous state at constant temperature. ee Latent heat of o4 oo vaporization e e 4 piel e ° eote %le = ef, e ° Water at 100 °c Steam at 100 °C

• Mass of Beaker= Mi ke
• Mass of Beaker + Water = Ms kg
• Time taken to Boil “ty Minutes

• Heat gained by steam = (Ms_M))L

Generally Heat gained by steam =tou Heat gained by Water (Mb = My, = %

(MyM) CWX 100

In this experiment the Heat gained by the Beaker may be Neglected. Latent heat of fusion is the amount of heat required to change a substance from soli to liquid at

Example 4

Calculate the amount of Heat required to melts 800g of Ice at 0°C The specifie Latent of fasion of le 33400¥%eu Data given: rrr

Specific Heat of fusion, L~ 33400 Jike

Heat gained, H= ML

He (0.8.x 334001/ kg)

Determination of the specific Latent Heat of fusion of lee

• Mass of Calorimeter + ster = Mi
• Mass of ealorimeter +Water=M:
• Mass of Calorimeter #Water= Ms
• Initial Temperature of Water=Q
• Final temperature of Water ~Q
• Mass of Water=(_M:-Mi )
• Mass of ee= (Ms. Mz)

The loc melts and forms Water at 0°C The Water formed warm up to Temperature Qf Heat

Where » Wis the specific heat eapecity of water.

• Heat lost by the original water in the calorimeter = (M>-M,) (Qi -Qr) Cu
• heat lost hy ealorimeter and stirrer= MI Ce (Q,- Qe),
• Ceis the specific heat capacity of the material ofthe calorimeter,

Applying the heat equation: (Heat gained by ice im Melting + Heat gained by the Water formed) ~(Heat lost by calorimeter and stirrer + Heat lost by original Water) us

(My = Ma) L+ (M3- M2) CW QF= MI Ce (Qi – Qr}+ (M2- MI CW QI- QF)

L = [MCc + (Mp – M) Cw IQ, – Qe] – (Ms – Mp) CwO; ‘M3 -Mp Specific Latent heat of Vaporisation is the amount of heat required to change a unit Mass of liquid into gaseous state ( Vapour) at constant temperature Specific latent Heat of fusion is the amount of heat required to change a unit Mass of solid substance into liquid at constant temperature

SUBSTANCE SPECIFIC LATENT HEATOF FUSION J/kg

lee 331400 Naphihalene 16300 Lead 24962 Copper 79740 Aluminum 317680 Gold 66880

Example S

016 kg of ice at – 10°C is dropped into 2ke of Water 49°C contained in a Copper calorimeter of mass 0. 15kg_. If the final temperature of the Mixture is 20°C fin d the specific latent Heat of fusion of ice, Where

• Specific Heat capacity of iee= 2.1 x 103 J/ Kg%C
• Specific Heat capacity of eopper= 4200/ Kg °C
• Specific Heat Capacity of Water= 4200 j/ Kg °C

Solution

Heat gained by ice during warming up form – 10 °C to 0°C (0.652. 1X 103 x10)

of ice Heat gained by cold Water in warming up from O°C to 20°C

50400 J Heat lost by Water during cooling from 49°C to 20°C

But Total Heat gained = Total Heat lost 12600 + 0.6 L + 50400 ~ 243600 + 1827

The Mechanism of Refrigeration

Describe tie mechanism of refrigeration Reffigerator is a machine which can enable Heat to flow fiom a cold Region to a Hot region The Basie principle used in Refrigeration is Cooling by absorption of latent Heat Waid

How it Works

A volatile liquid such as freon, evaporates inside the copper coils A surrounding the freezing cabinet or the refrigeration.

• The latent heat of Vaporization comes from the air surrounding the coil ie. from the

inside of the freezing g eabinet

• An ecleetically driven pump Premove the vapor from A and foree it into the heat

exchanger, which 1s made of eopper coils.

• The coils of the heat exchanger are filled with cool fins F
• Inthe heat exchanger, vapor is compressed by the pump and condensed back to liquid
• The conversion of vapour into liquid in (c) gives out the latent heat of vaporization,

which is conducted away by the fins,

• The condensed liquid is then retumed to the evaporator coil (A) through avalve (V) (in

this way a continuous circulation of vapour and liquid is set up).

• The rate of evaporation and the degree of cooling is controlled by @ thermostat, which

switches the pumps motor on and of at intervals.

• The thermostat ean be adjusted to give the desired low temperature inside the freezing

cabinet where food is preserved,