Definition of Heat Capacity

7/29/2019 Definition of Heat Capacity

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Definition of Heat Capacity

Heat capacity is an important property of materials that must be considered in

engineering devices, planning a scientific experiment and keeping your lunch

cold.

There are three main types of heat capacity, the understanding of which willexpand your ability to take advantage of this property of matter.

Definition

1. A material's heat capacity is a measure of how much energy must be exchanged

between an object and its environment to produce a change in temperature. In

other words, it is a measure of an object's "capacity" to hold "heat." Materials with

high heat capacities, such as water, require large amounts of energy to produce

a small temperature change.

Significance

2. Gardeners shield their tomatoes with water-walls, and during the Middle Ages, a

lady-in-waiting put hot rocks into her master's bed. Both practices rely on the

ability of water and stone to hold heat very well, a property owing to the fact that

they have a high heat capacity.

Heat capacity is used for much more than saving tomatoes or keeping feet warm.

It is an essential parameter considered in many fields of engineering from

architecture to aerodynamics.

Heat Capacity

Heat capacity is the amount of heat required to increase the

temperature of an object by 1 oC (or 1 K).


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Types of Heat Capacity

3. The general term "heat capacity" includes three more specific, often confused,

terms. These are molar heat capacity, specific heat and just heat capacity.

Molar Heat Capacity and Specific Heat

4. A substance's molar specific heat capacity is the amount of energy required to

raise one mole of that substance 1 degree Celsius. A mole specifies a number of

objects, as does a dozen, except a mole equals 6.02 x 10^23 objects instead of

12. Because of the way the mole is defined, one mole of a substance is equal to

that substance's atomic or molecular mass in grams.

The more general term "specific heat" usually refers to the amount of energy

required to raise a certain mass of material, usually a kilogram, 1 degree Celsius.

This is probably the most common use of the term "heat capacity."

Both molar heat capacity and specific heat are intrinsic properties of a substance;

that is, they are independent of the amount of material you have on hand.


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Heat Capacity

Mathematical definition of heat capacity

An object's heat capacity, C, is the product of its specific heat capacity (the

amount of heat required to raise 1 kg of the material 1 degree C) and its mass in

kg. Heat capacity is an extensive property of a substance. That is, its value

depends on how much substance you have. For example, the heat capacity of a

20 oz. soda is exactly double that of a 10 oz. soda.


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Mathematical Definition and Theory

5. Mathematically, heat capacity is defined as the first order partial derivative of the

change in heat of an object with respect to the object's change in temperature, all

else being held constant. The root of what determines a substance's heat

capacity is a course in statistical thermodynamics. However, there is a general

rule: the larger and more free molecules are, the greater their heat capacity will

be. In other words, the more degrees of freedom the atoms or molecules making

up a substance have, the greater the heat capacity will be.

Examples of Heat Capacity in Use

6. Metals tend to have very low heat capacities compared to nonmetals. For

example, 1 kg of copper has a heat capacity of 394 Joules per Celsius, while 1

kg of marble has a heat capacity of 880. Thus, stone can hold much more heat

than metal.

The definition of a calorie is based on the heat capacity of water. A dietary

Calorie (note the capital C) is the amount of heat required to raise the

temperature of 1 kg of water 1 degree. Therefore, if all of the Calories in a

Twinkie (150) went toward heating 1 liter of water at room temperature, say at 22

degrees Celsius, then the Twinkie would have enough energy to raise the

temperature of the water to boiling and then some.


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Heat capacity


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Specific Heat CapacityThe amount of heat required to change the temperature of 1 kg of a

substance by 1oC.

Formula of Specific Heat Capacity

Example 1

How much thermal energy is required to raise the temperature of a 2 kg aluminium

block from 25C to 30 C? [The specific heat capacity of aluminium is 900 Jkg-1

oC-1]

Answer:

Mass, m = 2kg

Specific heat capacity, c = 900 Jkg-1oC-1

Temperature change, = 30 - 25 = 5 oC

Thermal energy required,

Q = mc = (2)(900)(5) = 9000J.


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Conversion Of Electrical Energy Into Thermal Energy

Example 2

An electric heater supplies 5 kW of power to a tank of water. Assume all the energy

supplied is converted into heat energy and the energy losses to the surrounding is

negligible. How long will it take to heat 500 kg of water in the tank from 25 to 100

C? [Specific heat capacity of water = 4200 J kg-1oC-1]

Answer:

P = 5000W

m = 500kg

c = 4200 J kg-1oC-1

= 100 - 25 = 75oC

t = ?

We assume,

all the electrical energy supplied = heat energy absorbed by the water

Pt = mc

(5000) t = (500)(4200)(75)

t = 31500s = 525 minutes = 8 hours 45 minutes

(Practically the time can be much longer than this because a lot of heat may be

losses to the surrounding.)


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Conversion of Gravitational Energy into Thermal Energy

Example 3

A lead shot of mass 5g is placed at the bottom of a vertical cylinder that is 1m long

and closed at both ends. The cylinder is inverted so that the shot falls 1 m. By how

much will the temperature of the shot increase if this process is repeated 100 times?

[The specific heat capacity of lead is 130Jkg-1K-1]

Answer:

m = 5g

h = 1m 100 = 100m

g = 10 ms-2

c = 130Jkg-1K-1

= ?

In this case, the energy conversion is from potential energy to heat energy. We

assume that all potential energy is converted into heat energy. Therefore

mgh = mc

gh = c

(10)(100) = (130)

= 7.69

o

C


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Conversion Of Kinetic Energy Into Thermal Energy

Mixing 2 Liquid


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Example 4

What will be the final temperature if 500 cm3 of water at 0C is added to 200cm3 of

water at 90C? [Density of water = 1gcm-3]

Answer:

The density of water is 1g/cm3, which means the mass of 1 cm3 of water is equal to

1g.

Let the final temperature =

m1 = 500g = 0.5kg

c1 = c

1 = - 0 =

m2 = 200g = 0.2kg

c2 = c

2 = 90 -

m1c11 = m2c22

(0.5) c ( ) = (0.2) c ( 90 - )0.5 = 18 - 0.2

0.5 + 0.2 = 18

0.7 = 18

= 25.71 oC


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Specific heat capacity of water

If you give same amount of heat to different type of matters you observe that

changes in

their

temperatures

are different.

For instance,

all you

experience that

given an equal

amount of heat

to metal

spoons and

wooden spoons, metal spoon has greater change in its temperature. Thus, most of

the housewives use wooden or plastic spoons while cooking. These examples show

that each matter has its own characteristics to absorb heat. We call this concept as

specif ic heat capacityof the matters. It is the distinguishing property of matters. We

show it with the letter c and give the definition of it as, heat required to increasetemperature of unit mass 1 C.On the contrary, heat capacity of the system is

defined as heat required increasing the temperature of whole substance and we

show it with C.

C=m.c where m is the mass of the substance and c is the specific heat of the

matter.


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Specific heat capacity


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Measurement of the specific heat capacity of a metal by an electrical method

Apparatus

Joulemeter, block of metal, heating coil to match, beaker, lagging, thermometer

accurate to 0.1 C, glycerol, electronic balance and a low voltage a.c. supply.

Procedure

1. Find the mass of the metal block m .

2. Set up the apparatus as shown in the diagram.

3. Record the initial temperature q1 of the metal block.

4. Plug in the joulemeter and switch it on.

5. Zero the joulemeter and allow current to flow until there is a temperature rise of 10

C.

6. Switch off the power supply, allow time for the heat energy to spread throughout the

metal block and record the highest temperature q2.

7. The rise in temperature Dq is therefore q2 q1.

8. Record the final joulemeter reading Q .


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Results

Mass of metal block m =

Initial temperature of the block q1 =

Final temperature of the block q2 =

Rise in temperature Dq = q2 q1=

Final joulemeter reading Q =

Calculations

The specific heat capacity of the metal c can be calculated from the following

equation:

Energy supplied electrically = energy gained by the metal block

Q = mcDq .

Sample readings might be as follows.

Mass of aluminium block = 1 kg

Initial temperature of block = 15 C

Final temperature of block = 22 C

Joulemeter reading = 6350 J

Calculations

Electrical energy supplied = energy gained by aluminium block

W= mc Dq

6350 = 1 c 76350 = 7c

Specific heat capacity of aluminium, c = 9.1 102 J kg -1 K-1


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Measurement of specific heat capacity of water by an electrical method

Apparatus

Joulemeter, calorimeter, heating coil, beaker, lagging, thermometer reading to 0.1

C, electronic balance and a low voltage a.c. supply.

Procedure

1. Find the mass of the calorimeter m cal.

2. Find the mass of the calorimeter plus the water m1. Hence the mass of the water mw

is m1 mcal.

3. Set up the apparatus as shown. Record the initial temperature q1.

4. Plug in the joulemeter, switch it on and zero it.

5. Switch on the power supply and allow current to flow until a temperature rise of 10

C has been achieved.

6. Switch off the power supply, stir the water well and record the highest temperature

q2. Hence the rise in temperature Dq is q2 q1.

7. Record the final joulemeter reading Q .


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Results

Mass of the calorimeter m cal =

Mass of the calorimeter plus the water m1 =

Mass of the water mw = m1 mcal =

Initial temperature of water qSUB>1 =

Final temperature q2 =

Rise in temperature Dq = q2 q1 =

Final joulemeter reading Q =

CalculationsGiven that the specific heat capacity of the calorimeter ccal is known, the specific

heat capacity of water cw can be calculated from the following equation:

Energy supplied = energy gained by water + energy gained by calorimeter

Q = mwcwDq + mcal ccalDq

A typical set of results might look like this:

Mass of calorimeter = 0.080 kg

Mass of calorimeter + water = 0.150 kg

Mass of water = 0.070 kg

Initial temperature of water + calorimeter = 15 C

Final temperature of water + calorimeter = 24 C

Increase in temperature = 9 K

Specific heat capacity of copper = 390 J kg1 K1

Energy supplied = 2900 J

Calculations

Q = mwcw Dq+ mcccDq

2900 = 0.07 cw 9 + 0.08 390 9

2900 = 0.63 cw + 280.8

cw = 4157.46

= 4.2 103 Jkg1 K1


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Measurement of the specific heat capacity of a metal or water by a mechanical

method

Apparatus

Copper calorimeter, copper rivets, beaker, boiling tube, lagging, thermometer

accurate to 0.1 C, heat source and electronic balance.

Procedure

1. Place some copper rivets in a boiling tube. Fill a beaker with water and place the

boiling tube in it.

2. Heat the beaker until the water boils. Continue boiling for a further five minutes to

ensure that the copper pieces are 100 C.

3. Find the mass of the copper calorimeter mcal.

4. Add cold water to the calorimeter until it is quarter full. Find the combined mass of

the calorimeter and water m1. Hence the mass of the water mw is m1 mcal.

5. Record the initial temperature of the calorimeter plus water q1.

6. Quickly add the hot copper rivets to the calorimeter, without splashing.

7. Stir the water and record the highest temperature q2. The fall in temperature Dqc of

the copper rivets is 100 C 02. The rise in temperature Dqw of the calorimeter pluswater is q2 q1.


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8. Find the mass of the calorimeter plus water plus copper rivets m2 and hence find the

mass of the rivets mc.

Results

Mass of the calorimeter mcal =

Mass of the calorimeter plus the water m1 =

Mass of the water mw = m1 mcal =

Initial temperature of water q1 =

Initial temperature of rivets 100 C

Initial temperature of calorimeter q1 =

Final temperature of water Dq2 =

Rise in temperature of water Dq2 = q2 q1 =

Fall in temperature of rivets Dq1 = 100 C q2

Mass of calorimeter plus water plus rivets m2 =

Mass of rivets mco = m2 m1

Calculations

Assume that heat losses to the surroundings or heat gains from the surroundings are

negligible.

Given that either the specific heat capacity of water cw or the specific heat capacity

of copper cc is known, the other specific heat capacity can be calculated from the

following equation:

Energy lost by copper rivets = energy gained by copper calorimeter + the

energy gained by the water

mcoccDqc = mcalccDqw + mwcwDq2

Ifcw is known, then cc can be calculated or alternatively ifcc is known, cw can be

found.

Sample results

Mass of calorimeter 0.082 kg

Mass of calorimeter + water 0.158 kg

Mass of water 0.076 kg

Mass of copper pieces 0.033 kg


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Initial temperature of copper pieces 100 C

Initial temperature of water in calorimeter 16 C

Final temperature of water in calorimeter 19 C

Specific heat capacity of water 4180 J kg-1 K-1

mcoccDq1 = mcalccDq2 + mwcwDq2

0.033 cc 81 = 0.076 4180 3 + 0.082 cc 3

2.673 cc = 953.0 + 0.246 cc

cc = 3.9 102 J kg-1 K-1


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Application of Specific Heat Capacity

Car radiator

Water is pumped through the channels in the engine block to absorb

heat.

Water is used as the cooling agent due to its high specific heat

capacity.

The hot water flows to the radiator and is cooled by the air flows

through the fins of the radiator.

The cool water flows back to the engine again to capture more heat and

this cycle is repeated continuously.

Cooking utensils

Cooking utensils are made of metal which has low specific heat

capacity so that it need less heat to raise up the temperature.

Handles of cooking utensils are made of substances with high specific

heat capacities so that its temperature wont become too high even if it

absorbs large amount of heat.

Thermal Radiator

Thermal radiators are always used in cold country to warm the house.

Hot water is made to flow through a radiator. The heat given out from

the radiator is then warm the air of the house.

The cold water is then flows back to the water tank. This process is

repeated continuously.

Water is used in the radiator because it has high specific heat capacity.


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Phenomena of specific heat capacitysea breeze

Sea Breeze

Land has lower heat capacity than sea water. Therefore, in day time,

the temperature of the land increases faster than the sea.

Hot air (lower density) above the land rises. Cooler air from the sea

flows towards land and hence produces sea breeze.

Land Breeze

Land has lower heat capacity than sea water. During night time, the

temperature of the land drops faster than the sea.

Hot air (lower density) above the sea rises. Cooler air from the land

blows towards sea and hence produces land breeze.

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Molar Heat Capacities

Molar heat capacity is heat energy to raise one mole of a substance by 1 oC.

C = (molecular weight) x (specific heat capacity)

Dulong-Petit Law is approximately true for many solids at room temperatures:The molar heat capacity of a solid is roughly 6 cal/mole/oC.

The Molar Heat Capacity of gases (at constant volume) depends on thenumber of atoms in the molecule:

Monoatomic: C = 3 cal/mole/ oC (approximate)

Diatomic: C = 5 cal/mole/ oC (approximate)

Polyatomic: C = 6 cal/mole/ oC (approximate)

Phase Changes

In a phase change, heat is added to a system without changing itstemperature.

System separates into macroscopic phaseswith different physicalproperties (ice and water, for example)

On microscopic scale this means changes in ordering of atoms (change incrystal structure, distance between atoms, magnetization, etc.)

Latent Heat

Latent Heat is the amount of heat added per unit mass of a substance duringa phase change.

Latent heat of fusion = heat per unit mass added to melt (or removed tofreeze).

Latent heat of vaporization = heat per unit mass added (or removed) for liquid-vapor phase change.

http://physics.bgsu.edu/~stoner/p201/heat/tsld011.htm


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Phase changes of latent heat of fusion and latent heat of vapourisation

Change of Phase

Matters can be in four states like solid, liquid, gas and plasma. Distance between

the molecules or atoms of the matter shows its state or phase .

Temperature and pressure are the only factors that affect the phases of matter.

Under constant pressure, when you heat matter, its speed of motion increases

and as a result the distance between the atoms or molecules becomes larger.

If you give heat to a solid substance, its temperature increases up to a specific

point and after this point temperature of it is constant and it starts to change its

phase from solid to liquid.

Another example that all you in experience daily life, when you heat water it boils

and if you continue to give heat it starts to evaporate.

In this section we will learn these changes in the phases of substances and

learn how to calculate necessary heat to change the states of them.

Melting and Freezing

If solid matters gain enough heat they change state solid to liquid. Heat is a form

of energy and in this situation it is used for the break the bonds of the atoms and

molecules. Heated atoms and molecules vibrate more quickly and break their

bonds. We call this process melting changing state solid to liquid. Inverse of

melting is called freezing, changing state liquid to solid, in which atoms and

molecules lost heat and come together, their motion slows down and distance

between them decreases.


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Look at the given graph which shows the melting of the ice.

This is a phase of change of water from solid to liquid. As you can see at the

beginning ice is at -15 C, we give heat and its temperature becomes 0 C which is

the melting point of ice. During melting process temperature of the ice-water mixture

does not change. After all the mass of ice is melted its temperature starts to rise.

Every solid matter has its own melting point; we can say that melting point is a

distinguishing property of solids. Inverse of this process is called freezing in which

liquid lost heat and change phase liquid to solid. Freezing point and melting point are

the same for same matter and it is also distinguishing property of matter.

We find the heat necessary for melting the solid substance with following formula;

Lfusion, like specific heat, it shows how much heat you should give for

melting/freezing unit of mass. For example, 3,3105 joule/kg is the latent heat of

fusion for ice and liquid.


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Example: Find the amount of heat for melting the ice having mass 1,3kg at -10

C? (Lfusion =3, 3105 joule/kg cice=2,2X103j/kg.C)

We first increase the temperature of the ice from -10 C to 0 C (melting point)

Effects of Pressure and Impurity on Freezing and Melting Point

Pressure is the force exerting on the surface perpendicularly. Thus, it helps tokeep particles together. If volume of the matter increases after melting, pressure

decreases the melting point. On the contrary, if the volume of the substance

decreases after melting, pressure increases the melting point of the matter. For

example, when you walk on the snowy road you observe that snow under your

feet melt later than around, because you exert pressure on it with your feet. Ice

melting at 0 C can be melt at -3 C with the applied pressure on it. Impurity like

pressure affects the latent heat of fusion. For instance, salty water freezes under

0 C.


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Boiling Evaporation and Condensation

Evaporation is the change of phase from liquid to gas. Evaporation occurs only

at the surface of the water and at every temperature. However, evaporation is

directly proportional to the temperature, increasing in the temperature increasingin the rate of

evaporation.

Inverse of this

process is called

condensation in

which; gas

molecules/atoms

lost heat and

change phase from

gas to liquid. As in

the case of melting,

when you give heat to liquid, at one certain point its temperature does not

change. Gained heat spent on breaking the bonds between molecules and

atoms. At this temperature, vapor pressure of the liquid is equal to the pressure

of surrounding. During this process evaporation occurs in everywhere of the

liquid which is called boiling. Boiling point is a distinguishing property of liquids;

each matter has its own boiling point. For example, water boils at 100 C in

atmospheric pressure. We use the following formula to find required heat to boil

liquid matter.

Where; m is the mass of the liquid matter and Lvaporization is the latent heat of

vaporization that shows the necessary heat to evaporate unit of mass. For

example, you should give 2,3X106 joule heat to change the phase of water from

liquid to gas.


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Example: Find the amount of heat for evaporating 2,8kg of water at 45 C?

(Lvaporization =2, 3106 joule/kg cwater=4190j/kg.C)

Effects of Pressure and Impurity on Boiling Point

Boiling occurs only when the vapor pressure of liquid and pressure of outside

equals to each other. If the

pressure of outside increases then

the boiling point of the liquid also

increases. On the contrary, if the

pressure of the outside decreases,

then boiling point of the liquid also

decreases. For example, at the top

of a mountain atmospheric

pressure is lower than the

atmospheric pressure of the sea

level. In addition to this, impurity of the liquid matter also affects the boiling point

of that matter. For instance, if you mix water with a salt or sugar, you increase the

boiling point of the water


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Sublimation

Sublimation is the change of state from solid to gas. Some of the solid matters

change their states directly to the gas with the gained heat. For example, dry ice

(frozen CO2) sublimate when heat is given. Inverse of this process is calleddeposition, in which gas matters lost heat and change their phase to solid.

Example: Graph given below shows the relation of temperature and gained heat on

different matters. Which ones of them are possible?

Line of A shows the relation that, gained heat by the matter is constant however, its

temperature is increasing. Such a relation between heat and temperature of the

matter is not possible.


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Line B shows that, temperature of the substance increases with the gained heat. It is

possible.

Line C shows that, matter gains heat but its temperature stays constant. This is also

possible; C can be change of its phase.

Line D says that, matter gains heat however, its temperature decreases. This

situation is not possible.


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Does more heat always mean a higher temperature?

From what we've said so far, you might be forgiven for thinking that giving something

more heat always makes its temperature rise. Generally that's true, but not always.

Suppose you have a lump of ice floating in a pan of water and you place it on your

hot stove. If you stick a thermometer in the ice-water mixture, you'll find it's around

0C (32F)the normal freezing point of water. But if you keep heating, you'll find

the temperature stays the same until pretty much all the ice has melted, even though

you're supplying more heat all the time. It's almost as though the ice-water mixture is

taking the heat you're giving it and hiding it away somewhere. Oddly enough, that's

exactly what's happening!

When a substance changes from solid to liquid or from liquid to gas, it takes energy

to change its state. To turn solid ice into liquid water, for example you have to push

the water molecules inside further apart and break apart the framework (or

crystalline structure) that holds them together. So while ice is melting (in other words,

during the change of state from solid water to liquid ice), all the heat energy you

supply is being used to separate molecules and none is left over for raising the

temperature.


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The heat needed to change a solid into a liquid is called the latent heat of fusion.

Latent means hidden and "latent heat of fusion" refers to the hidden heat involved in

making a substance change state from solid to liquid or vice-versa. Similarly, you

need to supply heat to change a liquid into a gas, and this is called the latent heat of

vaporization.

Latent heat is a kind of energy and, although it may seem to be "hidden," it doesn't

vanish into thin air. When liquid water freezes and turns back to ice, the latent heat

of fusion is given off again. You can see this if you cool water systematically. To start

with, the temperature of the water falls regularly as you remove heat energy. But at

the point where liquid water turns to solid ice, you'll find water freezes without getting

any colder. That's because the latent heat of fusion is being lost from the liquid as it

solidifies and it's stopping the temperature from falling so quickly.


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Measuring Specific Latent Heat

There are many methods that can be used to do this and the best one for the

purpose will depend in part on the materials being tested. The method for measuring

the latent heat of fusion of ice would not work very well for iron or aluminium. Here

are two different specific latent heat experiments for you to carry out.

Latent heat of fusion

Method 1

This method works well for ice as well as other chemicals with a melting point at a

little below room temperature.


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Experimental Setup with Power Circuit

The two set ups should be identical other than the fact that the control is not

connected to a power supply.:

Place the two immersion heaters in the funnels and pack the crushed ice

around them. The element of the heaters should be positioned towards the

end of the metal barrel; check that this is in good contact with the ice. Connect

one heater to the power supply but do not turn it on.

Record the weights of the two beakers and place them under the funnels.

Wait until both funnels are dripping water into their beakers then turn on the

power supply and at the same time start the stop clock.

Record the current and potential difference of the supply circuit and do not

alter it (if data logging equipment is available the values can be constantly

tracked and the resultant power and energy delivered calculated

automatically).

Ensure that the ice is kept topped up and in good contact with the two

heaters.

The ideal duration of the experiment will depend on several factors including

the power of the heater, the size of the funnel and room temperature etc.

However as a guide, a duration of five minutes should be adequate.

Turn off the power and stop the clock but do not move the beakers.

After another minute or so (when the flow of melt water is back to the rate of

the control setup) remove the beakers and weigh them again.

Analysis 1

Calculate the total energy (in Joules) delivered to the heater from the

equation:

Calculate the weight of water deposited in each beaker. Subtract the weight of

water in the control beaker from that in the experimental beaker. This mass,

, is equal to the weight of ice melted due solely to the effect of the heater.


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Divide the energy delivered by the mass of the water (in kilograms). This is

the Specific latent heat of ice:

A cautious experimenter would then repeat the experiment but switching the

role of the experimental and control apparatus.


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Method 2

This method works well for substances with a melting point at a modest

distance above room temperature. Chemicals such as stearic or lauric acid are

ideal.

Experimental Setup with Power Circuit

Weigh the insulated beaker.

Measure out sufficient of the test material to well cover the element of the

immersion heater when it is placed in the insulated beaker. Weigh the

measuring beaker with it contents.

Turn on the heater and fully melt the test solid so that the heater and

temperature probe can be put in place and fixed with clamps. Allow the

material to resolidify and record the temperature at regular intervals.

Allow the material to cool a few degrees below the melting point. Turn on the

heater and record the current and potential difference readings.


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Record the temperature at regular intervals until the material is a few degrees

above the melting point.

Repeat the previous two steps several times but with a different settings for

the power supply.

Analysis 2

Plot a graph for each different power run of temperature against time and use

it to measure the melting time of the sample (from when the sample begins to

melt to when it is fully liquid.

The time to fully melt the sample depends on two factors, the latent heat of

the sample and the rate of heat loss to the surroundings. The analysis is as

follows:

The latent heat of fusion of the sample is given by:

Where L is the specific latent heat of fusion and is the mass of the sample. If we

divide by time we have the power , that is going solely into melting the sample.


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But the total power supplied is divided between this and the rate of heat loss to the

surroundings :

Writing this out in full we can put the equation into form:

So if we draw a graph of total power against the inverse of the melting time we

should produce a straight line graph with a y intercept of the rate of heat loss to the

surroundings and a gradient of the specific heat capacity multiplied by the mass of

the sample.

There is one additional point that can be added to the graph as the x intercept

corresponds to zero total power and the relevant time is therefore the time it took the

sample to solidify when the power had been turned off.


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Latent heat of vaporisation

Temperature and State Changes

At sea level, water freezes at 32F (0C) and boils at 212F (100C). These are the

temperatures at which water changes state.

When a liquid boils, changes to a gas, it absorbs heat. When a gas condenses,

changes back to a liquid, it gives off heat.

Water requires one BTU of heat per pound to rise one degree Fahrenheit. If you

place one pound of water at 32F in a container over a flame, its temperature rises

1F for each BTU of heat the water absorbs from the flame.

Once the water has reached a temperature of 212F, it has absorbed 180 BTUs of

heat.

As the flame continues to heat the water, it boils, changing from a liquid to a gas,

and it continues to boil until all of it has changed to a gas.


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If this gas is collected in a container and checked with a thermometer; it would also

have a temperature of 212 F. The temperature has not risen further, but the flame

has applied an additional 970 BTUs of heat. The liquid absorbs the heat as it boils. It

is "hidden" in the water vapor.


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If the vapor contacted cool air, the heat would flow into the cooler air as the vapor

condensed back into water. This hidden heat is called the "latent, hidden, heat of

vaporization".

Water has a latent heat of vaporization of 970 BTUs. This means one pound of water

at 212 F will absorb 970 BTUs of heat when it boils and becomes a vapor. In the

same way, the vapor will give off 970 BTUs of heat when it condenses back to water.


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Latent Heat of Vaporization


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The Effects of Pressure on Boiling Points

As the pressure on a liquid is increased, the boiling point rises.

As the pressure on a liquid is decreased, the boiling point drops.

At sea level, where the atmospheric pressure is 14.7 psi, the boiling point of water is

212F (100C). At any point higher than sea level, the atmospheric pressure is lower

and so is the boiling point. In Denver, Colorado (elevation 5,300 feet), water boils at

only 206F (97C).

Atmospheric pressure is approximately 14.7 psi (absolute) at sea level, and

somewhat lower at higher elevations. At sea level, the entire weight of a "column" of

air approximately 600 miles high, presses down on everything. At higher elevations,

the column of air is shorter and the air is thinner, so the pressure is lower.


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Of course, you don't

notice the 14.7 psi pressing

in on everything, and air

pressure gages are

calibrated to read 0 psi at

atmospheric pressure. But

this atmospheric pressure

exists, and you can feel its

effects, particularly at higher

elevations; for example, if

you exercise vigorously, at a

high elevation, you become

winded more quickly.

In an air conditioning system, the pressure in the evaporator is low, so that all the

refrigerant vaporizes. The pressure in the condenser is high, so that all the

refrigerant readily changes state to a liquid.


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In an automotive cooling system, an overpressure condition is maintained to raise

the boiling point. For example, a cooling system having a pressure cap rated at 22psi

(1.5 bar) would raise the boiling point of pure water 268F (131), a 56F (31C)

increase.


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Definition of Heat Capacity

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