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Chap 11 ThermalProperties 12 S

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9/22/20

Thermal Properties of Matter

Syllabus (Pure)

a. Describe a rise in temperature of a body in terms of an increase in its

internal energy (random thermal energy).

b. Define the terms heat capacity and specific heat capacity.

c. Recall and apply the relationship thermal energy = mass x specific heat

capacity x change in temperature to new situations or to solve related problems.

d. Describe melting/solidification and boiling/condensation as processes of

energy transfer without a change in temperature.

e. Explain the difference between boiling and evaporation. f. Define the term latent heat and specific latent heat.

g. Recall and apply the relationship thermal energy = mass x specific latent

heat to new situations or to solve similar problems.

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Internal Energy

 The internal energy of the body is the sum of all kinetic and

potential energies of its molecules.

Kinetic energy in the movement of the molecules

Potential energy in the form of the inter-molecular force

 When an object is heated, there is energy transfer from the

heat source to the object. The molecules of the object gains kinetic energy. The temperature of the object also rises.

 The temperature of the object is dependent of the average

kinetic energy of the molecules.

 The internal energy of the body is proportional to the

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Heat Capacity

Heat capacity:

• The amount of heat required to raise the temperature of

the object by 1K or 1°C.

SI unit: J/ K or J/ °C.

C = Q /

Q is heat absorbed or released, in J.

•  Is change in temperature in K or °C.

E.g.11.1 (pg.190)

Question:

When the same amount of heat is supplied to two metal

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Specific Heat Capacity

The amount of heat required to raise the temperature of 1kg

of the object by 1K or 1°C.

SI unit: J / (kg K) or J / (kg °C).

Specific heat capacity, c = C / mass = Q / m ,

Q is heat absorbed or released, in J.

•  Is change in temperature in K or °C.

C is the heat capacity

Q = mc

E.g. 11.2/11.3 (pg. 193 / 194)

Application of high specific heat capacity in cooling fluid and

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To Determine the SHC

Note that knowledge of this

experiment is out of the syllabus.

 Measuring specific heat capacity of

water.

 Here, the beaker contains 0.5 kg of

water. When the I00 W electric heater is switched on for 230 seconds, the

temperature of the water rises by 10

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To Determine the SHC

Energy transferred to water = mc

= 0.5 x c x 10.

Energy- supplied by heater = power x time.

Heat absorbed by water = heat supplied by heater

0.5 x c x 10 = 100 x 230 = 23 000 J.

c = 4600.

Hence, the specific heat capacity of water is 4600 J/

(kg K).

This method makes no allowance for any thermal

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Heat Transfer Calculations

Power x time = mc

(Assuming no

thermal energy losses).

For all calculations, always note the following:

Heat supplied, heat loss

Heat gained

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Melting and solidification

Melting is a process which a substance changes its state

from solid to a liquid, without a change in the temperature.

Pure substance melt at a definite temperature, known as its

melting point.

Although the solid (e.g. Ice) is absorbing the heat, the

temperature remains constant until all the solid has melted. Only then, will temperature of the melted solid start to rise.

Thus, the energy resulted in a change of state rather than a

change in temperature.

What happens to the energy?

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Temperature Time Graph

Graph show that the temperature of the object increases as

time increases. Can you identify:

The melting point

Time taken for the object to melt completely

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Cooling Curve

A cooling can be used to determine the melting point of a

substance.

A substance is heated until it has melted. It is then left to

stand and the temperature is taken at a regular interval. A temperature time graph is plotted.

Three clearly defined portions of the graph.

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Boiling and Condensation

 Boiling is the process in which a substance changes from the

liquid state to the gaseous state, without a change in temperature.

 Boiling point is the definite temperature at which a pure

substance boil.

 The heat gained by the substance is the latent heat of

vaporization.

 Condensation point is the temperature at which the substance

will change from a gaseous state to the liquid state.

 When steam condense, it releases its latent heat of

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Evaporation

Evaporation occurs when a liquid changes to a gas,

below its boiling point.

Some particles in the liquid moves faster than the

rest, the faster ones near the surface of the liquid

has enough energy to escape into the air.

Evaporation occurs at all temperature, and takes

place only on the exposed surface of the liquid.

The heat required is absorbed from the

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Evaporation

Factors affecting the rate of evaporation

Temperature

Higher temperature

Area of exposed

area

Larger area

Humidity of

surrounding air

Low humidity

Motion of air

Windy

Pressure

Lower pressure

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Evaporation and Boiling

Similarity:

both have the same change of state

both require latent heat of vaporisation

Differences

Factors

Boiling

Evaporation

Speed of process

Quick

Slow

Visible

observation

Location of

occurrence

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Impurities and Pressure

Summary of factors for water

Add

impurities

Pressure

Increase

Melting

point

Lowers

Lowers

Boiling

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Latent Heat of Fusion

 Latent heat of fusion: the energy supplied to the particles to

overcome the attractive forces in solids.

 Work done is converted to potential energy, there is no rise

in temperature.

 During solidification, energy is released.

 Specific latent heat of fusion (lf): the quantity of heat that is

needed to change unit mass of the substance from solid state to liquid state, without a change in temperature.

 SI unit: J/kg

lf = Q / m, Q is amount of heat & m is mass.

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Latent Heat of Vaporization

Latent heat of vaporization: energy needed to overcome the

attractive forces holding the particles together.

Specific latent heat of vaporization (lv): the quantity of heat

that is needed to change unit mass of the substance from liquid state to vapour state, without a change in temperature.

SI: J/kg

lv = Q /m

Eg. 11.6 (pg. 204)

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Summary

Gas Liquid

Solid

Latent heat Change in state

causes

melting freezing

boiling or

evaporation condensation

Specific latent heat l = Q/m

(in J kg-1)

Heat capacity C = Q/m (in J kg-1)

Change in temperature

depends on

Specific heat capacity

c = Q/ m

(in J kg-1 0C-1)

depends on

Specific latent heat of fusion lf

(in J kg-1)

Specific latent heat of vaporisation lV

(in J kg-1) for change from

liquid to gas

for change from solid to liquid

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Challenge

A glass of water of mass 500g is at 30°C. 40g of

ice is added to the glass of water. Assuming no

loss of heat to the glass or the surrounding,

calculate the final temperature of the mixture.

(Assume specific heat capacity of water is 4200 J/

kg K, and specific latent heat of ice is 336 000

References

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