4

4.1a TemperatureTemperature in Kelvin =Temperature (oC) +273

Eg:

27oC = 27 + 273

T = 300 K

1. Complete the following conversion

a. to Kelvin from degree Celsius

i.27oC. =

ii.-27oC =

iii.500 oC =

b.from the absolute scale( K) to degrees Celsius

i.0 K=

ii.500 K=

iii.1000 K=

b.Temperature and Kinetic Energy

Increasing the average kinetic energy of the atom of the molecules in the substance results in it becoming warmer, that is a rise in its temperature.Temperature is a measure of the average kinetic energy of the atoms or molecules in a substance.

4.1b HEATCONTENT: a) Thermal Equilibrium1. Heat is form of energy which involves warm and hot conditions while temperature is its measurement.

2.The S.I unit for heat is the Joule (J) while the S.I unit for temperature is Kelvin (K).

3.Heat is the energy that transfers from one object to another because of a temperature difference.4.Two objects are in thermal contact when heat energy can be transferred between them.

Thermal equilibrium can be summarised as:

There are no net flows of heat between two objects that are in thermal equilibrium. Two objects in thermal equilibrium have the same temperature.

4.A Thermometer is an instrument which is used to measure and it is constructed base on the principle of thermal equilibrium.

ASSESMENT

Section A

1. Thermal equilibrium is reached when

A temperature of P = temperature of Q

B force on P = force on Q

C density of P = density of Q

D volume of Q displaced = volume of P

2. Thermal equilibrium between two objects mean

Aboth objects are at the stable phase at the same temperature

Bnet heat is zero

Cthere is heat transfer from one object to another object.

Done object is cooler than the other object.

3. X Y

The Figure above shows two objects, X and Y, which are touching each other. At the beginning, object X is hotter than object Y. Thermal equilibrium is achieved when

A materials of X and Y are the same

B specific heat capacities of X and Y are the same

Cthe rate of heat transfer is the same for both X and Y

Dthermal equilibrium cannot be achieved because the temperature of object X is higher than that of object Y

4 Thermal characteristic is the physical characteristic of the

material that will

A change with temperature

Bdecrease with temperature

C increase with temperature

5 Which of the following is not a criteria in the making of a thermometer?

A obey Charles Law

Bhas physical characteristic that changes with temperature

C has two fixed temperature points D the range of temperature between

two points can be sub-divided into smaller scales

Section B

1. A liquid X of mass 250g was heated until almost boiling point and left to cool down as shown in figure 1 below. The timing is started when Bunsen burner was removed. Figure 2 shows the graph of temperature against time for the cooling of the liquid.

[Specific heat capacity of liquid X = 4200 JkgoC-1]

a)(i)What is meant by thermal equilibrium ?

(ii)From the graph in figure 2, what is the temperature of liquid X when it is the thermal equilibrium with the surroundings?

(iii) Explain in term of transfer of energy how liquid X achieves to thermal equilibrium with its surroundings.

(iv) State the temperature of the surroundings in his experiment.

b)(i)Calculate the amount of heat lost by liquid X before attaining thermal equilibrium.

(ii)Calculate the time taken by a 48 W heater to reheat liquid X to a

temperature of 95o C.

Marking scheme

Section A

1. A

2. B.

3. C.

4. A 5. A

Section B

1. (a) (i) Two objects in thermal equilibrium have the same temperature and there is no net flow of heat between them.

(ii)35oC

(iii) The initial temperature of liquid X is higher than the surroundings.

There is a net transfer of heat from liquid X to the surrounding.

This lost of heat results in the drop in temperature for liquid X.

When the temperature of liquid X is equal to the temperature of the

surrounding, thermal equilibrium is achieved and there is no net

transfer of heat between liquid X and surroundings.

(iv)35oC

(b)(i)Q = mc

= 0.250 x 4200 x (90- 35)

= 57 750 J

(ii) Time taken = t

Pt = mc

48 x t = 0.250 x 4200 x (90- 35)

t = 1312.5 s

4.2 Understanding Specific Heat Capacity

Situation

You can hold a freshly fried piece of pisang goreng and bite into the batter quite comfortably, but the hot banana filling will burn your tongue.

Definition of heat capacity

The heat capacity of an object is the amount of heat required to change the temperature of the object by 1oC or 1K. Unit of heat capacity is expressed in JoC-1 or JK-1.

The ability of a substances to heat up, or store heat (actually store internal energy), varies for different things. It is called heat capacity.

E.g.

A pot of rice has a larger heat capacity than a bowl of rice.

A piece of stone by the road side has a smaller heat capacity than a big rock at the side of the hill.

Definition of specific heat capacity

The specific heat capacity of a substance is the amount of heat that must be transferred to change the temperature by 1oC or 1K for a mass of 1 kg of the substance.

Q is the heat energy transferred to the substance in joules (J)

m is the mass of the material in kilograms (kg)

is the temperature change in oC or Kelvin

c is the specific heat capacity in Jkg-1oC-1 or Jkg-1K-1Applications:

(a) When two objects of equal masses are heated at equal rates, the object with the smaller specifics heat capacity will have a faster temperature increase.

(b) When two objects of equal masses are heated, to obtain the same temperature increase, more heat needs to be supplied to the object with a larger specific heat capacity.

(c) Cooking pot:

(i) The body, base and handle of the cooking pot in figure are made of materials

with different specific heat capacities.

(ii) Table below shows the characteristics of the parts of the cooking pot.

PartCharacteristics

Base-Copper base.

-Low specific heat capacity. Becomes hot very quickly. Enables

quick cooking of the food in the pot.

-High density. The heavier base ensures that the pot is stable and

Will not topple over easily.

Handle-Handle made of synthetic material or wood.

-Large specific heat capacity. Will not become too hot when heat is

absorbed.

-Poor conductor of heat. Very little heat from the body and contents of

the pot are transferred to the hand of the person holding the pot.

-Low density. Does not add very much to the total weight of the pot.

Body-Aluminium body.

-Relatively low specific heat capacity. Becomes hot quickly.

-Low density. Reduces the overall weight of the pot.

-Does not react with the food in the pot.

Example 4.21

How much heat energy is required to raise the temperature of a 3kg sheet of glass from 24oC to 36oC? [Specific heat capacity of glass = 840 Jkg-1oC-1]

Solution

Mass, m = 3kg

Change in temperature,

= 36 oC 24 oC

=12 oC

Heat energy required, Q

=

=3 x 840 x 12

=30240 J

Example 4.22

The water in the ice maker of a refrigerator has a mass of 0.4kg and temperature of 22oC. What is the temperature of the water after 33600J of heat has been removed from it?

[ Specific heat capacity of water = 4200 Jkg-1oC-1]

Solution

Mass, m = 0.4 kg

Heat removed, Q = 33600 J

Let be the drop in temperature of the water.

Q =

33600 = 0.4 x 4200 x

= = 20oC

Final temperature of the water = 22 20 = 2oC

Exercises

1. The specific heat capacity of aluminium is larger than the specific heat capacity of lead. This means that

A. a 1 kg block of aluminium contains more heat energy than a 1 kg block of lead.

B. a block of aluminium will heat up faster than a block of lead of equal mass when they are heated at the same rate.

C. More heat is required to raise the temperature by 1oC for 1 kg of aluminium than for 1 kg of lead.

D. a block of aluminium will cool down faster than a block of lead of equal mass when they lose heat at the same rate.

2. If you stand near a satay stall when the seller is grilling the sticks of satay, you will notice small fragments of red hot charcoal ambers flying up from the flames. When these ambers touch your arm you feel no pain or just a slight sensation of hotness even though the red hot ambers are at a few hundreds degrees Celsius. Can you explain why?

Answer

The charcoal fragments have very small mass. Thus the heat capacity is very small. Though there is a big temperature drop when the fragments come into contact with our skin, only a very small amount of heat is released.

3. At a certain section of the Victoria Falls in Africa, water drops vertically through a height of 480m.

(a) Explain why the water at the base of the waterfall could have a temperature slightly higher than the water at the top.

(b) Estimate the maximum possible difference in temperature between the water at the base and at the top of the waterfall. (Take g = 10 ms-1)

Solution

(a) As the water falls, it loses potential energy. Some of its potential energy is

converted to heat energy to increase the temperature of the water.

(b) Assume all the potential energy lost by the falling water is converted to heat energy. Therefore

mgh =

10 x 480 = 4200 x

= 1.14oC4. Figure below shows an immersion heater with a power rating of 240 V, 500 W

used to boil 500 g of water.

The initial temperature of the water is 25 o C and the specific heat capacity of the water is 4200 Jkg -1 o C-1 .

(a) Calculate the energy used by the heater in 15 minutes in the joule.

Energy= power (W) x time (s)

= 500 x 15 x 60

= 450 000 J(b) If the mass of the metal container is 100 g and its specific capacity is 1200 Jkg -1 o C-1 , calculate the total heat required to boil the water (assume the temperature of the water and container are the same).

Total energy

= Energy to heat metal container + Heat to boil water

= mccc + mwcw

= (0.1 X 1200 X 75) + (0.5 X 4200 X 75)

= 9000 + 157 500

= 166 500 J

(c) i)What is the minimum time required to boil the water?

Pt = Total energy

500t= 166 500

t=333 s

(ii) The actual time required is more than time calculate in (c) (i) . Explain.

This is due to heat lost to the surroundings.

(d) A student propose to replace the container with a similar container but with its external surface painted white. Will this reduce the time required to boil the water?. Explain.Yes, because by this method, heat loss to the surroundings can

be reduce.

(e) Figure 8 (b) shows a situation where the immersion heater is only half immersed in the water. Comment on the time taken to boil the water.

The time taken to boil the water will increase because part of the heat from the heater is lost to the surroundings by radiation.

Application

4.3 UNDERSTANDING SPECIFIC LATENT HEAT

By the of this subtopic, you will be able to

State that transfer of heat during a change of phase does not cause a change in temperature Define specific latent heat State that = Q/m Determine the specific latent heat of fusion and specific latent heat of vaporisation Solve problem involving specific latent heat.What is latent heat?

Latent heat is the total energy absorbed or released when a substance

completely changes its physical state at constant temperature

heat absorbed

melting

vaporisation

ice water

(solid) fusion (liquid) condensation steam

(gas )

heat released

There are 2 types of latent heat:

i. latent heat of fusion

solid liquid

ii. latent heat of vapourisation

liquid gas

*** both happening at constant temperature

Definition of specific latent heat of fusionIs the quantity of heat energy required to change 1kg of a substance from solid to liquid without change of temperature.

Definition of specific latent heat of vaporisation

Is the quatity of heat energy required to change 1kg of a substance from the liquid

phase to gaseous state, without change in temperature.

Application of Specific Heat Latent Heat in Daily Life

1. Water as a Coolant

When 1 kg of ice melts , because of its high latent heat (3.36 X 105 J), it will absorb a large amount of heat from its surrounding. This property allows ice to be used as a coolant to maintain other substances at cooler temperature.

Add ice to drinks e.g milo, its cooled down because ice absorps the heat from the drink and melts. The same applies to vegetables and other fruits when we place them in contact with ice. The ice absorbs large amount of heat and melts at the same time , foodstuff can be maintained at low temperature.

2. Steam for cooking

Water has a large specific latent heat of vaporisation. So when steam condenses to water, it releases large amount of heat (2.26 X 206 J). This heat is used in steaming fish, cakes, buns. This large amount of heat can be used to cook food at a faster rate.

*** Asians drink hot drink in the afternoon, so that we sweat . When we do sweat, large amount of heat from the body is absorbed by sweat when it evaporates. Naturally our body is cooled down.

TUTORIAL

1. A block of iron with mass, mi and temperature, w is out into a beaker containing water of mass, mw and temperature, w. . The iron block then cools down to an equilibrium temperature of . The specific heat capacities of iron and water are ci and cw respectively. What is the heat lost by the iron block?

Amwcw ( - w )

C mici(i - )

Bmwcw (i - )

D mici ( w - )

2. Which of the following liquid A, B ,C or D requires the most heat to change it totally from a liquid into a gas at the boiling point of the liquid?

m/gl/J kg -1

A2003.0 106

B1802.6 106

C5002.3 106

D4003.1 106

3.A block of aluminium with a mass of 5 kg is given 40.5 kJ of energy. Its temperature rises by 9C. What is the specific heat capacity of aluminium?

A 700 J kg-1C-1

C 900 J kg-1C-1 B 800 J kg-1C-1

D 1000 J kg-1C-1E 1100 J kg-1C-14.Scalding of the skin by boiling water is less serious then by steam. This is because

A the boiling point of water is less than the temperature of steam

B the heat of boiling water is quickly lost to the surroundings.

C steam has a high specific latent heat.

D steam has a high specific heat capacity.

5.The rate of evaporation of water in a glass depends on the following factors except..

A the temperature of the water

B the area of the water

C the depth of the water

D the moisture content of the atmosphere.

6.The latent heat of fusion of a substance is the quantity of heat required to..

A separate the molecules in the solid substances so that they are able to move further..

B increase the kinetic energy of the molecules in the solid substances.

C increase the temperature of the substance

7. The specific latent heat of fusion is the heat required to

A. change the temperature of 1 kg of water by 1 C

B. change the temperature of 1 kg of ice by 1 C

C. change 1 kg of water to steam at 100 C

D. change 1 kg of ice into water at 0

8. Specific latent heat of vapourisation is measured in

A. J C-1

B. J kg-1

C. J kg-1C

D. J kg-1C-19. Latent heat is absorbed when

A. water is heated

B. water is cooled

C. water changes into steam

D. water changes into ice

10. Water is used as a coolant in motor car engines because

A. water has low viscosity

B. water helps to clean the engine

C. water has a high specific heat capacity

D. water has a high specific latent heat of fusion

11. When the boiling temperature of water exceeds 100 C, it may be because

A water has been boiled for a long period.

B the volume of water is too large

C of the presence of impurities.

Answers.

1.C

2.D

3.C

4. C

5. C

6. B

7. D

8. B

9. D

10. D

11. C

TUTORIAL 2

1. A P watt immersion heater is used to melt ice at 0 oC. If 89.3 g of ice is melted in 5 minutes, what is the value of P? (Specific latent heat of fusion of ice = 3.36 x105 Jkg-1 ).

2. Calculate the total heat energy required to change 0.5 kg of water at room temperature( 27oC) to steam.

3. In an experiment to determine the specific latent heat of fusion of substance K, 1.5kg of substance K is heated at a rate of 250 W. Figure 1 below shows the graph of temperature against time.

/oC

60

35

X

Y Z

0 2 4 6 8 10 12 t/min

Figure1

(a) What are the states of matter in regions X,Y, and Z?

(b) State the melting point of K.

(c) What is the time taken to melt substance K completely?

(d) Calculate the total energy required to melt substance K completely.

(e) Calclute the specific latent heat of fusion of substance K.

(f) Explain in relation to the kinetic theory of matter why

(i) the temperature of the substance remains constant in the region Y eventhough it continues to absorb heat.

(ii) The temperature of the substance rises in region Z.

Answers:

1. P =100 W

2. Total heat required = 1283.3. kJ

3. (a) X solid, Y Liquid and solid, Z Liquid

(b) 35 oC

(c) 6 mins

(d) Q = 9.0 x104 J

(e) 6.0 x 104 J

(f) (i) In the region of Y, the heat absorbed. This is used to overcome the force of attraction between the molecules so that they can have a wider range of movement in the liquid state thus maintaining a constant temperature throughout the melting process.

(ii)In the region Z , the substance exist in the liquid state only. The heat absorbed in this region causes the average kinetic energy of the molecules to increase thereby causing its temperature to rise.The Gas Laws

Property and symbolSI UnitSymbol in SI unitOther Units

Pressure, P

pascal, (Nm-2)

Pa

N cm-2, cm Hg

Volume, V

(metre)3m3

mm3, cm3

Temperature, T

kelvin

K

C, F

For a gas in an enclosed container, the

(a)number of molecules is constant.

(b)mass of the gas is constant.

(c)behaviour of the gas depends on the volume, temperature and pressure of the gas.

Property of

a gasExplanation based on the kinetic theory

VolumeThe molecules are moving freely in random motion. Therefore the molecules fill up the whole space in the container. The volume of the gas is equal to the volume of the container

TemperatureThe molecules are in continuous random motion. They have an average kinetic energy that is proportional to the temperature.

PressureThe molecules are in continuous random motion. At every instant, molecules collide with the walls of the container and bounce back. There is a change in momentum and a force is exerted on the walls. The total force per unit area is the pressure of the gas.

The relationship between pressure, volume and temperature can be investigated in three ways.

Boyles Law

Boyles law gives the relationship between the pressure and volume of a fixed mass of gas at constant temperature.

The relationship between pressure and volume can be explained using the kinetic theory of gases:

(a)When the volume of a gas is decreased, the number of molecules per unit volume increases.

(b)The same number of molecules moves in a smaller space.

(c)The molecules collide more frequently with the walls of the container.

(d)The increase in the rate of collision results in an increase in the pressure exerted by the gas.

Boyles law states that for a fixed mass of gas, the volume of the gas is inversely proportional to its pressure when the temperature is kept constant.

The mathematical expression for Boyles law is,

that is PV = constant, when the temperature is kept constant.

Consider a gas with initial pressure P1 and volume V1. When the pressure is changed to P1, the volume of the gas changes to V2.

According to Boyles law, when temperature is constant.

The relationship between pressure and volume can also be expressed with the following graphs.

The SI unit for temperature is Kelvin (K).

The relationship between temperature in Kelvin and degrees Celsius for three common temperatures.

Temperatures in C can he converted to K by adding 273.

C = ( + 273) K

Examples:

An iron cylinder containing gas has a pressure of kPa when it is kept in a store at temperature 27oC What is the pressure of the gas when the cylinder is moved outdoors where the temperature is 40C?

Answer:

P1=360 kPaT1 = (27+273) = 300 K

P2 = Final pressure= (40 + 273) = 313 K

Using the pressure law,

P1=P2

T1T2

360=P2

300313

P2 =360x313

300

Therefore P2 = 375.6 kPa

Charles Law

1.Charles law gives the relationship between the volume and temperature of a fixed mass of gas at constant pressure.

2.The relationship between volume and temperature can be explained using the kinetic theory of gases:

(a)When a gas is heated, the average kinetic energy of the molecules increases. The temperature of the gas increases.

(b)The rate of collision between the molecules and the walls will increase if the volume is constant.

(c)If the gas is allowed to expand, the faster molecules now move in a bigger space.

(d)Therefore, the rate of collision between the molecules and the walls remain constant and thus the pressure is constant.

3. When the pressure is kept constant, Charles Law states that for a fixed mass of gas, the volume of the gas is directly proportional to its absolute temperature.The mathematical expression for Charles Law is

4.Consider a gas with initial volume V, and temperature T1. When the temperature changes to T2, the volume of the gas changes to V2.

5.According to the Charless law, when the pressure is constant, 6.The relationship between volume and absolute temperature can also be expressed with the following graphs in Figure 4.61.

Example:

A syringe in a refrigerator contains 4.5 ml of air at -3C. When the syringe was taken out and placed in a room where the temperature was 27C, the air in it expands. Calculate the final volume of the air in the syringe.

V1 = 4.5 mlT1 = (-3 + 273) = 270 K

V2 = Final volumeT2 = (27 + 273) = 300 K

Answer:

Using the Charles law,

4.5=V2

270300

V2 =4.5 x300

270

Therefore V2 = 5.0 ml

Short Note

A fixed mass of gas has pressure, volume and temperature.

Boyles law states that for a fixed mass of gas, the pressure of the gas is inversely proportional to its volume when the temperature is kept constant.

Charles law states that for a fixed mass of gas, the volume of the gas is directly proportional to its absolute temperature when its pressure is kept constant.

The pressure law states that for a fixed mass of gas, the pressure of the gas is directly proportional to its absolute temperature when its volume is kept constant.

Gas LawsObjective Question:

1.A flask contains air at a pressure of 150 kPa and temperature 27C.The flask is then immersed in a hot water with a steady temperature of 87C. Calculate the new value of the air pressure.

A 47 kPa

B 125 kPa

C 180 kPa

D 483 kPa

2..The figure shows air trapped in a glass tube. Which of the following ways can increase the length of the air column?

A The glass tube is heated.

BThe open end of the tube is inclined upwards.

C There is an increase in atmospheric pressure.

DThe mercury is replaced with concentrated sulphuric acid.5.The volume of an air bubble in water is V at a depth of 10 m and V2 at a depth of 5 m. If the atmospheric pressure at the surface of the water is 10 m of water, what is the ratio of V1 to V2?

A 4:3

B 3:4

C 2:1

D 1:2

6.Which physical quantity is a constant in all the three gas laws?

ADensity

BAtmospheric pressure

CMass

7. A fixed mass of gas in a closed container is heated. The following statements are true except

A the gas pressure increases.

Bthe average distance between molecules increases.

Cthe average kinetic energy of molecules increases.

Dthe frequency of collisions of the molecules increases3.The air in a tube has a pressure of 10 kPa at a temperature of 27 0C. When the pressure in the tube becomes 35 kPa, what will be its temperature?

A1 050 0C

B777 0C

C770 0C

D105 0C

E77.7 0C

4An air bubble has a volume of 10 cm3 at the bottom of a 50 m deep point. What will be its volume at the surface of the point? [Atmospheric pressure = 10 m water]

A6 cm3

B50 cm3

C60 cm3

D500 cm3

E600 cm3

Structure Questions:

1.A student wanted to use a flask as shown in Figure 1 to measure temperature. The flask was placed in pure melting ice and then in steam. The readings of the pressure gauge is given in Table 1.

.

Figure 1

Fixed pointTemperature/

CReading of pressure gauge/ kPa

Ice point0110.0

Steam point100150.3

Table 1

(a)(i)Why is a flask with a large volume was used?

[1 mark]

(ii)Explain why the pressure did not give a zero reading at temperature 0C.

[1 mark]

(b)(i)Sketch the graph of pressure against temperature.

[2 marks]

(ii)If the graph is extrapolated, at what temperature in 0 0C will the pressure become zero?

[1 mark]

(iii)Name the temperature in (b)(ii) when the pressure of the air becomes zero. [1 mark]

(iv)Name the temperature scale that starts with zero at zero air pressure

[1 mark]

(v)State the gas law that relates the pressure of a gas and the temperature measured with the scale in (b)(iv).

[1 mark]

(c)Using the gas law in (b)(v), calculate the pressure of the air when the flask is cooled to -23C.

[3 marks]

2.Figure 4 shows a manometer connected to a flask which is immersed in water bath. Liquid Y is 14 cm in height and the water is 20 cm in height. Before starting the experiment, the position of the level of liquid Y is shown in Figure 4.(a)Calculate the density of liquid Y.

[2 mark]

(b)When the flask is heated, the level of the mercury rises in the right arm of the manometer.

i.Based on the kinetic theory of matter, explain the change which occurs to the air in the flask.

[1 mark]

ii.State the principle of physics involved in the experiment.

[1 mark]

iii.What is the temperature of the air if the air pressure becomes 6 x 105 Pa when the initial temperature of air in the flask is 27C at a pressure of 5 x 105 Pa.

[1 mark]

(c)What is the change that occurs in the mercury level if a longer rubber tube is used? [1 mark](d)Why is an asbestos shield is placed between the manometer and the water bath? [1 mark]

Essay Questions

1.A diver near the bottom of the sea observed that air bubbles released by him increase in size as he moved up towards the surface.Using this observation;

(a)make one suitable inference,

[1 mark]

(b)state one appropriate hypothesis that could be investigated.

[1 mark]

(c)describe how you would design an experiment to test your hypothesis using a thick glass syringe and other apparatus. In your description, state clearly the following:

(i)aim of the experiment

(ii)variables in the experiment

(iii)list of apparatus and materials

(iv)arrangement of the apparatus

(v)the procedure of the experiment, which includes the method of controlling the manipulated variable and the method of measuring the responding variable

(vi)the way you would tabulate the data

(vii) the way you would analyse the data

[10 marks]

oC

Kelvin

273K

0 oC

-273oC

0 K

Aluminium body

Wooden handle

Aluminium

body

Copper base

Copper base

Wooden handle

(a) (b)

PAGE 15

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