Form 4 Physics Chapter 4 - Teacher's Copy

CHAPTER 4 : HEAT A student is able to: 4.1.1 Explain thermal equilibrium 4.1.2 Explain how a liquid-in-glass thermometer works 4.1

Understanding Thermal Equilibrium * Choose the correct word in the bracket.

1.

*( Heat , Temperature ) is the degree of hotness of a body.

2.

*( Heat , Temperature ) is a form of energy.

3.

A hot body has a high temperature whereas a cold body has a low temperature.

4.

The SI unit for heat is joule, J.

5.

The SI unit for temperature is kelvin, K.

6.

Temperature is a *( base , derived ) quantity.

7.

Heat is a *( base , derived ) quantity.

8.

The figure shows two metal blocks in thermal contact.

A 30 0 C

(a) (b) (c) (d) (e)

B 80 0 C

Energy is transferred from *( A , B ) to *( A , B ) at a faster rate. Energy is transferred from *( A , B ) to *( A , B ) at a slower rate. Temperature A will *( increase , decrease ). Temperature B will *( increase , decrease ). The net heat will flow from *( A , B ) to *( A , B ) until they are at the same temperature.

9.

Two bodies are said to be in thermal equilibrium when : (a) they are at the *( zero , same ) temperature. (b) the net rate of heat flow between the two bodies is *( zero , same ).

10.

Temperature is measured by a thermometer with works with the principle of thermal equilibrium.

11.

Name the physical property (thermometric property) which varies with temperature used in a liquid-in-glass thermometer. Expansion of a liquid.

12.

The liquids commonly used in liquid-in-glass thermometers are mercury and alcohol.

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13.

Comparison of mercury and alcohol as a liquid-in-glass thermometer. Mercury

14.

Alcohol

Freezing point : -39 0 C Boiling point : 357 0 C It does not wet the tube.

Freezing point : -115 0 C Boiling point : 78 0 C It wet the tube.

Opaque Easy to read. It is poisonous.

colourless It needs to be dyed. It is safe liquid.

It is expensive.

It is cheap.

Conducts heat well, responds faster to temperature changes.

Responds more slowly then mercury.

Complete the following table concerning a liquid-in-glass thermometer. Features

Explanation

The glass bulb is thin. The bulb is made small. The bore of the capillary tube is narrow and uniform.

The walls of the long tube above the bulb are made thick

15.

Temperature of liquid, θ =

Heat transfer by conduction is faster and thermal equilibrium can be reached faster A small amount of liquid will be more responsive to heat. The contraction and expansion of mercury is more obvious for a small change in temperature (more sensitive). The uniform tube ensures even expansion of liquid. This acts as a magnifying glass for easy reading of the mercury thread in the stem.

lθ − l 0 × 100 0 C , l100 − l 0

Where, l 0 = length of mercury at ice point.

l100 = length of mercury at steam point. lθ = length of mercury at θ point. 16.

An uncalibrated thermometer is attached to a centimetre scale and reads 5.0 cm in pure melting ice and 30.0 cm in steam. When the thermometer is immersed in the liquid y, the length of the mercury column is 15.0 cm. What is temperature of liquid y? 40 0 C

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4.2

Understanding Specific Heat Capacity

A student is able to: 4.2.1 Define specific heat capacity I 4.2.2 State that c= Q mθ 4.2.3 4.2.4 4.2.5 4.2.6

Determine the specific heat capacity of a liquid Determine the specific heat capacity of a solid Describe applications of specific heat capacity Solve problems involving specific heat capacity

1.

The specific heat capacity of a substance is the quantity of heat needed to increase the temperature of a mass of 1 kg by 1 0 C or 1 K.

2.

Specific heat capacity, c =

Q mθ

Where, m = mass Q = heat absorbed or released θ = change in temperature 3.

The unit of specific heat capacity is J kg −1

4.

The quantity of heat absorbed or lost from a body is given by, Q = mc θ

5.

How much heat energy is required to raise the temperature of 1.5 kg of water from 30 0 C to its boiling point? The specific heat capacity of water is 4200 J kg −1 0 C −1 .

0

C −1 or J kg −1 K −1 .

441000 J 6.

Conversion of energy (a)

Electrical energy from heater transformed into heat energy. Pt = m c θ

(b)

Potential energy of a falling object transformed into heat energy. mgh = m c θ

(c)

Kinetic energy of a moving object is transformed into heat energy when it is stop due to friction. ½ mv 2 = m c θ

7.

A 700 W electric heater is used to heat 2 kg of water for 10 minutes. Calculate the temperature rise of the water. The specific heat capacity of water is 4200 J kg −1 0 C −1 . 50 0 C

3

8.

A copper block weighing 2 kg is dropped from a height of 20 m. What is the rise in temperature of the copper block after it hits the floor. The specific heat capacity of copper is 400 J kg −1 0 C −1 . 0.5 0 C

9.

10.

A bullet traveling at 60 m s −1 hit a sand bag. The temperature of the bullet rises by 4.5 0 C. Calculate the specific heat capacity of the bullet. 400 J kg −1 0 C −1 100 g of hot water at 90 0 C is mixed with 200 g of cold water at 30 0 C. Assuming that no heat is lost, calculate the final temperature of the mixture. 50

11.

4.3

0

C

Complete the following table. Material has a high specific heat capacity It takes a longer time to be heated.

Material has a low specific heat capacity It becomes hot very quickly.

It does not lose heat easily.

It lose heat easily.

It is a heat insulator.

It is a good heat conductor.

Understanding Specific Latent Heat

A student is able to: 4.3.1 State that transfer of heat during a change of phase does not cause a change in temperature 4.3.2 Define specific latent heat (l) 4.3.3 State that l = Q m 4.3.4 4.3.5 4.3.6

Determine the specific latent heat of fusion Determine the specific latent heat of 4vaporization Solve problems involving specific latent heat

1.

Matter exists in three states, that is solid, liquid and gas.

2.

The heat released or absorbed at constant temperature during a change of state of matter is known as latent heat.

4

3.

Latent heat is released Melting

boiling

Solid

liquid ………

Freezing

Gas condensation

Latent heat is absorbed 4.

Specific latent heat of fusion is the quantity of heat that is needed to change 1 kg of a substance from solid state to liquid state, without a change in temperature.

5.

Specific latent heat of vaporization is the quantity of heat that is needed to change 1 kg of a substance from liquid state to vapour state, without a change in temperature.

6.

Specific latent heat, L =

Q m

Where, Q = latent heat absorbed or released by the substance m = mass of the substance. 7. 8.

The SI unit for specific latent heat is J kg −1 . What is the quantity of heat required to melt 2 kg of ice at 0 0 C? Specific latent heat of fusion of ice = 336000 J kg −1 . 6.72 X 10 5 J

9.

Temperature /

0

C F

90 B

C

5

15

D

E

20

35 Time/minute

70

30

0

A

5

Figure shows the temperature-time graph for a substance, S of mass 2.0 kg, being heated using a 500 W heater. (a)

Based on the graph state the physical condition of substance, S in (i)

AB : solid

(ii)

BC : solid and liquid

(iii)

CD : liquid

(iv)

DE : liquid and gas

(b)

Melting point : 70 0 C

(c)

Boiling point : 90 0 C

(d)

By using kinetic theory, explain why the temperature of substance, S in AB is increasing. The heat energy absorbed is used for increasing the kinetic energy of the molecules.

(e)

By using kinetic theory, explain why the temperature of substance, S is constant in BC even though heat is still been supplied to it. The heat energy absorbed is used for breaking up the bonds of the molecules, and not for increasing the kinetic energy of the molecules.

(f)

Calculate the specific heat capacity of the substance in solid state. 1875 J kg −1

(g)

0

C −1

Calculate the specific latent heat of vaporisation of substance, S. 225000 J kg −1

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4.4

Understanding The Gas Laws

A student is able to: 4.4.1 Explain gas pressure, temperature and volume in terms of the behavior of gas molecules 4.4.2 Determine the relationship between pressure and volume at constant temperature for a fixed mass of gas ie Pv = constant 4.4.3 Determine the relationship between volume and temperature at constant pressure for a fixed mass of gas ie V/T = constant 4.4.3 Determine the relationship between pressure and temperature at constant volume for a fixed mass of gas ie P/T = constant 4.4.4 Explain about zero 4.4.5 Explain the absolute/Kelvin scale of temperature 4.4.6 Solve problems involving the pressure, temperature and volume of a fixed mass of gas 1.

Complete the table below about gas laws’ Boyle’s Law

Charles’ Law

Pressure Law

1 V

Vα T

Constant Variable : 1. Mass of gas 2. Temperature

Constant Variable : 1. Mass of gas 2. Pressure of gas

Constant Variable : 1. Mass of gas 2. volume

Boyle’s law states that the pressure of a fixed mass of gas is inversely proportional to its volume at constant temperature.

Charles’ law states that the volume of a fixed mass of gas is directly proportional to the absolute temperature at constant gas pressure.

Pressure law states that the pressure of a fixed mass of gas is directly proportional to the absolute temperature at constant volume.

P

V

P

Pα

0

V

Pα T

T/ 0 C

0

7

T/ 0 C

P

V

1 V

0

P V

0

T/K

0

P

P V

0

V

0

V T

0

V

T/K

P T

V T

0

2.

P

P

P T

0

T

0

T

Figure (a) show 18 cm of air column trapped in a capillary tube by 4 cm of mercury. If the glass tube is inverted, what is the length, L, of the air column trapped in the capillary tube? (Atmospheric pressure = 76 cm Hg)

4 cm

L

18 cm

air

air

(a)

(b)

20 cm

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3.

An air bubble released by a diver has a volume of 4.0 cm 3 at depth of 15 m. What is the volume of the bubble at a depth of 10 m? (Atmospheric pressure = 10 m water) 5 cm 3

4.

The value -273 0 C is equivalent to 0 K. This temperature is known as the absolute zero.

5.

Convert 27 0 C to its equivalent temperature in Kelvin.

6.

300 K Convert 330 K to its equivalent temperature in degrees Celsius. 57

7.

0

C

The volume of a gas is 5 cm 3 at 27 0 C. The gas is heated at fixed pressure until the column becomes 6 cm 3 . Calculate the final temperature of the gas. 87 0 C

8.

A gas of volume 20 cm 3 at 47 0 C is heated until its temperature becomes 87 0 C at constant pressure. What is the final volume of the gas? 22.5 cm 3

9.

Before a journey from Parit Buntar to Ipoh, the air in a car tyre has a pressure of 200 kPa and a temperature of 27 0 C. After the journey, the air pressure in the tyre is 220 kPa. What is the temperature of the air in the tyre after the journey? 57 0 C

10.

The pressure of gas in a light bulb is 50.5 kPa at 30 0 C. Calculate the pressure of the gas when the temperature inside the bulb rises to 87 0 C after the bulb is lighted up. 60 kPa

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