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UNITY SECONDARY SCHOOL PHYSICS CHAPTER 11: THERMAL PROPERTIES OF MATTER ANSWERS NAME:___________________________(
) CLASS: ________ DATE: _________
Exercise 11A: Heat Capacity and Specific Heat Capacity 1.
Define heat capacity. The heat energy required to raise the temperature of a substance by 1 °C. ________________________________________________________________________ ________________________________________________________________________
2.
Define specific heat capacity. The heat energy required to raise the temperature of 1 kg of a substance ________________________________________________________________________ by 1 °C. ________________________________________________________________________
3.
Distinguish between the heat capacity C and the specific heat capacity c of a body. How are they related? C is heat energy required to raise the temperature of a substance by 1 °C but c is for 1 ________________________________________________________________________ kg of the substance. C is directly proportional to c or C=mc ________________________________________________________________________
4.
Explain what is meant by “the specific heat capacity of water = 4.2 kJ kg-1 0C-1 “. Heat energy of 4200 J is required to raise the temperature of 1 kg of water by 1 °C. ______________________________________________________________________ ______________________________________________________________________
Materials
c / J kg-1 K-1
Materials
c / J kg-1 K-1
Plain water
4200
Lead
460
Brass
380
Mercury
140
Copper
400
Methylated spirit
2400
Glass
670
Sea water
3900
Ice
2100
Aluminium
900
Iron
460
Zinc
390
Chapter 11: Thermal Properties of Matter
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Science Department
5.
A kettle is rated at 25 W. Calculate: (a) the quantity of heat generated in 2 s, Q = Pt = 25 × 2 = 50 J
Heat generated in 2 s = 50 J (b) the rise in temperature of 150 g of water if the electric kettle is switched on for 5 minutes. The specific heat capacity of water is 4 J (g°C)-1. heat gained = heat sup plied mcθ = Pt 50W × 5 min× 60 θ = 150 g × 4 J ( g °C ) = 12.5°C
Rise in temperature = 12.5°C
6.
A 800 W heater is totally immersed in a huge block of mass 10 kg which is at a temperature of 30 °C. In 7 minutes, the temperature of the mass rises to 80 °C. Calculate the specific heat capacity of the mass.
Heat gained = Heat sup plied mcθ = Pt Pt c= mθ 800 × 7 × 60 = 10 × ( 80 − 30 ) = 672 J /( kg °C )
Specific heat capacity of the mass = 672J/(kg°C) Chapter 11: Thermal Properties of Matter
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7.
The arrangement shown in the diagram is used to determine the specific heat capacity of an unknown metal block. The circuit is switched on for a time interval of 500 s and the following readings are obtained: thermometer V A
heater
metal block
Change in thermometer reading = 50 °C Mass of metal block = 1 kg Ammeter reading, I = 5 A Voltmeter reading, V = 8 V (a) Using the above data, calculate the specific heat capacity of the unknown metal. (Note: Power = V x I) Heat gained = heat sup plied
mcθ = Pt Pt c= mθ VIt = mθ 8 × 5 × 500 = 1 × 50 = 400 J /( kg °C ) Specific heat capacity of unknown metal = 400J/(kg°C) (b) Is it advisable to take the thermometer reading immediately after switching on the current? Why? No. Some time is needed for the heat to be evenly distributed through the metal ______________________________________________________________ block before an accurate reading of the final temperature can be taken. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ Chapter 11: Thermal Properties of Matter
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8.
A lump of metal of mass M x and initial temperature T x is dropped into water of M w and temperature T w .
X Water Mass = Mw Initial Temp = Tw Final Temp = ?
Mass = Mx Initial Temp = Tx
Based on the following assumptions, 1. T x is higher than T w 2. There is no change in state for the water. 3. There is no heat loss. determine an expression for the final temperature of the mixture. Use c x and c w for specific heat capacity of X and water respectively.
M x c x ( Tx − T f ) = M w c w ( T f − Tw ) M x c xTx − M x c xT f = M w c wT f − M w c wTw M x c xTx + M w c wTw = M w c wT f + M x c xT f = T f ( M w cw + M x c x ) Tf =
9.
M x c xTx + M w c wTw M w cw + M x c x
A block of zinc has a mass of 2 kg. It is dropped into water of mass 3 kg and temperature 10 °C. The final temperature of the water and the zinc block is 30 °C. Find the initial temperature of the block of zinc.
Heat loss by zinc = Heat gain by water m z c z ( T − 30 ) = m w c w ( 30 − 10 ) m z c zT − 30 m z c z = 20 m w cw T =
20 m w c w + 30 m z c z mz cz
20 ( 3 )( 4200 ) + 30 ( 2 )( 390 ) ( 2 )( 390 ) = 353°C =
Initial temperature of zinc = 353°C
Chapter 11: Thermal Properties of Matter
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10.
The initial temperature of a container of 3 kg water is 90 °C. When a block of metal of temperature 25 °C is added into the water, the temperature of the water drops to 60 °C. If the metal is aluminium, find its mass.
Heat loss by water = Heat gain by Alu min ium m w c w ( 90 − 60 ) = m Al c Al ( 60 − 25 ) 30 m w c w = 35 m Al c Al m Al =
30 m w c w 35 c Al
30 ( 3 )( 4200 ) ( 35 )( 900 ) = 12 kg =
11.
Mass of aluminium block = 12 kg
The initial temperature of a container of 5 kg water is 40 °C. When a block of metal of temperature 500 °C is added into the water, the temperature of the water rises to T f . If the block is 1 kg of iron, find T f .
Tf =
M Fe c FeTFe + M w c wTw M w c w + M Fe c Fe
( 1 )( 460 )( 500 ) + ( 5 )( 4200 )( 40 ) ( 5 )( 4200 ) + ( 1 )( 460 ) = 49.9°C =
T f = 49.9°C
12.
The initial temperature of a container of 1500 g water is 32 °C. When a 1 kg block of temperature 300 °C is added into the water, the temperature of the water rises to 58 °C. What material is the block?
Heat loss by block = Heat gain by water mb cb ( 300 − 58 ) = m w c w ( 58 − 32 ) 242 mb cb = 26 m w c w cb =
26 m w c w 242 mb
26 ( 1.5 )( 4200 ) ( 242 )( 1 ) = 677 J /( kg °C ) =
The block is made of glass Chapter 11: Thermal Properties of Matter
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UNITY SECONDARY SCHOOL PHYSICS CHAPTER 11: THERMAL PROPERTIES OF MATTER ANSWERS NAME:__________________________(
)
CLASS: ________ DATE: _________
Exercise 11B: Melting and Freezing 1.
What are the three states of matter? Solid, liquid and gas _________________________________________________________________
2.
Complete the table below: Process Melting
Freezing
Change of state
Energy (absorbed or removed)
From solid to liquid
absorbed
From liquid to solid
removed
3.
The temperature of a substance (decreases / remain constant / increases) during melting.
4.
Define the melting point of a substance. The melting point of a substance is the definite temperature where a substance _________________________________________________________________ changes its state from solid to liquid. _________________________________________________________________
5.
Latent heat of fusion is needed to change a substance from solid to liquid without a temperature change.
Chapter 11: Thermal Properties of Matter
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6.
The figure below shows how the temperature of a piece of ice changes as it is heated. Temperature/ °C
100
M
80 60 40 20
-20
J 1
(a)
7.
2
3
4
5
6
7
8
9
Time/ min
What are the states of matter at region J-K, K-L and L-M? J – K: solid
(b)
L
K
0
K – L: solid + liquid
L-M: liquid
What is the melting point of ice? 0°C
The figure below shows how the temperature of a cup of liquid changes as it is cooled. Temperature/ °C
140
J
120 100 80
K
L
60 40
M
20
0
(a)
2
4
6
8
10
12
14
16
18
Time/ min
What are the states of matter at region J-K, K-L and L-M? J – K: liquid
K – L: liquid + solid
(b)
What is the melting point of the liquid? 70°C
(c)
What is the freezing point of the liquid? 70°C
Chapter 11: Thermal Properties of Matter
L-M: solid
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8.
The figure below shows how the temperature of 0.5 kg of a solid substance changes as it is heated at a rate of 250 W. Temperature/ °C
70 60 50 40 30 20 10
0
(a)
1
2
3
4
5
6
7
8
9
Time/ min
What is the relationship between power, energy and time? Power x time = energy _____________________________________________________________
(b)
What is the melting point of the substance? 35°C
(c)
What is the time taken to melt the solid completely? 6 – 2 = 4 min
(d)
What is the total energy required to melt the substance?
Energy = 4 x 60s x 250 W = 60,000 J
Total energy required to melt the substance =60,000J (e)
What is the energy required to melt 1 kg of the solid? Energy = 60,000 J x 2 = 120,000 J
Energy required to melt 1 kg of the solid = 120,000J
Chapter 11: Thermal Properties of Matter
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UNITY SECONDARY SCHOOL PHYSICS CHAPTER 11: THERMAL PROPERTIES OF MATTER ANSWERS NAME:___________________________(
) CLASS: ________ DATE: _________
Exercise 11C: Boiling and Condensation 1.
Complete the table below: Energy (absorbed or removed)
Process
Change of state
Boiling
From liquid to gas
Absorbed
Condensation
From gas to liquid
Removed
2.
The temperature of a substance (decreases / remain constant / increases) during boiling.
3.
Define the boiling point of a substance. The boiling point of a substance is the definite temperature where a substance _________________________________________________________________ changes its state from liquid to gas. _________________________________________________________________
4.
Latent heat of vaporisation is needed to change a substance from liquid to gas without a temperature change.
Chapter 11: Thermal Properties of Matter
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Science Department
5.
The figure below shows how the temperature of a piece of ice changes as it is heated. Temperature/ °C
M
100
N
80 60 40 20
-20
L
K
0
J 1
2
3
4
5
6
7
8
9
Time/ min
(a)
What are the states of matter at region J-K, K-L, L-M and M-N? J – K: solid K – L: solid+liquid L-M: liquid M-N: liquid+gas
(b)
What is the boiling point of ice? 100°C
Chapter 11: Thermal Properties of Matter
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Science Department
UNITY SECONDARY SCHOOL PHYSICS CHAPTER 11: THERMAL PROPERTIES OF MATTER ANSWERS NAME:___________________________(
) CLASS: ________ DATE: _________
Exercise 11D: Evaporation and Boiling 1.
What is the similarity between evaporation and boiling? In both process, a liquid is changed to a gas. _________________________________________________________________
2.
3.
Complete the table below to distinguish between boiling and evaporation: Boiling
Evaporation
Quick
Slow
Bubbles formed
Bubbles not formed
Occurs throughout
Occurs at surface
Occurs at a fixed temperature
Occurs at all temperature
Heat from point source
Heat from surrounding
What are the six factors that affect the rate of evaporation?
↑ Temperature ↑ evaporation (a) ____________________________________________ ↑ Area of the exposed surface ↑ evaporation (b) ____________________________________________ ↑ Motion of the air ↑ evaporation (c) ____________________________________________ ↑ Pressure ↓ evaporation (d) ____________________________________________ ↑ Humidity ↓ evaporation (e) ____________________________________________ ↑ Boiling point of substance ↓ evaporation (f) ____________________________________________
Chapter 11: Thermal Properties of Matter
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Science Department
4.
Why does washing dry faster (a) on a warm, windy day? On a warm and windy day, the temperature is higher, and there is moving air ______________________________________________________________ which increases the rate of evaporation. ______________________________________________________________ (b) when it is spread out rather than left in a pile? Spreading out the washing increases the area of exposed surface, thus ______________________________________________________________ increasing the rate of evaporation. ______________________________________________________________
5.
Why is it advisable to sponge a feverish baby with cool water? Cool water takes in heat from baby to evaporate. Temperature of baby can be _________________________________________________________________ lowered. _________________________________________________________________
6.
Using the kinetic theory of matter, explain why evaporation of a liquid causes the temperature of the liquid to fall. Particles at the surface are exposed to the surrounding. _________________________________________________________________ They gain heat easier from the surrounding and move faster. _________________________________________________________________ They escape from the surface of the water. _________________________________________________________________ Leaving behind particles of slower speed. Therefore temperature falls _________________________________________________________________
7.
Why do you feel cool when some alcohol is applied onto your skin? Alcohol takes in heat from skin to evaporate. _________________________________________________________________ Therefore skin lose heat and feels cool. _________________________________________________________________
Chapter 11: Thermal Properties of Matter
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Science Department
UNITY SECONDARY SCHOOL PHYSICS CHAPTER 11: THERMAL PROPERTIES OF MATTER ANSWERS NAME:___________________________(
) CLASS: ________ DATE: _________
Exercise 11E: Specific Latent Heat 1.
Explain what is meant by “the specific latent heat of fusion of ice = 336 kJ kg-1 “. 1 kg of ice needs 336 kJ of heat energy to melt completely at 0°C. ________________________________________________________________
2.
A cup contains 200 g of tea at 60 oC. What is the minimum amount of ice at 0 oC needed to cool the drink to 0 oC? Heat loss by tea = Heat gain by ice to melt
m w c w ( 60 − 0 ) = m i l f mi =
60 m w c w lf
( 60 )( 0.2 )( 4200 ) 3.36 × 10 5 = 0.15 kg =
3.
Mass of ice required = 0.15 kg
A red-hot iron of mass 0.5 kg and temperature 1100 °C is added into water which has a temperature of 100 °C. What is the mass of water lost as steam? [specific heat capacity of iron = 460 J kg-1 K-1 ]
Heat loss by iron = Heat gain by water to change to steam m Fe c Fe ( 1100 − 100 ) = m s lv ms =
1000 m Fe c Fe lv
( 1000 )( 0.5 )( 460 ) 2.26 × 10 6 Mass of water lost as steam = 0.102 kg = 0.102 kg A hot piece of brass of mass 2 kg is added into water of temperature 100 °C. What is the initial temperature of the brass if 0.3 kg of water is lost as steam? [specific heat capacity of brass = 380 J kg-1 K-1 ] Heat loss by brass = Heat gain by water to change to steam =
4.
m B c B ( TB − 100 ) = m s l v m B c BTB − 100 m B c B = m s l v TB = =
Chapter 11: Thermal Properties of Matter
m s l v + 100 m B c B mBcB 0.3( 2.26 × 10 6 ) + ( 100 )( 2 )( 380 ) = 992°C ( 2 )( 380 ) Initial temperature of brass = 992 °C 13
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5.
A block of metal Y of mass M and initial temperature T is added into water at 100 °C. Based on the following assumption: 1. T is higher than 100 °C 2. There is no heat loss. Determine an expression for the mass of water that will be changed into steam. Use c for specific heat capacity of Y and l v for specific latent heat of vaporization of water.
Heat loss by Y = Heat gain by water to change to steam Mc( T − 100 ) = m s lv ms =
6.
Mc( T − 100 ) lv
Adding lumps of ice to it can cool limejuice. Andy discovered that 80 g of ice at a 0 oC temperature of cools 0.30 kg of limejuice from 25 oC to 4 oC. Determine: (a) The energy gained by the ice in melting, Heat gain by ice to change to water at 0°C = m i l f
= ( 0.08 )( 3.36 × 10 5 ) = 26880 J Energy gained by ice in melting = 26,880J (b) The energy gained by the melted ice, Heat gain by melted ice to rise to 4°C = m i cθ
= ( 0.08 )( 4200 )( 4 − 0 ) = 1344 J Energy gained by melted ice = 1,344J (c) The energy lost by the limejuice, Heat loss by lim ejuice = Total heat gain by ice
= 26880 + 1344 = 28224 J Energy lost by limejuice = 28,224J (d) The specific heat capacity of the limejuice.
Heat loss by lim ejuice = m J c J ( 25 − 4 ) 28224 = 21m J cJ cJ =
Chapter 11: Thermal Properties of Matter
28224 28224 = = 4480 J /( kgK ) 21m J 21 × 0.3 Specific heat capacity of limejuice = 4480 J/(kg K)
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Science Department
7.
Ice of mass 0.5 kg at temperature 0°C is added into 2 kg of water at 5°C. What is the mass of ice remaining if the final temperature of the mixture is 0°C?
Heat loss by water = Heat gain by ice to change to water m w cw ( 5 − 0 ) = mi l f mi =
5 m w cw lf
( 5 )( 2 )( 4200 ) = 0.125 kg 3.36 × 10 5 Mass of ice remaining = 0.5 − 0.125 = 0.375 kg =
8.
Mass of ice remaining = 0.375 kg
Ice of mass 0.5 kg at temperature 0 °C is added into 2 kg water at 25 °C. What is the final temperature of the mixture if all the ice has melted?
Heat loss by water = Heat gain by ice to change to water + Heat gain by melted ice to ↑ T to T f m w c w ( 25 − T f ) = m i l f + m i c w ( T f − 0 ) m w c w 25 − m w c wT f = m i l f + m i c wT f m w c wT f + m i c wT f = m w c w 25 − m i l f Tf =
m w c w 25 − m i l f mw cw + mi c w
( 2 )( 4200 )( 25 ) − ( 0.5 )( 3.36 × 10 5 ) = = 4°C ( 2 )( 4200 ) + ( 0.5 )( 4200 ) Final temperature of mixture = 4 °C *9.
Calculate (a) the amount of heat required to melt 50 g of ice at 0 oC,
Heat gain by ice to change to water at 0°C = ml f = (0.05 )( 3.36 × 10 5 ) = 16800 J Amount of heat required = 16,800 J (b) the amount of heat required to change 50 g of ice at 0 oC to water at 30 oC
Heat gain by ice to ↑ T to 30°C = 16800 + mcθ = 16800 + (0.05 )( 4200 )( 30 − 0 ) = 23100 J Amount of heat required = 23,100 J (c) the amount of heat required to change 50 g of ice at 0 oC to steam at 100 oC.
Heat gain by ice to ↑ T to 100°C steam = 23100 + mc ( 100 − 30 ) + ml v = 23100 + (0.05 )( 4200 )( 70 ) + ( 0.05 )( 2.26 × 10 6 ) = 150800 J
Amount of heat required = 150,800 J Chapter 11: Thermal Properties of Matter
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Science Department
*10.
Heat was supplied at a constant rate of 2000 W to ice of mass 10 kg and at an initial temperature of –10 oC. [Specific heat capacity of ice = 2100 J kg-1 0C-1] (a) Assuming no heat is lost, calculate the amount of heat required to convert the ice into steam at 100 0C. [Hint: find heat required for these changes: ice from –10 oC to ice at 0 oC, ice from 0 oC to water at 0 oC, water from 0 oC to water at 100 oC, and then from water at 100 oC to steam at 100 0C.]
Heat gain by ice( −10°C → 0°C ) + Heat gain by ice( melt at 0°C ) + Heat gain by water ( 0°C → 100°C ) + Heat gain by water ( → steam at 100°C ) = mc i ( 0 − ( −10 )) + ml f + mc w ( 100 − 0 ) + ml v = ( 10 )( 2100 )( 10 ) + ( 10 )( 3.36 × 10 5 ) + (10 )( 4200 )( 100 ) + ( 10 )( 2.26 × 10 6 ) = 30 ,370 ,000 J Amount of heat required = 30.4 MJ ≈ 30 MJ (b) Find the time taken to change ice at –10 oC into steam at 100 0C entirely.
Q = Pt Q t= P 30370000 = = 15185 s = 4.22 h 2000 *11.
Time taken = 15185 s or 4.22 h
A hot piece of copper of mass 3 kg and temperature 900 °C is put into 3 kg of water which temperature is 50 °C. What is the mass of water remaining after some water is lost as steam? [specific heat capacity of copper = 400 J kg-1 K-1]
Heat loss by copper = Heat gain by water ( 50°C → 100°C ) + Heat gain by water ( → steam at 100°C ) mcuccu ( 900 − 100 ) = mc ( 100 − 50 ) + m s lv ms = =
mcuccu ( 800 ) − mc ( 50 ) lv
(3 )( 400 )( 800 ) − ( 3 )( 4200 )( 50 ) 2.26 × 10 6
= 0.146 kg mass of water remaining = 3 − 0.146 = 2.854 kg *12.
Mass of water remaining = 2.854 kg
A piece of 5 kg aluminium at temperature -40 °C is added into 1 kg of water at 5 °C. How much ice is formed around the aluminium when the mixture has reached thermal equilibrium? [specific heat capacity of aluminium = 900 J kg-1 K-1] Heat gain by Al = Heat loss by water ( 5°C → 0°C ) + Heat loss by water ( → ice at 0°C )
m Al c Al ( 0 − ( −40 )) = mc ( 5 − 0 ) + m i l f mi = =
m Al c Al ( 40 ) − mc ( 5 ) lf
(5 )( 900 )( 40 ) − ( 1 )( 4200 )( 5 ) 3.36 × 10 5
Mass of ice formed = 0.473 kg
= 0.473 kg Chapter 11: Thermal Properties of Matter
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