Exe-11B Calorimetry ICSE Class-10 Concise Physics Selina Solutions Chapter-11. We Provide Step by Step Answer of Exercise – 11(A), MCQs – 11(A), Numericals – 11(A) , , Exercise-11(B), MCQ -11(B), Numericals -11(B) Questions of Exercise-11 Calorimetry ICSE Class-10 Concise . Visit official Website CISCE for detail information about ICSE Board Class-10.
Board | ICSE |
Publications | Selina Publication |
Subject | Physics |
Class | 10th |
Chapter-11 | Calorimetry (Exercise – 11B) |
Book Name | Concise |
Topics | Solution of Exercise – 11(A), MCQs – 11(A), Numericals – 11(A) ,, Exercise-11(B), MCQ -11(B), Numericals -11(B) |
Academic Session | 2021-2022 |
Calorimetry ICSE Class-10 Concise Physics Selina Solutions Chapter-11
–: Select Exercise :-
Exercise – 11(A), MCQs – 11(A), Numericals – 11(A) ,
Exercise-11(B), MCQ -11(B), Numericals -11(B)
How to Solve Numericals of Calorimetry ICSE Class-10
Note:- Read the chapter Calorimetry carefully and solve All the example given in your text book with Numericals Problems before starting Solutions of Chapter – 11 “Calorimetry” for ICSE Class-10 exercise. Get ICSE Class 10 Physics Formula Chapter-Wise to Solve Calorimetry Numericals problems.
ICSE Class-10″Calorimetry” Selina Physics Solution, Exercise – 11 (B)
Page 280
Question 1
(a)What do you understand by the change of phase of substance?
(b)Is there any change in temperature during the change of phase?
(c)Does the substance absorb or liberate any heat during the change of phase?
(d) what is the name given to the energy absorbed during a phase change?
Answer 1
(a) The process of change from one state to another at a constant temperature is called the change of phase of substance.
(b)There is no change in temperature during the change of phase.
(c)Yes, the substance absorbs or liberates heat during the change of phase.
(d) Latent heat
Question 2
A substance changes from its solid state to the liquid state when heat is supplied to it
(a) Name the process.
(b) What name is given to heat observed by the substance.
(c) How does the average kinetic energy of the molecules of the substance change.
Answer 2
(a) Melting.
(b) Latent heat of melting.
(c) As their no change in temperature the average kinetic energy of the molecules does not change.
Question 3 (Exe-11B Calorimetry ICSE Class-10 )
A substance undergoes (i) a rise in its temperature, (ii) A change in its phase without change in its temperature. In each case, state the energy change in the molecules of the substance.
Answer 3
(i)Average kinetic energy of molecules changes.
(ii)Average potential energy of molecules changes.
Question 4
How does the (a) average kinetic energy (b) average potential energy of molecules of a substance change during its change in phase at a constant temperature, on heating?
Answer 4
(a) Average kinetic energy does not change.
(b) Average potential energy increases.
Explanation: When a substance is heated at constant temperature (i.e. during its phase change state), the heat supplied makes the vibrating molecules gain potential energy to overcome the inter molecular force of attraction and move about freely. This means that the substance changes its form.
However, this heat does not increase the kinetic energy of the molecules, and hence, no rise in temperature takes place during the change in phase of a substance.
This heat supplied to the substance is known as latent heat and is utilized in changing the state of matter without any rise in temperature.
Question 5
State the effect of presence of impurity on the melting point of ice. Give one use of it.
Answer 5
The melting point of ice decreases by the presence of impurity in it.
Use: In making the freezing mixture by adding salt to ice. This freezing mixture is used in preparation of ice creams.
Question 6 (Exe-11B Calorimetry ICSE Class-10 )
State the effect of increase of pressure on the melting point of ice.
Answer 6
The melting point of ice decreases by the increase in pressure. The melting point of ice decrease by 0.0072oC for every one atmosphere rise in pressure.
Question 7
The diagram shows the change of phases of a substance on temperature time graph.
(a)What do parts AB, BC, CD and DE represent?
(b) What is the melting point of the substance?
(c) What is the boiling point of the substance?
Answer 7
(a) AB part shows rise in temperature of solid from 0 to T1oC, BC part shows melting at temperature T1oC, CD part shows rise in temperature of liquid from T1oC to T3oC , DE part shows the boiling at temperature T3oC.
(b) T1oC.
(c) T3oC.
Question 8 (Exe-11B Calorimetry ICSE Class-10 )
The melting point of naphthalene is 800C and the room temperature is 250C. A sample of liquid naphthalene at 900 is cooled down to room temperature. Draw a temperature-time graph to represent this cooling. On the graph mark the region which corresponds to the freezing process.
Answer 8
The melting point of naphthalene , a crystalline solid is 80 degree C and the room temperature is 25 °C
The temperature – time graph is as follows: –
Question 9
1 kg of ice at 0o is heated at constant rate and its temperature is recorded after every 30 s till steam is formed at 100o C. Draw a temperature time graph to represent the change of phase.
Answer 9
Question 10
Explain the terms boiling and boiling point. How is the volume of water affected when it boils at 100oC?
Answer 10
Boiling: The change from liquid to gaseous phase on heating at a constant temperature is called boiling.
Boiling point: The particular temperature at which vaporization occurs is called the boiling point of liquid.
Volume of water wills increases when it boils at 100oC.
Question 11
How is the boiling point of water affected when some salt is added to it?
Answer 11
The boiling point of water increases on adding salt.
Question 12
What is the effect of increase in pressure on the boiling point of a liquid?
Answer 12
The boiling point of a liquid increases on increasing the pressure.
Question 13 (Exe-11B Calorimetry ICSE Class-10 )
Water boils at 120 °C in a pressure cooker. Explain the reason
Answer 13
- The boiling point of a liquid increases with the increase in pressure and decreases with the decrease in pressure.
- The boiling point of pure water at one atmospheric pressure (= 760 mm of Hg) is 100 °C.
- In a pressure cooker, the water boils at about 120 °C to 125 °C due to increase in pressure, as the steam is not allowed to escape out of it.
Question 14
Write down the approximate range of temperature at which the water boils in a pressure cooker.
Answer 14
In a pressure cooker, the water boils at about 120oC to 125oC.
Question 15
It is difficult to cook vegetables on hills and mountains. Explains the reason.
Answer 15
This is because at high altitudes atmospheric pressure is low; therefore boiling point of water decreases and so it does not provide the required heat energy for cooking.
Question 16
Complete the following sentences:
(a)When ice melts, its volume………….
(b)Decrease in pressure over ice ………….. its melting point.
(c)Increase in pressure ………..the boiling point of water.
(d)A pressure cooker is based on the principle that boiling point of water increases with ……………….
(e)The boiling point of water is defined as ………………………..
(f) water can be made to boil at 115°C by …………….. pressure over its surface.
Answer 16
(a) When ice melts, its volume decreases.
(b) Decrease in pressure over ice increases its melting point.
(c) Increase in pressure increases the boiling point of water.
(d)A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure.
(e) The boiling point of water is defined as the constant temperature at which water changes to steam.
(f) water can be made to boil at 115°C by increasing pressure over its surface.
Question 17 (Exe-11B Calorimetry ICSE Class-10 )
What do you understand by the term latent heat?
Answer 17
Latent heat: The heat energy exchanged in change of phase is not externally manifested by any rise or fall in temperature, it is considered to be hidden in the substance and is called the latent heat.
Page 281
Question 18
Define the term specific latent heat of fusion of ice. State its S.I. unit.
Answer 18
The quantity of heat required to convert unit mass of ice into liquid water at (melting point) is called the specific latent heat of fusion of ice.
Question 19
Write the approximate value of specific latent heat of ice.
Answer 19
Specific latent heat of ice: 336000 J kg-1.
Its S.I. unit is Jkg-1.
Question 20
‘The specific latent heat of fusion of ice is 336 J g-1 ‘. Explain the meaning of this statement.
Answer 20
It means 1 g of ice at 0oC absorbs 360 J of heat energy to convert into water at 0oC.
Question 21
1 g ice at 0o C melts to form 1 g water at 0o. State whether the latent heat is absorbed or given out by ice.
Answer 21
Latent heat is absorbed by ice.
Question 22
Which has more heat: 1 g of ice at 0o C or 1 g of water at 0oC? Give reasons.
Answer 22
1 g of water at 0oC has more heat than 1 g of ice at 0oC. This is because ice at 0oC absorbs 360 J of heat energy to convert into water at 0oC.
Question 23 (Exe-11B Calorimetry ICSE Class-10 )
(a) Which requires more heat: 1 g ice at 0o C or 1 g water at 0oC to raise its temperature to 10oC?
(b) Explain your answer in part (a).
Answer 23
(a) 1 g ice at 0oC requires more heat because ice would require additional heat energy equal to latent heat of melting.
(b) 1 g ice at 0oC first absorbs 336 J heat to convert into 1 g water at 0oC.
Question 24
Ice cream appears colder to the mouth than water at 0oC. Give reasons.
Answer 24
This is because 1 g of ice at 0oC takes 336 J of heat energy from the mouth to melt at 0oC. Thus mouth loses an additional 336 J of heat energy for 1 g of ice at 0oC than for 1g of water at 0oC. Therefore cooling produced by 1 g of ice at 0oC is more than for 1g of water at 0oC.
Question 25
The soft drink bottles are cooled by (i) ice cubes at 0°C, and (ii) iced-water at 0°C. Which will cool the drink quickly? Give reason.
Answer 25
This is because 1 g of ice at 0oC takes 336 J of heat energy from the bottle to melt into water at 0oC. Thus bottle loses an additional 336 J of heat energy for 1 g of ice at 0oC than for 1 g iced water at 0oC. Therefore bottled soft drinks get cooled, more quickly by the ice cubes than by iced water.
Question 26 (Exe-11B Calorimetry ICSE Class-10 )
It is generally cold after a hailed storm than during and before the hail storm. Give reasons.
Answer 26
The reason is that after the hail storm, the ice absorbs the heat energy required for melting from the surrounding, so the temperature of the surrounding falls further down and we feel colder.
Question 27
The temperature of surroundings starts falling when ice in a frozen lake starts melting. Give reasons.
Answer 27
The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it became very cold.
Question 28
Water in lakes and ponds do not freeze at once in cold countries. Give reason.
Answer 28
The specific latent heat of fusion of ice is sufficiently high, about 336 J g-1. Before freezing the water in the lakes and ponds will have to release a large quantity of heat to the surrounding. If there is any layer of ice formed on water than water being a poor conductor of heat will also prevent the loss of heat from water of lake. Hence, in cold countries water in lakes and ponds do not freeze.
Question 29 (Exe-11B Calorimetry ICSE Class-10 )
Explain the following:
(a) The surroundings become pleasantly warm when water in a lake starts freezing in cold countries.
(b) The heat supplied to a substance during its change of state, does not cause any rise in its temperature.
Answer 29
(a) The reason is that the specific latent heat of fusion of ice is sufficiently high, so when the water of lake freezes, a large quantity of heat has to be released and hence the surrounding temperature becomes pleasantly warm.
(b) Heat supplied to a substance during its change of state, does not cause any rise in its temperature because this is latent heat of phase change which is required to change the phase only.
Selina Solution of ICSE Physics “Calorimetry’ Multiple Choice Type – 11(B)
Page 281
Question 1
The S.I. unit of specific latent heat is :
(a) cal g-1
(b) cal g-1 K-1
(c) J kg-1
(d) J kg -1 K-1
Answer 1
J kg-1
Question 2
The specific latent heat of fusion of water is :
(a) 80 cal g-1
(b) 2260 J g-1
(c) 80 J g-1
(d) 336 J kg-1
Answer 2
80cal g-1
“Calorimetry” Concise Physics Solution, Numericals:- 11-(B) ICSE Class-10
Page 281
Question 1
10g of ice at 0oC absorbs 5460J of heat energy to melt and change to water at 50oC. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200Jkg-1K-1.
Answer 1
Mass of ice=10g = 0.01kg
Amount of heat energy absorbed, Q=5460J
Specific latent heat of fusion of ice=?
Specific heat capacity of water = 4200Jkg-1K-1
Amount of heat energy required by 10g (0.01kg) of water at 0oC to raise its temperature by 50oC= 0.01X4200X50=2100J.
Let Specific latent heat of fusion of ice=L Jg-1.
Then,
Q = mL + mcT
5460 J =10 x L + 2100J
L=336Jg–
Question 2 (Exe-11B Calorimetry ICSE Class-10 )
How much heat energy is released when 5.0 g of water at 20oC changes into ice at 0oC? Take specific heat capacity of water =4.2 J g-1 K-1, specific latent heat of fusion of ice =336 J g-1.
Answer 2
Mass of water m = 5.0 g
specific heat capacity of water c = 4.2 J g-1 K-1
specific latent heat of fusion of iceL =336 J g-1
Amount of heat energy released when 5.0 g of water at 20oC changes into water at 0oC = 5 x 4.2 x 20 = 420 J.
Amount of heat energy released when 5.0g of water at 0oC changes into ice at 0oC=5x336J=1680J.
Total amount of heat released =1680 J + 420 J = 2100 J.
Question 3
A molten metal of mass 150 g is kept at its melting point 800oC. When it is allowed to freeze at the same temperature, it gives out 75000 J of heat energy.
(a)What is the specific latent heat of the metal?
(b) If the specific heat capacity of metal is 200 J kg-1 K-1, how much additional heat energy will the metal give out in cooling to -50oC?
Answer 3
Mass of metal =150 g
Specific latent heat of metal
Specific heat capacity of metal is 200 J kg-1 K-1.
Change in temperature= 800-(-50) = 850oC (or 850 K).
Question 4
A solid metal of mass 150g melts at its melting point of 800°C by providing heat at the rate of 100W.The time taken for it to completely melt at the same temperature is 4 min. What is the specific latent heat of fusion of the metal?
Answer 4
Question 5 (Exe-11B Calorimetry ICSE Class-10 )
A refrigerator converts 100g of water at 20oC to ice at -10oC in 73.5 min. calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 Jg-1K-1, specific latent heat of ice is 336Jg-1 and the specific heat capacity of ice is 2.1Jg-1K-1.
Answer 5
heat released when 100g of water cools from 20o to 0oC =100X20X4.2=8400J.
Amount of heat released when 100g of water converts into ice at 0oC =100X336=33600J.
and Amount of heat released when 100g of ice cools from 0oC to -10oC =100X10X2.1=2100J.
Total amount of heat=8400+33600+2100=44100J.
Time taken= 73.5min=4410s.
Average rate of heat extraction (power)
.
Question 6
In an experiment, 17 g of ice is used to bring down the temperature of 40 g of water at 34oC to its freezing temperature. The specific heat capacity of water is 4.2 J g-1 K-1. Calculate the specific latent heat of ice. State one important assumption made in the above calculation.
Answer 6
Mass of ice m1 =17 g
Mass of water m2 =40 g.
Change in temperature =34-0=34K
Specific heat capacity of water is 4.2Jg-1K-1.
Assuming there is no loss of heat, heat energy gained by ice (latent heat of ice), Q= heat energy released by water
Q = 40 x 34 x 4.2 = 5712 J.
Specific latent heat of ice=
Question 7
The temperature of 170g of water at 500C to be lowered to 5oC by adding certain amount of ice to it. Find the mass of ice added. Given: Specific heat capacity of water=4200Jkg-1C-1 and specific latent het of ice =336000JKg-1.
Answer 7
Question 8
Find the result of mixing 10g of ice at -10oC with 10g of water at 10oC. Specific heat capacity of ice is 2.1Jg-1K-1, specific latent heat of ice is 336Jg-1, and specific heat capacity of water is 4.2Jg-1 K-1.
Answer 8
Let whole of the ice melts and let the final temperature of the mixture be ToC.
Amount of heat energy gained by 10g of ice at -10oC to raise its temperature to 0oC= 10x10x2.1=210J
The Amount of heat energy gained by 10g of ice at 0oC to convert into water at 0oC=10×336=3360 J
And Amount of heat energy gained by 10g of water (obtained from ice) at 0oC to raise its temperature to ToC = 10×4.2x(T-0)=42T
Hence Amount of heat energy released by 10g of water at 10oC to lower its temperature to ToC = 10×4.2x(10-T)=420-42T
Heat energy gained = Heat energy lost
210 + 3360 + 42T = 420-42T
T = -37.5oC
This cannot be true because water cannot exist at this temperature.
So whole of the ice does not melt. Let m gm of ice melts. The final temperature of the mixture becomes 0oC.
Therefore, amount of heat energy gained by 10g of ice at -10oC to raise its temperature to 0oC= 10x10x2.1=210J
Because Amount of heat energy gained by m gm of ice at 0oC to convert into water at 0oC=mx336=336m J
Then Amount of heat energy released by 10g of water at 10oC to lower its temperature to 0oC = 10×4.2x(10-0)=420
Heat energy gained = Heat energy lost
210 + 336m = 420
m = 0.625 gm
Question 9 (Exe-11B Calorimetry ICSE Class-10 )
A piece of ice of mass 40 g is added to 200 g of water at 50oC. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water is 4200 J kg-1 K-1, specific latent heat of fusion of ice =336 x 103 J kg-1.
Answer 9
Let final temperature of water when all the ice has melted =ToC.
Amount of heat lost when 200g of water at 50oC cools to ToC=
200X4.2X(50-T) = 42000-840T
Amount of heat gained when 40g of ice at 0oC converts into water at 0oC.= 40X336J = 13440 J
Amount of heat gained when temperature of 40g of water at 0oC rises to ToC= 40X4.2X(T-0) = 168T
We know that
Amount of heat gained=amount of heat energy lost.
13440+168T= 42000-840T
168T+840T= 42000-13440
1008T= 28560
T=28560/1008=28.33oC.
Question 10
Calculate the mass of ice needed to cool 150g of water contained in a calorimeter of mass 50g at 32oC such that the final temperature is 5oC. Specific heat capacity of calorimeter =0.4Jg-1C-1, Specific heat capacity of water =4.2Jg-1C-1, latent heat capacity of ice=330Jg-1
Answer 10
Heat loss by (water + Calorimeter) = Heat gain by ice
Heat loss by (water + calorimeter) = mw Cpw ΔT + mC Cpc ΔT = mi ( L + Cpw δT ) ………………….(i)
where, mw = mass of water = 50 g
Cpw = Specific heat of water = 4.2 J/( g °C )
mC = mass of calorimeter = 50 g
Cpc = Specific heat capacity of calorimeter = 0.4 J/( g °C )
ΔT = fall in temperature of water and calorimeter = 32-5 = 27°C
mi = mass of ice in gram
L = latent heat capacity of ice = 330 J/g
δT = rise in temperature = 5 °C
by substituting all the values in eqn.(1)we get the mass of ice as
Question 11 (Exe-11B Calorimetry ICSE Class-10 )
250 g of water at 30o C is contained in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of vessel and its contents to 5o C. specific latent heat of fusion of ice = 336 x 103 J kg-1, specific heat capacity of copper = 400 J kg-1 K-1, specific heat capacity of water is 4200 J kg-1 K-1.
Answer 11
Given Mass of copper vessel m1 = 50 g.
and Mass of water contained in copper vessel m2 = 250 g.
Mass of ice required to bring down the temperature of vessel = m
Final temperature = 5o C.
Amount of heat gained when ‘m’ g of ice at 0o C converts into water at 0o C = m × 336 J
Amount of heat gained when temperature of ‘m’ g of water at 0o C rises to 5o C = m × 4.2 × 5
Total amount of heat gained = m × 336 + m × 4.2 × 5
Amount of heat lost when 250 g of water at 30o C cools to 5o C =
250 × 4.2 x 25 = 26250 J
Amount of heat lost when 50 g of vessel at 30o C cools to 5o C =
50 x 0.4 × 25 = 500 J
Total amount of heat lost = 26250 + 500 = 26750 J
We know that amount of heat gained = amount of heat lost
m × 336 + m × 4.2 × 5 = 26750
357 m = 26750
m = 26750/357 = 74.93 g
Hence, mass of ice required is 74.93 g.
Question 12
2 kg of ice melts when water at 100oC is poured in a hole drilled in a block of ice. What mass of water was used? Specific heat capacity of water is 4200 J kg-1 K-1, specific latent heat of fusion of ice = 336 J g-1.
Answer 12
Since the whole block does not melt and only 2 kg of it melts, so the final temperature would be 0 oC.
Amount of heat energy gained by 2 kg of ice at 0oC to convert into water at 0oC=2X336000=672000 J
Let amount of water poured=m kg.
Initial temperature of water =100oC.
Final temperature of water =0oC.
Amount of heat energy lost by m kg of water at 100oC to reach temperature 0oC =mX4200X100 = 420000m J
We know that heat energy gained =heat energy lost.
672000J= mX420000J
m=672000/420000=1.6kg
Question 13
Calculate the total amount of heat energy required to convert 100 g of ice at -10o C completely into water at 100o C. Specific heat capacity of ice 2.1 J g-1 K-1, specific heat capacity of water is 4.2 J g-1 K-1, specific latent heat of fusion of ice = 336 J g-1.
Answer 13
Amount of heat energy gained by 100 g of ice at -10o C to raise its temperature to 0o C =
100 × 2.1 × 10 = 2100 J
Amount of heat energy gained by 100 g of ice at 0o C to convert into water at 0o C =
100 × 336 = 33600 J
Amount of heat energy gained when temperature of 100 g of water at 0o C rises to 100o C =
100 × 4.2 × 100 = 42000 J
Total amount of heat energy gained is = 2100 + 33600 + 42000 = 77700 J = 7.77 × 104 J
Question 14 (Exe-11B Calorimetry ICSE Class-10 )
The amount of heat energy required to convert 1 kg of ice at -10oC completely into water at 100oC is 777000 J. calculate the specific latent heat of ice. Specific heat capacity of ice = 2100 J kg-1 K-1, Specific heat capacity of water is 4200 J kg-1 K-1.
Answer 14
Total Amount of heat energy gained by 1kg of ice at -10oC to raise its temperature to 0oC= 1 x 2100 x 10 = 21000 J
Therefore Amount of heat energy gained by 1kg of ice at 0oC to convert into water at 0oC=L
So Amount of heat energy gained when temperature of 1kg of water at 0oC rises to 100oC= 1 x 4200 x 100 = 420000 J
Hence Total amount of heat energy gained = 21000+420000+L=441000 +L.
Given that total amount of heat gained is =777000J.
So,
441000+L=777000.
L=777000-4410
Question 15
200 g of ice at 0 °C converts into water at 0 °C in 1 minute when heat is supplied to it at a constant rate. In how much time, 200 g of water at 0 °C will change to 20 °C? Take specific latent heat of ice = 336 J g-1.
Answer 15
Mass of ice, mice = 200 g
Time for ice to melt, t1 = 1 min = 60 s
Mass of water, mw = 200 g
Temperature change of water, ΔT = 20 °C
Rate of heat exchange is constant. So, power required for converting ice to water is same as the power required to increase the temperature of water.
Hence, the time required is 15 seconds.
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