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## Access ICSE Selina Solutions for 10 Physics Chapter 11 - Calorimetry

1. Define the term heat.

Ans: The kinetic energy due to random motion of the molecules of a substance is known as its heat energy.

2. Name the S.I unit of heat.

Ans: S.I. unit of heat is joule (symbol J)

3. Define the term calorie. How is it related to joule?

Ans: One calorie of heat is heat energy required to raise the temperature of 1 g of water from 14.5 oC to 15.5 oC.

1 calorie = 4.186 J

4. Define one kilo-calorie of heat.

Ans: One kilo-calorie of heat is heat energy required to raise the temperature of 1 kg of water from 14.5o C to 15.5o C.

5. Define temperature and name its S.I. unit.

Ans: The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature.

S.I. unit kelvin (K).

6. State three differences between heat and temperature.

Ans:

Heat | Temperature |

The kinetic energy due to random motion of the molecules of a substance is known as its heat energy. | The quantity which determines the direction of flow of heat between two bodies kept in contact is called temperature |

S.I. unit joule (J). It is measured by the principle of calorimetry. | S.I. unit kelvin (K).It is measured by a thermometer. |

7. Define calorimetry.

Ans: The measurement of the quantity of heat is called calorimetry.

8. Define the term heat capacity and state its S.I. unit.

Ans: The heat capacity of a body is the amount of heat energy required to raise its temperature by 1℃ or 1K. S.I. unit is joule per kelvin (JK^{-1})

9. Define the term specific heat capacity and state its S.I. unit.

Ans: The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of the unit mass of that substance by 1℃ (or 1K). S.I. unit is joule per kilogram per kelvin (J kg-1 K-1).

10. How is the heat capacity of a body related to the specific heat capacity of its substance?

Ans: Heat capacity = Mass × specific heat capacity

11. State three differences between the heat capacity and specific heat capacity.

Ans: Heat capacity of the body is the amount of heat required to raise the temperature of (the whole) body by 1 ℃ whereas specific heat capacity is the amount of heat required to raise the temperature of unit mass of the body by 1 ℃. Heat capacity of a substance depends upon the material and mass of the body. Specific heat capacity of a substance does not depend on the mass of the body. S.I. units of heat capacity are J K-1 and S.I. The unit of specific heat capacity is J kg-1 K-1.

12. Name a liquid which has the highest specific heat capacity.

Ans: Water has the highest specific heat capacity.

13. Write the approximate value of specific heat capacity of water in S.I unit.

Ans: Specific heat capacity of water = 4200 J kg-1 K-1

14. What do you mean by the following statement :

(i) the heat capacity of a body is 50 J k-1 ?

Ans: The heat capacity of a body is 50J K-1 means to increase the temperature of this body by 1K then we have to supply 50 joules of energy.

(ii) the specific heat capacity of copper is 0.4 J kg^{-1 }K^{-1} ?

Ans: The specific heat capacity of copper is 0.4 J kg-1 K-1 means to increase the temperature of one gram of copper by 1K then we have to supply 0.4 joules of energy.

15. Specific heat capacity of a substance A is 3.8 J kg-1 K-1 and of the substance B is 0.4 J kg-1 K-1. Which substance is a good conductor of heat ? How did you arrive at your conclusion?

Ans: B. for the heat energy and same mass, the rise in temperature of B will be more hence, B is a good conductor of heat.

16. Name two factors on which the heat energy liberated by a body on cooling depends.

Ans: Heat liberated by a body depends on the mass of the body, specific heat capacity of the body and change in temperature experienced by the body.

17. Name three factors on which the heat energy absorbed by a body depends and state how it depends on them.

Ans: The quantity of heat energy absorbed by the body depends on three factors: (i) Mass of the body - The amount of heat energy required by the body is directly proportional to the mass of the substance.

(ii) Nature of material of the body - The amount of heat energy required by the body depends on the nature of the substance and it is expressed in terms of its specific heat capacity c.

(iii) Rise in temperature of body - The amount of heat energy required by the body is directly proportional to the rise in temperature.

18. Write the expression for the heat energy Q received by m kg of a substance of specific heat capacity c J kg-1 K-1 when it is heated through ∆t ℃.

Ans: The expression for the heat energy Q

Q= mc ∆ t (in joule)

19. Same amount of heat is supplied to two liquids A and B. The liquid A shows a greater rise in temperature. What can you say about the heat capacity of A as compared to that of B?

Ans: Heat capacity of liquid A is less than that of B

20. Two blockers P and Q of different metals having their mass in the ratio 2:1 are given the same amount of heat. Their temperature rises by the same amount. Compare their specific heat capacities.

Ans: Let CP and CP be the specific heat capacities of blocks P and Q respectively, We know that, c = Q / m × ∆ P

∴ (CP/Cq) = (Q / 2m × ∆ P) / (Q / m × ∆ P) = 1/ 2

Hence, the required ratio is 1 : 2

21. What is the principle of method of mixture? What other name is given to it? Name the law on which this principle is based.

Ans: The principle of method of mixture is:

Heat energy lost by hot body = Heat energy gained by cold body.

This principle is based on the law of conservation of energy.

22. A mass m1 of a substance of specific heat capacity c1 at temperature t1 is mixed with a mass m2 of another substance of specific heat capacity c2 at a lower temperature t2. Deduce the expression for the temperature t of the mixture. State the assumption made, if any.

Ans: A mass m1 of a substance A of specific heat capacity c1 at temperature T1 is mixed with mass m2 of other substance B of specific heat capacity c2 at a lower temperature T2 and the final temperature of the mixture becomes T.

Fall in temperature of substance A = T1 – T

Rise in temperature of substance B = T – T2

Heat energy lost by A = m1 × c1 × fall in temperature = m1c1 (T1 – T)

Heat energy gained by B = m2 × c2 × rise in temperature = m2c2 (T – T2)

If no energy lost in the surrounding, then by the principle of mixtures,

Heat energy lost by A = Heat energy gained by B

m1c1(T1- T) = m2c2(T – T2)

After rearranging this equation, we get

T = m1c1T1+m2c2T2 / m1c1+m2c2

Here we have assumed that there is no loss of heat energy.

23. Why do the farmers fill their fields with water on a cold winter night?

Ans: In the absence of the water, if on a cold winter night the atmospheric temperature falls below 0o C, the water in small capillaries of the plant will freeze, so the veins will burst due to the increase in the volume of the water on freezing. As a result, plants will die and the crop will be destroyed. In order to save the crop on such cold nights, farmers fill their fields with water because water has a high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0o C.

24. Discuss the role of high specific heat capacity of water with reference to climate in coastal areas.

Ans: The specific heat capacity of water is very high. It is about five times as high as that of sand. Hence the heat energy required for the same rise in temperature by a certain mass of water will be nearly five times that required by the same mass of sand. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature. As such, sand gets heated or cooled more rapidly as compared to water under similar conditions. So a large difference in temperature is developed between the land and sea due to which land and sea breezes are formed. These breezes make the climate near the sea shore moderate.

25. Water is used in hot water bottles for fomentation. Give a reason.

Ans: The reason is that water does not cool quickly due to its large specific heat capacity, so a hot water bottle provides heat energy for fomentation for a long time period.

26. What property of water makes it an effective coolant?

Ans: By allowing the water to flow in pipes around the heated parts of a machine, heat energy from such parts is removed. Water in pipes extracts more heat from the surrounding without much increase in its temperature because of its large specific heat capacity. So, Water is used as an effective coolant.

27. Give one example each where a high specific heat capacity of water is used (i) as coolant, (ii) as heat reservoir.

Ans: (1) Radiator in car. (2) To avoid freezing wine & juice bottles.

28. A liquid X has a specific heat capacity higher than liquid Y. Which liquid is useful as (i) coolant in car radiators.

Ans: As the specific heat capacity of liquid X is higher than liquid Y, the rise in temperature for X will be less than that of Y.

(i)The liquid needs to absorb more energy without much change in temperature as a coolant in car radiators. Hence liquid X is useful for this function.

(ii) Which liquid is useful as a heat reservoir to keep juice bottles without freezing?

Ans: The liquid needs to give out a large amount of heat before reaching freezing temperatures as a heat reservoir to keep juice bottles without freezing. Hence, liquid X is useful for this function.

29. (a) What is a calorimeter?

Ans: A calorimeter is a cylindrical device that is used to measure the amount of heat gained or lost by a body when it is mixed with another body.

(b) Name the material of which it is made of. Give two reasons for using the material satiated by you.

Ans: Calorimeter is made up of a thin copper sheet because:

(i) Copper is a good conductor of heat, thus the vessel soon acquires the temperature of its contents.

(ii) Copper has low specific heat capacity thus the heat capacity of the calorimeter is low and the amount of heat energy taken by the calorimeter from its contents to acquire the temperature of its contents is negligible.

(c) Out of the three metals A,B, and C specific heat capacity 900 J kg^{-1} ℃^{-1},380 J kg^{-1} ℃^{-1} and 460 J kg^{-1} ℃^{-1} respectively, which will you prefer for a calorimeter ? Give a reason.

Ans: Metal B with specific heat capacity 380 J kg-1 ℃-1 should be selected to make a calorimeter. By selecting this metal, the heat capacity of the calorimeter is low and the amount of heat energy taken by the calorimeter from its contents to acquire the temperature of its contents is negligible.

(d) How is the loss of heat due to radiation minimised in a calorimeter?

Ans: Heat loss due to radiation will be minimized by polishing the box in order to smoothen it.

30. Why is the base of a cooking pan made thick and heavy?

Ans: By making the base of a cooking pan thick, because its thermal capacity becomes large and it imparts sufficient heat energy at a low temperature to the food for its proper cooking. Further it keeps food warm for a long time, after cooking.

### Multiple Choice Type

1. The S.I. unit of heat capacity is:

(a) J kg-1

(b) J K-1

(c) J kg-1K-1

(d) cal ℃-1

Ans: Correct option is (b) J K-1

2. The S.I. unit of specific heat capacity is :

(a) J kg-1

(b) J K-1

(c) J kg-1K-1

(d) kcal kg-1 ℃-1

Ans: Correct option is (c) J kg-1 K-1

3. The specific heat capacity of water is :

(a) 4200 J kg-1 K-1

(b) 420 J g-1 K-1

(c) 0.42 J g-1 K-1

(d) 4.2 J kg-1 K-1

Ans: Correct option is (a) 4200 J kg-1 K-1

### Numericals

1. By imparting heat to a body, its temperature rises by 15℃. What is the corresponding rise in temperature on the kelvin scale?

Ans: The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on Celsius scale. So, the difference in temperature is the same on both the Celsius and Kelvin scales. Therefore, the corresponding rise in temperature on the Kelvin scale will be 15 K.

2. (a) Calculate the heat capacity of a copper vessel of mass 150 g if the specific heat capacity of copper is 410 J kg-1 K-1

Ans: (i) Mass of copper vessel = 150 g = 0.15 kg

The specific heat capacity of copper = 410 J kg-1 K-1

Heat capacity = Mass × specific heat capacity

= 0.15 kg × 410 J kg-1 K-1 = 61.5 J K-1

Change in temperature = (35 − 25)o C = 10oC = 10K

(b) How much heat energy will be required to increase the temperature of the vessel in part (a) from 25°C to 35°C?

Ans: Energy required to increase the temperature of vessel ∆Q = mc ∆T

= 0.15 × 410 × 10 = 615 J

3. A piece of iron of mass 2.0 kg has a heat capacity of 966 J K -1 .

(i) Find heat energy needed to warm it by 15℃, and

Ans: We know that heat energy needed to raise the temperature by 15 ℃ is = heat capacity × change in temperature.

Heat energy required = 966 J K-1 × 15 K = 14490 J.

(ii) Find its specific heat capacity in the S.I. unit.

Ans: We know that specific heat capacity is = heat capacity/ mass of substance, So specific heat capacity is = 966 / 2 = 483 J kg-1 K-1

4. Calculate the amount of heat energy required to raise the temperature of 100 g of copper from 20oC to 70 oC. Specific heat capacity of copper = 390 J kg -1K-1

Ans: Mass of copper m = 100 g = 0.1 kg

Change of temperature ∆t = (70 − 20) oC

Specific heat of capacity of copper = 390 J kg-1 K-1

Amount of heat required to raise the temperature of 0.1 kg of copper is

Q = m × the ∆t × c = 0.1 × 50 × 390 = 1950 J

5. 1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20oC to 40 oC. Calculate the specific heat capacity of lead.

Ans: Heat energy supplied = 1300 J

Mass of lead = 0.5 kg

Change in temperature = (40-20) oC = 20oC (or 20 K)

Specific heat capacity of lead c= ∆Q / m∆t = 1300 / 0.5 × 20

c = 130 J kg-1 K-1

6. Find time taken by a 500W heater to raise the temperature of 50 kg of material of specific heat capacity 960 from 18oC to 38 oC. Assume that all the heat energy supplied by the heater is given to the material.

Ans: Specific heat capacity of material c = 960 J kg-1 K-1

Change in temperature ∆T = (38 − 18)oC = 20oC (or 20 K)

Power of heater P = 500 W

∆Q = mc ∆T ∆Q = 50 × 960 × 20

Time taken by a heater to raise the temperature of material

t=∆Q/P

t= 50 × 960 × 20 / 500

t = 1920 seconds

t = 32 min

7. An electric heater of power 600 W raises the temperature of 4 kg of a liquid from 10oC to 15oC in 100 s.

(i) Calculate the heat capacity of 4kg of liquid.

Ans: Power of heater P = 600 W

Mass of liquid m = 4.0 kg

Change in temperature of liquid = (15 − 10)o C = 5oC(or 5 K)

Time taken to raise its temperature = 100s

Heat energy required to heat the liquid

∆Q = mc ∆T

And

∆Q = P × t = 600 × 100 = 60000J

c= ∆Q / m∆t

=6000 / 4 × 5

= 3000 J kg-1 K-1

Heat capacity = c × m

(ii) Calculate the specific heat capacity of liquid.

Ans: Heat capacity = 4 × 3000J kg-1 K-1

= 1.2 × 104 J/K

8. 0.5 kg lemon squash at 30oC is placed in a refrigerator which can remove heat at an average rate of 30 J s-1. How long will it take to cool the lemon squash to 5oC ? Specific heat capacity of squash = 4200 J kg-1 K-1

Ans: Change in temperature= 30 − 5 = 25 K.

∆Q= mc ∆T

= 0.5× 4200×25

=52500J

t=∆Q/p

= 52500/ 30

=1750 s

t = 29 min 10 sec

9. A mass of 50g of a certain metal at 150oC is immersed in 100 g of water at 11oC .The final temperature is 20oC . Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 J kg-1 K-1

Ans: Mass of certain metal (m1) = 200 g

Temperature (T1) = 83o C

Mass of water (m2) = 300 g

Temperature of water (T2) = 30o C

Final temperature (T) = 33o C

Specific heat capacity of water c2= 4.2 J/g/K

The specific heat capacity of the metal c2 is given by this formula

m1c1(T1- T) = m2c2(T – T2)

c1 =m2c2(T – T2) / m1(T1 – T)

c1 =(33-30)×300×4.2 / 200×(83-33)

c1 =0.378JK-1

10. 45 g of water at 50oC in a beaker is cooled when 50g of copper at 18oC is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39J kg-1 K-1 and that of water is 4.2 J kg-1 K-1. State assumption used

Ans: Mass of water (m1) = 45 g

Temperature of water (T1) = 50o C

Mass of copper (m2) = 50 g

temperature of copper(T2) = 18o C

Final temperature (T) = ?

The specific heat capacity of the copper c2 = 0.39 J/g/K

The specific heat capacity of water c1 = 4.2 J/g/K

m1c1(T1- T) = m2c2(T – T2)

T = m1c1T1 + m2c2T2 / m1c1+ m2c2

= 45 ×4.2 ×50 +50× 0.39× 18 / 45× 4.2 + 50 ×0.39

T = 9801 / 208.5

T = 47o C

11. 200 g of hot water at 80oC is added to 300 g of cold water at 10oC. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg-1 K-1.

Ans: Mass of hot water (m1) = 200g

Temperature of hot water (T1) = 80oC

Mass of cold water (m2) = 300g

Temperature of cold water (T2) = 10oC

Final temperature (T) = ?

m1c1(T1- T) = m2c2(T – T2)

c1 = c2

T = m1T1 + m2T2 / m1+ m2

T = 200× 80 + 300× 10/ 500

T = 38oC.

12. The temperature of 600 g of cold water rises by 15oC when 300 g of hot water at 50oC is added to it. What was the initial temperature of the cold water?

Ans: Mass of hot water (m1) = 300 g

Temperature (T1) = 50oC

Mass of cold water (m2) = 600 g

Change in temperature of cold water (T − T2) = 15oC

Final temperature = ToC

The specific heat capacity of water is c = m1c1(T1- T) = m2c2(T – T2)

300(50 – T) = 600 (15)

T = 20oC

Final temperature = 20oC

Change in temperature = 15oC

Initial temperature of cold water = 20oC − 15oC = 5oC.

13. 1kg of water is contained in a 1.25 kW kettle. Calculate the time taken for the temperature of water to raise from 25oC to its boiling point 100oC.Specific heat capacity of water =4.2 J kg-1 K-1.

Ans: Mass of water = 1000 g.

Change in temperature=100oC − 25 oC = 75 oC (or 75K)

Amount of heat energy required to raise temperature = 1000 × 4.2 × 75 = 315000 J. Time taken to raise the temperature, T = 4 min 12 seconds.

### Exercise. 11 B

1. (a) What do you understand about the change of phase of a substance ?

Ans: The process of change from one state to another at a constant temperature is called the change of phase of substance.

(b) Is there any change in temperature during the change of phase?

Ans: No change in temperature during the change of phase.

(c) Does the substance absorb or liberate any heat energy during the change of phase?

Ans: Yes, the substance absorbs or liberates heat during the change of phase.

(d) What is the name given to the energy absorbed during a phase change ?

Ans: Latent heat

2. A substance changes from its solid state to the liquid state when heat is supplied to it.

(a) Name the process.

Ans: Melting: The change from solid to liquid phase on heating at a constant temperature is called melting.

(b) What name is given to heat absorbed by the substance?

Ans: Latent heat of melting.

(c) How does the average kinetic energy of molecules of the substance change?

Ans: The average kinetic energy of the molecules does not change as there is no change in temperature.

3. A substance on heating undergoes (i) a rise in its temperature,(ii) a change in its phase without change in its temperature. In each case, state the change in energy of molecules of the substance.

Ans: (i) Average kinetic energy of the molecules changes.

(ii) Average potential energy of the molecules changes.

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?

Ans: (a) it does not change

(b) it increases

5. State the effect of the presence of impurity on the melting point of ice. Give one use of it.

Ans: The melting point of ice will decrease by the presence of an impurity in it.

Use: In making a freezing mixture by adding salt to ice. This freezing mixture is to be used in the preparation of ice cream.

6. State the effect of the increase of pressure on the melting point of ice.

Ans: The melting point of the ice decreases by the increase in pressure. The melting point of the ice decreases by 0.0072oC for every one atmosphere's rise in pressure.

7. The diagram in below shows the change of phases of a substance on a temperature time graph on heating the substance at a constant rate.

(a) What do parts AB, BC, CD and DE represent?

Ans: In part AB the temperature of solid rises from 0 to T1oC, In part BC shows melting at the temperature T1oC, In CD part shows rise in temperature of liquid from T1oC to T3oC and in part DE shows the boiling at the temperature T3oC.

(b) What is the melting point of the substance?

Ans: T1oC.

(c) What is the boiling point of the substance?

Ans: T3oC.

8. The melting point of naphthalene is 80oC and the room temperature is 25oC. A sample of liquid naphthalene at 90o C 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.

Ans: Temperature-time graph is as follows: -

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.

Ans:

10. Explain the terms boiling and boiling point. How is the volume of water affected when it boils at 100oC ?

Ans: Boiling: Change from liquid to gaseous phase on heating at a constant temperature is called boiling.

Boiling Point: Particular temperature at which vaporization occurs is called the boiling point of liquid.

The Volume of water will increase when it boils at 100oC.

11. How is the boiling point of water affected when some salt is added to it?

Ans: Boiling point of water increases on adding salt.

12. What is the effect of an increase in pressure on the boiling point of a liquid?

Ans: Boiling point of water increases on adding salt.

13. Water boils at 120 °C in a pressure cooker. Explain the reason

Ans: Boiling point of a liquid increases with the increase in the pressure and decreases with the decrease in the pressure. 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 the pressure, as the steam is not allowed to escape out from it.

14. Write down the approximate range of temperature at which the water boils in a pressure cooker.

Ans: In a pressure cooker, the water boils at about 120o C to 125o C

15. It is difficult to cook vegetables on hills and mountains. Explain the reason.

Ans: Because at high altitudes atmospheric pressure is low; therefore, the boiling point of the water decreases and so it doesn't provide required heat energy for the cooking.

16. Complete the following sentences:

(a) When ice melts, its volume………….

Ans: Its volume decreases.

(b) Decrease in pressure over ice ………….. its melting point.

Ans: Decrease in pressure over ice increases its melting point.

(c) Increase in pressure ………..the boiling point of water.

Ans: Increase in pressure increases the boiling point of water.

(d) A pressure cooker is based on the principle that the boiling point of water increases with....…

Ans: A pressure cooker is based on the principle that the boiling point of water increases with the increase in pressure.

(e) The boiling point of water is defined as ………………………..

Ans: The boiling point of water is defined as the constant temperature at which water changes to the steam.

(f) water can be made to boil at 115°C by ................. pressure over its surface.

Ans: Water can be made to boil at 115°C by increasing pressure over its surface.

17. What do you understand by the term latent heat?

Ans: Latent heat: Heat energy exchanged in change of phase is not externally manifested by any rise or fall in the temperature, it is considered to be hidden in the substance and is called the latent heat.

18. Define the term specific latent heat of fusion of ice. State its S.I. unit.

Ans: The quantity of heat required to convert a unit mass of ice into liquid water at 0° C (melting point) is called the specific latent heat of fusion of ice.

It's S.I. unit is Jkg-1.

19. Write the approximate value of specific latent heat of ice.

Ans: The specific latent heat of ice: 336000 J kg-1.

20. 'The specific latent heat of fusion of ice is 336 J g-1 '. Explain the meaning of this statement.

Ans: It means 1 g of ice at 0o C absorbs 360 J of heat energy to convert into water at 0o C.

21. 1 g ice at 0oC melts to form 1 g water at 0oC. State whether the latent heat is absorbed or given out by ice.

Ans: Latent heat will absorbed by ice

22. Which has more heat: 1 g of ice at 0oC or 1 g of water at 0oC? Give reasons.

Ans: 1 g of water at temperature 0oC has more heat than 1 g of ice at temperature 0oC. This is because ice at 0oC absorbs 360 J of heat energy to convert into water at 0oC.

23. (a) Which requires more heat: 1 g ice at 0o C or 1 g water at 0oC to raise its temperature to 10oC?

Ans: 1 g ice at temperature 0oC requires more heat because ice would require additional heat energy equal to latent heat of melting.

(b) Explain your answer in part (a).

Ans: 1 g ice at temperature 0oC first absorb heat 336 J to convert into 1 g water at 0oC

24. Ice cream appears colder to the mouth than water at 0oC. Give reasons.

Ans: Because 1 g ice at temperature 0oC takes 336 J of heat energy from the mouth to melt at 0oC. Thus the mouth loses an additional 336 J of the 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 the water at 0oC.

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 a reason

Ans: Because 1 g ice at temperature 0oC takes 336 J of heat energy from the bottle to melt into water at 0oC. Thus the bottle loses an additional 336 J of heat energy for 1 g of ice at 0oC than for 1 g of iced water at 0oC. So bottled soft drinks get cooled, more quickly by the ice cubes than by iced water.

26. It is generally colder after a hail storm than during and before the hail storm. Give reasons.

Ans: 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 feels colder.

27. The temperature of surroundings starts falling when ice in a frozen lake starts melting. Give reasons.

Ans: Because heat energy requires form, melting the frozen lake is absorbed from the surrounding atmosphere. so the temperature of the surrounding falls and it becomes very cold.

28. Water in lakes and ponds does not freeze at once in cold countries. Give a reason.

Ans: Specific latent heat of fusion for ice is sufficiently high, approx 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 area. If there is any layer of ice formed on water then water being a poor conductor of heat will also prevent the loss of heat from water of the lake. so, in cold countries water in lakes and ponds does not freeze.

29. Explain the following:

(a) The surroundings become pleasantly warm when water in a lake starts freezing in cold countries.

Ans: Reason is that the specific latent heat of fusion of ice is sufficiently high, so when water of a lake freezes, a large quantity of heat will be released hence the surrounding temperature becomes pleasantly warm.

(b) The heat supplied to a substance during its change of state, does not cause any rise in its temperature.

Ans: Heat supplied to a substance during its change of state, doesn’t cause any rise in its temperature because this is latent heat of phase change which is required to change phase only.

### Multiple Choice Type

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

Ans: Correct option is (c) J kg-1

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

Ans: Correct option is (c) 80cal g-1

### Numericals

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.

Ans: Mass of ice=10g = 0.01kg

Heat energy absorbed, Q=5460J

Specific latent heat of fusion of ice=?

Specific heat capacity of water = 4200Jkg-1K-1

Heat energy requires 10g of water at 0oC to raise its temperature by 50oC is = 0.01×4200×50=2100J.

Let, Specific latent heat of fusion of ice is L Jg-1.

Then,

Q = mL + mc $\Delta $ T

5460 J =10 × L + 2100J

L = 336Jg-1.

2. 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.

Ans: Given, 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 ice L =336 J g-1

Heat energy will be released when 5.0 g of water at 20oC changes into water at 0oC is, = 5×4.2×20 = 420 J.

This energy is released when 5.0g of water at 0oC changes into ice at

0oC = 5×336J = 1680J.

So, total Heat released = 1680 J + 420 J = 2100 J.

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?

Ans: (a) Given, mass of metal =150 g

Now, specific latent heat of metal

L=Q/m

= 75000/150

= 500Jg-1

(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 -50°C?

Ans: Specific heat capacity of metal=200 J kg-1 K-1.

Now, change in temperature= 800-(-50) = 850oC (or 850 K).

ΔQ = mcΔT

= 0.15×200×850

= 25500J

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?

Ans: Given that, m=150g=.15 kg,power=100W,time=4 min=60second

Total heat supplied in 4 min=100×4×60 =24000J

Let L= latent specific heat

mL = 24000

L = 24000/0.15

= 1.6×105 Jkg-1

5. 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 is2.1Jg-1K-1.

Ans: heat is released when 100g of water cools from 20o to 0oC

= 100X20X4.2 = 8400J.

heat is released when 100g of water converts into ice at 0oC

= 100X336 = 33600J.

heat is released when 100g of ice cools from 0oC to -10oC

= 100X10X2.1= 2100J.

Total heat = 8400+33600+2100 = 44100J.

Time taken = 73.5min = 4410s.

Now, average rate of heat extraction (power)

P = E/t

= 44100

= 10W

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.

Ans: Given, 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.

Assume that there is no loss of heat, heat energy gained by ice (latent heat of ice), Q= heat energy release by water

Q = 40 × 34 × 4.2 = 5712 J.

Specific latent heat of ice, L=Q/m

= 5712/12

=336J g-1

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.

Ans:

Given mass of water = 170 g = 0.17 kg

T1 = 50°C, T2 = 5°C,

Specific heat capacity of water = 4200 J kg –1°C –1

Specific Latent heat of ice = 336000 J kg –1

Let ‘y’ mass of ice add then,

Heat lost by water = heat used by ice.

Water at 50°C to 5°C = ice at °C to water at °C + water at 0°C to 5°C

0.17 × 4200 ×(50-5)= y × 336000 + y × 5

= 32130

= 357000y

Now, y = 32130/357000

= 0.09 Kg or 90 g

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.

Ans: Assuming all of the ice melt and the final temperature of mixture be ToC.

heat energy gain by ice of 10g at -10oC to raise its temperature to 0oC= 10x10x2.1=210J

heat energy gain by ice of 10g at 0oC to convert into water at 0oC=10x336=3360 J

heat energy gain by 10g of water (obtained from ice) at 0oC to raise its temperature to ToC = 10x4.2×(T-0)=42T

heat energy release by 10g of water at 10oC to lower its temperature to ToC = 10x4.2×(10-T)=420-42T

Heat energy gain = Heat energy lost

210 + 3360 + 42T = 420-42T

T = -37.5oC

This will not be possible because water cannot exist at this temperature.

Thus, the whole ice will not melt. Assume m gm of ice melt then the final temperature of the mixture becomes 0oC.

Thus, heat energy gain by 10g of ice at -10oC to raise its temperature to 0oC

= 10×10×2.1=210J

heat energy gain by m gram of ice at 0oC to convert into water at 0oC

= m × 336=336m J

then, released heat energy by 10g of water at 10oC to lower its temperature to 0oC is, = 10×4.2×(10-0)=420

Heat energy gain = Heat energy lost

210 + 336m = 420

m = 0.625 gm

9. 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.

Ans: Let the final temperature of water when it completely melts =ToC.

heat lost when 200g of water at 50oC cools to ToC is,

200 × 4.2×(50-T) = 42000-840T

heat gain when 40g of ice at 0oC converts into water at 0oC.is

40 ×336J

=13440 J

heat gain by water of 40g at 0oC when temperature rises to ToC= 40×4.2×(T-0)

= 168T

We know,

Amount of heat gain = amount of heat energy lost.

13440+168T= 42000-840T

168T+840T= 42000-13440

1008T= 28560

T=28560/1008=28.33oC.

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

Ans: We know that, Heat energy loss by (water+Calorimeter) = Heat energy gain by ice

So, heat energy loss by (water+calorimeter) = mw Cpw ΔT + mC Cpc ΔT = mi ( L + Cpw δT ) ........(1)

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 placing all the values in equation (1) and will get the mass of ice,

mi =150×4.2×27 + 50×0.4x27 / 330×4.2×5

= 50g

11. 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.

Ans: Given, mass of copper vessel (m1 ) is 50 g.

The mass of water contained in a copper vessel (m2) is 250 g.

mass of ice required to bring down the temperature of vessel is m

And the final temperature is 5o C.

heat gain when 'm' g of ice at 0o C melts into water at 0o C = m × 336 J

heat gained when temperature of 'm' g of water at 0o C rises to 5o C = m× 4.2× 5

Total heat gain = m × 336 + m × 4.2 × 5

heat lost when 250 g of water at 30o C cool down to 5o C

=250 × 4.2 × 25 = 26250 J

heat lost when 50 g of vessel at 30o C cools to 5o C

=50 × 0.4 × 25 = 500 J

Total heat lost = 26250 + 500 = 26750 J

Now, heat gained = heat lost

m × 336 + m × 4.2 × 5 = 26750

357 m = 26750

m = 26750/357 = 74.93 g

So, the mass of ice will be 74.93 g.

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.

Ans: Till the whole block does not melt and only 2 kg of it melts, final temperature will be 0 oC.

heat energy gained by ice of 2 kg at 0oC to convert into water at 0oC=2×336000=672000 J

Assume that amount of water poured is m kg.

Initial temperature of water =100oC.

Final temperature of water =0oC.

heat energy loss by water of m kg at 100oC to reach temperature 0oC =

m ×4200×100 = 420000m J

Now, we know

heat energy gained =heat energy lost.

672000J= m×420000J

m=672000/420000

=1.6 kg

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.

Ans: heat energy gain by 100 g of ice at -10o C to raise its temperature to 0o C is,

100 × 2.1 × 10 = 2100 J

heat energy gain by 100 g of ice at 0o C to convert into water at 0o C is,

100 × 336 = 33600 J

heat energy gain when temperature of 100 g of water at 0o C rises to 100o C is,

100 × 4.2 × 100 = 42000 J

So, total heat energy gained is = 2100 + 33600 + 42000 = 77700 J = 7.77 × 104 J.

14. 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.

Ans: heat energy gain by 1kg of ice at -10oC to raise its temperature to 0oC is,

= 1 × 2100 × 10 = 21000 J

Let, heat energy gain by 1kg of ice at 0oC to convert into water at 0oC is L

So, heat energy gained when temperature of 1kg of water at 0oC rises to 100oC is,

= 1 × 4200 × 100 = 420000 J

Total heat energy gained is = 21000+420000+L=441000 +L.

Given that total heat gained is =777000J.

So,

441000+L=777000.

L=777000-441000.

L=336000 J Kg-1

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.

Ans: Given that, mass of ice (mice) is = 200 g

Time taken by ice to melt is, t1 = 1 min = 60 s

Mass of the water, mw = 200 g

And temperature change of water is, ΔT = 20 °C

We know that, Rate of heat exchange is constant. So, power require for converting ice into water will same as the power require to rice the temperature of water

So,, P Ice= P Water

E Ice / t1 = EWater / t2

miceL/t1 = mWcw ΔT/t2

t_{2}=mWcw ΔTxt1/miceL

t2=200×4.2x20x60 / 200×336

t2=15 s

So, time will be 15 seconds.

### Chapters Included in Class 10 Mathematics Textbook (Concise – Selina Publication)

The book is divided into a total of 7 sections, and each section contains a particular number of chapters. Check the below-given list to find out more.

### Unit 1. Commercial Arithmetic

Value Added Tax.

It is followed by Banking (Recurring Deposit Account).

And then, Shares and Dividends.

### Unit 2. Algebra

Linear inequations (In one variable).

Quadratic equations.

Solving (simple) problems (Based on Quadratic Equations).

Ratio and Proportions (Including Properties and Uses).

Remainder and Factor Theorems.

Matrices.

Arithmetic Progression.

Geometric Progression.

### Unit 3. Coordinate Geometry

Reflection in x-axis, y-axis, x = a, Y = a and the origin; Invariant Points is the first.

It is followed by Section and Midpoint Formula.

And then Equation of a line.

### Unit 4. Geometry.

The similarity is the first chapter here.

It is followed by, Loci (Locus and its constructions)

Then comes the Circles.

It is followed by Tangents and Intersecting Chords.

Lastly, Constructions (Circles).

### Unit 5. Mensuration

Cylinder, Cone, and Sphere (Surface area and volume) is the only chapter here.

### Unit 6. Trigonometry

Trigonometric Identities (Including Trigonometric Ratios of Complementary Angles and use of four-figure Trigonometrical Tables) and Heights and Distances, are the two chapters here.

### ICSE Board Class 10 Mathematics Chapter 12 (Concise – Selina Publication).

Chapter number 12 of the class 10 Mathematics textbook published by Concise – Selina Publication is “Reflection (In x-axis, y-axis, x = a, y = a and the origin; Invariant points)”. It is the first chapter of Unit 3, which is Coordinate Geometry.

The chapter contains the following topics, Introduction, Co-ordinate Axis, Co-ordinates, Reflection, Reflection in the line Y = 0 i.e., in the x-axis, Reflection in the line x = 0 i.e., in the y-axis, Reflection in the origin, Invariant Point, Reflection of a point in the lines x = a and y = a, using graph paper.

Exercises of Class 10 Mathematics Chapter 12 Reflection (In x-axis, y-axis, x = a, y = a and the origin; Invariant points).

The first exercise which is 12 (A) consists of 19 questions in total and the next exercise which is 12 (B) consists of 17 questions, which are to be solved on the graph paper.

In Mathematics, practicing is the only way to master it. There are no shortcuts but to practice, if there is a mantra to be the master of Mathematics, then it is “practice”. The only magic word in mathematics is “practice”. Therefore, after learning the concepts, students are required to practice as much as possible and the exercise provided at the end of the chapter is the best way to start practice.

## FAQs on Chapters included in Class 10 Mathematics Textbook (Concise – Selina Publication)

**1. I have learned the concepts of the chapters but it is taking me too long to solve the questions. What should I do?**

As said earlier you should do practice, by attempting as many questions as possible. As you practice more and more questions your speed at solving the problem will gradually increase and eventually you will be able to solve the problems in no time. The principle of practice can be applied to anything in which one wants to become a master. Therefore, if you wish to be a master of Mathematics you should keep on practicing, as much as possible without any distractions and detours.

**2. Why should I follow the books published by Concise – Selina Publications.**

Concise – Selina publication books are the most reliable and sought after books for class 10 ICSE boards. The books are designed in a concise manner sought-after places and illustrations, yet include not a single example and illustration which are not required; therefore, you can also solve the examples, in the same manner, you will be solving the exercise questions, just for the practice and to be better on the topic.

**3. I have solved many exercise questions but some of them are incorrect and I am not able to figure out what am I doing wrong, what should I do?**

You should visit the Vedantu website and download the Solution for Class 10 Mathematics Chapter 12 Reflection (In x-axis, y-axis, x = a, y = a and the origin; Invariant points) solution, which is totally free for you. In the solution provided by Vedantu, you will find everything you need. It is a step by step guide, so you will easily find out what you were doing wrong. After finding out your mistake, work on that area, and solve those questions again, in this manner you will become better at the topic.

**4. Do Vedantu Solutions provide the graphs for the questions which require graphical representation?**

Yes, the solutions provided by Vedantu for Class 10 Chapter 12, Reflection, (In the x-axis, y-axis, x = a, y = a, and the origin; Invariant points) includes the graphs. Since all the questions of Exercise 12 (B) require graphs, Vedantu solutions provide the same for each of those questions. Graphs are an important part of this chapter therefore Vedantu expert teachers provide the same with precision and visibility in a clear manner, so that students can understand it easily without any problem.

**5. Why choose Vedantu Solutions?**

The solutions that Vedantu provides are prepared for you by expert teachers, who excel in the particular field. Solutions provided are well formatted and are step by step so that it becomes easy for the students to follow the solutions, everything will be easy for the eyes to see and for the mind to understand, all the resource material available at the Vedantu is prepared in such a manner, so students will not have to go through any sort of trouble at all.