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Class-X Question Bank 1

Question Bank

Calorimetry

1. Define the term heat.

Ans. It is the sum total of kinetic and potential energy of all the molecules of a

given substance.

2. Name the SI unit of heat.

Ans. Joule (J) is the SI unit of heat.

3. Define temperature :

Ans. The average kinetic energy of all the molecules of a given substance is

called temperature.

4. State the SI unit of temperature.

Ans. Kelvin (K) is the SI unit of temperature.

5. The temperature of a substance rises by 25°C. How much is the rise on

kelvin scale and why?

Ans. The rise on kelvin scale is 25 K. It is because, rise in temperature in

°C = rise in temperature in kelvin.

6. What is a calorimeter? Why is it made of copper? Give two reasons.

Ans. A copper vessel along with copper stirrer used in the measurement of heat

energy is called calorimeter. Calorimeter is made from copper because,

(i) it takes the temperature of the contents within it very quickly.

(ii) it has a very low specific heat capacity and hence takes very little amount

of heat energy from the contents within it.


7. (a) Define : (i) Calorie, (ii) Kilocalorie.

(b) How many calories are in one kilocalorie?

(c) How many joules of energy are in : (i) one calorie, (ii) one kilocalorie?

(d) State the SI unit for measuring heat energy.

Ans. (a) (i) Calorie : It is the amount of heat energy required to raise the

temperature of 1 g of water through 1°C.

(ii) Kilocalorie : It is the amount of heat energy required to raise the

temperature of 1 kg of water, through 1°C.

(b) 1000 calories = 1 kilocalorie

(c) (i) 1 calorie = 4.2 J

(ii) 1 kilocalorie = 4200 J

(d) The SI unit for measurement of heat energy is joule.

8. State three factors, which determine the heat absorbed by a cold body.

Ans. (i) Heat absorbed is directly proportional to the mass of the body.

(ii) Heat absorbed is directly proportional to the specific heat capacity of the

body.

(iii) Heat absorbed is directly proportional to the rise in temperature of the body.


9. (a) Define the following terms :

(i) specific heat capacity

(ii) thermal capacity or heat capacity.

(b) State the SI units of : (i) specific heat capacity (ii) thermal capacity.

(c) Which material is calorimeter made of?

Ans.(a) (i) The amount of heat energy required to raise the temperature of one kg

of a substance through 1 K is called its specific heat capacity.

(ii) The amount of heat energy required to raise the temperature of a given

mass of a substance through 1 K is called its thermal capacity.

(b) (i) The SI unit of specific heat capacity is J kg–1 K–1.

(ii) The SI unit of thermal capacity is JK–1.

(c) Calorimeter is made from copper. Recently copper has been replaced by

aluminium.

10. Specific heat capacity of water is 4200 Jkg–1 k–1. What do you understand

by the statement?

Ans. From the above statement, it implies that, if the temperature of 1 kg of

water is raised through 1 K, then the amount of heat energy required to

do so is 4200 J.


11. The thermal capacity of a vessel is 428 J K–1. What do you understand by

the statement?

Ans. From the above statement, it implies that, if the temperature of the given vessel

is raised through 1 K, then heat energy required to do so is 428 K.

12. The heat capacity of a vessel is 42 J °C–1. How much heat energy is required

to raise its temperature from 10°C to 20°C?

Ans. Rise in temperature = (20 – 10)°C = 10°C

∴ Heat energy required to raise its temperature through 10°C =

Heat capacity × Rise in temperature = 42 J°C–1 × 10°C = 420 J.

13. Some amount of heat energy is supplied to two liquids A and B. The liquid A

shows a greater rise in temperature. What can you say about heat capacity of

A as compared to that of B?

Ans. Heat capacity of A is less than that of B.

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

Ans. Cooking pans are made from metals such as copper, brass, stainless steel, etc.

They have low specific heat capacity and hence rapidly attain the temperature

of flame used for heating. This in turn can burn the food.

To avoid burning of food the base is made heavy and thick, so that the

temperature within the pan is less than the flame and hence the food cooks

without burning.


15. A substance of mass m1, specific heat capacity c1 at a temperature T1 is mixed

with another substance of mass m2, specific heat capacity c2 at a lower

temperature T2. Deduce the expression for the final temperature of the mixture.

State any assumption made, if any.

Ans. Let the final temperature of mixture be T

∴ Rise in temperature (θR) = (T – T2)

Fall in temperature (θ f) = (T1 – T)

Now, heat gained by substance at lower temp.

= Heat lost by substance at higher temp.

m2c2 (T – T2) = m1c1 (T1 – T)

m2c2T – m2c2T2 = m1c1T1 – m1c1T

m2c2T + m1c1T = m1c1T1 + m2c2T2

(m2c2 + m1c1)T = m1c1T1 + m2c2T2

T = 1 1 1 2 2 2

2 2 1 1

m c T m c Tm c m c

++

16. Why are hot water bottles very efficient for fomentation?

Ans.Water has the highest sp. heat capacity of 4.2 Jg–1 °C–1. Furthermore, its

temperature does not rise beyond 100 °C, under normal conditions. Thus, it

can store large amount of heat energy at quite a bearable temperature and

hence it is ideal for fomentation purposes.


17. What are land and sea breezes? Explain how are they caused in the coastal

regions.

Ans. Local winds which blow from the sea to the land during the day-time and from

the land to the sea during night-time are called sea and land breezes.

Formation of sea breeze : Specific latent heat capacity of land is about five time

less, as compared to water. Thus, when the sun shines equally on the sea, as well

as on the land, the land gets comparatively hot as compared to the sea. The air

above the land gets hot and hence, rises, thereby causing a drop in pressure. To

make up for the drop in pressure, the cold air from the sea, blows towards the

land, thereby giving rise to sea breeze.

Formation of land breeze : At night the sea water, as well as the land radiate

heat energy. However, the land radiates heat energy, at much faster rate as

compared to the sea, owing to its five times less specific heat capacity, as

compared to the sea water. Thus, on the whole, the sea is comparatively warmer

as compared to the land. The air above the sea gets warm and rises up.

This causes a fall in pressure. To make up for this fall in pressure, the cold air

from the land blows towards the sea, thereby giving rise to land breeze.


18. Discuss the role of high specific heat capacity of water, with reference to the

climate in coastal regions.

Ans. The climate in coastal regions remains moderate due to the blowing in of the cold

air from sea during day-time (sea breeze) and blowing out of the cold air from

land to sea (land breeze), during night-time. The land and sea breeze are formed

as under :

Write here formation of land and sea breeze as in Q 17.

19. Why is water used as a coolant in motor-car radiators?

Ans. Water has the highest specific heat capacity of 4.2 Jg–1 °C–1 and its

temperature does not rise beyond 100 °C. Thus, it can absorb large amount

of heat from a working engine, which is then radiated out through the radiator.

Thus, on this low temperature the engine works efficiently.

20. Why is water sprayed on roads in the evening in hot summer?

Ans.Water has the highest specific heat capacity of 4.2 Jg–1 °C–1. Thus, water

absorbs large amount of heat from the roads, but its own temperature does not rise

much. Thus, on the whole the roads get cooled.

21. Why the temperature in hot summer falls sharply, after a sharp shower?

Ans. Water has the highest specific heat capacity of 4.2 Jg–1 °C–1. Thus, when

there is a sharp shower, the water evaporates. In doing so, it absorbs large

amount of heat from surroundings, hence temperature falls sharply.


22. Why do sandy soils get heated up quickly as compared to wet soils?

Ans. The specific heat capacity of the sand is about five times less than that of the

water. Thus, when sun shines equally on sandy soil and the wet soil, the sandy

soil gets heated up rapidly as compared to the wet soil. It is because, water

absorbs large amount of heat energy, but its temperature does not rise

sufficiently.

23. Why is water considered the best liquid for quenching thirst?

Ans. Water has the highest specific heat capacity of 4.2 Jg–1 °C–1 and hence it

can absorb large amount of heat energy, without rising in temperature

sufficiently. Thirst is the natural signal, when body produces more heat energy

than required. Thus, water is ideal for quenching thirst, because it can absorb

large amount of heat energy.

24. Why is it advisable to pour cold water over burns, caused on the human body,

by hot solids?

Ans. Water has the highest specific heat capacity of 4.2 Jg–1 °C–1. Thus, it can

extract large amount of heat energy rapidly from the site of burn and hence

gives a lot of relief.


25. Why does a wise farmer waters his fields, if the forecast says it is going to be

frost?

Ans. The frost can seriously damage the leaves and fruits of the plants.

When a farmer waters his fields, during night, this water gives large amount

of heat energy, because 1 g of water will liberate 4.2 J of energy for every

1 °C fall in temperature. Thus, the air around the field is saturated with heat

energy and its temperature does not fall below 0 °C. Thus, no frost is formed.

26. Why are big tubs of water kept in underground cellars for storing fresh fruit

and vegetables in cold countries?

Ans. During sub-zero temperatures, the water within the cells of fruits and vegetables

freezes and hence damages them. To avoid such a damage the huge water tubs

are kept. The water gives off 4200 J of heat energy for every one kg for 1 °C fall

in temperature. Thus, it does not allow the temperature of cellar to fall below

0 °C, and hence, their is no damage to fresh fruits or vegetables.

27. Define the following terms :

(i)Fusion (ii) Melting point

Ans. Fusion : The process of changing of a solid in liquid state, at some fixed

temperature, by the absorption of heat from some external source is called

fusion.

Melting point : It is the temperature at which a solid changes into liquid state.


28. What do you understand by the following terms :

(i) latent heat of fusion?

(ii) specific latent heat of fusion?

(iii) specific latent heat of fusion of ice?

(iv) state the value of sp. latent heat of fusion of ice in : (1) the SI system

(2) calories?

Ans. (i) Latent heat of fusion : It is the amount of heat energy required to change

a given mass of solid at its m.p. into its liquid state, without any rise in

temperature.

(ii) Specific latent heat of fusion : It is the amount of heat energy required to

change 1 kg of solid, at its melting point, into its liquid state, without any

rise in temperature.

(iii) Specific latent heat of fusion of ice : It is the amount of heat energy

required to change 1 kg of ice at 0 °C, into 1 kg of water at 0 °C, without

any rise in temperature.

(iv) Specific latent heat of fusion of ice is

(i) 336,000 Jkg–1 in the SI system (ii) 80 calories g–1 in CGS system.


29. Bottled drinks are cooled more effectively, when surrounded by lumps of ice

than iced water.

Or

Why do bottled soft drinks get cooled more quickly by ice cubes than by iced

water.

Ans. Every 1 kg of ice at 0 °C absorbs 336,000 J of heat energy to form water at

0 °C. As ice can extract 336,000 J of heat energy, more than water at 0 °C,

therefore, it cools the bottled drinks more effectively.

30 . Why does atmospheric temperature fall after a hailstorm?

Ans. Ice has the highest sp. latent heat of fusion of 336,000 Jkg–1. Thus, when the

hail melts, it absorb large amount of heat energy from the surroundings.

Thus, the atmospheric temperature drops.

31. Why does weather become pleasant when it starts freezing in cold countries?

Ans. Ice has the highest sp. latent heat of fusion of 336,000 Jkg–1. Thus, every 1 kg

of water at 0 °C, on freezing releases 336,000 J of heat energy. As enormous

amount of heat energy is released in atmosphere, therefore weather becomes

pleasant.

32. Why does it gets warm during a snow storm?

Ans. Same as Q.31.


33. Why does it become bitterly cold when ice melts in cold countries?

Ans. Same as Q.30.

34. Specific latent heat of fusion of lead is 27 × 103 Jkg –1. What do you

understand by the statement?

Ans. It means one kg of lead at its melting point, on solidification at the same

temperature will release 27 × 103 J of heat energy.

35. Given figure shows variation in temperature with time when some wax cools

from liquid phase to solid phase.

(i) In which part of the curve the wax is in liquid phase?

(ii) What does the part QRS represent?

(iii) At which point on the curve, the wax will be in the liquid as well as solid

phase?

(iv) In which part of curve the wax is in solid phase.

Ans. (i) Wax is in liquid phase in region PQ.

(ii) QRS represents change in state of liquid wax to solid wax at constant

temperature.

(iii) At point R the wax will be in the solid as well as liquid phase.

(iv) Wax is in the solid phase in region ST.


36. 1 kg of ice at 0°C is heated at a constant rate and its temperature is recorded

after every 30 s, till steam is formed at 100°C. Draw a temperature-time graph

to represent the change of phase.

Ans.

37. The melting point of naphthalene, a crystalline solid, is 80°C and room

temperature is 30°C. Liquid naphthalene at 100°C is cooled down to room

temperature. Draw a temperature-time graph to represent cooling curve.

Ans.

38. Ice-cream appears colder to mouth than water at 0°C. Give a reason.

Ans. Ice has the highest sp. latent heat of fusion, i.e., 336 Jg–1. Thus, every one

gram of ice on melting in mouth extracts out 336 J of extra heat energy, as

compared to water at 0°C. Thus, ice-cream appears colder than water at 0°C.


39. What do you understand by the following terms?

(i) Boiling, (ii) Boiling point.

Ans. (i) Boiling : The process of rapid change of a liquid into gaseous state, without any

change in temperature, and by the absorption of heat energy from an external

source is called boiling.

(ii) Boiling point : The constant temperature at which a liquid rapidly changes

to its gaseous state is called boiling point.

40. Graph shows change of phases of a substance on temperature-time graph.

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

(b)What is the melting point of substance?

(c) What is the boiling point of substance?

Ans.(a) (i) AB represents the solid being heated still it attains a temp T1.

(ii) BC represents changes in the state of solid to liquid at constant temperature.

(iii) CD represents the liquid being heated till it attains a temperature of T3.

(b) The melting point of the substance is T1.

(c) The boiling point of the substance is T2.


Numerical Problems

1. Calculate the amount of heat energy required to raise the temperature of 100 g

of copper from 20°C to 70°C (sp. heat capacity of copper = 390 J kg–1 K–1)

Ans. Mass of copper (m) = 100 g = 0.1 kg

Rise in temperature = (70 – 20) = 50°C R(θ )

Sp. heat capacity of copper (c) = 390 J kg–1 K–1

= 390 J kg–1 °C–1

Heat required = mcθR

= 0.1 kg × 390 J kg–1°C × 50°C

= 1950 J

2. 1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from

20°C to 40°C. Calculate the specific heat capacity of lead.

Ans. Mass of lead (m) = 0.5 kg

Rise in temperature = (40 – 20)°C = 20°C R(θ )

Heat required (H) = 1300 J

H = mc Rθ

∴ c = R

H 1300 Jθ 0.5 kg × 20°Cm

= 130 J kg–1 °C–1


3. Find the time taken by a 500 W heater to raise the temperature of 50 kg of a

material of specific heat capacity 960 J kg–1 K–1 from 18°C to 38°C. Assume

all heat energy supplied by the heater is given to the material.

Ans. Power of the heater (P) = 500 W = 500 Js–1

Let the time for which the heater works = t

Energy supplied by the heater = P × t

= 500 J s–1 × t

Rise in temperature of the material R(θ )

= (38 – 18)

= 20°C = 20 K

Mass of the material (m) = 50 kg

sp. heat capacity of the material (c) = 960 J kg–1 K–1

∴ Heat energy absorbed by the material

= mcθR = 50 kg × 960 J kg–1 K–1 × 20 K

= 960000 J

Energy supplied by the heater = Heat energy absorbed by the material

500 J s–1 × t = 960000 J

∴ t = –196000 J500 Js

= 1920 s

= 1920

60 min = 32 min


4. An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid

from 10.0°C to 15.0°C in 100 s. Calculate : (i) heat capacity of 4.0 kg of liquid,

(ii) the specific heat capacity of liquid.

Ans. Power of heater (P) = 600 W = 600 Js–1

Time (t) = 100 s

Energy supplied by heater = P × t

= 600 Js–1 × 100 s

= 60000 J

Rise in temperature = (15.0 – 10.0)°C R(θ )

= 5°C = 5 K

(i) For 5 K rise in temperature energy required

= 60000 J

For 1 K rise in temperature energy required

= 60000 J

5 K = 12000 JK–1

Heat capacity of liquid = 12000 JK–1

(ii) Specific heat capacity of liquid Heat capacity

=Mass

–112000 JK4.0 kg

–1 –1= 3000 Jkg K


5. A piece of iron of mass 2.0 kg has thermal capacity 966 J °C–1. Find : (i) the

heat energy needed to raise it to 15°C and (ii) its sp. heat capacity in SI unit.

Ans. (i) Thermal capacity = 966 J°C–1 = 966 JK–1

Rise in temperature = 15°C = 15 K

Heat energy required to warm iron

= Thermal capacity × Rise in temperature

= 966 JK–1 × 15 K = 14490 J

(ii) Specific heat capacity of iron

–1

Thermal capacity=

Mass of iron

966 JK2.0 kg

–1 –1483 Jkg K

6. 0.5 kg of lemon squash at 30°C is placed in a refrigerator, which can remove

heat at an average of 30 Js–1. How long will it take to cool lemon squash to

5°C? (sp heat capacity of lemon squash is 4200 J kg–1°C–1).

Ans. Mass of lemon squash (m) = 0.5 kg

sp. heat capacity of lemon squash (c) = 4200 Jkg–1°C–1

Fall in temperature (θ f) = (30 – 5)°C = 25°C

Heat extracted by refrigerator (H) = mc f θ

= 0.5 kg × 4200 Jkg–1°C–1 × 25°C = 52500 J


Rate of extraction of heat = 30 Js–1

Time in which lemon squash cools

= –15250030 Js

= 1750 s

= 1750 s

60 = 29.167 min

7. 200 g mass of a certain metal at 83°C is immersed in 300 g water at 30°C. The

final temperature is 33°C. Calculate the specific heat capacity of the metal.

(Assume that specific heat capacity of water is 4.2 Jg–1K–1).

Ans. DATA

Substances Metal Water

Mass 200 g 300 g Sp. heat capacity ? 4.2 Jg–1K–1 Initial temp. 83°C 30°C Final temp. = 33°C R = (33 – 30)

= 3°C = 3 K θ θ f = (83 – 33)

= 50°C = 50 K Heat lost by metal = mcθ f

= 200 g × c × 50 K

Heat gained by water = mcθR

= 300 g × 4.2 Jg–1K–1 × 3 K

= 3780 J

Heat lost = Heat gained


200 g × c × 50 K = 3780 J

c = 3780 J

200 50 gk×

= 0.378 Jg–1K–1

8. You are required to make a water bath of 40 kg, at 40 °C, by mixing hot water

at 100 °C and cold water at 15 °C. Calculate the ratio of hot and cold water,

which should be mixed.

Ans. Let the mass of cold water = x.

∴ Mass of hot water = (40 kg – x).

Substance Mass S.H.C Initial temp.

Final temp. = 40 °C

Cold water Hot water

x 40 kg – x

4200 Jkg–1°C–14200 Jkg–1°C–1

15 °C 100 °C

θR = (40 – 15) = 25 °C θF = (100 – 40) = 60 °C

Heat lost by hot water = Heat gained by cold water.

mc θF = mc θR.

(40 – x) × 4200 × 60 = x × 4200 × 25

∴ (40 – x) × 12 = 5x.

480 = 17x

∴ x = 48017

= 28.235 kg.

∴ Mass of cold water = 28.235 kg.

Mass of hot water = (40 – 28.235) = 11.765 kg.

∴ Mass of hot water : Mass of cold water = 11.765 : 28.235 = 1 : 2.4.


9. Temperature of 600 g cold water rose by 15 °C, when 300 g of hot water at

50 °C was added to it. What is the initial temp. of cold water?

Ans.

Substance Mass S.H.C Initial temp. Final temp. = ? (x)

Cold water

Hot water

600 g

300 g 4.2 Jg–1 °C–1

4.2 Jg–1 °C–1

?

50 °C

θR = 15°C

θF = (50°C – x)

Heat lost by hot water = Heat gained by cold water

300 × 4.2 × (50 – x) = 600 × 4.2 × 15

(50 – x) = 30

∴ x = 20 °C (Final temp.)

∴ Initial temp. of water = (Final temp. – θR) = (20 – 15) = 5 °C.

10. 200 g of hot water at 80°C is added to 400 g of cold water, when the temperature

of cold water rises by 18 °C. Calculate the initial temperature of cold water.

Ans. Proceed as in Q 9. [Ans. 26 °C]


11. Some amount of heat is given out to 120 g of water and its temperature rises by

10 K. When same amount of heat is given to 60 g of oil, its temperature rises by

40 K. The specific heat capacity of water is 4200 Jkg–1 K–1. Calculate :

(i) amount of heat in joules given to water.

(ii) specific heat capacity of oil.

Ans. (i) Heat absorbed by water = mcθR = 1201000

× 4200 × 10

= 5040 J.

(ii) Heat absorbed by water = Heat absorbed by oil

∴ 5040 J = 60 g × c × 40.

c = 5040 1000

60 40××

= 2100 Jkg–1K–1.

12. A metal drill of power 500 W, drills a hole in a metal cube of mass 0.25 kg in

6.5s. The specific heat capacity of lead is 130 J kg–1 °C–1.

(i) How much heat is generated by metal drill in one second?

(ii) Calculate the heat generated in 6.5 s.

(iii) If t °C, is the rise in temperature of metal cube, state the amount of heat

energy absorbed by it in terms of t.

(iv) Write down an equation and calculate the value of t.


Ans.(i) Energy generated by drill in 1 s = P × t = 500 × 1 s = 500 J.

(ii) Energy generated by drill in 6.5 s = 500 × 6.5 = 3250 J.

(iii) Energy absorbed by metal = mcθR = 0.25 × 130 × t = 32.5 t.

(iv) Energy absorbed by metal = Energy generated by drill

32.5 t = 3250

∴ t = 325032.5

= 100 °C.

13. An electric heater is of power 1000 W. It heats 0.4 kg of water for 2 minutes.

The specific heat capacity of water is 4200 J kg–1K–1.

(i) How much heat is liberated by heater in 1 s?

(ii) How much heat is liberated by heater in 2 minutes?

(iii) If θ °C, is the rise in temperature of water, state the amount of heat energy

absorbed by it, in terms of θ.

(iv) Write down an equation and calculate the value of θ.

Ans. Proceed as in Q. 12. [Ans. (i) 1000 J; (ii) 120,000 J; (iii) 1680 θ; (iv) 71.428 °C]

14. A copper calorimeter, weighing 57.5 g, contains 60 g of water at 12 °C. 55 g of

iron nails at 100 °C, are dropped into the calorimeter and stirred rapidly. The

final temperature attained by the calorimeter and its contents is 20 °C. Calculate

the specific heat capacity of iron. (Sp. heat capacity of copper is 0.4 Jg–1 °C–1

and that of water is 4.2 Jg–1 °C–1.


Ans.

Substance Mass S.H.C Initial temp. Final temp. = 20 °C

Calorimeter

Cold water

Iron

57.5 g

60 g

55 g

0.4 Jg–1 °C–1

4.2 Jg–1 °C–1

?

12°C

12°C

100°C

θR = (20 – 12) = 8°C

θF = (100 – 20) = 80°C

Heat gained by calorimeter = mcθR = 57.5 × 0.4 × 8 = 184 J

Heat gained by cold water = mcθR = 60 × 4.2 × 8 = 2016 J

Heat lost by iron = mcθF = 55 × c × 80 = 4400 cg °C.

Heat lost by iron = Heat gained by calorimeter + Heat gained by water.

4400 cg °C = (184 + 2016) = 2200 J

∴ c = 22004400

= 0.5 Jg–1 °C–1.

15. A molten metal weighing 150 g, is kept at its melting point, at 800°C. When

allowed to solidify at the same temperature, it gives off 7500 J of heat energy.

What is the specific latent heat of fusion of metal?

If its specific heat capacity is 200 Jkg–1 K–1, how much additional heat will it

give out in cooling to –50 °C?


Ans. (i) Heat given off by metal during change of state

= mass × sp. latent heat of metal

7500 J = 150 g × sp. latent heat of fusion of the metal.

Sp. latent heat of fusion of the metal = 7500150

= 50 Jg–l

(ii) Heat given off by metal on cooling = mcθF

= 1501000

× 200 × [800 – (– 50)] J = 25,500 J.

16. A solid weighing 200 g, is kept at its melting point at 0°C. When allowed to

melt at the same temperature, it absorbs 67200 J of heat energy.

Calculate the specific latent heat of fusion of solid. If its specific heat

capacity is 4200 Jkg–1K–1, how much additional heat will it absorb in

attaining a temperature of 80 °C?

Ans Proceed as in [Ans. (i) 336 Jg–l, (ii) 67200 J]

17. One kg of molten lead at its melting point 327 °C, is dropped in 1 kg of water at

20 °C. Assuming no heat is lost to surroundings, calculate final temperature of

water. [Sp. heat capacity of lead and water is 130 Jkg–1 K–1 and 4200 Jkg–1 K–1

respectively. Sp. latent heat of fusion of lead = 27000 J kg–1]


Ans.

Substance Mass S.H.C/S.L.H Initial temp.

Final temp. = ? (x)

Lead Water

1 kg 1 kg

130 Jkg–1 °C–1/27000 Jkg–1

4200 Jkg–1 °C–1

327°C 20°C

θR = (x – 20°C)

θF = (327°C – x) Heat lost by lead to change its state = mL = 1 × 27000 = 27000 J.

Heat lost by solid lead from 327 °C to final temp. = mcθF = 1 × 130 × (327 – x).

Heat gained by water = mcθR = 1 × 4200 × (x – 20).

Heat gained = Heat lost.

⇒ 4200 (x – 20) = 130 (327 – x) + 27000.

⇒ 4200x – 84000 = 42510 – 130x + 27000

⇒ 4330x = 153510

∴ x = 35.45 °C.

18. 0.6 kg of molten metal at its melting point 420 °C, is dropped into 2 kg of water

at 25 °C. Assuming no heat is lost to surroundings, calculate the final

temperature of water.[sp. heat capacity of metal and water are 250 Jkg–1 K–1

and 4200 Jkg–1 K–1 respectively. Sp. latent heat of fusion of metal = 29000

Jkg–1]

Ans. Proceed as Q : 17 [Ans. 30.98°C]


19. A block of lead of mass 1 kg at 27 °C was heated in a furnace, till it melts

completely. Find the quantity of heat required : (i) To bring the lead block to

its melting point, (ii) To completely melt the lead block at its melting point.

[m.p of lead = 327 °C; sp. heat capacity of lead = 130 Jkg–1 K–1 and sp.

latent heat of fusion of lead = 26,000 Jkg–1].

Ans.(i) Heat required to bring the lead to its melting point = mcθR

= 1 × 130 × (327 – 27) = 39000 J

(ii) Heat required to melt lead = mL = 1 × 26,000 = 26,000 J.

20. A block of a metal of mass 0.8 kg at 20 °C is heated in a furnace, till it

completely melts. Find the quantity of heat required : (i) to bring the metal to its

melting point, (ii) to completely melt the metal at its melting point [m.p of metal

= 420 °C; sp. heat capacity of metal 250 Jkg–1 K–1 and sp. latent heat of fusion

of metal = 29000 Jkg–1]

Ans. Proceed as Q : 19 [Ans. (i) 80,000 J, (ii) 23200 J]

21. A refrigerator converts 100 g of water at 20 °C to ice at –10 °C in 73.5

minutes. Calculate the average rate of heat extraction from water in watts.

[Specific heat capacity of ice = 2100 Jkg–1 K–1; sp. heat capacity of water

is 4200 J kg–1 K–1 and sp. latent heat of fusion of ice is 336,000 Jkg–1].

Ans. Heat lost by water at 20 °C to attain temp. of 0 °C

= mcθF = 1001000

kg × 4200 × 20 °C = 8400 J


Heat lost by water at 0 °C to form ice at 0 °C = mLice = 1001000

kg × 336,000

Jkg–1 = 33600 J

Heat lost by ice to attain temp. of – 10 °C

= mcθF = 1001000

kg × 2100 J

kg. C × 10 °C = 2100 J

Total heat lost= (8400 + 33600 + 2100) J = 44,100 J

∴ Rate of extraction of heat = 44,100 J

73.5 60 s× = 10 Js–1 = 10 W.

22. A cooling machine freezes 1.2 kg of water at 20°C, into ice at – 40°C in

12 minutes. Calculate the average rate of extraction of heat energy in watts.

[S.H.C of ice = 2100 Jkg–1 K–1, S.H. C of water = 4200 Jkg–1 K–1, S.L.H of ice

= 336000 Jkg–1]

Ans. Proceed as in Q. 21 [Ans. 336 W]

23. An ice cube of mass 30 g, is dropped in 200 g of water at 30 °C, such that all the

ice melts. Calculate final temperature of water. [Latent heat of ice = 80 cal g–1;

Specific heat capacity of water = 1 cal g–1 °C–1]

Ans.

Substance Mass S.H.C/S.L.H Initial temp. Final temp. = ? (x)

Ice

Water

30 g

200 g

80 cal g–1

1 cal g–1 °C–1

0°C

30 °C

θR = (x – 0) = x

θF = (30 – x)


Heat gained by ice to form water at 0 °C = mL = 30 × 80 = = 2400 calories

Heat gained by water at 0 °C to attain temp. x = mcθR = 30 × 1 × x = 30x.

Heat lost by water at 30 °C = mcθF = 200 × 1 × (30 – x) = 6000 – 200x.

Heat gained = Heat lost.

2400 + 30x = 6000 – 200x.

∴ 230 x = 3600

∴ x = 15.65 °C.

24. 50 g of ice cubes are placed in 500 g of water at 40 °C, such that all the ice melts.

Calculate final temperature of water. [Latent heat of ice = 80 cal g–1; specific heat

capacity of water = 1 cal g–1 °C–1]

Ans. Proceed as in Q. 23 [Ans. 29.09°C]

25. 10 g of ice at 0°C absorbs 5460 J of heat energy to melt and change to water

at 50°C. Calculate the specific latent heat of fusion of ice. Specific heat

capacity of water is 4200 J kg–1 K–1.

Ans.Mass of ice = 10 g.

Heat absorbed by ice to form water at 0°C = m Lice = 10 g × Lice

Rise in temperature = (50 – 0) = 50°C. R(θ )

∴ Heat absorbed by water at 0°C to attain temp. of 50°C

= mcθR = 10 × 4.2 × 50 = 2100 J

∴ Total heat absorbed by ice = 10 g × Lice + 2100 J


Also total heat absorbed by ice = 5460 J

10 g × Lice + 2100 J = 5460 J

⇒ 10 g × Lice = (5460 – 2100) J = 3360 J

∴ Lice = 3360 J10 g

= 336 Jg–1.

26. How much heat is released when 5.0 g of water at 20°C changes to ice at 0°C?

[Take sp. heat capacity of water = 4.2 J g–1°C–1, specific latent heat of

fusion of ice = 336 Jg–1]

Ans. Fall in temperature (θ )f = (20 – 0) = 20°C.

∴ Heat lost by water to attain temp. of 0°C = mcθf = 5 × 4.2 × 20 = 420 J

Heat lost by water to form ice at 0°C = mLice = 5 × 336 = 1680 J

∴ Total amount of heat released by water = (420 + 1680) J = 2100 J

27. In an experiment, 17 g of ice is used to bring down the temperature of 40 g

of water at 34°C to its freezing point. The specific heat capacity of water is

4.2 J kg–1 K–1. Calculate the specific latent heat of ice. State one

assumption made in the above calculation.


Ans. Fall in temperature of water (θ )f = (34 – 0) = 34°C

∴ Heat lost by water = mcθ f = 40 × 4.2 × 34 = 5712 J

Heat gained by ice to form water at 0°C = mLice = 17 g × Lice.

Heat gained by ice = Heat lost by water

17 g × Lice = 5712 J

∴ Lice = 5712 J17 g

= 336 Jg–1.

28. Find the result of mixing 10 g of ice at –10°C with 10 g of water at 10°C.

[sp. heat capacity of ice = 2.1 Jg–1K–1; sp. latent heat of ice = 336 Jg–1

and specific heat capacity of water = 4.2 J g–1 K–1]

Ans. Let the amount of ice which melts = x g

Heat gained by ice at –10°C to form ice at 0°C = mc = 10 × 2.1 × 10 = 210 J Rθ

Heat gained by ice at 0°C to melt= mLice = x × 336

∴Total heat gained = 210 J + 336x

Heat lost by water at 10°C = mc = 10 × 4.2 × 10 = 420 J Rθ

Heat gained = Heat lost

210 J + 336 x = 420 J 336x = (420 – 210) = 210 J

x = 210336

= 0.625 g

Thus, 0.625 g of ice melts.


29. What will be the result of mixing 400 g of copper chips at 500°C with 500 g of

crushed ice at 0°C.

[sp. heat capacity of copper = 0.42 Jg–1K–1 and sp.

latent heat of fusion of ice = 340 Jg–1].

Ans. Let the amount of ice which melts = x g ∴ Heat absorbed by ice on melting = mLice = x × 340

Fall in temperature of copper chips (θ f ) = (500 – 0) = 500°C

∴ Heat given out by copper chips = mcθf = 400 × 0.42 × 500 = 84000 J

Heat absorbed by ice = Heat given out by copper chips

x × 340 = 84000

∴ x = 84000

340 = 247.05 g

∴ Thus, 247.05 g of ice will melt.

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