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Chapter 11

Calorimetry

Class 10 - Concise Physics Selina


Exercise 11(A)

Question 1

Define the term heat.

Answer

Heat is the internal energy of molecules constituting the body. It flows from a hot body to a cold body when they are kept in contact.

Question 2

Name the S.I. unit of heat.

Answer

The S.I. unit of heat is joule (J).

Question 3

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

Answer

One calorie is the quantity of heat energy required to raise the temperature of 1 g of water from 14.5°C to 15.5°C.

1 calorie (or 1 cal) = 4.186 J or 4.2 J (nearly)

Question 4

Define one kilo-calorie of heat.

Answer

One kilo calorie is the heat energy required to raise the temperature of 1 kg of water from 14.5°C to 15.5°C

Question 5

Define temperature and name it's S.I. unit.

Answer

Temperature is a parameter which tells the thermal state of a body (i.e., the degree of hotness or coldness of body). It determines the direction of flow of heat when two bodies at different temperatures are placed in contact.

The S.I. unit of temperature is Kelvin (K).

Question 6

State three differences between heat and temperature.

Answer

HeatTemperature
Heat is that form of energy which flows from a hot body to a cold body when they are kept in contact.Temperature is a quantity which determines the direction of flow of heat on keeping the two bodies at different temperatures in contact.
The S.I. unit of heat is joule (J)The S.I. unit of temperature is kelvin (K).
The amount of heat contained in a body depends on mass, temperature and substance of body.The temperature of a body depends on the average kinetic energy of it's molecules due to their random motion.

Question 7

Define calorimetry.

Answer

The measurement of quantity of heat is known as calorimetry.

Question 8

Define the term heat capacity and state it's S.I. unit.

Answer

The term heat capacity of a body is the amount of heat energy required to raise it's temperature by 1° C (or 1 K).

The S.I. unit of heat capacity is joule per kelvin (J K-1)

Question 9

Define the term specific heat capacity and state it's S.I. unit.

Answer

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through 1°C (or 1 K). i.e.,

Specific heat capacity c=Amt of heat energy suppliedmass×rise in temperature\text{Specific heat capacity c} = \dfrac{\text{Amt of heat energy supplied}}{\text{mass} \times \text{rise in temperature}}

The S.I. unit of specific heat capacity is joule per kilogram per kelvin (J kg-1 K-1).

Question 10

How is the heat capacity of a body related to the specific heat capacity of it's substance?

Answer

The equation which relates the heat capacity of a body to the specific heat capacity is —

Heat capacity (C') = Mass (m) × Specific heat capacity (c)

Question 11

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

Answer

Heat capacitySpecific heat capacity
It is the amount of heat energy required to raise the temperature of entire body by 1°CIt is the amount of heat energy required to raise the temperature of unit mass of the body by 1°C.
It depends both on the substance and mass of the body. More the mass of the body more is it's heat capacity.It does not depend on the mass of the body, but it is the characteristic property of the substance of the body.
S.I. unit is J K-1S.I. unit is J kg-1 K-1.

Question 12

Name a liquid which has the highest specific heat capacity.

Answer

A liquid which has the highest specific heat capacity is water.

Question 13

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

Answer

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

Question 14

What do you mean by the following statements —

(i) The heat capacity of a body is 50 J K -1?

(ii) The specific heat capacity of copper is 0.4 J g-1 K-1 ?

Answer

(i) "The heat capacity of a body is 50 J K-1" means 50 J of heat energy is required to raise the temperature of that body by 1 K.

(ii) "The specific heat capacity of copper is 0.4 J g-1 K-1" means that the heat energy required to raise the temperature of 1 g of copper by 1 K is 0.4 J.

Question 15

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

Answer

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

Question 16

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

Answer

The factors on which the heat energy liberated by a body on cooling depends are:

  1. Mass of the body
  2. Temperature of the body.

Question 17

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

Answer

Three factors on which the heat energy absorbed by a body depends are:

  1. Mass of the body — Heat energy absorbed by a body is directly proportional to the mass of the body i.e., Q ∝ m.
  2. Increase in temperature of the body — Heat energy absorbed is directly proportional to the rise in temperature i.e., Q ∝ △t.
  3. The material of the body — Heat energy absorbed by a body depends on the substance of the object which is expressed in terms of it's specific heat capacity c i.e., Q ∝ c.

Question 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° C.

Answer

The expression for the heat energy Q is given by:

Q = m x c x △t joules,

where, c is specific heat capacity

m is mass of substance

△t is change in temperature

Question 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?

Answer

As the substance having low specific heat capacity will show a rapid and high rise in temperature and since liquid A shows a greater rise in temperature as compared to B when same amount of heat is supplied to both. Hence, heat capacity of liquid A is less than that of B.

Question 20

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

Answer

Let,

Specific heat capacity of block P = Cp

Specific heat capacity of block Q = CQ

From relation,

c=Qm×tc = \dfrac{Q}{m \times △t} \\[0.5em]

where, c = specific heat capacity

m = mass

Q = heat energy

△t = change in temperature

Now,

CpCQ=Q2m×tQm×tCpCQ=m×t2m×tCpCQ=12\dfrac{C_p}{C_Q} = \dfrac{\dfrac{Q}{2m × △t}}{\dfrac{Q}{m × △t}} \\[0.5em] \dfrac{C_p}{C_Q} = \dfrac{m × △t}{2m × △t} \\[0.5em] \Rightarrow \dfrac{C_p}{C_Q} = \dfrac{1}{2} \\[0.5em]

Hence, the required ratio is 1 : 2

Question 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.

Answer

Heat energy lost by the hot body = Heat energy gained by the cold body. This is called the principle of method of mixture.

The other name given to it is the principle of calorimetry.

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

Question 22

A mass m1 of a substance of specific heat capacity c1 at temperature t1 is mixed with a mass m2 of other 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.

Answer

Let a substance A of mass m1, specific heat capacity c1 at temperature t1 is mixed with another substance B of mass m2, specific heat capacity c2 at a lower temperature t2.

If the final temperature of the mixture becomes t, then

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 heat energy is 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)

m1 c1 t1 - m1c1 t = m2 c2 t - m2 c2 t2

m1c1 t1 + m2c2 t2 = m1c1 t + m2c2 t

m1c1 t1 + m2c2 t2 = t (m1c1 + m2c2)

Therefore,

t=m1c1t1+m2c2t2m1c1+m2c2\text{t} = \dfrac{\text{m}_1 \text{c}_1 \text{t}_1 + \text{m}_2 \text{c}_2 \text{t}_2}{\text{m}_1\text{c}_1 + \text{m}_2\text{c}_2}

The assumption made here is that there is no loss of heat energy.

Question 23

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

Answer

On a cold winter night, if the atmospheric temperature falls below 0° C, water in the fine capillaries of plants will freeze, so the veins will burst due to the increase in volume of water on freezing. As a result, plants will die and the crop will get destroyed. In order to save 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 0° C.

Question 24

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

Answer

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 than that required by the same mass of sand. Similarly, a certain mass of water will impart nearly five times more heat energy than that given by the same mass of sand for the same fall in temperature.

As such sand (or earth) gets heated or cooled more rapidly as compared to water under the similar conditions (exposure to the Sun). Thus, near the sea shore, there becomes a large difference in temperature between the land and sea due to which convection currents are set up. The cold air from the land blows towards the sea during the night (i.e., land breeze) and during the day cold air blows from the sea towards the land (i.e., sea breeze). These breezes near the sea makes the climate in coastal areas moderate.

Question 25

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

Answer

Hot water bottles are used for fomentation because water does not cool quickly due to it's high specific heat capacity, so a hot water bottle provides more heat energy for fomentation over a longer period. Hence, water is used in hot water bottles for fomentation.

Question 26

What property of water makes it an effective coolant?

Answer

Water is an effective coolant. By allowing water to flow in pipes around the heated parts of a machine, heat energy from such parts is removed. Water in pipes can extract more heat from the surroundings without much rise in it's temperature because of it's high specific heat capacity. This is why radiators in car and generator use water for cooling.

Question 27

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

Answer

(i) Used as a coolant — Radiators in car and generator use water for cooling. Water in car radiator can absorb more heat without much rise in temperature due to it's high specific heat capacity. Hence, in acts as an effective coolant.

(ii) Used as heat reservoir — In cold countries, water is used as heat reservoir for wine and juice bottles to avoid their freezing. The reason is that water due to it's high specific heat capacity can impart a large amount of heat before reaching to it's freezing point. Hence, bottles kept in water remain warm and do not freeze when there is considerable fall in temperature.

Question 28

A liquid X has specific heat capacity higher than the liquid Y. Which liquid is useful as (i) coolant in car radiators and, (ii) heat reservoir to keep juice bottles without freezing?

Answer

(i) Coolant in car radiators — Liquid X is used as a coolant in car radiators because liquid X has specific heat capacity higher than the liquid Y, hence it will absorb more heat energy without much change in temperature.

(ii) Heat reservoir to keep juice bottles without freezing — The liquid needs to give out large amount of heat before reaching freezing temperatures and as liquid X has specific heat capacity higher than the liquid Y, hence liquid X will be used.

Question 29

(a) What is calorimeter?

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

(c) Out of the three metals A, B and C of specific heat 900 J kg-1 °C-1 , 380 J kg-1 °C-1 and 460 J kg-1 °C-1 respectively, which will you prefer for calorimeter? Given reason.

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

Answer

(a) A calorimeter is a cylindrical vessel which is used to measure the amount of heat gained (or lost) by a body when it is mixed with the other body.

(b) It is made up of a thin sheet of copper. The reasons for using copper are:

  1. Copper is a good conductor of heat, so the vessel soon acquires the temperature of it's contents.
  2. Copper has the low specific heat capacity so the heat capacity of calorimeter is low and the amount of heat energy taken by the calorimeter from the contents to acquire it's temperature, is very small.

(c) Heat capacity of the calorimeter should be low. Hence, B should be preferred for making the calorimeter.

(d) The outer and inner surfaces of the vessel should be polished so as to reduce the loss of heat due to radiation in a calorimeter.

Question 30

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

Answer

The base of a cooking pan is made thick and heavy because it's heat capacity becomes large due to which it gets heated slowly and it imparts sufficient heat energy at a slow rate to the food for it's proper cooking and after cooking it keeps the food warm for a long time.

Multiple Choice Type

Question 1

The S.I. unit of heat capacity is:

  1. J kg-1
  2. J K-1
  3. J kg-1 K-1
  4. cal °C-1

Answer

The term heat capacity of a body is the amount of heat energy required to raise it's temperature by 1°C (or 1 K).
The S.I. unit of heat capacity is joule per kelvin (J K-1).

Question 2

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

  1. J kg-1
  2. J K-1
  3. J kg-1 K-1
  4. kcal kg-1 °C-1

Answer

The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of unit mass of that substance through 1°C (or 1 K). i.e.,

Specific heat capacity c=Amt of heat energy suppliedmass×rise in temperature\text{Specific heat capacity c} = \dfrac{\text{Amt of heat energy supplied}}{\text{mass} \times \text{rise in temperature}}

The S.I. unit of specific heat capacity is joule per kilogram per kelvin (J kg-1 K-1).

Question 3

The specific heat capacity of water is:

  1. 4200 J kg-1 K-1
  2. 420 J g-1 K-1
  3. 0.42 J g-1 K-1
  4. 4.2 J kg-1 K-1

Answer

Water has an unusually high specific heat capacity of 4200 J kg-1 K-1

Numericals

Question 1

By imparting heat to a body, it's temperature rises by 15°C. What is the corresponding rise in temperature on the Kelvin scale?

Answer

A degree (or temperature difference) is same on both the celsius and kelvin scales i.e., △t° C = △T K. Therefore, the corresponding rise in temperature on the Kelvin scale will be 15 K.

Question 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.

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

Answer

(a) Given,

mass (m) = 150 g = 0.15 kg

specific heat capacity (c) = 410 J kg-1 K-1

We know that,

Heat capacity (C') = Mass (m) × specific heat capacity (c)

Substituting the values in the formula above we get,

C' = 0.15 x 410 = 61.5 J K-1

Hence, heat capacity of a copper vessel = 61.5 J K-1

(b) Change in temperature = 35°C – 25°C = 10°C = 10 K

Energy required to increase the temperature of vessel Q = ?

From relation,

Q = m x c x △T

Substituting the values in the formula above we get,

Q = 0.15 x 410 x 10 = 615 J

Hence, heat energy required to increase the temperature of the vessel = 615 J

Question 3

A piece of iron of mass 2.0 kg has a heat capacity of 966 J K-1 . Find (i) heat energy needed to warm it by 15° C, and (ii) it's specific heat capacity in S.I unit.

Answer

(i) Given,

mass (m) = 2.0 kg

heat capacity (C') = 966 J K-1

rise in temperature (△t) = 15°C

Heat energy needed (Q) = ?

From relation,

Q = C' × △t

Substituting the values in the formula above we get,

Q = 966 x 15 = 14,490 J

Hence, heat energy needed to warm it by 15°C = 14,490 J

(ii) specific heat capacity (c) = ?

We know that,

c=Cmc = \dfrac{C'}{m} \\[0.5em]

Substituting the values in the formula above we get,

c=9662=483 J kg1K1c = \dfrac{966}{2} = 483 \text{ J kg}^{-1} \text{K}^{-1}

Hence, Specific heat capacity = 483 J kg-1 K-1

Question 4

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

Answer

Given,

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

Change of temperature (△t) = (70 – 20)°C = 50°C

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

Amount of heat required to raise the temperature of 0.1 kg of copper = ?

From relation,

Q = m x c x △t

Substituting the values in the formula above we get,

Q = 0.1 x 390 x 50 = 1950 J

Hence, the amount of heat energy required to raise the temperature of 100 g of copper from 20° C to 70° C = 1950 J

Question 5

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.

Answer

Given,

Heat energy supplied (Q) = 1300 J

Mass of lead (m) = 0.5 kg

Change in temperature (△t) = (40 – 20)°C = 20° C

Specific heat capacity (c) = ?

From relation,

Q=m×c×tQ = m \times c \times △t \\[0.5em]

Substituting the values in the formula above we get,

1300=0.5×c×20c=13000.5×20c=130 J kg1K11300 = 0.5 \times c \times 20 \\[0.5em] \Rightarrow c = \dfrac{1300}{0.5 \times 20} \\[0.5em] \Rightarrow c = 130 \text{ J kg}^{-1} \text{K}^{-1}

Hence, specific heat capacity of lead = 130 J kg-1 K-1

Question 6

Find the time taken by a 500 W heater to raise the temperature of 50 kg of material of specific heat capacity 960 J kg-1 K-1, from 18°C to 38°C. Assume that all the heat energy supplied by heater is given to the material.

Answer

Given,

Power of heater (P) = 500 W

mass of material (m) = 50 kg

Specific heat capacity of material (c) = 960 J kg-1 K-1

Change in temperature △t = (38 – 18)°C = 20° C = 20 K

From relation,

Q=mctQ = m c △t

Substituting the values in the formula above we get,

Q=50×960×20Q=960,000Q = 50 \times 960 \times 20 \\[0.5em] \Rightarrow Q = 960,000

Now,

Q=Power×timeQ = \text {Power} \times \text{time}

Substituting the values in the formula above we get

960,000=500×timetime=960,000500time=1920 s=32 min960,000 = 500 \times \text{time} \\[0.5em] \Rightarrow \text{time} = \dfrac{960,000}{500} \\[0.5em] \Rightarrow \text{time} = 1920 \text{ s}= 32 \text{ min} \\[0.5em]

Hence, time taken = 32 min

Question 7

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) the heat capacity of 4.0 kg of liquid, and (ii) the specific heat capacity of liquid.

Answer

(i) Power of heater (P) = 600 W

Mass of liquid (m) = 4.0 kg

Change in temperature of liquid = (15 – 10)°C = 5° C (or 5 K)

Time taken to raise it's temperature (t) = 100 s

heat capacity = ?

From relation,

Q=Power×timeQ = \text {Power} \times \text{time}

Substituting the values in the formula above we get,

Q=600×100Q=60000JQ = 600 \times 100 \\[0.5em] \Rightarrow Q = 60000 J \\[0.5em]

Now,

C=QTC^{'} = \dfrac{Q}{△T} \\[0.5em]

Substituting the values in the formula above we get,

C=60,0005C=1.2×104 JK1C^{'} = \dfrac{60,000}{5} \\[0.5em] C^{'} = 1.2 \times 10^{4} \text{ J} \text{K}^{-1} \\[0.5em]

Hence, heat capacity = 1.2 x 10 J K-1

(ii) specific heat capacity {c} = ?

c=Qm×Tc = \dfrac{Q}{m \times △T} \\[0.5em]

Substituting the values in the formula above we get,

c=600004×5c=600004×5c=3×103 J kg1K1c = \dfrac{60000}{4 \times 5} \\[0.5em] c = \dfrac{60000}{4 \times 5} \\[0.5em] c = 3 \times 10^{3} \text{ J kg}^{-1} \text{K}^{-1}

Hence, specific heat capacity = 3 x 103 J Kg-1 K-1

Question 8

0.5 kg of lemon squash at 30°C 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 5°C ? Specific heat capacity of squash = 4200 J kg-1 K-1.

Answer

Given,

mass (m) = 0.5 kg

Change in temperature = (30 – 5)°C = 25° C = 25 K

Specific heat capacity of squash (c) = 4200 J kg-1 K-1

From relation,

Q=m×c×TQ = m \times c \times △T \\[0.5em]

Substituting the values in the formula above we get,

Q=0.5×4200×25Q=52,500JQ = 0.5 \times 4200 \times 25 \\[0.5em] \Rightarrow Q = 52,500 J

Let time taken to remove 52500 J of heat be t.

Now it is given that,

30 J of heat is removed in 1 sec

So, 52500 J of heat ⇒ t = ?

t=130×52,500t=1750 st=29 min 10 sect = \dfrac{1}{30} \times 52,500 \\[0.5em] \Rightarrow t = 1750 \text{ s} \\[0.5em] \Rightarrow t = 29 \text{ min } 10 \text{ sec} \\[0.5em]

Hence, time taken = 29 min 10 sec

Question 9

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

Answer

Given,

Mass of metal (m) = 50 g

Fall in temperature of metal = (150 – 20) = 130°C

Rise in temperature of water = (20 - 11) = 9°C

Heat energy given by metal = mc△t
= 50 x c x 130
= 6500 x c     [Equation 1]

Heat energy taken by water = 100 × 4.2 × 9
= 3780     [Equation 2]

Assuming that there is no loss of heat energy,

Heat energy given by metal = Heat energy taken by water.

Equating equations 1 & 2, we get,

6500×c=3780c=37806500c=0.582 J g1K16500 \times c = 3780 \\[0.5em] c = \dfrac{3780}{6500} \\[0.5em] c = 0.582 \text{ J g}^{-1} \text{K}^{-1} \\[0.5em]

Hence, specific heat capacity of the metal = 0.582 J g-1 K-1

Question 10

45 g of water at 50°C in a beaker is cooled when 50 g of copper at 18°C 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 0.39 J g-1 K-1 and that of water is 4.2 J g-1 K-1. State the assumptions used.

Answer

Given,

Mass of water = 45 g

Let the final constant temperature reached be t°C

Fall in temperature of water = (50 – t)°C

Mass of copper = 50 g

Rise in temperature of copper = (t - 18)°C

The specific heat capacity of the copper cc = 0.39 J g-1 K-1

The specific heat capacity of water cw = 4.2 J g-1 K-1

Heat energy given by water = mc△t
= 45 x 4.2 x (50 – t)     [Equation 1]

Heat energy taken by copper = 50 x 0.39 x (t - 18)     [Equation 2]

Assuming that there is no loss of heat energy

Heat energy given by water = Heat energy taken by copper

Equating equations 1 & 2, we get,

45×4.2×(50t)=50×0.39×(t18)189×(50t)=19.5×(t18)(189×50)(189×t)=(19.5×t)(19.5×18)(9450)(189×t)]=(19.5×t)(351)9450+351=(19.5t)+(189t)9801=(208.5t)t=9801208.5t=47.0072°C47°C45 \times 4.2 \times (50 - t) = 50 \times 0.39 \times (t - 18) \\[0.5em] \Rightarrow 189 \times (50 - t) = 19.5 \times (t - 18) \\[0.5em] \Rightarrow (189 \times 50) - (189 \times t) = (19.5 \times t) - (19.5 \times 18) \\[0.5em] \Rightarrow (9450) - (189 \times t)] = (19.5 \times t) - (351) \\[0.5em] \Rightarrow 9450 + 351 = (19.5 t) + (189 t) \\[0.5em] \Rightarrow 9801 = (208.5t) \\[0.5em] \Rightarrow t = \dfrac{9801}{208.5} \\[0.5em] \Rightarrow t = 47.0072°C \approx 47°C

Hence, final temperature = 47°C

Question 11

200 g of hot water at 80°C is added to 300 g of cold water at 10°C. 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.

Answer

Mass of hot water = 200 g

Temperature of hot water = 80° C

Mass of cold water = 300 g

Temperature of cold water = 10° C

Let final temperature be t

Fall in temperature of hot water = (80 – t)°C

Rise in temperature of cold water = (t - 10)°C

The specific heat capacity of water cw = 4200 J kg-1 K-1 = 4.2 J g-1 K-1

Heat energy given by hot water = mc△t
= 200 x 4.2 x (80 – t)     [Equation 1]

Heat energy taken by cold water = 300 x 4.2 x (t - 10)     [Equation 2]

Assuming that there is no loss of heat energy,

Heat energy given by hot water = Heat energy taken by cold water

Equating equations 1 & 2, we get,

200×4.2×(80t)=300×4.2×(t10)2×(80t)=3×(t10)1602t=3t30160+30=3t+2t190=5tt=1905t=38°C200 \times 4.2 \times (80 - t) = 300 \times 4.2 \times (t - 10) \\[0.5em] 2 \times (80 - t) = 3\times (t - 10) \\[0.5em] 160 - 2t = 3t - 30 \\[0.5em] 160 + 30 = 3t + 2t \\[0.5em] \\[0.5em] 190 = 5t \\[0.5em] \Rightarrow t = \dfrac{190}{5} \\[0.5em] \Rightarrow t = 38°C

Hence, final temperature = 38°C

Question 12

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

Answer

Given,

Mass of cold water = 600 g

Mass of hot water = 300 g

Temperature of hot water = 50° C

Let initial temperature of cold water be ti

Let the final temperature be t

Gain in temperature of cold water = (t - ti) = 15° C

Loss of heat from hot water = (50 - t)

Heat energy given by hot water = mc△t
= 300 x 4.2 x (50 – t)     [Equation 1]

Heat energy taken by cold water = 600 x 4.2 x (t - ti)     [Equation 2]

Assuming that there is no loss of heat energy

Heat energy given by hot water = heat energy taken by cold water

Equating equations 1 & 2, we get,

300×4.2×(50t)=600×4.2×153×(50t)=6×1550t=2×1550t=30t=5030=20°C300 \times 4.2 \times (50 - t) = 600 \times 4.2 \times 15 \\[0.5em] \Rightarrow 3 \times (50 - t) = 6 \times 15 \\[0.5em] \Rightarrow 50 - t = 2 \times 15 \\[0.5em] \Rightarrow 50 - t = 30 \\[0.5em] \Rightarrow t = 50 - 30 = 20° C \\[0.5em]

final temperature (t) = 20°C

initial temperature = ?

We know,

t - ti = 15

20 - ti = 15

Therefore, ti = 20 - 15 = 5° C

Hence, initial temperature = 5° C

Question 13

1.0 kg of water is contained in a 1.25 kW kettle. Calculate the time taken for the temperature of water to rise from 25° C to it's boiling point 100° C. Specific heat capacity of water = 4.2 J g-1 K-1.

Answer

Given,

mass = 1 kg = 1000 g

Heat energy taken by water (Q) = mc△t
= 1000 x 4.2 x (100 – 25)
= 1000 x 4.2 x 75
= 315000 J

Now,

Q=Power×timeQ = \text {Power} \times \text{time}

Substituting the values in the formula above we get,

315000=1250×tt=3150001250t=252 sec=4 min 12 sec315000 = 1250 \times t \\[0.5em] \Rightarrow t = \dfrac{315000}{1250} \\[0.5em] \Rightarrow t = 252 \text{ sec} = 4 \text{ min } 12 \text{ sec} \\[0.5em]

Hence, time = 4 min 12 sec

Exercise 11(B)

Question 1

(a) What do you understand by the change of phase of a 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

(a) The process of change from one state to another at a constant temperature is called the change of phase.

(b) No, there is no change in temperature during the change of phase.

(c) Yes, the substance absorb or liberates heat during the change of phase. The heat is absorbed during melting and it is liberated during freezing at a constant temperature.

(d) The name given to the energy absorbed during a phase change is latent heat.

Question 2

A substance changes from it's solid state to the liquid state when heat is supplied to it

(a) Name the process.

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

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

Answer

(a) The process is known as melting.

(b) The heat absorbed by the substance is called the latent heat of melting.

(c) The average kinetic energy of the molecules does not change as there is no change in temperature.

Question 3

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

Answer

(i) When a substance on heating undergoes a rise in it's temperature then average kinetic energy of molecules increases.

(ii) When a substance on heating undergoes a change in it's phase without change in it's temperature then average potential energy of molecules increases.

Question 4

How does the (a) average kinetic energy (b) average potential energy of molecules of a substance change during it's change in phase at a constant temperature, on heating?

Answer

(a) The average kinetic energy of molecules does not change during it's change in phase at a constant temperature, on heating.

(b) Average potential energy of molecules increases during it's change in phase at a constant temperature, on heating.

Explanation: During the change in phase of the substance at a constant temperature on heating, the heat supplied is utilized in increasing the separation against the attractive forces between the molecules. This increases the potential energy of the molecules.
As the temperature of the substance remains constant, the average kinetic energy of the molecules does not change. The heat energy supplied during melting is utilised only in increasing the potential energy of the molecules and is called the latent heat of melting.

Question 5

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

Answer

The melting point of a substance decreases by the presence of impurities in it. The melting point of ice decreases from 0 ° C to -22 ° C on mixing salt to it in proper proportion. This fact is utilized in making the freezing mixture by adding salt to ice. The freezing mixture is used in preparing 'kulfis'.

Question 6

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

Answer

The melting point of the substances which contract on melting (like ice) decreases by the increase in pressure. For example, the melting point of ice decreases by 0.0072°C for every one atmosphere rise in pressure.

Question 7

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

The diagram below shows the change of phases of a substance on a temperature-time graph on heating the substances at a constant rate. What do parts AB, BC, CD and DE represent? What is the melting point of the substance? What is the boiling point of the substance? Calorimetry, Concise Physics Solutions ICSE Class 10.

(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

(a) The different parts represent the following:

  • Part AB — It shows the rise in temperature of solid from 0°C to t1°C.
  • Part BC — It shows melting at temperature t1°C.
  • Part CD — It shows the rise in temperature of liquid from t1°C to t3°C.
  • Part DE — It shows boiling at temperature t3°C

(b) The melting point of the substance is t1° C.

(c) The boiling point of the substance is t3° C.

Question 8

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

Temperature-time graph representing the cooling of naphthalene from 90°C to room temperature of 25°C is shown below:

The melting point of naphthalene is 80° C and the room temperature is 25°. A sample of liquid naphthalene at 90° 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. Calorimetry, Concise Physics Solutions ICSE Class 10.

Question 9

1 kg of ice at 0°C is heated at a constant rate and it's temperature is recorded every 30 s till steam is formed at 100°C. Draw a temperature-time graph to represent the change of phases.

Answer

Temperature-time graph representing the change of phases when ice is heated till steam is formed at 100°C is shown below:

1 kg of ice at 0°C is heated at a constant rate and it's temperature is recorded every 30 s till steam is formed at 100°C. Draw a temperature-time graph to represent the change of phases. Calorimetry, Concise Physics Solutions ICSE Class 10.

Question 10

Explain the terms boiling and boiling point. How is the volume of water affected when it boils at 100°C.

Answer

The change from liquid to gas (or vapour) phase on absorption of heat at a constant temperature is called boiling or vaporisation.

The particular temperature at which vaporisation occurs is known as the boiling point of liquid.

Volume of water increases when it boils at 100° C. 1 cm3 of water at 100 ° C becomes 1760 cm3 of steam at 100 ° C.

Question 11

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

Answer

The boiling point of water increases by addition of salt to it. If common salt is added to water, it boils at a temperature higher than 100° C.

Question 12

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

Answer

The boiling point of a liquid increases with the increase in pressure.

Question 13

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

Answer

The boiling point of liquid increases with the increase in pressure and decreases with the decrease in pressure. The boiling point of pure water at one atmospheric pressure is 100° C.

In a pressure cooker, steam is not allowed to escape out. The vapour pressure on water inside the cooker becomes nearly 1.75 times the atmospheric pressure, so water boils in it at about 120° C to 125° C due to increased pressure.

Question 14

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

Answer

The water boils at about 120° C to 125° C in a pressure cooker.

Question 15

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

Answer

At high altitudes, such as hills and mountains, the atmospheric pressure is low (less than one atmospheric pressure), therefore at these places, water boils at temperature lower than 100° C and so it does not provide the required heat energy to it's contents for cooking. Thus, cooking there becomes very difficult and it takes a much longer time.

Question 16

Complete the following sentences:

  1. When ice melts, it's volume ..........
  2. Decrease in pressure over ice .......... it's melting point.
  3. Increase in pressure .......... the boiling point of water.
  4. A pressure cooker is based on the principle that boiling point of water increases with the ..........
  5. The boiling point of water is defined as ..........
  6. Water can be made to boil at 115°C by .......... pressure over it's surface.

Answer

  1. When ice melts, it's volume decreases.
  2. Decrease in pressure over ice increases its melting point.
  3. Increase in pressure increases the boiling point of water.
  4. A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure
  5. The boiling point of water is defined as the constant temperature at which water changes to steam
  6. Water can be made to boil at 115°C by increasing pressure over it's surface.

Question 17

What do you understand by the term latent heat?

Answer

Heat energy absorbed or liberated in change of phase that is not externally manifested by any rise or fall in temperature is called the latent heat.

Question 18

Define the term specific latent heat of fusion of ice. State it's S.I. unit.

Answer

The specific latent heat of fusion of ice is the heat energy required to melt unit mass of ice at 0° C to water at 0° C without any change in temperature.

S.I. unit of latent heat of fusion is J kg-1

Question 19

Write the approximate value of specific latent heat of ice.

Answer

The approximate value of specific latent heat of ice is 336 x 103 J kg-1

Question 20

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

Answer

'The specific latent heat of fusion of ice is 336 J g-1' means 1 g of ice at 0° C absorbs 336 J of heat energy to convert into water at 0° C.

Question 21

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

Answer

Latent heat is absorbed by ice when 1 g ice at 0° C melts to form 1 g water at 0° C.

Question 22

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

Answer

1 g of water at 0° C has more heat because 1 g of water at 0° C liberates 80 cal heat to form 1 g of ice at 0° C.

Question 23

(a) Which requires more heat: 1 g ice at 0° C or 1 g water at 0° C to raise it's temperature to 10° C ?

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

Answer

(a) 1 g ice at 0° C requires more heat to raise it's temperature to 10° C.

(b) 1 g ice at 0° C requires more heat to raise it's temperature to 10° C because 1 g ice at 0° C first absorbs 336 J heat to convert into 1 g water at 0° C and then the water absorbs heat to raise it's temperature from 0° to 10° C.

Question 24

Ice cream appears colder to the mouth than water at 0° C. Give reasons.

Answer

Ice cream absorbs heat energy as well as the latent heat while water absorbs only heat energy. Therefore, ice cream absorbs more amount of energy from the mouth as compared to water. Hence, ice cream appears colder to the mouth than water at 0° C.

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

Ice cubes at 0°C will cool the soft drink bottles more quickly than iced-water at 0°C.

1 g of ice at 0° C takes 336 J of heat energy from the drink to melt into water at 0° C. Thus, the drink liberates an additional 336 J of heat energy to 1 g ice at 0° C than to 1 g ice-cold water at 0° C. Therefore, cooling produced by 1 g ice at 0° C is much more than that by 1 g water at 0° C.

Question 26

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

Answer

It is generally cold after a hail storm than during and before it because after the hail storm, ice absorbs the heat energy required for it's melting from the surroundings, so the temperature of the surrounding falls further down and we feel more cold.

Question 27

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

Answer

The temperature of surroundings starts falling when ice in a frozen lake starts melting because quite a large amount of heat energy is required for melting the frozen lake which is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it becomes very cold.

Question 28

Water in lakes and ponds do not freeze at once in cold countries. Give reason.

Answer

Water in lakes and ponds do not freeze at once in cold countries because the specific latent heat of fusion of ice is sufficiently high (= 336 J g-1). The water in lakes and ponds will have to liberate a large quantity of heat to the surrounding before freezing. The layer of ice formed over the water surface, being a poor conductor of heat, will also prevent the loss of heat from the water of lake, hence the water does not freeze all at once.

Question 29

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 it's change of state, does not cause any rise in it's temperature.

Answer

(a) The surroundings become pleasantly warm when water in a lake starts freezing in cold countries because the specific latent heat of fusion of ice is very high, hence large quantity of heat is released when the water in the lake freezes. Therefore, the temperature of the surrounding becomes pleasantly warm.

(b) During the change in phase of the substance at a constant temperature on heating, the heat supplied is utilized in increasing the separation against the attractive forces between the molecules. This increases the potential energy of the molecules.
The average kinetic energy of the molecules does not change. Hence, the temperature of the substance remains constant. The heat energy supplied during melting is utilised only in increasing the potential energy of the molecules and is called the latent heat of melting.

Multiple Choice Type

Question 1

The S.I. unit of specific latent heat is:

  1. cal g-1
  2. cal g-1 K-1
  3. J kg-1
  4. J kg-1 K-1

Answer

The S.I. unit of specific latent heat is J kg-1

Question 2

The specific latent heat of fusion of water is:

  1. 80 cal g-1
  2. 2260 J g-1
  3. 80 J g-1
  4. 336 J kg-1

Answer

The specific latent heat of fusion of water is 80 cal g-1

Numericals

Question 1

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.

Answer

Given,

Mass (m) = 10 g

Heat energy absorbed (Q) = 5460 J

Specific heat capacity of water = 4200 J kg-1 K-1 = 4.2 J g-1 K-1

Specific latent heat of fusion of ice (L) = ?

(i) Heat energy required to melt the ice at 0° C to water at 0° C (Q1) = m x L

Substituting the values in the formula we get,

Q1=10×LQ_1 = 10 \times L \\[0.5em]

(ii) Heat energy required to raise temperature from 0° C to 50° C = m x c x rise in temperature

Substituting the values in the formula we get,

Q2=10×4.2×50=2100JQ_2 = 10 \times 4.2 \times 50 \\[0.5em] = 2100 J \\[0.5em]

From relation,

Q=Q1+Q25460=(10×L)+210054602100=10×L3360=10×LL=336010L=336 J g1Q = Q_1 + Q_2 \\[0.5em] 5460 = (10 × L) + 2100 \\[0.5em] 5460 - 2100 = 10 × L\\[0.5em] 3360 = 10 × L\\[0.5em] \Rightarrow L = \dfrac{3360}{10} \\[0.5em] \Rightarrow L = 336 \text{ J g}^{-1} \\[0.5em]

Hence, Specific latent heat of fusion of ice = 336 J g-1

Question 2

How much heat energy is released when 5.0 g of water at 20° C changes into ice at 0° C? 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

Given,

Mass (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

(i) Heat energy released when water lowers it's temperature from 20° C to 0° C

= m x c x change in temperature

Substituting the values in the formula we get,

Q1=5×4.2×(200)Q1=5×4.2×20Q1=420JQ_1 = 5 \times 4.2 \times (20 - 0) \\[0.5em] Q_1 = 5 \times 4.2 \times 20 \\[0.5em] \Rightarrow Q_1 = 420 J \\[0.5em]

(ii) Heat energy released when water at 0° C changes into ice at 0° C = m x L

Substituting the values in the formula we get,

Q2=5×336JQ2=1680JQ_2 = 5 × 336 J \\[0.5em] \Rightarrow Q_2 = 1680 J \\[0.5em]

From relation,

Q=Q1+Q2Q=420+1680Q=2100Q = Q_1 + Q_2 \\[0.5em] Q = 420 + 1680 \\[0.5em] \Rightarrow Q = 2100 \\[0.5em]

Total heat released = 2100 J

Question 3

A molten metal of mass 150 g is kept at it's melting point 800° C. When it is allowed to freeze at the same temperature, it gives out 75,000 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 - 50° C?

Answer

(a) Given,

Mass (m) = 150 g

Heat energy given out (Q) = 75,000 J

Specific latent heat of the metal = ?

From relation Q = m x L

Substituting the values in the formula we get,

75000=150×LL=75000150L=500 J g175000 = 150 \times L \\[0.5em] \Rightarrow L = \dfrac{75000}{150} \\[0.5em] \Rightarrow L = 500 \text{ J g}^{-1} \\[0.5em]

Hence, the specific latent heat of the metal = 500 J g-1

(b) Specific heat capacity of metal is 200 J kg-1 K-1

Change in temperature
= 800 – (-50) = 800 + 50 = 850° C = 850 K

From relation,

Q = m x c x change in temperature

Substituting the values we get,

Q=0.15×200×850Q=25,500JQ = 0.15 \times 200 \times 850 \\[0.5em] Q = 25,500 J \\[0.5em]

Hence, 25,500 J of heat energy will be given out

Question 4

A solid metal of mass 150 g melts at it's melting point of 800° C by providing heat at the rate of 100 W. 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

Given,

m = 150 g = 0.15 kg

P = 100 W

t = 4 min = 240 s

specific latent heat of fusion of the metal = ?

Heat supplied = P x t

Substituting the values we get,

Heat supplied=100×240Heat supplied=24000J\text{Heat supplied} = 100 \times 240 \\[0.5em] \text{Heat supplied} = 24000 J

We know,

Q = m x L

Substituting the values we get,

24000=0.15×LL=240000.15L=160,000L=1.6×105 J kg124000 = 0.15 \times \text{L} \\[0.5em] \Rightarrow \text{L} = \dfrac{24000}{0.15} \\[0.5em] \Rightarrow \text{L} = 160,000 \\[0.5em] \Rightarrow \text{L} = 1.6 \times 10^{5} \text{ J kg}^{-1}

Hence, specific latent heat of fusion of the metal = 1.6 x 105 J kg-1

Question 5

A refrigerator converts 100 g of water at 20° C to ice at -10° C in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J g-1 K-1 , specific latent heat of ice is 336 J g-1 and the specific heat capacity of ice is 2.1 J g-1 K-1.

Answer

Given,

mass (m) = 100 g

time (t) = 73.5 min

specific heat capacity of water = 4.2 J g-1 K-1

specific latent heat of ice = 336 J g-1

specific heat capacity of ice = 2.1 J g-1 K-1

Heat energy released by water in fall of it's temperature from 20° to 0° C (Q1)
= mass x specific heat capacity x fall in temperature
= 100 × 4.2 x (20 - 0)
= 100 × 4.2 x 20
= 8400 J

Hence, Q1 = 8400 J

Heat energy released by water when it converts into ice at 0° C (Q2) = m x Lice
= 100 × 336
= 33600 J

Heat energy released when ice cools from 0° C to -10° C (Q3) = m x c x change in temperature
= 100 × 2.1 x [0 - (-10)]
= 100 x 2.1 x 10
= 2100 J

Hence,

Total heat energy = Q1 + Q2 + Q3
= 8400 + 33600 + 2100
= 44100 J

Time taken = 73.5 min = 4410 s

Average rate of heat extraction (P)

P=EtP = \dfrac{E}{t}

Substituting the values in the formula we get,

P=441004410P=10WP = \dfrac{44100}{4410} \\[0.5em] P = 10 W

Hence, average rate of heat extraction = 10 W

Question 6

In an experiment, 17 g of ice is used to bring down the temperature of 40 g of water at 34° C to it's 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

Given,

Mass of ice (m1) = 17 g

Mass of water (m2) = 40 g

Change in temperature = 34 – 0 = 34° C = 34 K

Specific heat capacity of water (c) = 4.2 J g-1 K-1

specific latent heat of ice = ?

Assuming that no heat energy is lost,

heat energy required by ice to melt = heat energy given by water

So,

m1 x L = m2 x c x change in temperature

Substituting the values in the relation above we get,

17×L=40×4.2×3417×L=5712L=571217L=336 J g117 \times L = 40 \times 4.2 \times 34 \\[0.5em] 17 \times L = 5712 \\[0.5em] \Rightarrow L = \dfrac{5712}{17} \\[0.5em] \Rightarrow L = 336 \text{ J g}^{-1} \\[0.5em]

Hence, specific latent heat of ice = 336 J g-1

Assumption — There is no loss of energy.

Question 7

The temperature of 170 g of water at 50° C is lowered to 5° C by adding certain amount of ice to it. Find the mass of ice added. Given : specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of ice = 336000 J kg-1.

Answer

Given,

mw = 170 g = 0.17 kg

specific heat capacity of water = 4200 J kg-1 K-1

specific latent heat of ice = 336000 J kg-1

mi = ?

Heat energy given out by water in lowering it's temperature from 50° C to 5° C
= m x c x change in temperature
= 0.17 x 4200 x (50 - 5)
= 0.17 x 4200 x 45
= 32,130

Heat energy taken by m kg ice to melt into water at 0° C
= mi x L
= mi x 336000

Heat energy taken by water at 0° C to raise it's temperature to 5° C
= mi x c x change in temperature
= mi x 4200 x (5 - 0)
= mi x 4200 x 5
= mi x 21000

heat energy released = heat energy taken

Substituting the values we get,

32130=(mi×336000)+(mi×21000)32130=mi×357000mi=32130357000mi=0.09 Kg=90 g32130 = (m_i \times 336000 ) + (m_i \times 21000 ) \\[0.5em] 32130 = m_i \times 357000 \\[0.5em] \Rightarrow m_i = \dfrac{32130}{357000} \\[0.5em] \Rightarrow m_i = 0.09 \text{ Kg} = 90 \text{ g}

Hence, the mass of ice added = 90 g

Question 8

Find the result of mixing 10 g of ice at - 10° C with 10 g of water at 10° C. Specific heat capacity of ice = 2.1 J g-1 K-1, specific latent heat of ice = 336 J g-1, and specific heat capacity of water = 4.2 J g-1 K-1.

Answer

Given,

mass of ice = 10 g

mass of water = 10 g

Specific heat capacity of ice = 2.1 J g-1 K-1

specific latent heat of ice = 336 J g-1,

specific heat capacity of water = 4.2 J g-1 K-1

Let the final temperature be t° C.

Heat energy taken by ice at – 10° C to raise it's temperature to 0° C (Q1)
= m x c x change in temperature
= 10 x 2.1 x [0 - (-10)]
= 10 x 2.1 x 10
= 210 J

Hence, Q1 = 210 J

Heat energy taken by ice at 0° C to convert into water at 0° C (Q2)
= m x Lice
= 10 × 336
= 3360 J

Hence, Q2 = 3360 J

Heat energy taken by water at 0° C to raise it's temperature to t° C (Q3) = m x c x change in temperature
= 10 × 4.2 × (t – 0)
= 10 × 4.2 × t
= 42 t

Hence, Q3 = 42 t

Heat energy released by water at 10° C to lower it's temperature to t° C (Q4) = m x cwater x change in temperature
= 10 × 4.2 × (10 – t)
= 42 × (10 – t) = 420 – 42t

Hence, Q4 = 420 – 42t

If there is no loss of heat,

Heat energy gained = Heat energy lost

210+3360+42t=42042t3570420=42t42t3150=84tt=315084t=37.5°C210 + 3360 + 42 t = 420 - 42t \\[0.5em] 3570 - 420 = - 42 t - 42 t \\[0.5em] 3150 = - 84 t \\[0.5em] t = - \dfrac{3150}{84} \\[0.5em] t = - 37.5° C

This cannot be true because water cannot exist at -37.5° C.

Hence, we can say that the whole of the ice did not melt.

Let amount of ice which melts = m g

Final temperature of the mixture = 0° C

Heat energy gained by ice at -10° C to raise it's temperature to 0° C

= m x cice x change in temperature
= 10 × 2.1 x [0 - (-10)]
= 10 × 2.1 x 10
= 210 J

Heat energy gained by m gm of ice at 0° C to change into water at 0° C
= m × Lice
= m x 336
= 336m J

Heat energy released by 10 g of water at 10° C to lower it's temperature to 0° C
= m x cwater x change in temperature
= 10 × 4.2 × (10 – 0)
= 10 × 4.2 × 10 = 420 J

If there is no loss of heat,

Heat energy gained = Heat energy lost

210+336m=420336m=420210336m=210m=210336m=0.625g210 + 336 m = 420 \\[0.5em] \Rightarrow 336 m = 420 - 210 \\[0.5em] \Rightarrow 336 m = 210 \\[0.5em] \Rightarrow m = \dfrac{210}{336} \\[0.5em] \Rightarrow m = 0.625 g

Hence, 0.625 g of ice will melt and temperature will remain at 0° C

Question 9

A piece of ice of mass 40 g is added to 200 g of water at 50° C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J kg-1 K-1 and specific latent heat of fusion of ice = 336 x 103 J kg-1.

Answer

Given,

mi = 40 g

mw = 200 g

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

specific latent heat of fusion of ice = 336 x 103 J kg-1

final temperature of water = ?

Let final temperature = t

Heat energy given out by water when it cools from 50° C to t° C
= m x c x change in temperature
= 200 × 4.2 × (50 – t)
= 42000 – 840t

Heat energy taken by ice when it converts from ice into water at 0° C
= m x L
= 40 × 336 J
= 13440 J

Heat energy taken by water when it raises it's temperature from 0° to t° C
= m x c x change in temperature
= 40 × 4.2 × (t – 0)
= 40 × 4.2 × t
= 168t

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

13440+168t=42000840t168t+840t=42000134401008t=28560t=285601008t=28.330°C13440 + 168t = 42000 – 840t \\[0.5em] 168t + 840t = 42000 – 13440 \\[0.5em] 1008t = 28560 \\[0.5em] t = \dfrac{28560}{1008} \\[0.5em] t = 28.330° C

Hence, final temperature = 28.330° C

Question 10

Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 32°C such that the final temperature is 5°C. Specific heat capacity of calorimeter = 0.4 J g-1 °C-1, Specific heat capacity of water = 4.2 J g-1 °C-1, Latent heat capacity of ice = 330 J g-1.

Answer

Given,

mass of water mw = 150 g

mass of calorimeter mc = 50 g

Specific heat capacity of calorimeter = 0.4 J g-1 °C-1,

Specific heat capacity of water = 4.2 J-1 °C-1,

Latent heat capacity of ice = 330 J g-1

mass of ice mi = ?

Heat energy imparted by calorimeter and water contained in it in cooling from 32° C to 5° C is used in melting ice and then raising the temperature of melted ice from 0°C to 5°C.

Heat energy imparted by water (Q1)
= m x c x change in temperature
= 150 x 4.2 x (32 - 5)
= 150 x 4.2 x 27
= 17,010 J

Heat energy imparted by calorimeter (Q2)
= m x c x change in temperature
= 50 x 0.4 x (32 - 5)
= 50 x 0.4 x 27
= 540 J

Heat energy taken by ice to melt (Q3)
= mi x L
= mi x 330

Heat energy gained by water from melted ice to reach from 0°C to 5°C (Q4)
= mi x c x change in temperature
= mi x 4.2 x (5 - 0)
= mi x 4.2 x 5
= mi x 21

From the principle of calorimetry, if the system is fully insulated then,

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

Q1 + Q2 = Q3 + Q4

Substituting the values we get,

17010+540=(mi×330)+(mi×21)17550=mi×351mi=17550351mi=50g17010 + 540 = (m_i \times 330) + (m_i \times 21) \\[0.5em] 17550 = m_i \times 351 \\[0.5em] \Rightarrow m_i = \dfrac{17550 }{351} \\[0.5em] \Rightarrow m_i = 50 g

Hence, mass of ice = 50 g

Question 11

250 g of water at 30° C is contained in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and it's contents to 5° C. Given 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

Given,

mcopper = 50 g

mwater = 250 g

Final temperature = 5° C.

Let mass of ice required be mi.

Heat energy gained by ice at 0° C to convert into water at 0° C
= mi × L
= mi x 336 J

Heat energy gained by (m) g of water at 0° C to rise it's temperature to 5° C
= m x c x change in temperature
= mi x c x (5 - 0) = mi × 4.2 × 5
= 21 x mi

Heat energy lost by water at 30° C in cooling to 5° C
= m x c x change in temperature
= 250 × 4.2 × (30 - 5)
= 250 × 4.2 × 25
= 26250 J

Heat energy lost by vessel at 30° C to cool down to 5° C =
= m x c x change in temperature
= 50 × 0.4 × (30 - 5)
= 50 × 0.4 × 25
= 500 J

If there is no loss of heat,

Heat energy gained = Heat energy lost

Substituting the values in the relation above we get,

(336×mi)+(21×mi)=26250+500357×mi=26750mi=26750357mi=74.93g(336 \times m_i ) + (21 \times m_i) = 26250 + 500 \\[0.5em] 357 \times m_i = 26750 \\[0.5em] \Rightarrow m_i = \dfrac{26750}{357} \\[0.5em] \Rightarrow m_i = 74.93 g

Hence, required mass of ice = 74.93 g

Question 12

2 kg of ice melts when water at 100° C is poured in a hole drilled in a block of ice. What mass of water was used? Given: Specific heat capacity of water = 4200 J kg-1 K-1, specific latent heat of ice = 336 × 103 J Kg-1.

Answer

Given,

mi = 2 kg

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

Specific latent heat of ice = 336 × 103 J Kg-1.

mwater = ?

Since the whole block does not melt and only 2 kg of it melts, so final temperature would be 0° C.

Heat energy taken by ice at 0° C to convert into water at 0° C
= m x L
= 2 × 336000
= 672000 J

Initial temperature of water = 100° C

Final temperature of water = 0° C

Heat energy lost by (mw) kg at 100° C to reach temperature 0° C
= m x c x change in temperature
= m × 4200 × (100 - 0)
= m × 4200 × 100
= 420000 m J

If there is no loss of energy,

heat energy gained = heat energy lost

Substituting the values in the relation above we get,

672000=m×420000m=672000420000m=1.6kg672000 = m × 420000 \\[0.5em] m = \dfrac{672000}{420000} \\[0.5em] m = 1.6 kg \\[0.5em]

Hence, mass of water = 1.6 kg

Question 13

Calculate the total amount of heat energy required to convert 100 g of ice at -10° C completely into water at 100° C. Specific heat capacity of ice = 2.1 J g-1 K-1, specific heat capacity of water = 4.2 J g-1 K-1, specific latent heat of ice = 336 J g-1.

Answer

Given,

m = 100 g

Specific heat capacity of ice = 2.1 J g-1 K-1

specific heat capacity of water = 4.2 J g-1 K-1

specific latent heat of ice = 336 J g-1

total amount of heat energy required = ?

Heat energy taken by ice to raise it's temperature from at –10° C to 0° C
= m x c x change in temperature
= 100 × 2.1 × [0 - (-10)]
= 100 x 2.1 x 10
= 2100 J

Heat energy taken by ice at 0° C to convert into water at 0° C
= m x L
= 100 × 336
= 33600 J

Heat energy taken by water to raise the temperature from 0° C to 100° C
= m x c x change in temperature
= 100 × 4.2 × (100 - 0)
= 100 × 4.2 × 100
= 42000 J

Total heat energy gained is
= 2100 + 33600 + 42000
= 77700 J
= 7.77 × 104 J

Hence, total amount of heat energy required = 7.77 × 104 J

Question 14

The amount of heat energy required to convert 1 kg of ice at -10° C completely to water at 100° C is 7,77,000 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

Given,

m = 1 kg

heat energy required = 7,77,000 J

Specific heat capacity of ice = 2100 J kg-1 K-1

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

specific latent heat of ice (L) = ?

Heat energy taken by ice to raise temperature from – 10° C to to 0° C
= m x c x change in temperature
= 1 × 2100 × [0 - (-10)]
= 21000 J

Heat energy gained by ice at 0° C to convert into water at 0° C
= m x L
= 1 x L
= L

Heat energy taken by water to raise it's temperature from 0° C to 100° C
= m x c x change in temperature
= 1 x 4200 x (100 - 0)
= 1 × 4200 × 100
= 4,20,000 J

Total heat energy gained
= 21,000 + L + 4,20,000 = 4,41,000 + L

As,

4,41,000 + L = 7,77,000
L = 7,77,000 – 4,41,000
L = 3,36,000 J kg-1

Hence, specific latent heat of ice = 3,36,000 J kg-1

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

Given,

mi = 200 g

mw = 200 g

Time for ice to melt (t1) = 1 min = 60 s

Change in temperature of water (Δt) = 20° C

specific latent heat of ice = 336 J g-1

Rate of heat exchange is constant.

Therefore, power required for converting ice to water = power required to increase the temperature of water.

Pi=PwEit1=Ewt2mi×Lt1=mw×cw×Δtt2200×33660=200×4.2×20t2t2=5040336t2=15sP_i = P_w \\[0.5em] \dfrac{E_i}{t_1} = \dfrac{E_w}{t_2} \\[0.5em] \dfrac{ m_i \times L}{t_1} = \dfrac{ m_w \times c_w \times Δt }{t_2} \\[0.5em] \dfrac{ 200 \times 336}{60} = \dfrac{ 200 \times 4.2 \times 20 }{t_2} \\[0.5em] \Rightarrow t_2 = \dfrac{5040}{336} \\[0.5em] \Rightarrow t_2 = 15 s

Hence, time taken = 15 s

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