Calorimetry ICSE Class-10 Concise Physics Selina Solutions

Calorimetry ICSE Class-10 Concise Physics Selina Solutions Chapter-11. We Provide Step by Step Answer of Exercise – 11(A), MCQs – 11(A), Numericals – 11(A) , Exercise-11(B), MCQ -11(B), Numericals -11(B)  Questions of Exercise-11 Calorimetry ICSE Class-10 Concise . Visit official Website CISCE  for detail information about ICSE Board Class-10.

Board ICSE
Publications Selina Publication
Subject Physics
Class 10th
Chapter-11 Calorimetry (Exercise – 11A)
Book Name Concise
Topics Solution of Exercise – 11(A), MCQs – 11(A), Numericals – 11(A) , Exercise-11(B), MCQ -11(B), Numericals -11(B)
Academic Session 2021-2022

Calorimetry ICSE Class-10 Concise Physics Selina Solutions Chapter-11


: Select Exercise :-

Exercise – 11(A), MCQs – 11(A), Numericals – 11(A) ,

Exercise-11(B), MCQ -11(B), Numericals -11(B)


How to Solve Numericals of Calorimetry ICSE Class-10

Note:- Read the chapter Calorimetry carefully and solve All the example given in your text book with Numericals Problems before starting Solutions of Chapter – 11 “Calorimetry” for ICSE Class-10 exercise.  Get ICSE Class 10 Physics Formula Chapter-Wise  to Solve Calorimetry Numericals problems.


“Calorimetry” Selina Physics Solution, Exercise -11(A)  ICSE Class-10

Page 269

Question 1

Define the term heat.

Answer 1

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

Question 2

Name the S.I. unit of heat.

Answer 2

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

Question 3

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

Answer 3

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

1 calorie = 4.186 J

Question 4

Define one kilo-calorie of heat.

Answer 4

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

Question 5

Define temperature and name the S.I. unit.

Answer 5

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

S.I. unit kelvin (K).

Question 6

State three differences between heat and temperature.

Answer 6

Heat Temperature
The kinetic energy due to random motion of the molecules ofa 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). S.I. unit kelvin (K).
It is measured by the principle of calorimetry. It is measured by a thermometer

Question 7

Define calorimetry.

Answer 7

The measurement of the quantity of heat is called calorimetry.

Question 8

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

Answer 8

The heat capacity of a body is the amount of heat energy required to raise its temperature by 1oC or 1K.

S.I. unit is joule per kelvin (JK-1).

Question 9

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

Answer 9

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 by 1oC (or 1K).

S.I. unit is joule per kilogram per kelvin (Jkg-1K-1).

Question 10

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

Answer 10

Heat capacity = Mass x specific heat capacity

Question 11

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

Answer 11

Heat capacity of the body is the amount of heat required to raise the temperature of (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. unit of heat capacity is      and S.I. unit of specific heat capacity is .


Page 270

Question 12

Name a liquid which has the highest specific heat capacity.

Answer 12

Water has the highest specific heat capacity.

Question 13

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

Answer 13

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

Question 14

What do you mean by the following statements:

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

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

Answer 14

(i) The heat capacity of a body is 50JK-1 means to increase the temperature of this body by 1K we have to supply 50 joules of energy.

(ii) The specific heat capacity of copper is 0.4Jg-1K-1 means to increase the temperature of one gram of copper by 1K we have to supply 0.4 joules of energy.

Question 15

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

Answer 15

The specific heat capacity of substance B is lesser than that of A. So, for same mass and same heat energy, the rise in temperature for B will be more than that of A. Hence, substance 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 16

Change in temperature and the nature of material

Question 17

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

Answer 17

The quantity of heat energy absorbed by a body depends on three factors :

(i)Mass of the body – The amount of heat energy required is directly proportional to the mass of the substance.

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

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

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 incrementoC.

Answer 18

The expression for the heat energy Q

Q= mc increment t (in joule)

Question 19

Same amount of heat is supplied to two liquid 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 19

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

As the substance with low heat capacity shows greater rise in temperature.

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 temperatures rise by same amount, compare their specific heat capacities.

Answer 20

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 21

The principle of method of mixture:

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

This principle is based on 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 of mixture. State assumption made, if any.

Answer 22

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

Fall in temperature of substance A = T– T

Rise in temperature of substance B = T – T2

Heat energy lost by A = mcross times  c1cross times   fall in temperature

= m1c1(T1 – T)

Heat energy gained by B= m2  cross times   c2  cross times  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

Here we have assumed 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 23

In the absence of water, if on a cold winter night the atmospheric temperature falls below 0oC, the water in the fine capillaries of plant will freeze, so the veins will burst due to the increase in the volume of 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 high specific heat capacity, so it does not allow the temperature in the surrounding area of plants to fall up to 0oC.

Question 24

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

Answer 24

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 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 the similar conditions. Thus a large difference in temperature is developed between the land and the sea due to which land and sea breezes are formed. These breezes make the climate near the sea shore moderate.

Question 25

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

Answer 25

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.

Question 26

Water property of water makes it an effective coolant?

Answer 26

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

Question 27

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

Answer 27

(i)Radiator in car.

(ii)To avoid freezing of wine and juice bottles.

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 28

The specific heat capacity of liquid X is higher than that of Y. So, for same mass and same heat energy, the rise in temperature for X will be less than that of Y.

(i) As a coolant in car radiators, the liquid needs to absorb more energy without much change in temperature. So, liquid X is ideal for this function.

(ii) As a heat reservoir to keep juice bottles without freezing, the liquid needs to give out large amount of heat before reaching freezing temperatures. Hence, liquid X is ideal for this function.

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 29

(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 other body.

(b)  It is made up of thin copper sheet because:

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

   (ii) Copper has low specific heat capacity so the heat capacity of 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) Heat capacity of the calorimeter should be low so out of three metals the one which have lowet specific heat capacity should be preferred.

(d) By polishing the outer and inner surface of the vessel the loss due to radiation can be minimized.

Question 30

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

Answer 30

By making the base of a cooking pan thick, 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 the food warm for a long time, after cooking.


“Calorimetry” ICSE Class-10 Physics Solution, Multiple Choice Type – 11(A)

Page 270

Question 1

The S.I. unit of heat capacity is :

(a) J kg-1

(b) J K-1

(c) J kg-1 K-1

(d) cal 0C-1

Answer 1

(b) J K-1

Question 2

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

(a) J kg-1

(b) J K-1

(c) J kg-1 K-1

(d) kilocal kg-10C-1

Answer 2

J kg-1 K-1

Question 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

Answer 3

4200 J kg-1 K-1


“Calorimetry” Selina ICSE Physics Solution, Numericals – 11(A)

Page 270

Question 1

By imparting heat to a body, its temperature rises by 150C. What is the corresponding rise in temperature on the Kelvin scale?

Answer 1

The size of 1 degree on the Kelvin scale is the same as the size of 1 degree on the Celsius scale. Thus, the difference (or change) in temperature is the same on both the Celsius and Kelvin scales.

Therefore, the corresponding rise in temperature on the Kelvin scale will be 15K.

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 25oC to 35oC?

Answer 2

(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 X specific heat capacity

=0.15 kg X 410Jkg-1K-1

=61.5JK-1

Change in temperature= (35-25)oC=10oC=10K

(ii) Energy required to increase the temperature of vessel

=0.15 X 410 X 10

=615 J


Page 271

Question 3

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

(i) Heat energy needed to warm it by 15oC, and

(ii) Its specific heat capacity in S.I unit.

Answer 3

(i)We know that heat energy needed to raise the temperature by 15o is = heat capacity x change in temperature.

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

(ii)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.

Question 4

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

Answer 4

Mass of copper m = 100 g = 0.1 kg

Change of temperature increment 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 = begin mathsize 14px style straight m space straight x space increment straight t space straight x space straight c end style

= 0.1 x 50 x 390

= 1950 J

Question 5

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

Answer 5

Heat energy supplied = 1300 J

Mass of lead = 0.5 kg

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

Specific heat capacity of lead

c = 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 18oC to 38oC. Assume that all the heat energy supplied by the heater is given to the material.

Answer 6

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

Time taken by a heater to raise the temperature of material

t= 1920 seconds

t=32 min

Question 7

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10oC to 15oC in 100 s. calculate :

(i)the specific heat capacity of the liquid.

(ii)the heat capacity of 4.0 kg of liquid.

Answer 7

Power of heater P= 600 W

Mass of liquid m=4.0 kg

Change in temperature of liquid = (15-10)oC = 5oC(or 5 K)

Time taken to raise its temperature =100s

Heat energy required to heat the liquid

and

 =600X100=60000J

  =

Heat capacity= c x m

Heat capacity= 4 x 3000JKg-1 K-1 = 1.2 x 104 J/K

Question 8

0.5 kg of 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.

Answer 8

Change in temperature= 30 – 5 = 25 K.

=

t=29 min 10 sec.

Question 9

A mass of 50g of a certain metal at 150° C is immersed in 100g 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 9

Heat liberated by metal=

Heat absorbed by water=

Heat energy lost= heat energy gained

= =

S=0.52 J g-1 K

Question 10

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

Answer 10

Mass of water (m1)=45 g

Temperature of water (T1) =50oC

Mass of copper (m2) =50 g

temperature of copper(T2) =18oC

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

                                                             

         oC

T= 47oC

Question 11

200g 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 the water. Specific heat capacity of water = 4200 J kg-1 K-1.

Answer 11

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) =?

c1=c2

T=38oC.

Question 12

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

Answer 12

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.

T = 20oC.

Final temperature = 20o c

Change in temperature = 15oC

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

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 its boiling point 100°C. Specific heat capacity of water = 4.2 J g-1 K-1.

Answer 13

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