Calorimetry Class-10 Goyal Brothers ICSE Physics Solutions Ch-4

Calorimetry Class-10 Goyal Brothers  ICSE Physics Solutions Chapter-4 Calorimetry . We Provide Step by Step Answer of Exercise, MCQs, Numericals Practice Problem Questions of Exercise-1 Calorimetry ICSE Class-10 . Visit official Website CISCE  for detail information about ICSE Board Class-10 Physics.

Calorimetry Class-10 Goyal Brothers ICSE Physics Solutions Ch-4


-: Select Topics :-

Exercise-1

 MCQ -1

Numericals Practice Problems-1

Exercise-2

 MCQ -2

Numericals Practice Problems-2

 


Exe-1 Calorimetry Class-10 Goyal Brothers ICSE Physics Solutions Ch-4

Page-54,55

Question 1.
Define calorie.
Answer 1:
CALORIE : “Is the amount of heat energy required to raise the temperature of 1 g of water from 14.5 °C to 15.5°C”.

Question 2.
State the modern unit of heat energy. How is this unit related to calorie ?
Answer 2:
Modern unit of heat energy is joule
Relation of joule with calorie :
1 J = 2.4 calorie
or 1 calorie = 4 .186 J = 4.2 J (approx)

Question 3.
What do you understand by the term thermal capacity ? State its unit is SI system.
Answer 3:
Thermal capacity : “The amount of heat energy required to raise the temperature of a given mass of substance through 1°C (1k) is called thermal capacity”.

Question 4.
Define specific heat capacity and state its SI and CGS units.
Answer 4:
Specific heat capacity : “Is the amount of heat energy required to raise the temperature of unit mass of a substance through 1°C or 1K is called specific heat capacity.”
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 3

Question 5.
Is the specific heat capacity of ice greater, equal to or less than water ?
Answer 5:
Specific heat capacity of ice is less than specific heat capacity of water.

Question 6.
Explain the following :

(a) Water is used in hot water bottles for fomentation purposes.
Answer 6:
Water provides heat energy for longer time and does not cool quickly as specific heat capacity of water is large.

(b) Water is used as a coolant in motor car radiators.
Answer:
When water is circulated in the pipes, it absorbs more amount of heat from surroundings (removes heat) without much rise in its temperature because of its high specific heat capacity.

(c) A wise farmer always waters his fields in the evening, if there is a forecast for frost.
Answer:
To save the crops on such cold nights farmers fill their fields with water as water has high sp. heat capacity. So water does not allow the temp, in the surrounding area of plants to fall upto 0°C. Other wise when temp, falls below 0°C water in the fine capillaries of plants will freeze, so the veins will burst due to increase in volume of water on freezing.

(d) Wet soil does not get as hot as dry soil in the sun.
Answer:
Water has high sp. heat capacity as compared to soil (dry) and absorbs heat from surrounding for longer time and takes longer time to set as compared to dry soil.

(e) Water is sprinkled on the roads in the evening during hot summer.
Answer:
Water has high sp. heat capacity and removes heat from the hot soil and decreases its temperature during hot summer evening.

(f) Water is used for internal heating in cold countries.
Answer:
In cold countries water is used for internal heating as it can carry large amount of heat energy from the furnace to the rooms at a fairly moderate temperature.

(g) Cold water is poured on the burns caused on the skin by some hot object.
Answer:
Water has high sp. heat capacity and can remove more heat from the bums caused on the skin by hot object and releives of the pain.

(h) Water rubs are kept in warehouses storing fruits and vegetables in cold countries during winter.
Answer:
Water has high sp. heat capacity and water kept in tubs lose heat for a longer time and keep the surrounding hot for longer time and save the vegetables from busting due to increase in volume at low temp, of water present vegetables.

Question 7.
Explain how is land breeze caused ?
Answer 7:
Land breeze : Blowing of cold air from land towards sea. During night temp, of land falls more rapidly as compared to water. Since water has high sp. heat capacity Pressure over sea water decrease and hence air blows from land (high pressure) towards sea (low pressure)

Question 8.
Explain the formation of sea breeze.
Answer 8:
Sea breeze : Blowing of cool air from sea towards land. During day time land gets heated up rapidly due to low sp. heat of land as compared to water. Pressure at land decreases. Hence air blows from sea (high pressure) towards land (low pressure).

Question 9.
Why is the weather in coastal regions moderate ?
Answer 9:
The climate near coastal regions moderate : The sp. heat capacity of water is very high or sp. heat capacity of land is much low as compared to water. As such land (or sand) gets cooled more rapidly as compared to water under similar conditions. Thus, a large difference in temperature is developed between the land and the sea, due to which cold air blows from land towards sea during night (i.e. land breeze) and during the day cold air blows from sea towards land (i.e. sand breeze). These make the climate near coastal region moderate.


 Multiple Choice Questions-1

Calorimetry Class-10 Goyal Brothers  ICSE Physics Solutions

1. The specific heat capacity of a substance :
(a) changes with the mass of given substance.
(b) changes with the area or volume of substance.
(c) changes with rise or fall in temperature.
(d) is a constant quantity for a given substance.
Answer:
(d) is a constant quantity for a given substance.

2. Land and sea breezes are formed in coastal regions because :
(a) water has very high specific heat capacity than the land.
(b) land has very high specific heat capacity than the water.
(c) sea water cools the cooler regions. .
(d) all the above.
Answer:
(a) water has very high specific heat capacity than the land.

3. The base of cooking pans is made thicker and heavy because:
(a) it lowers the heat capacity of pan
(b) it increases the heat capacity of pan
(c) the food does not get charred and keeps hot for long time
(d) both (a) and (c)
Answer:
(d) both (a) and (c)

4. The S.I. unit of specific heat capacity is :
(a) JKg-1
(b) JK-1
(c) JKg-1 K-1
(d) kJkg-1 K-1
Answer:
(b) Jk-1

5. The specific heat capacity of water in S.I. system is :
(a) 4.2 Jkg-1 K-1
(b) 42 JKg-1 K-1
(c) 4200 JKg-1 k-1
(d) 420 JKg-1 K-1
Answer:
(c) 4200 JKg-1 K-1

6. S.I. unit of thermal capacity is :
(a) Jkg-1
(b) kJ Kg-1
(c) Jkg-1 K-1
(d) cal oC-1
Answer:
(c) Jkg-1 K-1


Numerical Problems on Calorimetry -1

Calorimetry Class-10 Goyal Brothers  ICSE Physics Solutions

Page-55,56,57,58,59,60,61

Question 1.
A solid of mass 0.15 kg is heated from 10°C to 90°C. If the specific heat capacity of the solid is 390 Jkg-10 C-1, find the heat absorbed by the solid.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 6
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 7

Question 2.
A liquid of mass 0.2 kg and temperature 135°C is cooled to 25°C. If the specific heat capacity of liquid is 750 Jkg-10 C-1, find the heat energy given out.
Hint : OF = (135 – 25) = 110°C
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 7.1

Practice Problems 2

Question 1.
0.08 kg of a substance is heated from 30°C to 130°C when 2000 calories of energy is supplied to it Calculate the specific heat capacity of the substance in (a) calories, (b) joules.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 7.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 8

Question 2.
0.50 kg of lead at 327°C is cooled to 27°C, when it gives off 22500 calories of energy. Calculate the specific heat 1 capacity of lead in (a) calories, (b) joules.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 8.1

Practice Problems 3 

Question 1.
272 calories of heat is required to heat 0.02 kg of a metal of specific heat capacity 170 cal kg-10 C-1 to a temperature T. If the initial temperature of the metal is 20°C, calculate the final temperature T.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 8.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 9

Question 2.
3.75 × 105 calories of heat is given out by 5 kg of water at 100°C. Calculate the temperature of cooled water. Specific heat capacity of water is 1000 cal kg-1 °C-1.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 9.1

Question 3.
A burner, supplies heat energy at a rate of 20 Js-1 Find the specific heat capacity of a solid of mass 25 g, if its temperature rises by 80°C in one minute.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 9.2

Question 4.
A liquid of mass 100 g loses heat at a rate of 200 Js-1 for 1 minute. If the temperature of liquid drops by 100°C, calculate the specific heat capacity of the liquid.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 10
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 10.1

Practice Problems 4

Question 1.
A heater, rated 1000 W, is used to heat 1.5 kg of water at 40°C to its boiling point. Calculate the time in which the water starts to boil Specific heat capacity of water is 4200. J kg-10 C-1.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 10.2

Question 2.
400 g of mercury of specific heat capacity 0.14 Jg-1 °C-1 is heated by a 200 W heater for 1 min. and 40 s. If initially mercury is at 0°C, calculate its final temperature.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 10.3
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 11

Question 3.
A power drill of 400 W makes a hole in a lead cube of specific heat capacity 0.13 Jg-1  °C-1 in 80 s. If the temperature of lead rises from 27°C to 327°C, calculate the mass of the lead cube.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 11.1

Practice Problems 5

Question 1.
A solid of mass 150 g at 200°C is placed in 0.4 kg of water at 20°C till a constant temperature is attained. If the S.H.C. of the solid is 0.5 Jg-1 K-1, find the resulting temperature of the mixture.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 11.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 12

Question 2.
A liquid of mass 100 g at 120°C is poured in water at 20°C, when the final temperature recorded is 40°C. If the specific heat capacity of the liquid is 0.8 Jg-1 °C-1, calculate the initial mass of water.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 12.1

Question 3.      Calorimetry Class-10 Goyal Brothers
A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C, when the final temperature recorded is 20°C. Find the specific heat capacity of the solid.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 12.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 13

Practice Problems 6

Question 1.
20 g of hot water at 80°C is poured into 60 g of cold water, when the temperature of cold water rises by 20°C. Calculate the initial temperature of cold water.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 13.1

Question 2.
50 g of a hot solid of specific heat capacity 0.25 Jg-10 C-1 and at 100°C is placed in 80 g of cold water, when the temperature of cold water rises by 3°C. Find the initial temperature of cold water.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 13.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 14

Practice Problems 7

Question 1.      Calorimetry Class-10 Goyal Brothers
What mass of a solid of specific heat capacity 0.75 Jg-10 C-1 will have heat capacity 93.75 Jg-1 °C-1 ?
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 14.1

Question 2.
A solid of mass 1.2 kg has sp. heat capacity of 1.4 Jg-1 °C-1. Calculate its heat capacity in SI units.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 14.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 15

Practice Problems 8

Question 1.
A solid of mass 0.15 kg and at 100°C is placed in 0.25 kg of water, contained in a copper calorimeter of mass 0. 12 kg at 10°C. If the final temperature of the mixture is 20°C, calculate the sp. heat capacity of the solid.
(given, H.C of water = 4200 Jkg-1 k-1, SHC of copper = 400 J Kg-1 k-1)
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 15.1

Question 2.
A piece of brass of mass 200 g and 100°C, is placed in 400 g of turpentine oil, contained in a copper calorimeter of mass 50 g at 15°C. The final temperature recorded is 23CC. Find the sp. heat capacity of turpentine oil.
[SHC for brass = 370 J kg-1 k-1 ; SCH of copper = 390 J Kg-1 k-1 ] Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 16
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 16.1

Practice Problems 9

Question 1.
A copper vessel contains 200 g of water at 24°C. When 112 g of water at 42°C is added, the resultant temperature of water is 30°C. Calculate the thermal capacity of the calorimeter.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 17

Question 2.          Calorimetry Class-10 Goyal Brothers
A copper calorimeter contains 50 g of water at 16°C. When 40 g of water at 36°C is added, the resulting temperature of the mixture is 24°C. Calculate the heat capacity of the calorimeter.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 17.1
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 18

Practice Problems 10

Question 1.
A liquid X of specific heat capacity 1050 J kg-1 K-1 and at 90°C is mixed with a liquid Y of specific heat capacity 2362,5 J kg-1 K-1 and 20°C, when the final temperature recorded is 50°C. Find in what proportion the weights of the liquids are mixed.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 18.1

Question 2.
Your are required to make a water bath of 50 kg at 45°C, by mixing hot water at 90°C, with cold water at 20°C. Calculate the amount of hot water required.
Hint : Let amt. of hot water = x
 Amount of cold water = (50 – x) kg
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 18.2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 19

Practice Problems 11

Question 1.
Heat energy is given to 100 g of water, such that its temperature rises by 10 K. When the same heat energy is given to a liquid L of mass 50 g its temperature rises by 50 K. Calculate

  1. heat energy given to water
  2. the specific heat capacity of liquid L.
[Take sp. heat capacity of water = 4200 J Kg-1 k-1] Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 19.1

Question 2.      Calorimetry Class-10 Goyal Brothers
Heat energy is given to 80 g of alcohol (sp. heat capacity 2200 J kg-1 K-1) when its temperature rises by 20 K. If the same heat energy is given to 200 g of mercury of sp. heat capacity 140 J kg-1 K-1, what is the rise in temperature.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 20

Practice Problems 12

Question 1.
A copper ball is dropped from a vertical height of 1200 m. If the initial temperature of copper ball at the height is 12°C, what is its temperature of copper is 400 Jkg-1 °C-1 and g = 10 ms-2.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 20.1
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 21

Question 2.
A waterfall is 1.5 km high. If the temperature of water at its top is 20°C find its temperature at the bottom of waterfall, assuming all the kinetic energy is converted into heat energy.
[Take g – 10 ms-2 and sp. heat capacity of water = 4200 J Kg-1 c-1] Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Calorimetry 21.1


Exe-2 Calorimetry Class-10 Goyal Brothers  ICSE Physics Solutions Ch-4

Page-64,65

Question 1.
(a) What do you understand by the term latent heat of fusion?
Answer :
Latent heat of fusion : When a solid is heated change in phase from solid to liquid takes place at a constant temp. “The heat supplied to change solid to liquid at constant temp, is called latent heat of fusion.”

(b) Why does the temperature remain constant during the fusion of a substance ?
Answer:
“The heat absorbed by solid is utilised in increasing the potential energy of the molecules.”

Question 2.
What do you understand by the term specific latent heat of fusion ? State its C.GS. and S.I. unit.
 Specific latent heat of fusion : “The heat energy required to convert unit mass of the substance from solid to liquid state without change in temperature.”
Units :
C.G.S. → Cal g-1
S.I. J kg-1

Question 3.
Define specific latent heat of fusion of ice. State its magnitude in calories and joules.
Answer:
Specific latent heat of fusion of ice : “Is the heat energy required to convert unit mass of ice to water without the change in temp, (or ice at 0°C to water at 0°C).”
Specific latent heat of fusion of ice = 80 cal g-1 or 336000 Jkg-1

Question 4.
The specific heat of fusion of lead is 27 Jg-1. What do you understand from the statement ?
Answer:
The specific latent heat of lead is 27 Jg-1 means 1 g of lead will absorb 27 J of heat in changing from solid to liquid at constant temperature.

Question 5.          Calorimetry Class-10 Goyal Brothers
Why should bits of ice to wiped dry before adding them to the calorimeter during the determination of specific latent heat of fusion of ice ?
Answer:
If bits of ice are not wiped dry waterdrops are already in liquid state will absorb less heat and result will not be correct.

Question 6.
Explain the following :

(a) Why does the weather become moderate in cold countries when the freezing of lakes and other water bodies start ?
(b) Why does it become very cold when ice starts melting in the cold countries ?
(c) Why is melting of ice a better coolant than water at zero degree Celsius ?
(d) Why does ice-cream feel more colder than water at 0°C ?
(e) Why does the weather become warm, when it snows ?
(f) Why does the weather become very cold after a hail storm ?
(g) Why are icebergs carried thousands of kilometers away without melting substantially ?
(h) Why does snow/ice not melt rapidly on the mountains during summer ?

Answer:    
Reasons used : 1 kg of ice on meting absorbs 336000 J of heat energy and 1 kg of water to freeze will absorb 336000 J of heat energy.
(a) When freezing of lakes and other water bodies start in cold countries every 1 kg of water gives out 336000 J of heat and temp, of atmosphere increase making the weather moderate.
(b) When ice start meting heat is absorbed from the atmosphere (336000 J for every 1 kg of ice) and temp, falls in the surrounding and it becomes very cold.
(c) Sp. Latent heat of ice is 336000 J for every 1 kg ice. Hence to change ice at 0°C to water at 0°C, it will extract 336000 J of heat from the hot engine and will cool the engine for longer time.
(d) Sp. latent heat of ice is very high and it is 336000 J Hence ice will absorb more heat from mouth and temp, of mouth will fall considerably and ice cream feels more colder than water.
(e) When it snows, water evolves heat i.e. it gives out 336000 J for every 1 kg, in the surrounding and it becomes warm.
(f) After hail storm, to melt ice balls very large amount of heat is extracted from surroundings (sp. heat capacity of ice 336000 J) hence temp, falls and it becomes very cold.
(g) Specific latent heat of ice and also density of ice (less than water) makes it flow in water and ice bergs lose heat slowly and are carried to large distance.
(h) It is the high latent heat of ice (336000 J) for every 1 kg to change into under at 0°C. Snow melts slowly on the mountains in summer and water is available in the rivers.

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