Selina Concise Class 10 Physics Solution Chapter No. 11 – ‘Calorimetry’ For ICSE Board Students.
Selina Concise Class 10 Physics Chapter 11 Calorimetry Exercise All Questions and Answers by Physics Teacher here in this post.
Exercise 11(A)
Question: 1
Define the term heat.
Solutions: The energy that flows from the hot body to the cold body when kept in contact is called as heat.
Question: 2
Name the S.I. unit of heat.
Solution: The S.I. unit of heat is joule.
Question: 3
Define the term calorie. How is it related to joule?
Solution: The quantity of heat required to raise the temperature of 1g of water by 1oC.
1 calorie = 4.186 joule.
Question: 4
Define one kilo-calorie of heat.
Solution: One kilo-calorie is the heat energy required to raise temperature of 1kg of water by 1oC.
Question: 5
Define temperature and name its S.I. unit.
Solution: Temperature is a parameter which tells the thermal state of body.
S.I. unit of temperature is kelvin (K).
Question: 6
State three differences between heat and temperature.
Solution:
Heat | Temperature |
Heat is the energy that flows from the ot body to cold body when kept in contact. | Temperature is the quantity which determines the direction of flow of heat. |
S.I. unit of heat is joule. | S.I. unit of temperature is kelvin. |
Heat is measured by principle of calorimetry. | Temperature is measured by thermometer. |
Question: 7
Define calorimetry.
Solution: The measurement of amount of heat is called as calorimetry.
Question: 8
Define the term heat capacity and state its S.I. unit.
Solution: The amount of heat energy required for a body to raise its temperature by 1 K is called as heat capacity of the body.
S.I. unit of heat capacity is joule per kelvin.
Question: 9
Define the term specific heat capacity and state its S.I. unit.
Solution: The amount of heat energy required to raise the temperature of unit mass of substance by 1 K is called as specific heat capacity of the body.
S.I. unit of specific heat capacity is joule per kilogram per kelvin (J/kg x K).
Question: 10
How is the heat capacity of a body related to the specific heat capacity of its substance?
Solution:
Heat capacity = Mass x Specific heat capacity
Question: 11
State three differences between the heat capacity and specific heat capacity.
Solution:
Heat capacity | Specific heat capacity |
The amount of heat energy required to raise the temperature of body by 1 K. | The amount of heat energy required to raise the temperature of unit mass of a body by 1 K. |
It depends upon the mass of body. | It does not depend upon the mass of body. |
S.I. unit is J/K | S.I. unit is J/kgK. |
Question: 12
Name a liquid which has the highest specific heat capacity.
Solution: Water has the highest specific heat capacity.
Question: 14
What do you mean by the following statements:
(i) The heat capacity of a body is 50 JK-1?
(ii) The specific heat capacity of copper is 0.4 J g-1 K-1?
Solution:
i.) The heat capacity of body is 50 J/K, means 50J heat energy is required to raise the temperature of body by 1 K.
ii.) The specific heat capacity of copper is 0.4 J/g K, means 0.4J energy is required to raise the temperature of 1 g of copper by 1 K.
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?
Solution: The heat conductivity of body is inversely proportional to the specific heat capacity of body. Hence, as the specific heat capacity of B is lower than A it will act as a good conductor of heat. As the heat capacity of B is less than A rise in temperature of B is more than A for same mass and same energy provided.
Question: 16
Name two factors on which the heat energy liberated by a body on cooling depends.
Solution: Mass of the body and specific heat capacity of the body are the two factors on which the heat energy liberated by a body on cooling depends.
Question: 17
Name three factors on which the heat energy absorbed by a body depends and state how does it depend on them.
Solution: The three factors on which the heat energy absorbed by the body depends are, mass of body as the mass of the body is directly proportional to the amount of heat energy required. Temperature rise, as the amount of heat energy required is also directly proportional to the rise in the temperature of the body. Lastly specific heat capacity of the body, as the amount of heat absorbed by a body depends upon its specific heat capacity.
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 △toC.
Solution: expression for heat energy
Q = m c ∆t joule
Where,
c = specific heat capacity
m = mass of substance
∆t = rise 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?
Solution: As the temperature of liquid A shows a greater rise in the temperature, the specific heat capacity of A is lower than that of specific heat capacity of B.
Question: 20
Two blocks P and Q of different metals having their mass in the ratio 2:1 are given the same amount of heat. Their temperatures rise by the same amount, compare their specific heat capacities.
Solution: Let us assume p and q be the specific heat capacities of two blocks P and Q respectively. E be the heat energy provided.
As we know,
Specific heat capacity = heat energy / m x ∆t
p = E / 2m x ∆t
q = E / m x ∆t
p/q = (E / 2m x ∆t)/ (E / m x ∆t)
hence,
p/q = ½
the ratio of specific heat capacities of two block P and Q is 1:2
Question: 21
What is the principle of the method of mixture? What other name is given to it? Name the law on which this principle is based.
Solution: Principle of the methods of mixture is heat energy lost by the hot body is equal to heat energy gained by the cold body. It is also called as principle of calorimetry. This principle is based of 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.
Solution:A mass m1 of a substance P of specific heat capacity c1 at temperature T1 is mixed with a mass m2 of other substance Q of specific heat capacity c2 at a lower temperature T2. And the final temperature becomes T.
For substance P, loss in temperature = T1 – T
For substance Q, gain in temperature = T – T2
Heat energy loss of P = m1 x c1 x (T1 – T)
Heat energy gain of Q = m2 x c2 x (T – T2)
In ideal situation, by principle of mixtures
Heat loss by P = heat gained by Q
m1 x c1 x (T1 – T) = m2 x c2 x (T – T2)
by rearranging equation,
T = (m1 x c1 x T1 + m2 x c2 x T2)/ (m1 x c1 + m2 x c2).
Question: 23
Why do the farmers fill their fields with water on a cold winter night?
Solution: The specific heat of water is very high. Hence, it is used as a heat reservoir keep the plants from freezing as freezing can kill plants. The farmers fill their fields with water in cold night to keep the temperature from falling below 0oC.
Question: 24
Discuss the role of high specific heat capacity of water with reference to climate in coastal areas.
Solution: the specific heat capacity of water is very high than compared to the sand by 5 times. Hence, the heat required to raise the temperature of 1 kg of water by 1 K will be 5 times more than the heat required to raise the temperature of same amount sand by same degree. As a result, sand gets heated and cooled more rapidly than water under similar conditions. Due to which near costal regions the temperature difference between the water and sand is very high which sets up the convection of air currents. The cold air blows from sea towards land in the day time and cold air blows from the land to sea in the night. These sir current make the climate near sea shore moderate.
Question: 25
Water is used in hot water bottles for fomentation. Give reason.
Solution: Water has the highest specific heat capacity. Hence, water does not get cooled easily. When hot water bottles are used for fomentation as they provide heat energy for extended period of time.
Question: 26
What property of water makes it an effective coolant?
Solution: specific heat capacity of water is very high. Hence, it is used as very effective coolant as it can remove heat from a system without having much rise in the temperature.
Question: 27
Give one example each where high specific heat capacity of water is used (i) as coolant, (ii) as heat reservoir.
Solution:
i.) Water is used as coolant in radiators in car and generators.
ii.) In cold countries, water is used as heat reservoir for wine and juice bottles to avoid freezing.
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?
Solution: The liquid X has higher specific heat capacity than the liquid Y
i.) Liquid X is used as coolant in car radiators as it absorbs more heat without much rise in temperature due to its high specific heat.
ii.) As the liquid X higher specific heat capacity it requires large amount of heat before reaching its freezing point. Hence it can be used as heat reservoir to keep juice bottles without freezing.
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 minimised in a calorimeter?
Solution:
a.) A calorimeter is a cylindrical vessel which is used to ensure the amount of heat gained by a body when it is mixed with another body or substance.
b.) The calorimeter is made up of thin sheet of copper as copper is very good conductor of heat, so vessel can be in equilibrium with temperature of the content. The specific heat capacity of copper is very low so it doesn’t affect reading of calorimeter.
c.) The specific capacity of material used must be very low. Hence, metal B can be used for calorimeter.
d.) The loss of heat due to radiation can be minimised in a calorimeter can be minimised by polishing outer and the inner surface of the vessel.
Question: 30
Why the base of a cooking pan made thick and heavy?
Solution: The base of a cooking pan made thick and heavy as heat capacity of coking pan increases with increase in the thickness. The higher heat capacity allows the pan to get heated slowly and evenly for proper cooking and after that pan gets cooled slowly keeping the food warm for longer time.
MULTIPLE CHOICE TYPE
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
Solution: (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) kcal kg-10C-1
Solution:(c) Jkg-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
Solution:(a) 4200 J kg-1 K-1
NUMERICAL
Question: 1
By imparting heat to a body, its temperature rises by 150C. What is the corresponding rise in temperature on the Kelvin scale?
Solution: Given
Temperature rise = 150C
Temperature rise in kelvin is also 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 25oC to 35oC?
Solution: Given
Mass = 150g = 0.15 kg
Specific heat capacity = 410 J/kgK.
a.) Heat capacity = specific heat capacity x mass
Heat capacity = 410 x 0.15
Heat capacity = 61.5 J/K
b.) Change in temperature, ∆t = 35 – 25 = 100C
Heat energy required to raise 10oC = ∆t x specific heat capacity x mass
Heat energy = 10 x 410 x 0.15
Heat energy = 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 15oC, and
(ii) Its specific heat capacity in S.I unit.
Solution: Given
Mass= 2 kg
Heat capacity = 966 J/K
i.) Heat energy required to raise 15oC = ∆t x heat capacity
= 15 x 966
= 14490 J
ii.) Specific heat capacity = heat capacity / mass
Specific heat capacity = 966 / 2
Specific heat capacity = 483 J/kgK
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.
Solution: Given
Mass = 100g = 0.1 kg
Change in temperature ∆t = 70 -20 = 50 0C.
Specific heat capacity = 390 J/kgK
Heat energy = ∆t x mass x specific heat capacity
Heat energy = 50 x 0.1 x 390
Heat energy = 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.
Solution: Given
Heat energy = 1300 J
Mass = 0.5kg
Rise in temperature, ∆t = 40 – 20 = 20 oC.
Specific heat capacity = heat energy / mass x ∆t
Specific heat capacity = 1300 / 20 x 0.5
Specific heat capacity = 130 J/kgK
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 heater is given to the material.
Solution: Given
Power = 500 W
Mass = 50 kg
Specific heat capacity = 960 J/kgK
Change in temperature, ∆t = 38 -18 = 20oC
Heat energy = mass x ∆t x specific heat capacity
Heat energy = 50 x 20 x 960
Heat energy = 96000 J
As we know,
Power = energy / time
Time = energy / power
Time = 96000 / 500
Time = 1920 s = 32 min.
Question: 7
An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0oC to 15.0oC in 100 s. Calculate:
(i.) the heat capacity of 4.0 kg of liquid, and
(ii.) the specific heat capacity of liquid
Solution: Given
Power = 600 W
Mass = 4 kg
Rise in temperature ∆t = 15 – 10 = 5oC.
Time = 100s
i.) As we know,
Power = heat energy / time
Heat energy = power x time
= 600 x 100
= 60000 J
Heat capacity = heat energy / ∆t
= 60000 / 5
= 12000 J/K
ii.) Specific heat capacity = heat capacity / mass
= 12000 / 4
= 3000 J/kgK
Question: 8
0.5 kg of lemon squash at 30o 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 5o C? Specific heat capacity of squash = 4200 J kg-1 K-1.
Solution: Given
Mass = 0.5 kg
Specific heat capacity of lemon = 4200 J/kgK
Rate of removal of heat = 30 J/s
heat removed by fridge = mass x specific heat capacity x change in temperature
= 0.5 x 4200 x 25
= 52500 J
Time taken = heat removed / rate of removal of heat
= 52500 / 30
= 1750 s = 29 min 10 s
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.
Solution: Given
Mass of metal = 50g
Mass of water = 100g
Temperature of water = 11 oC
Change in temperature of metal ∆t = 130 oC
Change in temperature of water = 9 oC
Specific heat capacity of water = 4.2 J/gK
Heat lost by metal = mass x specific heat capacity x ∆t
= 50 x s x 130
Heat gained by water = mass x specific heat capacity x ∆t
= 100 x 4.2 x 9
By principle of mixtures
Heat lost = heat gained
50 x s x130 = 100 x 4.2 x 9
Specific heat of metal = 3780 / 6500 = 0.582 J/gK
Specific heat of metal is 0.582 J/gK
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 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.
Solution: Given
Mass of water, m1= 45g
Mass of copper, m2 = 50 g
Change in temperature of water, ∆t1 = (50 -t) oC
Change in the temperature of copper,∆t2 = (t- 18) oC
Specific heat capacity of water, c1 = 4.2 J/gK
Specific heat capacity of copper, c2 = 0.39 J/gK
According to principle of mixture
Heat gained = heat lost
m1 x c1 x ∆t1 = m2 x c2 x ∆t2
50 x 0.39 x (t – 18) = 45 x 4.2 x (50 -t)
19.5t – 351 = 9450 – 189t
19.5t + 189t = 9450 + 351
208.5t = 9801
t = 47.007 oC
The assumption used is the whole process is ideal means there is no loss of heat in any other way
Question: 11
200 g of hot water at 80o C is added to 300 g of cold water at 10oC. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg-1 K-1.
Solution: Given
Mass of hot water, m1= 200 g = 0.2 kg
Temperature of hot water, t1= 80oC
Mass of cold water, m2= 300 g = 0.3 kg
Temperature of cold water, t2 = 10 oC
According to principle of mixture
Heat gained = heat lost
m1 x c1 x ∆t1 = m2 x c1 x ∆t2
0.2 x (80 – t) = 0.3 x (t -10)
16 – 0.2t = 0.3t – 3
16 + 3 = 0.2t + 0.3t
19 = 0.5t
t = 38 oC.
Question: 12
The temperature of 600 g of cold water rises by 15oC when 300 g of hot water at 50oC is added to it. What was the initial temperature of the cold water?
Solution: given
Mass of cold water, m1 = 600g
Mass of hot water, m2 = 300 g
Change in temperature of cold water = 15oC
Change in temperature of hot water = (50 – t)
According to principle of mixture
Heat gained = heat lost
m1 x c1 x ∆t1 = m2 x c2 x ∆t2
600 x 15 = 300 x (50 – t)
30 = 50- t
t = 20oC
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.
Solution: Given
Mass of water = 1 kg
Power of kettle = 1.25 kW = 1250 W
Change in temperature of water, ∆t = 100 – 25 = 75 OC.
Specific heat capacity = 4.2 J/gK = 4200 J/kgK
Heat energy = mass x specific heat capacity x ∆t
= 1 x 4200 x 75
= 315000 J
As we know,
Power = energy / time
Time = energy / power
= 315000 / 1250
= 252 s
= 4 min 12s.
Exercise 11 (B)
Question: 1
(a.) What do you understand by the change of phase of 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?
Solution:
a.) When a substance changes its state from one to another at constant temperature that process is called as change of state.
b.) No, change of phase is carried out at constant temperature.
c.) Yes, the substance absorbs or liberate heat during change of state.
d.) Latent heat.
Question: 2
A substance changes from its solid state to the liquid state when heat is supplied to it
a.) Name the process.
b.) What name is given to heat observed by the substance.
c.) How does the average kinetic energy of the molecules of the substance change.
Solution:
a.) Changing of substance from solid to liquid is called melting.
b.) Latent heat of melting.
c.) As there is no change in temperature average kinetic energy of molecules doesn’t change.
Question: 3
A substance on heating undergoes (i) a rise in its temperature, (ii) A change in its phase without change in its temperature. In each case, state the change in energy of molecules of the substance.
Solution:
i.) If the substance undergo increase in temperature the average kinetic energy of molecules increase.
ii.) If the substance undergoes change of state without change in temperature the molecule gains potential energy.
Question: 4
How does the (a) average kinetic energy (b) average potential energy of molecules of a substance change during its change in phase at a constant temperature, on heating?
Solution:
A.)The average kinetic energy of molecules does no change.
b.) If the substance undergoes change of state without change in temperature the molecule gains potential energy, to overcome the intermolecular forces and change state with causing rise in temperature.
Question: 5
State the effect of the presence of impurity on the melting point of ice. Give one use of it.
Solution: The melting point of a substance decreases with the presence of impurity.
Question: 6
State the effect of the increase of pressure on the melting point of ice.
Solution: With increase in pressure, the melting point of ice decreases.
Question: 7
The diagram shows the change of phases of a substance on a temperature-time graph on heating the substance at a constant rate.
(a.) What do parts AB, BC, CD and DE represent?
(b.) What is the melting point of the substance?
(c.) What is the boiling point of the substance?
Solution:
a.) Part AB shows increase in temperature of solid from o to t1, part BC shows its melting at temperature t1, part CD shows increase in temperature of liquid from t1 to t3 and part DE shows boiling of the substance at temperature t3.
b.) Melting point of the substance is t1oC.
c.) Boiling point of substance is at t3oC.
Question: 10
Explain the terms boiling and boiling point. How is the volume of water affected when it boils at 100oC?
Solution: When a liquid undergoes change of phase and becomes vapour that process is called boiling
Temperature at which vapour pressure of the liquid becomes equal to atmospheric pressure that temperature is called boiling point.
The volume of water increases when it boils.
Question: 11
How is the boiling point of water affected when some salt is added to it?
Solution: The boiling point of water increases when salt is added to it.
Question: 12
What is the effect of increase in pressure on the boiling point of a liquid?
Solution: When pressure is increased the boiling point of a liquid also increases.
Question: 13
Water boils at 120°C in a pressure cooker. Explain the reason
Solution: As the definition of boiling point, it is a temperature at which the vapour pressure of liquid becomes equal to the surrounding pressure. Hence in pressure cooker as the pressure in the surrounding is more the boiling point of water reaches to 120 °C.
Question: 14
Write down the approximate range of temperature at which water boils in a pressure cooker.
Solution: 120 °C to 125 °C.
Question: 15
It is difficult to cook vegetables on hills and mountains. Explains the reason.
Solution: As atmospheric pressure in the hills or mountains is less the water boils at lower temperature than it boils at ground level. Hence, it is difficult to cook vegetables on hills and mountains.
Question: 16
Complete the following sentences:
(a.) When ice melts, its volume………….
(b.) Decrease in pressure over ice ………….. its melting point.
(c.) Increase in pressure ………..the boiling point of water.
(d.) A pressure cooker is based on the principle that boiling point of water increases with the …………..
(e.)The boiling point of water is defined as ………………………..
(f.) Water can be made to boil at 115°C by…………….. pressure over its surface.
Solution:
(a) decreases.
(b) increases
(c) increases
(d) pressure.
(e) temperature at which vapour pressure of the water becomes equal to atmospheric pressure.
(f) increasing
Question: 17
What do you understand by the term latent heat?
Solution: Heat absorbed or liberated by substance to change its state without changing its temperature is called latent heat.
Question: 18
Define the term specific latent heat of fusion of ice. State its S.I. unit.
Solution: Specific latent heat of fusion of ice is defined as the energy required to change unit mass of ice into water without change of its temperature.
S.I. unit of specific latent heat of fusion of ice is J/kg.
Question: 19
Write the approximate value of specific latent heat of ice.
Solution: Approximate value of specific latent heat of ice is 336 KJ/kg.
Question: 20
‘The specific latent heat of fusion of ice is 336 J g-1 ‘. Explain the meaning of this statement.
Solution:The specific latent heat of fusion of ice is 336 J/g meaning 1 g of ice at 0 °C absorbs 336 J amount of energy to form 1g of 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.
Solution: As 1g of ice at 0 °C melts to 1 g of water at 0 °C, the latent heat is absorbed by the ice.
Question: 22
Which has more heat: 1 g of ice at 0°Cor 1 g of water at 0°C? Give reasons.
Solution: 1 g of water at 0 °C has more heat as it gives out 80 cal heat while changing to 1g of ice at 0°C.
Question: 23
(a) Which requires more heat: 1 g ice at 0°Cor 1 g water at 0°C to raise its temperature to 10oC? (b) Explain your answer in part (a).
Solution:
a.) 1g of ice requires more heat to raise its temperature by 10°C.
b.) The conversion of 1g of ice at 0 °C first absorbs 336 J of energy to converted into 1 g of water at 0 °C. Then that water starts heating up so 1g of ice at 0°C requires more heat.
Question: 24
Ice cream appears colder to the mouth than water at 0°C. Give reasons.
Solution: Ice cream at 0 °C first absorbs latent heat to convert into water at 0°C. on the other hand water at 0 °C only absorbs heat to rise its temperature. Hence the ice creams feels more cold in mouth than water at the same temperature.
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.
Solution: Soft drink bottles will get cooled quickly in ice at ice cubes at 0 °C as ice at 0 °C absorbs more 336 J/g heat from soft drink bottles than water at 0 °C.
Question: 26
It is generally cold after a hail storm than during and before the hail storm. Give reasons.
Solution: Ice absorbs more heat toget converted into water and they absorb that heat from surrounding. Hence, it gets colder after and before than during the hail storm.
Question: 27
The temperature of surroundings starts falling when ice in a frozen lake starts melting. Give reasons.
Solution: To ice to melt it requires the heat which is absorbed by the ice from the surrounding hence temperature of surrounding starts falling when ice in a frozen lake starts melting.
Question: 28
Water in lakes and ponds do not freeze at once in cold countries. Give reason.
Solution: To freeze water has to give out large amount of energy to change its phase to ice. When a layer of ice forms as water being bad conductor of heat its not easy for the rest of water to loose that much of heat. Hence, the lakes or pond do not freeze at once in the cold countries.
Question: 29
Explain the following:
(i.) The surroundings become pleasantly warm when water in a lake starts freezing in cold countries.
(ii.) The heat supplied to a substance during its change of state, does not cause any rise in its temperature.
Solution:
i.) When water freezes it gives out large amount of heat to be converted into ice. Hence, the surrounding becomes pleasantly warm when water in a lake starts freezing in cold countries.
ii.) Latent heat is required to change the phase only. Hence, the heat supplied it only causes change in state without rise in temperature.
MULTIPLE CHOICE TYPE
Question: 1
The S.I. unit of specific latent heat is:
(a) cal g-1
(b) cal g-1 K-1
(c) J kg-1
(d) J kg -1 K-1
Solution: (c) J kg-1
Question: 2
The specific latent heat of fusion of water is:
(a) 80 cal g-1
(b) 2260 J g-1
(c) 80 J g-1
(d) 336 J kg-1
Solution: (a) 80 cal g-1
NUMERICAL
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.
Solution: Given
Mass of ice = 10g = 0.01kg
Heat absorbed, Q = 5460J
Specific heat capacity of water = 4200 J/kgK.
Heat required for water to raise the temperature from 0 °C to 50 °C = 0.01 x 4200 x 50
= 2100 J
Let specific latent heat of ice be l
As we know,
Q = ml + mc∆t
5460 = 10 x l + 2100
l = 5460 – 2100 / 10
l = 336 J/g
Question: 2
How much heat energy is released when 5.0 g of water at 20°Cchanges into ice at 0°CTake specific heat capacity of water =4.2 J g-1 K-1, specific latent heat of fusion of ice = 336 J g-1.
Solution: Given
Mass of water = 5g
Change in temperature = 20 °C
specific heat capacity of water =4.2 J/gK
specific latent heat of fusion of ice = 336 J/g.
heat energy released by water to change its temperature from 20 °C to 0 °C = 5 x 4.2 x 20
= 420 J
Heat released by water to get converted into ice at 0 °C = 5 x 336J
= 1680 J
Total heat energy released = 420 + 1680 = 2100 J
Question: 3
A molten metal of mass 150 g is kept at its melting point 800oC. When it is allowed to freeze at the same temperature, it gives out 75000 J of heat energy.
(a)What is the specific latent heat of the metal?
(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 – 50oC?
Solution: Given
Mass of metal = 150g
Heat energy released = 75000 J
a.) Specific latent heat of metal = 75000 / 150
= 500 J/ g
b.) Specific heat capacity of metal = 200 J/kgK
Change in temperature = 800 – (-50)
= 850 °C
Heat energy required = mass x specific heat capacity x change in temperature
= 0.15 x 200 x 850
= 25500 J
Question: 5
A refrigerator converts 100 g of water at 20oC to ice at -10oC in 73.5 min. calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 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.
Solution:
Heat released by water when 100 g of water is cooled from 20 °C to 0 °C = 100 x 20 x 4.2 = 8400 J
Heat released by water when 100 g of water gets converted into ice at 0 °C = 100 x 336 x = 33600 J
Heat released by ice when 100 g of ice cools from 0 °C to -10 °C = 100 x 10 x 2.1 = 2100 J
Total amount of heat = 8400 + 33600 + 2100 = 44100 J
Time taken by fridge is 73.5 min = 4410s
Power = heat / time
Power = 44100 / 4410
Power = 10 W
Power that is also called as rate of heat extraction is 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 its freezing temperature. The specific heat capacity of water is 4.2 J g-1 K-1. Calculate the specific latent heat of ice. State one important assumption made in the above calculation.
Solution: Given
Mass of ice = 17 g
Mass of water = 40 g
Change in temperature = 34 °C
Specific heat capacity of water = 4.2 J/gK
Latent heat = energy released by water
Energy released = mass of water x specific heat capacity x change in temperature
= 40 x 4.2 x 34
= 5712 J
Specific latent heat of ice = energy released / mass of ice
= 5712 / 17
= 336 J/g
One key assumption is that there is no loss of energy.
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 is 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.
Solution:
Let m gm of ice melts. The final temperature of the mixture becomes 0°C.
So, amount of heat energy gained by 10 g of ice at -10°Cto raise its temperature to 0°C= 10 × 10 × 2.1 = 210 J
heat energy gained by m gm of ice at 0°Cto convert into water at 0°C= m ×336 = 336 m heat energy released by 10 g of water at 10°Cto lower its temperature to 0°C= 10 × 4.2 × (10 – 0) = 420
Heat energy gained = Heat energy lost
210 + 336 m = 420
m = 210 / 336
m = 0.625 gm
Question: 9
A piece of ice of mass 40 g is added to 200 g of water at 50oC. Calculate the final temperature of the 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.
Solution:
Let the final temperature of the water when all the ice has melted ist °C
Amount of heat lost when 200 g of water at 500 C cools to t°C = 200 × 4.2 × (50 – t) = 42000 – 840t
Heat gained when 40 g of ice at 0°C converts into water at 0°C =mass x specific heat of fusion = 40 × 336 J
= 13440 J
Heat gained when temperature of 40 g of water at 0°Crises to t°C= 40 × 4.2 × (t – 0) = 168t
According to law of mixture,
Heat gained = Heatlost
13440 + 168t = 42000 – 840t
168t + 840t = 42000 – 13440
1008t = 28560
t = 28560 / 1008
t = 28.330 °C
Question: 11
250 g of water at 30oC 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 its contents to 5oC. 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, the specific heat capacity of water is 4200 J kg-1 K-1.
Solution: Given
Mass of copper vessel, m1 = 50 g
Mass of water contained in copper vessel, m2 = 250 g
Mass of ice required to bring down the temperature of vessel = M
Final temperature = 50 °C.
Heat gained when ‘M’ g of ice at 0°C converts into water at 0 °C = M × 336 J
Heat gained when temperature of ‘M’ g of water at 0°C rises to 50 °C = M × 4.2 × 5
Total heat gained = M × 336 + M × 4.2 × 5
Heat lost when 250 g of water at 30 °C cools to 50 °C = 250 × 4.2 × 25
= 26250 J
Heat lost when 50 g of vessel at 30°C cools to 50 °C = 50 × 0.4 × 25
= 500 J
Total heat lost = 26250 + 500
= 26750 J
According to principle of mixtures,
Heat gained = Heat lost
M × 336 + M × 4.2 × 5
= 26750
357 M = 26750
M = 26750 / 357
M = 74.93g
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.
Solution: Given
Mass of ice = 2kg
Temperature change = 100 – 0 = 100 °C
Specific heat capacity of water = 4200 J/kgK
Specific latent heat of ice = 336000 J/kg
Heat energy gained by ice = 2 x 336000 = 672000 J
Let the amount of water poured be M kg
Heat energy lost water = M x 4200 x 100
= 420000M Joule
According to principle of mixtures,
Heat gained = Heat lost
672000 = 420000M
M = 672000/420000
M =1.6 kg
Amount of water poured is 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 fusion of ice = 336 J g-1.
Solution: Given
Mass of ice = 100g
Heat energy gained by ice to raise the temperature from -10 °C to 0 °C= 100 x 2.1 x 10 = 2100 J
Heat energy gained by ice to convert water at 0 °C = 100 x 336 = 33600 J
Heat energy gained by water to raise the temperature from 0 °C to 100 °C= 100 x4.2 x 100 = 42000 J
Total heat energy gained = 2100 + 33600 + 42000 = 77700 J
Question: 14
The amount of heat energy required to convert 1 kg of ice at -10 °C completely into water at 100 °C is 777000 J. calculate the specific latent heat of ice. Specific heat capacity of ice = 2100 J kg-1 K-1, Specific heat capacity of water is 4200 J kg-1 K-1.
Solution: Given
Mass of ice = 1 kg
Total energy = 777000 J
Heat energy gained by ice to raise temperature from -10°C to 0°C = 1 x 2100 x 10 = 21000 J
Let heat energy gained by ice to convert into water at 0 °C be L.
Heat energy gained by water to raise the temperature from 0 °C to 100 °C= 1 x 4200 x 100 = 420000 J
Total energy = 21000 + 420000 + L
777000 = 441000 + L
L = 336000 J/kg
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.
Solution: Given
Mass of ice, m1=200 g
Mass of water, m2 = 200 g
Change in temperature ∆t= 20 °C
Time taken = 1 min = 60 s
As the rate of exchange is constant, hence power required is the same.
Power to convert ice = power to raise temperature of water
Energy gained by ice/time = energy gained by water/ time
m1 x specific latent heat of ice / time = m2 x specific heat capacity of water x ∆t/ time
200 x 336 /60 = 200 x 4.2 x 20/time
1120 = 16800/ time
Time = 16800/1120
Time = 15 s
Time taken by 200g of water at 0 °C to change to 20°C is 15s.