# SELINA Solutions for Class 10 Physics Chapter 11 - Calorimetry

## Chapter 11 - Calorimetry Exercise Ex. 11A

Define the term heat.

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

The S.I. unit of heat capacity is :

- J kg
^{-1} - J K
^{-1} - J kg
^{-1}K^{-1} - cal
^{0}C^{-1}

_{}

By imparting heat to a body,
its temperature rises by 15^{0}C. What is the corresponding rise in
temperature on the Kelvin scale?

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.

Name the S.I. unit of heat.

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

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

- J kg
^{-1} - J K
^{-1} - J kg
^{-1}K^{-1} - kilocal kg
^{-1}^{0}C^{-1}

J kg^{-1} K^{-1}

(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^{o}C to 35^{o}C?

(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^{-1}K^{-1}

=61.5JK^{-1}

Change in temperature= (35-25)^{o}C=10^{o}C=10K

(ii) Energy required to increase the temperature of vessel

=0.15 X 410 X 10

=615 J

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

One calorie of heat is the heat energy
required to raise the temperature of 1 g of water from 14.5^{o}C to
15.5^{ o}C.

1 calorie = 4.186 J

The specific heat capacity of water is :

- 4200 J kg
^{-1}K^{-1} - 420 J g
^{-1}K^{-1} - 0.42 J g
^{-1}K^{-1} - 4.2 J kg
^{-1}K^{-1}

4200 J kg^{-1} K^{-1}

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 15^{o}C, and

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

(i)We know that heat energy needed to raise the temperature by 15^{o} 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}.

Define one kilo-calorie of heat.

One
kilo-calorie of heat is the heat energy required to raise the temperature of
1 kg of water from 14.5^{o}C to 15.5^{o}C.

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

Mass of copper m = 100 g = 0.1 kg

Change of temperature t = (70-20)^{o}C

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 =

= 0.1 x 50 x 390

= 1950 J

Define temperature and name the S.I. unit.

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

S.I. unit kelvin (K).

1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20^{o}C to 40^{o}C. Calculate the specific heat capacity of lead.

Heat energy supplied = 1300 J

Mass of lead = 0.5 kg

Change in temperature = (40-20)^{o}C = 20^{ o}C (or 20 K)

Specific heat capacity of lead

c = 130 J kg^{-1 }K^{-1}

State three differences between heat and temperature.

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. |

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^{o}C to 38^{o}C. Assume that all the heat energy supplied by the heater is given to the material.

Specific heat capacity of material c =960 J kg^{-1 }K^{-1}

Change in temperature T=(38-18)^{o}C = 20^{o}C (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

Define calorimetry.

The measurement of the quantity of heat is called calorimetry.

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

(i)the specific heat capacity of the liquid.

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

Power of heater P= 600 W

Mass of liquid m=4.0 kg

Change in temperature of liquid = (15-10)^{o}C = 5^{o}C(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 10^{4} J/K

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

The heat capacity of a body is the amount
of heat energy required to raise its temperature by 1^{o}C or 1K.

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

0.5 kg of lemon squash at 30^{o}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^{o}C? Specific heat capacity of squash = 4200 J kg^{-1 }K^{-1}.

Change in temperature= 30 - 5 = 25 K.

=

t=29 min 10 sec.

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

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 1^{o}C (or 1K).

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

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}.

Heat liberated by metal=

Heat absorbed by water=

Heat energy lost= heat energy gained

=

S=0.52 J g^{-1}
K^{-1}.

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

Heat capacity = Mass x specific heat capacity

45 g of water at 50^{o}C in a beaker is cooled when 50 g of copper at 18^{o}C 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}.

Mass of water (m_{1})=45 g

Temperature of water (T_{1}) =50^{o}C

Mass of copper (m_{2}) =50 g

temperature of copper(T_{2}) =18^{o}C

Final temperature (T) =?

The specific heat capacity of the copper c_{2} = 0.39 J/g/K

The specific heat capacity of water c_{1} = 4.2 J/g/K

^{o}C

T= 47^{o}C.

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

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_{}.

200g of hot water at 80^{o}C is added to 300 g of cold water at 10^{o}C. 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}.

Mass of hot water (m_{1}) = 200g

Temperature of hot water (T_{1}) = 80^{o}C

Mass of cold water (m_{2}) = 300g

Temperature of cold water (T_{2}) =10^{o}C

Final temperature (T) =?

c_{1}=c_{2}

T=38^{o}C.

Name a liquid which has the highest specific heat capacity.

Water has the highest specific heat capacity.

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

Mass of hot water (m_{1}) = 300 g

Temperature (T_{1}) = 50^{o}C

Mass of cold water (m_{2}) = 600 g

Change in temperature of cold water (T-T_{2}) =15^{o}C

Final temperature =T^{o}C

The specific heat capacity of water is c.

T = 20^{o}C.

Final temperature = 20^{o} c

Change in temperature = 15^{o}C

Initial temperature of cold water = 20^{o}C -15^{o}C = 5^{o}C.

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

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

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}.

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^{-1}K^{-1}?

(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^{-1}K^{-1} means to increase the temperature of one
gram of copper by 1K we have to supply 0.4 joules of energy.

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?

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.

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

Change in temperature and the nature of material

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

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.

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^{o}C.

The expression for the heat energy Q

Q= mct (in joule)

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?

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

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

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.

What is the principle of method of mixture? What other name is given to it? Name the law on which this principle is based.

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.

A mass m_{1} of a substance of specific heat capacity c_{1} at temperature T_{1} is mixed with a mass m_{2} of other substance of specific heat capacity c_{2} at a lower temperature T_{2}. Deduce the expression for the temperature of mixture. State assumption made, if any.

A mass m_{1} of a substance A of specific heat capacity c_{1} at temperature T_{1} is mixed with a mass m_{2} of other substance B of specific heat capacity c_{2} at a lower temperature T_{2} and final temperature of the mixture becomes T.

Fall in temperature of substance A = T_{1 }- T

Rise in temperature of substance B = T – T_{2}

Heat energy lost by A = m_{1} c_{1} fall in temperature

= m_{1}c_{1}(T_{1} - T)

Heat energy gained by B= m_{2 } c_{2} rise in temperature

= m_{2}c_{2}(T – T_{2})

If no energy lost in the surrounding, then by the principle of mixtures,

Heat energy lost by A = Heat energy gained by B

m_{1}c_{1}(T_{1} - T)= m_{2}c_{2}(T – T_{2})

After rearranging this equation, we get

Here we have assumed that there is no loss of heat energy.

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

In the absence of water, if on a cold winter night the atmospheric temperature falls below 0^{o}C, 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 0^{o}C.

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

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.

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

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.

Water property of water makes it an effective coolant?

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.

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

(i)Radiator in car.

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

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?

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.

(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?

(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 minimised.

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

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.

## Chapter 11 - Calorimetry Exercise Ex. 11B

(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?

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

(b)There is no change in temperature during the change of phase.

(c)Yes, the substance absorbs or liberates heat during the change of phase.

(d) Latent heat

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

- cal g
^{-1} - cal g
^{-1}K^{-1} - J kg
^{-1} - J kg
^{-1}K^{-1}

J kg^{-1}

10g
of ice at 0^{o}C absorbs 5460J of heat energy to melt and change to
water at 50^{o}C. Calculate the specific latent heat of fusion of
ice. Specific heat capacity of water is 4200Jkg^{-1}K^{-1}.

Mass of ice=10g = 0.01kg

Amount of heat energy absorbed, Q=5460J

Specific latent heat of fusion of ice=?

Specific
heat capacity of water = 4200Jkg^{-1}K^{-1}

Amount
of heat energy required by 10g (0.01kg) of water at 0^{o}C to raise
its temperature by 50^{o}C= 0.01X4200X50=2100J.

Let
Specific latent heat of fusion of ice=L Jg^{-1}.

Then,

Q = mL + mcT

5460 J =10 x L + 2100J

L=336Jg^{-1}.

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.

a. Melting.

b. Latent heat of melting.

c. As their id no change in temperature the average kinetic energy of the molecules does not change.

The specific latent heat of fusion of water is :

- 80 cal g
^{-1} - 2260 J g
^{-1} - 80 J g
^{-1} - 336 J kg
^{-1}

80cal g^{-1}

How much heat energy is released when 5.0 g of water at 20^{o}C changes into ice at 0^{o}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}.

Mass of water m = 5.0 g

specific heat capacity of water c = 4.2 J g^{-1 }K^{-1}

specific latent heat of fusion of iceL =336 J g^{-1}

Amount of heat energy released when 5.0 g of water at 20^{o}C changes into water at 0^{o}C = 5 x 4.2 x 20 = 420 J.

Amount of heat energy released when 5.0g of water at 0^{o}C changes into ice at 0^{o}C=5x336J=1680J.

Total amount of heat released =1680 J + 420 J = 2100 J.

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 energy change in the molecules of the substance.

(i)Average kinetic energy of molecules changes.

(ii)Average potential energy of molecules changes.

A molten metal of mass 150 g is kept at its melting point 800^{o}C. 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 -50^{o}C?

Mass of metal =150 g

Specific latent heat of metal

Specific heat capacity of metal is 200 J kg^{-1 }K^{-1}.

Change in temperature= 800-(-50) = 850^{o}C (or 850 K).

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?

(a) Average kinetic energy does not change.

(b) Average potential energy increases.

Explanation: When a substance is heated at constant temperature (i.e. during its phase change state), the heat supplied makes the vibrating molecules gain potential energy to overcome the intermolecular force of attraction and move about freely. This means that the substance changes its form.

However, this heat does not increase the kinetic energy of the molecules, and hence, no rise in temperature takes place during the change in phase of a substance.

This heat supplied to the substance is known as latent heat and is utilized in changing the state of matter without any rise in temperature.

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

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

The melting point of ice decreases by the presence of impurity in it.

Use: In making the freezing mixture by adding salt to ice. This freezing mixture is used in preparation of ice creams.

A refrigerator converts 100g of water at 20^{o}C to ice at -10^{o}C in 73.5 min. calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 Jg^{-1}K^{-1}, specific latent heat of ice is 336Jg^{-1} and the specific heat capacity of ice is2.1Jg^{-1}K^{-1}.

Amount of heat released when 100g of water cools from 20^{o} to 0^{o}C =100X20X4.2=8400J.

Amount of heat released when 100g of water converts into ice at 0^{o}C =100X336=33600J.

Amount of heat released when 100g of ice cools from 0^{o}C to -10^{o}C =100X10X2.1=2100J.

Total amount of heat=8400+33600+2100=44100J.

Time taken= 73.5min=4410s.

Average rate of heat extraction (power)

.

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

The melting point of ice decreases by the increase in pressure. The melting point of ice decrease by 0.0072^{o}C for every one atmosphere rise in pressure.

In an experiment, 17 g of ice is used to bring down the temperature of 40 g of water at 34^{o}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.

Mass of ice m_{1} =17 g

Mass of water m_{2} =40 g.

Change in temperature =34-0=34K

Specific heat capacity of water is 4.2Jg^{-1}K^{-1}.

Assuming there is no loss of heat, heat energy gained by ice (latent heat of ice), Q= heat energy released by water

Q = 40 x 34 x 4.2 = 5712 J.

Specific latent heat of ice=.

The diagram shows the change of phases of a substance on temperature time graph.

(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?

(a) AB part shows rise in temperature of solid from 0 to T_{1}^{o}C, BC part shows melting at temperature T_{1}^{o}C, CD part shows rise in temperature of liquid from T_{1}^{o}C to T_{3}^{o}C , DE part shows the boiling at temperature T_{3}^{o}C.

(b) T_{1}^{o}C.

(c) T_{3}^{o}C.

The
temperature of 170g of water at 50^{0}C to be lowered to 5^{o}C
by adding certain amount of ice to it. Find the mass of ice added. Given:
Specific heat capacity of water=4200Jkg^{-1}C^{-1} and
specific latent het of ice =336000JKg^{-1}.

The melting point of naphthalene is 800C and the room temperature is 250C. A sample of liquid naphthalene at 900 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.

The melting point of naphthalene , a crystalline solid is 80 degree C and the room temperature is 25 °C

The temperature - time graph is as follows: -

Find the result of mixing 10g of ice at -10^{o}C with 10g of water at 10^{o}C. Specific heat capacity of ice is 2.1Jg^{-1}K^{-1}, specific latent heat of ice is 336Jg^{-1}, and specific heat capacity of water is 4.2Jg^{-1} K^{-1}.

Let whole of the ice melts and let the final temperature of the mixture be T^{o}C.

Amount of heat energy gained by 10g of ice at -10^{o}C to raise its temperature to 0^{o}C= 10x10x2.1=210J

Amount of heat energy gained by 10g of ice at 0^{o}C to convert into water at 0^{o}C=10x336=3360 J

Amount of heat energy gained by 10g of water (obtained from ice) at 0^{o}C to raise its temperature to T^{o}C = 10x4.2x(T-0)=42T

Amount of heat energy released by 10g of water at 10^{o}C to lower its temperature to T^{o}C = 10x4.2x(10-T)=420-42T

Heat energy gained = Heat energy lost

210 + 3360 + 42T = 420-42T

T = -37.5^{o}C

This cannot be true because water cannot exist at this temperature.

So whole of the ice does not melt. Let m gm of ice melts. The final temperature of the mixture becomes 0^{o}C.

So, amount of heat energy gained by 10g of ice at -10^{o}C to raise its temperature to 0^{o}C= 10x10x2.1=210J

Amount of heat energy gained by m gm of ice at 0^{o}C to convert into water at 0^{o}C=mx336=336m J

Amount of heat energy released by 10g of water at 10^{o}C to lower its temperature to 0^{o}C = 10x4.2x(10-0)=420

Heat energy gained = Heat energy lost

210 + 336m = 420

m = 0.625 gm

1 kg of ice at 0^{o} is heated at constant rate and its temperature is recorded after every 30 s till steam is formed at 100^{o} C. Draw a temperature time graph to represent the change of phase.

A piece of ice of mass 40 g is added to 200 g of water at 50^{o}C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water is 4200 J kg^{-1 }K^{-1}, specific latent heat of fusion of ice =336 x 10^{3} J kg^{-1}.

Let final temperature of water when all the ice has melted =T^{o}C.

Amount of heat lost when 200g of water at 50^{o}C cools to T^{o}C=

200X4.2X(50-T) = 42000-840T

Amount of heat gained when 40g of ice at 0^{o}C converts into water at 0^{o}C.= 40X336J = 13440 J

Amount of heat gained when temperature of 40g of water at 0^{o}C rises to T^{o}C= 40X4.2X(T-0) = 168T

We know that

Amount of heat gained=amount of heat energy lost.

13440+168T= 42000-840T

168T+840T= 42000-13440

1008T= 28560

T=28560/1008=28.33^{o}C.

Explain the terms boiling and boiling point. How is the volume of water affected when it boils at 100^{o}C?

Boiling: The change from liquid to gaseous phase on heating at a constant temperature is called boiling.

Boiling Point: The particular temperature at which vaporization occurs is called the boiling point of liquid.

Volume of water wills increases when it boils at 100^{o}C.

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

Heat loss by (water+Calorimeter) = Heat gain by ice

Heat
loss by (water+calorimeter) = m_{w} Cp_{w} ΔT + m_{C}
Cp_{c} ΔT = mi ( L + Cp_{w} δT ) ......................(1)

where, m_{w} = mass of water = 50 g

C_{pw} = Specific heat of water = 4.2 J/( g °C )

m_{C}
= mass of calorimeter = 50 g

C_{pc}
= Specific heat capacity of calorimeter = 0.4 J/( g °C )

ΔT = fall in temperature of water and calorimeter = 32-5 = 27°C

m_{i} =
mass of ice in gram

L = latent heat capacity of ice = 330 J/g

δT = rise in temperature = 5 °C

by substituting all the values in eqn.(1)we get the mass of ice as

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

The boiling point of water increases on adding salt.

250 g of water at 30^{o }C is contained in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of vessel and its contents to 5^{o }C. specific latent heat of fusion of ice = 336 x 10^{3} 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}.

Mass of copper vessel m_{1 }= 50 g.

Mass of water contained in copper vessel m_{2 }= 250 g.

Mass of ice required to bring down the temperature of vessel = m

Final temperature = 5^{o }C.

Amount of heat gained when 'm' g of ice at 0^{o }C converts into water at 0^{o }C = m × 336 J

Amount of heat gained when temperature of 'm' g of water at 0^{o }C rises to 5^{o }C = m × 4.2 × 5

Total amount of heat gained = m × 336 + m × 4.2 × 5

Amount of heat lost when 250 g of water at 30^{o }C cools to 5^{o }C =

250 × 4.2 x 25 = 26250 J

Amount of heat lost when 50 g of vessel at 30^{o }C cools to 5^{o }C =

50 x 0.4 × 25 = 500 J

Total amount of heat lost = 26250 + 500 = 26750 J

We know that amount of heat gained = amount of heat lost

m × 336 + m × 4.2 × 5 = 26750

357 m = 26750

m = 26750/357 = 74.93 g

Hence, mass of ice required is 74.93 g.

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

The boiling point of a liquid increases on increasing the pressure.

2 kg of ice melts when water at 100^{o}C is poured in a hole drilled in a block of ice. What mass of water was used? Specific heat capacity of water is 4200 J kg^{-1 }K^{-1}, specific latent heat of fusion of ice = 336 J g^{-1}.

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

Amount of heat energy gained by 2 kg of ice at 0^{o}C to convert into water at 0^{o}C=2X336000=672000 J

Let amount of water poured=m kg.

Initial temperature of water =100^{o}C.

Final temperature of water =0^{o}C.

Amount of heat energy lost by m kg of water at 100^{o}C to reach temperature 0^{o}C =mX4200X100 = 420000m J

We know that heat energy gained =heat energy lost.

672000J= mX420000J

m=672000/420000=1.6kg

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

- The boiling point of a liquid increases with the increase in pressure and decreases with the decrease in pressure.
- The boiling point of pure water at one atmospheric pressure (= 760 mm of Hg) is 100 °C.
- In a pressure cooker, the water boils at about 120 °C to 125 °C due to increase in pressure, as the steam is not allowed to escape out of it.

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

Amount of heat energy gained by 100 g of ice at -10^{o }C to raise its temperature to 0^{o }C =

100 × 2.1 × 10 = 2100 J

Amount of heat energy gained by 100 g of ice at 0^{o }C to convert into water at 0^{o }C =

100 × 336 = 33600 J

Amount of heat energy gained when temperature of 100 g of water at 0^{o }C rises to 100^{o }C =

100 × 4.2 × 100 = 42000 J

Total amount of heat energy gained is = 2100 + 33600 + 42000 = 77700 J = 7.77 × 10^{4} J

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

In a pressure cooker, the water boils at about 120^{o}C to 125^{o}C.

The amount of heat energy required to convert 1 kg of ice at -10^{o}C completely into water at 100^{o}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}.

Amount of heat energy gained by 1kg of ice at -10^{o}C to raise its temperature to 0^{o}C= 1 x 2100 x 10 = 21000 J

Amount of heat energy gained by 1kg of ice at 0^{o}C to convert into water at 0^{o}C=L

Amount of heat energy gained when temperature of 1kg of water at 0^{o}C rises to 100^{o}C= 1 x 4200 x 100 = 420000 J

Total amount of heat energy gained = 21000+420000+L=441000 +L.

Given that total amount of heat gained is =777000J.

So,

441000+L=777000.

L=777000-441000.

L=336000JKg^{-1}

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

This is because at high altitudes atmospheric pressure is low; therefore boiling point of water decreases and so it does not provide the required heat energy for cooking.

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}.

Mass of ice, mice = 200 g

Time for ice to melt, t1 = 1 min = 60 s

Mass of water, mw = 200 g

Temperature change of water, ΔT = 20 °C

Rate of heat exchange is constant. So, power required for converting ice to water is same as the power required to increase the temperature of water.

Hence, the time required is 15 seconds.

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

……………….

(e)The boiling point of water is defined as ………………………..

(f) water can be made to boil at 115°C by ................. pressure over its surface.

(a) When ice melts, its volume decreases.

(b) Decrease in pressure over ice increases its meltingpoint.

(c) Increase in pressure increases the boiling point of water.

(d)A pressure cooker is based on the principle that boiling point of water increases with the increase in pressure.

(e) The boiling point of water is defined as the constant temperature at which water changes to steam.

(f) water can be made to boil at 115°C by increasing pressure over its surface.

What do you understand by the term latent heat?

Latent heat: The heat energy exchanged in change of phase is not externally manifested by any rise or fall in temperature, it is considered to be hidden in the substance and is called the latent heat.

Define the term specific latent heat of fusion of ice. State its S.I. unit.

The quantity of heat required to convert unit mass of ice into liquid water at _{} (melting point) is called the specific latent heat of fusion of ice.

Its S.I. unit is Jkg^{-1}.

Write the approximate value of specific latent heat of ice.

Specific latent heat of ice: 336000 J kg^{-1}.

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

It means 1 g of ice at 0^{o}C absorbs 360 J of heat energy to convert into water at 0^{o}C.

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

Latent heat is absorbed by ice.

Which has more heat: 1 g of ice at 0^{o} C or 1 g of water at 0^{o}C? Give reasons.

1 g of water at 0^{o}C has more heat than 1 g of ice at 0^{o}C. This is because ice at 0^{o}C absorbs 360 J of heat energy to convert into water at 0^{o}C.

(a) Which requires more heat: 1 g ice at 0^{o} C or 1 g water at 0^{o}C to raise its temperature to 10^{o}C? (b) Explain your answer in part (a).

(a) 1 g ice at 0^{o}C requires more heat because ice would require additional heat energy equal to latent heat of melting.

(b) 1 g ice at 0^{o}C first absorbs 336 J heat to convert into 1 g water at 0^{o}C.

Ice cream appears colder to the mouth than water at 0^{o}C. Give reasons.

This is because 1 g of ice at 0^{o}C takes 336 J of heat energy from the mouth to melt at 0^{o}C. Thus mouth loses an additional 336 J of heat energy for 1 g of ice at 0^{o}C than for 1g of water at 0^{o}C. Therefore cooling produced by 1 g of ice at 0^{o}C is more than for 1g of water at 0^{o}C.

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.

This is because 1 g of ice at 0^{o}C takes 336 J of heat energy from the bottle to melt into water at 0^{o}C. Thus bottle loses an additional 336 J of heat energy for 1 g of ice at 0^{o}C than for 1 g iced water at 0^{o}C. Therefore bottled soft drinks get cooled, more quickly by the ice cubes than by iced water.

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

The reason is that after the hail storm, the ice absorbs the heat energy required for melting from the surrounding, so the temperature of the surrounding falls further down and we feel colder.

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

The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of the surrounding falls and it became very cold.

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

The specific latent heat of fusion of ice is sufficiently high, about
336 J g^{-1}. Before freezing the water in the lakes and ponds will
have to release a large quantity of heat to the surrounding. If there is any layer of ice formed on water
than water being a poor conductor of heat will also prevent the loss of heat
from water of lake. Hence, in cold countries water in lakes and ponds do not
freeze.

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.

(i) The reason is that the specific latent heat of fusion of ice is sufficiently high, so when the water of lake freezes, a large quantity of heat has to be released and hence the surrounding temperature becomes pleasantly warm.

(ii) Heat supplied to a substance during its change of state, does not cause any rise in its temperature because this is latent heat of phase change which is required to change the phase only.

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