# 5 Heat Exercise Solutions Science 1 – Maharashtra board

5 Heat Exercise, Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5 Heat Exercise, 5 Heat Class 10 question and answers, 5 Heat class 10 important questions and answers PDF

# 5 Heat Exercise Solutions Science 1 – Maharashtra board

## Question No. 1]

### Fill in the blanks and rewrite the sentences:

a. The amount of water vapour in air is determined in terms of its………..

The amount of water vapour in air is determined in terms of its absolute humidity.

b. If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their………….

If objects of equal masses are given equal heat, their final temperature will be different. This is due to difference in their specific heat capacities.

c. When a liquid is getting converted into solid, the latent heat is……….

When a liquid is getting converted into solid, the latent heat is released.

Heat Exercise

## Question No. 2]

### Observe the following graph. Considering the change in volume of water as its temperature is raised from 0 °C, discuss the difference in the behaviour of water and other substances. What is this behaviour of water called?

If the temperature of water is raised from 0 °C to 10 °C, its volume goes on decreasing in the range 0 °C to 4 °C. It is minimum at 4 °C. The volume of water goes on increasing in the range 4 °C to 10 °C.

In general, when a substance is heated, its volume goes on increasing with temperature. Thus, in the range 0 °C to 4 °C, behaviour of water is different from other substances. It is called anomalous behaviour of water.

Maharashtra Board Class 10 Science Solutions Part 1 Chapter 5

## Question No. 3]

### What is meant by specific heat capacity? How will you prove experimentally that different substances have different specific heat capacities?

The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C is called the specific heat capacity of that object.

5 Heat Class 10 question and answers

## Question No. 4]

### While deciding the unit for heat, which temperatures interval is chosen? why?

While deciding the unit for heat, the temperature interval chosen is 14.5 °C to 15.5 °C. For the reason, see the information given in the following box.

Heat Exercise

## Question No. 5]

### Explain the following temperature vs time graph:

The graph shows what happens when a mixture of ice and water is heated continuously. The temperature of the mixture remains constant (0 °C) till all the ice melts as shown by the line AB. This temperature is the melting point of ice. On further heating, the temperature rises steadily from 0 °C to 100 °C as shown by the line BC, At 100 °C water starts converting into steam. This temperature is the boiling point of water. Further heating does not change the temperature and the conversion waters steam continues as shown by the line CD.

5 Heat Class 10 question and answers

## Question No. 6]

### Explain the following:

a. the role of anomalous behaviour of water in preserving aquatic life in regions of cold climate?

In cold regions, during winter, the temperature of the atmosphere falls well below 0 °C. As the temperature decreases, the water at the surfaces of lakes and ponds starts contracting. Hence, its density increases and it sinks to the bottom. This process continues till the temperature of all the water in a lake falls to 4 °C. As the water at the surface cools further, i.e., its temperature falls below 4 °C, it starts expanding instead of contracting. Therefore, its density decreases and it remains at the surface. The temperature of the water at the surface continues to fall to 0 °C. Finally, the water at the surface is converted into ice, but the water below the layer of ice is at 4 °C. Ice is a bad conductor of heat. Hence, the layer of the ice at the surface does not allow transfer of heat from the water to the atmosphere. As the water below the layer of ice remains at 4 °C, fish and other aquatic animals and plants can survive in it.

b. How can you relate the formation of water droplets on the outer surface of a bottle taken out of a refrigerator with formation of dew?

At a given temperature, there is a limit on how much water vapour the given volume of air can hold. The lower the temperature, the lower is the capacity of air to hold water vapour.

The temperature of a bottle kept in a refrigerator is lower than room temperature. Hence, when the bottle is taken out of the refrigerator, the temperature of the air surrounding the bottle is lowered. Therefore, the capacity of the air to hold water vapour becomes less. Hence, the excess water vapour condenses to form water droplets (like dew) on the outer surface of the bottle.

c. In cold regions in winter, the rocks crack due to anomalous expansion of water.

Sometimes water enters into crevices of the rocks. When the temperature of the atmosphere falls below 4 °C, water expands. Even when water freezes to form ice, there is increase in its volume. As there is no room for expansion, it exerts a tremendous pressure on the rocks which crack and break up into small pieces.

Heat Exercise

## Question No. 7]

a. What is meant by latent heat? How will the state of matter transform if latent heat is given off?

When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.

In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).

(Note: change of state = change of phase)

b. Which principle is used to measure the specific heat capacity of a substance?

The principle of heat exchange is used to measure the specific heat capacity of a substance. This principle is as follows: If a system of two objects is isolated from the environment by keeping it inside a heat resistant box, then no energy can leave the box or enter the box. In this situation, heat energy lost by the hot object = heat energy gained by the cold object.

c. Explain the role of latent heat in the change of state of a substance?

When a solid is heated, initially, its temperature increases. Here, the heat absorbed by the body (substance) is used in increasing the kinetic energy of the particles (atomic, molecules, etc.) of the body as well as for doing work against the forces of attraction between them. As the heating is continued, at a certain temperature (melting point), solid is converted into liquid. In this case, the temperature remains constant and the heat absorbed is used for weakening the bonds and conversion into liquid phase (liquid state). This heat is called the latent heat of fusion.

When a liquid is converted into the gaseous phase (gaseous state), at the boiling point, the heat absorbed is used for breaking the bonds between the atoms or molecules. This heat is called the latent heat of vaporization. Some solids, under certain conditions, are directly transformed into the gaseous phase. Here the heat is absorbed but the temperature remains constant. The absorbed heat is used for breaking the bonds between atoms or molecules. This heat is called the latent heat of sublimation.

In general, latent heat is the heat absorbed or given out by a substance during a change of state at constant temperature.

In transformations from liquid to solid, gas to liquid and gas to solid, latent heat is given out by the body (substance).

(Note: change of state = change of phase)

d. what basis and how will you determine whether air is saturated with vapour or not?

Whether the air is saturated with water vapour or not is determined on the basis of the extent of water vapour present in the air. If the relative humidity is 100%, the air is saturated with water vapour. In that case, we can see the formation of water droplets on the leaves of plants/grass.

If the relative humidity is less than 100%, the air is not saturated with water vapour.

5 Heat Class 10 question and answers

## Question No. 8]

If heat is exchanged between a hot and cold object, the temperature of the cold object goes on increasing due to gain of energy and the temperature of the hot object goes on decreasing due to loss of energy.

The change in temperature continues till the temperatures of both the objects attain the same value. In this process, the cold object gains heat energy and the hot object loses’ heat energy. If the system of both the objects is isolated from the environment by keeping it inside a heat resistant box (meaning that the energy exchange takes place between the two objects only), then no energy can flow from inside the box or come into the box.

(1) Heat is transferred from where to where?

(2) Which principle do we learn about from this process?

(3) How will you state the principle briefly?

(4) Which property of the substance is measured using this principle?

(1) Heat is transferred from a hot object to a cold object.

(2) This process shows the principle of heat exchange.

(3) In this process, the cold object gains heat energy and the hot object loses energy. If a system of two objects is isolated from the surroundings, heat energy lost by the hot object = heat energy gained by the cold object.

(4) This principle is used to measure the specific heat capacity of a substance.

Heat Exercise

## Question No. 9]

### Solve the following problems:

a. Equal heat is given to two objects A and B of mass 1 g. The temperature of A increases by 3 °C and B by 5°C. Which object has more specific heat? And by what factor?

Solution:

Data: m = 1 g, Δ T1 = 3 °C, Δ T2 = 5 °C,

Q same

Here, Q = mc1 ΔT1 = mc2 ΔT2 Thus, c1 > c2

The specific heat of A is more than that of B and b. Liquid ammonia is used in ice factory for making ice from water. If water at 20 °C is to be converted into 2 kg, ice at 0 °C, how many grams of ammonia is to be evaporated?

(Given: The latent heat of vaporization of 1 ammonia = 341 cal/g)

Solution:

Data : m1 = 2kg, ΔT1=20 °C – 0 °C

= 20 °C, c1 = 1 kcal/kg·°C, L1 (ice) = 80 kcal/kg,

L2 (vaporization of ammonia) = 341 cal/g = 341 kcal/kg, m2 =?

Q1 (heat lost by water) = m1c1 ΔT1 + m1L1

= 2kg × 1 kcal/kg·°C × 20 °C + 2 kg × 80 kcal/kg

=40 kcal + 160 kcal = 200 kcal

Q2 (heat absorbed by ammonia) = m2L2

= m2 × 34l kcal/kg

According to the principle of heat exchange, Q1 = Q2

∴ 200 kcal = m2 × 341 kcal/kg

∴ m2 = 200/341 kg = 0.5864 kg = 586.4 g

586.4 g of ammonia are to be evaporated.

c. A thermally insulated pot has 150 g ice at temperature 0 °C. How much steam of 100 °C has to he mixed to it, so that water of temperature 50 °C will be obtained?

(Given: Latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g °C)

Solution:

Data: m1 = 150 g, ΔT1 = 50 °C – 0 °C

= 50 °C, cw = 1 cal/g.°C, L1 = 80 cal/g, L2 = 540 cal/g,

Δ T2 = 100°C – 50 °C = 50 °C, m2 = ?

Q1 (heat absorbed by ice) = m1L1

= 150 g × 80 cal/g = 12000 cal

Q2 (heat absorbed by water formed on melting of ice) =m1 cw ΔT1

= 150 g × 1 cal/g·°C × 50 °C = 7500 cal

Q3 (heat given out by steam) = m2L2

= m2 × 540 cal/g

Q4 (heat given out by water formed on condensation of steam)

= m2 cw ΔT2 = m2 × 1 cal/g·°C × 50 °C

According to the principle of heat exchange,

Q1 + Q2 = Q3 + Q4

∴ 12000 cal + 7500 cal = m2 × 540 cal/g + m2 × 50 cal/g

∴ 19500 cal = m2 (540 + 50) cal/g

∴ m2 = 19500/590 g

33.5 g of steam is to be mixed.

d. A calorimeter has mass 100 g and specific heat 0.1 kcal/kg ·°C. It contains 250 g of liquid at 30 °C having specific heat of 0.4 kcal/kg·°C. If we drop a piece of ice of mass 10 g at 0 °C into the liquid, what will be the temperature of the mixture?

Solution:

Data: m1 = 100 g, c1 = 0.1 kcal/kg·°C,

= 0.1 cal/g·°C, T1 = 30 °C, m2 = 250 g,

c2 = 0.4 kcal/kg·°C = 0.4 cal/g·°C, T2 = 30 °C,

m3 = 10 g, T3 = 0 °C, L = 80 cal/g,

c (water) = 1 cal/g·°C, T = ?

Q1 (heat lost by calorimeter) = m1c1 (T- T1),

Q2 (heat lost by liquid) = m2c2 (T – T2),

Q3 (heat absorbed by ice) = m3 L,

Q4 (heat absorbed by water formed on melting of ice) = m3c (T – 0 °C)

According to the principle of heat exchange,

Q1 + Q2 = Q3 + Q4

∴ m1c1 (T1 – T) + m2c2 (T2 – T) = m3L + m3c (T – 0 °C)

∴ m1c1T1 – m1c1T + m2c2T2 – m2c2T = m3L + m3c (T – 0°C)

∴ m1c1T1 + m2c2T2 = m3L + (m1c1 + m2c2 + m3c)T

∴ 100g × 0.1 cal/g°C × 30 °C + 250g × 0.4 cal/g.°C × 30 °C J

= 10 g x× 80 cal/g + (100 g × 0.1 cal/g.°C + 250 g × 0.4 cal/g.°C + 10 g × 1 cal/g.°C) T

∴ (10 + 100 + 10) T = (300 + 3000 – 800)°C

∴ 120 T = 2500 °C

∴ T = 2500120 °C = 1256 °C = 20.83 °C

This is the temperature of the mixture.

6 Refraction of Light Exercise Solutions Science 1 – Maharashtra board 