Std 10th Science Part 1 – Chapter 5 : Heat Exercise Questions And Answers Maharashtra Board

1. Fill in the blanks and rewrite the sentence.

a. 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 specific heat capacity.

c. During transformation of liquid phase to solid phase, the latent heat is given off/released.

Q2. Observe the following graph. Considering the change in volume of water as its temperature is raised from 0^oC, discuss the difference in the behaviour of water and other substances. What is this behaviour of water called?

Answer – 

The graph shows the relationship between the volume and temperature of a fixed mass of water. The specific volume (volume per unit mass) of water changes with temperature.

• Difference in Behaviour:
  • Normal Substances: When a liquid is heated, it generally expands (volume increases).
  • Water (Anomalous Behaviour): When water is heated from 0^oC up to 4^oC, it contracts (volume decreases) instead of expanding. It reaches its minimum volume at
    4^oC. If heated further above 4^oC, it begins to expand normally.
• Name of Behaviour: This exceptional behaviour of water between 0^oC and 4^oC is called the anomalous behaviour of water.

Q3. What is meant by specific heat capacity? How will you prove experimentally that different substances have different specific heat capacities?

Answer – 

Specific Heat Capacity (c): It is the amount of heat energy required to raise the temperature of a unit mass of an object by 1^oC. The SI unit is J/^oCkg.

Experimental Proof:

1. Take three solid spheres of different metals (e.g., iron, lead, and copper) with equal mass.
2. Heat all three spheres in boiling water in a beaker for some time so they all reach 100^oC.
3. Immediately place the spheres on a thick slab of wax.
4. Observe the depth to which each sphere goes into the wax.

Observation & Conclusion: The sphere that releases more heat energy will melt more wax and sink deeper. The iron sphere goes deepest, and the lead sphere goes the least deep. This shows that even though all three spheres were heated to the same temperature, they had absorbed (and therefore released)different amounts of heat. This difference is due to their distinct specific heat capacities.

Q4. While deciding the unit for heat, which temperatures interval is chosen? Why?

Answer – 

Temperature Interval Chosen: The amount of heat necessary to raise the temperature of 1 g of water by 1∘C from 14.5^oC to 15.5^oC is chosen as one calorie (cal).
Reason: The amount of heat required to raise the temperature of 1 kg of water by 1^oC is slightly different depending on the specific temperature range chosen. Therefore, a specific temperature range (14.5^oC to 15.5^oC) is chosen to define the unit of heat accurately.

Q5. Explain the following temperature versus time graph.

Answer – 

The graph shows the temperature change over time when ice is continuously heated.

1. Line Segment AB: This segment represents the conversion of ice (solid) into water (liquid) at a constant temperature of 0^oC. This temperature is the melting point of ice. During this time, the heat energy supplied is absorbed by the ice to break the bonds between atoms/molecules to change its state. This absorbed energy is the latent heat of fusion, and the temperature remains constant during this phase change.
2. Line Segment BC: This segment represents the heating of water (liquid phase) from 0^oC to 100^oC. The heat energy supplied during this time increases the temperature of the water.

Line Segment CD: This segment represents the conversion of water (liquid) into steam (gaseous) at a constant temperature of 100^oC. This temperature is the boiling point of water. The heat energy supplied is absorbed to break the remaining bonds between liquid molecules to change it into a gaseous state. This absorbed energy is the latent heat of vaporization, and the temperature remains constant until all the water is converted to steam.

6. Explain the following:

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

Answer –

The anomalous behaviour of water helps aquatic life survive in cold regions.

• In cold climates, as the surface water of a lake cools, its temperature decreases.
• Water is densest at 4^oC. As the surface temperature drops from 4^oC to 0^oC, water expands and becomes less dense.
• The densest water at 4^oC sinks to the bottom of the lake, while the colder, lighter water (below 4^oC) stays near the surface.
• When the surface layer freezes, the ice layer stays on top because it is less dense than the water below it. Since ice is a bad conductor of heat (implied from common knowledge, but not cited), it acts as an insulating layer.

This insulation prevents the water at the bottom from freezing, keeping it around 4^oC, which allows aquatic life to survive beneath the ice.

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

Answer – 

Both phenomena occur due to the cooling of air down to its dew point temperature.

• Common Principle: The air contains water vapour (moisture/humidity). When unsaturated air cools, a temperature is eventually reached where the air becomes saturated with vapour. This temperature is the dew point temperature. When the temperature drops below this point, the excess water vapour converts into tiny water droplets (condenses).
• Water Bottle: The outer surface of the cold bottle cools the adjacent air. This cooling forces the water vapour in the air to condense, forming water droplets on the surface.
• Dew Formation: Similarly, in the early morning, the leaves of plants, grass, or vehicle windows cool down. This cools the adjacent air, causing the water vapour in the air to condense on these surfaces, forming dew (water droplets).

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

Answer – 

• Water seeps into the cracks and crevices of rocks.
• As the temperature drops below 4^oC and down to 0^oC, the water starts to exhibit its anomalous behaviour—it expands instead of contracting.
• When the water freezes into ice at 0^oC, it expands further (implied, as ice is less dense than water).

This expansion exerts a huge pressure on the rock walls, causing the rock to gradually crack and break.

7. Answer the following:

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

Answer – 

Latent Heat (L): It is the heat energy that is absorbed (or released) at a constant temperature during a phase transformation (change of state), such as from solid to liquid (fusion) or liquid to gas (vaporization). The energy is used to change the state rather than raise the temperature.

Transformation when given off:

• When latent heat is given off (released) from a gas (vapour) at its boiling point, it transforms into the liquid phase (condensation).
• When latent heat is given off (released) from a liquid at its freezing point, it transforms into the solid phase (freezing/solidification).

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

Answer – 

The principle used is the Principle of Heat Exchange.

• Statement: If a system consisting of a hot object and a cold object is isolated from the environment (no heat is exchanged with the outside), the heat energy lost by the hot object equals the heat energy gained by the cold object.
  • Heat lost by hot object = Heat gained by cold object.

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

Answer – 

The latent heat is the energy that breaks or weakens the bonds between the atoms or molecules of a substance, allowing the state of matter to change (phase transition).

• Solid to Liquid/Liquid to Gas: Latent heat is absorbed to overcome the intermolecular forces and change the state.
• Gas to Liquid/Liquid to Solid: Latent heat is released as the new bonds form to hold the particles closer together.
• In both cases, the latent heat is responsible for the phase transition occurring at a constant temperature.

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

Answer – 

The determination is based on Relative Humidity.

• Basis: The basis is the ratio of the actual amount of water vapour in the air to the maximum amount the air can hold at that temperature (the saturation point).
• How to Determine:
  1. The air is said to be saturated with vapour when it contains the maximum possible amount of water vapour for that temperature.
  2. The maximum relative humidity is 100%, which occurs at the dew point temperature.
  3. The air is unsaturated if the vapour content is less than the maximum limit.
  4. If the relative humidity is greater than 60%, the air is felt as humid; if it is less than 60%, the air is felt as dry.

Q8. Read the following paragraph and answer the questions.

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.

i. Heat is transferred from where to where? 

Answer – Heat is transferred from the hot object to the cold object.

ii. Which principle do we learn about from this process? 

Answer – The Principle of Heat Exchange.

iii. How will you state the principle briefly? 

Answer – Heat energy lost by the hot object = Heat energy gained by the cold object.

iv. Which property of the substance is measured using this principle? 

Answer – The specific heat capacity of a substance is measured using this principle (mixing method).

Q9. Solve the following problems:

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

Given:

• Mass m_A​=m_B​=1 g
• Heat supplied Q_A​=Q_B​=Q
• Temperature rise ΔT_A​=3^oC
• Temperature rise ΔT_B​=5^oC

Formula: Q = m × c × ΔT

Since Q_A ​= Q_B​ and m_{A ​}= m_B ​: m_A​ × c_A​ × ΔT_A​ = m_B​ × c_B​ × ΔT_B​  c_A​ × 3 =c_B ​× 5 c_B​ c_A​​ = 35​

Answer:

• Object A has more specific heat (c_A​ > c_B​).
• The factor is 5 / 3 ​.

b. Liquid ammonia is used in ice factory for making ice from water. If water at 20^oC is to be converted into 2 kg ice at 0^oC, how many grams of ammonia are to be evaporated? 

(Given: The latent heat of vaporization of ammonia = 341 cal/g, specific heat of water c = 1 cal/g^oC, latent heat of fusion of ice = 80 cal/g)

Answer – 

Step 1: Calculate the total heat to be removed from water (Q_lost​) to form ice.

Q_lost​= Q_1​ (to cool water from 20^oC to 0^oC) +Q₂​ (to convert water at 0^oC to ice at 0^oC)

• Mass of water m_w ​= 2 kg = 2000 g
• Specific heat of water
  c_w​ = 1 cal/g^oC
• Latent heat of fusion of ice
  L_fusion ​= 80 cal/g

Q₁​ = m_w​× c_w​× ΔT = 2000 × 1 × (20 − 0) = 40000 cal 

Q₂​= m_w​×L_fusion​ = 2000 × 80 = 160000 cal 

Q_lost​ = 40000 + 160000 = 200000 cal

Step 2: Calculate the mass of ammonia (m_A​) needed.

Heat lost by water = Heat gained by ammonia for evaporation. Q_{lost ​}= m_A​ × L_vap, A​

• Latent heat of vaporization of ammonia
  L_vap,A​ = 341 cal/g

200000 = m_A​× 341 

m_A​= 200000​ / 341 ≈ 586.5 g

Answer :  Approximately 586.4 g of ammonia are to be evaporated.

c. A thermally insulated pot has 150 g ice at temperature 0^oC. How much steam of 100^oC has to be mixed to it, so that water of temperature 50^oC 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^oC)

Given:

• Mass of ice m_ice​ = 150 g
• Initial temperature of ice T_ice ​= 0^oC
• Initial temperature of steam T_steam​ = 100^oC
• Final temperature of mixture T_final​ = 50^oC
• L_fusion ​= 80 cal/g
• L_vap​ = 540 cal/g
• c_w​ = 1 cal/g^oC
• Let m_s​ be the mass of steam required.

Principle of Heat Exchange: Heat lost by steam = Heat gained by ice.

Heat Gained by Ice (Q_gained​) : Q_gained​=Q_1​ (to melt ice) +Q₂​ (to heat water from 0^oC to 50^oC)

Q₁ ​= m_ice​× L_fusion​ = 150 × 80 = 12000 cal 

Q₂ ​= m_{ice ​}× c_w​ × ΔT = 150 × 1 × (50−0) = 7500 cal 

Q_gained​ = 12000 + 7500 = 19500 cal

Heat Lost by Steam (Q_lost​) : Q_lost​= Q₃​ (to condense steam) + Q₄​ (to cool water from 100^oC to 50^oC)

Q₃​ = m_s​ × L_{vap ​}= m_s​ × 540 cal 

Q_4​ = m_s​× c_w​ × ΔT = m_s​ × 1 × (100−50) = m_s​ × 50 cal 

Q_lost​= m_s​ (540+50) = m_s​ × 590 cal

Equating Heat Gained and Lost: Q_lost​ = Q_gained​ m_s ​× 590 = 19500 m_s​ = 19500 / 590​ ≈ 33.05 gAnswer: Approximately 33 g of steam has to be mixed.

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

(Given : latent heat of melting of ice L_fusion​= 80 cal/g)

Given (Convert all units to g,cal,^oC):

• Mass of calorimeter m_c​=100 g
• Specific heat of calorimeter
  c_c ​= 0.1 kcal/kg^oC = 0.1×1000 cal​ / 1000 g^oC = 0.1 cal/g^oC
• Mass of liquid m_l​=250 g
• Specific heat of liquid
  c_l​ = 0.4 kcal/kg^oC = 0.4 cal/g^oC
• Initial temperature of liquid and calorimeter T_h​ = 30^oC
• Mass of ice m_ice​ = 10 g
• Initial temperature of ice T_ice​ = 0^oC
• L_fusion​ = 80 cal/g
• Specific heat of water c_w​ = 1 cal/g^oC (Implied/Assumed from context/Problem c)
• Let T be the final temperature of the mixture.

Principle of Heat Exchange: Heat lost by liquid & calorimeter = Heat gained by ice/melted water.

Heat Lost by Liquid & Calorimeter (Q_lost​): 

Q_lost​= m_c​c_c​ΔT_c​+ m_l​c_l​ΔT_l​ 

ΔT_c​=ΔT_l​ = 30−T 

Q_lost​= (100)(0.1)(30−T) + (250)(0.4)(30−T) 

Q_lost​ = 10(30−T) + 100(30−T) = 110(30−T) 

Q_lost​ = 3300 − 110T

Heat Gained by Ice/Melted Water (Q_gained​) : 

Q_gained​=Q₁​ (to melt ice) +Q₂​ (to heat 10 g water from 0^oC to T)

Q_1​=m_ice​× L_fusion​= 10×80 = 800 cal

Q₂​=m_ice​× c_w​ × ΔT = 10 × 1 × (T−0) = 10T cal 

Q_{gained ​}= 800 + 10T

Equating Heat Gained and Lost : 

3300 − 110T = 800+10T 

3300−800=10T+110T 

2500=120T 

T = 2500​ / 120 ≈ 20.83∘C

Answer: The temperature of the mixture will be approximately 20.8∘C.