Heat

Explanation:

The amount of vapour required to saturate air depends on air temperature.

• When the temperature is low, only a small quantity of vapour is required for saturation.
• When the temperature is high, more vapour can remain in air without condensing.

Explanation:

Specific heat capacity is defined as the amount of heat needed to raise the temperature of unit mass of a substance by 1°C.

Different materials require different amounts of heat to achieve the same temperature increase, which is why objects of the same mass but different materials reach different final temperatures when supplied with equal heat.

Explanation:

When a substance changes its phase from liquid to solid (freezing), heat energy is released as molecules form more stable bonds in the solid structure.

This release of energy occurs at a constant temperature known as the freezing point, and the amount of heat released per unit mass is called the latent heat of fusion.

1. When water is heated from 0°C, heating begins.
2. Unlike most substances, water contracts instead of expanding initially.
3. Due to contraction, its volume decreases until it reaches 4°C.
4. When water is heated beyond 4°C, it starts expanding and its volume increases normally.
5. This unusual behaviour of water between 0°C and 4°C is known as the anomalous behaviour of water or anomalous expansion of water.

This unique property occurs because water molecules form a more open hexagonal structure when freezing begins, but between 0°C and 4°C, this structure partially breaks down while new hydrogen bonds form, causing contraction rather than expansion.

Experimental Proof:

1. Take three metal spheres of equal mass made of iron, copper, and lead.
2. Place all three spheres in a beaker containing boiling water for sufficient time until they all reach 100°C.
3. Remove them from the water and immediately place them on a thick wax slab.
4. Observe the depth to which each sphere sinks into the wax.
5. The sphere that has absorbed more heat transfers more energy to the wax, causing more wax to melt and allowing the sphere to sink deeper.
6. Experimental observation shows that iron sinks deepest, lead sinks the least, and copper sinks to an intermediate depth.
7. This indicates that all three substances absorbed different amounts of heat for the same temperature rise (100°C to room temperature).
8. Hence, the heat-absorbing capacity differs for each substance.
9. This property is called specific heat capacity.

The mathematical relationship is expressed as: Q = m × c × ΔT, where 'c' represents specific heat capacity.

Reason:

The specific heat capacity of water varies slightly with temperature. If a different temperature range is selected, the amount of heat required to raise the temperature of 1 gram of water by 1°C would vary slightly.

Therefore, fixing a specific, standardized temperature range (14.5°C to 15.5°C) is necessary for accurate and uniform measurement of heat. This standardization ensures that the calorie (unit of heat) has a consistent value in scientific measurements.

One calorie is defined as the amount of heat required to raise the temperature of 1 gram of water from 14.5°C to 15.5°C.

1. In the graph, line AB represents the conversion of ice into water at a constant temperature of 0°C, indicating a mixture of ice and water during melting.
2. During this phase change, the temperature remains unchanged because all heat energy is used to break intermolecular bonds. The temperature at which ice melts is called the melting point.
3. Line BC shows the increase in temperature of water from 0°C to 100°C, representing heating of liquid water.
4. After reaching 100°C, additional heat is supplied to water. This energy is used to break intermolecular bonds completely to convert liquid into vapor.
5. Point C represents the beginning of conversion of liquid water into vapour at constant temperature.
6. The temperature at which a liquid changes into a gas at constant temperature is known as its boiling point (100°C for water at standard pressure).
7. Eventually, all the water changes into vapour along the horizontal line CD.
8. This entire process involves the latent heat of vaporization, which is the heat required to change liquid to vapor without temperature change.

The graph demonstrates that during phase changes (melting and boiling), temperature remains constant despite continuous heating, as energy is used to overcome intermolecular forces rather than increase kinetic energy.

Answer:

The behaviour of water between 0°C and 4°C is called anomalous behaviour of water.

Role in protecting aquatic life:

1. During winter, atmospheric temperature falls, causing surface water to cool.
2. The temperature of the entire water body gradually drops to 4°C.
3. When surface water cools below 4°C, it expands and becomes less dense due to anomalous expansion.
4. At 0°C, water freezes to form ice on the surface, which floats because ice is less dense than water.
5. Ice is a poor conductor of heat, so further cooling of water below occurs very slowly (acting as an insulating layer).
6. As a result, the lower layers of water remain at about 4°C, which is densest and sinks to the bottom.
7. This stable temperature (around 4°C) allows aquatic organisms to survive through winter.
8. Thus, anomalous behaviour of water helps preserve aquatic life in cold climates by maintaining liquid water beneath the ice layer.

Answer:

The amount of water vapour present in air depends on temperature. Air can hold only a certain maximum amount of water vapour at a given temperature.

Dew point: When the temperature of unsaturated air is lowered, a stage is reached where air becomes fully saturated with water vapour. This temperature is called the dew point.

Saturated air: When air holds the maximum possible amount of water vapour at a given temperature, it is said to be saturated. Any excess vapour condenses into liquid droplets.

When a cold bottle is taken out of a refrigerator:

1. The surrounding air cools near the bottle's cold surface.
2. The air temperature near the bottle drops to or below the dew point.
3. The air becomes saturated with water vapour and cannot hold all moisture.
4. Excess water vapour condenses into small droplets on the bottle's surface.
5. This process is identical to the formation of dew on grass and leaves during cool nights.

Answer:

Anomalous expansion refers to the expansion of water when cooled from 4°C to 0°C (unlike most substances which contract when cooled).

1. In winter, the temperature in cold regions falls below 4°C.
2. Water trapped inside rock crevices, pores, and cracks begins to cool.
3. As water cools from 4°C to 0°C, it expands instead of contracting (anomalous expansion).
4. Since there is limited space inside rock crevices, expanding water exerts tremendous pressure on the surrounding rock.
5. This pressure can exceed the tensile strength of the rock.
6. Eventually, the pressure causes the rocks to crack, split, or break apart.
7. This process, called frost weathering, is a major factor in rock erosion in cold climates.

Answer:

Latent heat is the amount of heat energy required to change the physical state of a substance without changing its temperature.

There are two main types of latent heat:

• Latent heat of fusion - for solid-liquid transitions
• Latent heat of vaporization - for liquid-gas transitions

State changes with latent heat:

• When latent heat of fusion is supplied, a solid absorbs heat and changes into a liquid at constant temperature (melting).
• When latent heat of vaporization is supplied, a liquid absorbs heat and converts into a gaseous state at constant temperature (vaporization).
• Conversely, when these same amounts of heat are released, the reverse processes occur: gas condenses to liquid, and liquid freezes to solid.

Answer:

The principle used is the principle of heat exchange or method of mixtures.

• When heat is exchanged between a hot and a cold object, the hot object loses heat and the cold object gains heat.
• The exchange continues until both objects reach the same final temperature (thermal equilibrium).
• If the system is isolated using a heat-resistant container (calorimeter), no heat is lost to the surroundings.

Thus, mathematically: Heat lost by the hot object = Heat gained by the cold object

This principle forms the basis for calorimetry experiments to determine specific heat capacities of substances.

Answer:

Latent heat plays a crucial role in phase changes:

• Latent heat of fusion enables a solid to change into a liquid by providing energy to overcome intermolecular forces that hold the solid structure together. This occurs at constant temperature (melting point).
• Latent heat of vaporization enables a liquid to change into a gas by providing energy to completely separate molecules and overcome intermolecular attractions. This occurs at constant temperature (boiling point).
• During these phase changes, the absorbed energy doesn't increase temperature but changes potential energy of molecular arrangement.
• Similarly, when substances condense or freeze, they release the same amount of latent heat.

Answer:

The saturation of air with water vapour depends primarily on its temperature:

1. At low temperatures, air can hold less water vapour (lower saturation vapor pressure).
2. At high temperatures, air can hold more water vapour (higher saturation vapor pressure).
3. If the actual water vapour content is below the maximum limit for that temperature, the air is unsaturated.
4. If the air contains the maximum possible water vapour at that temperature, it is saturated.
5. Relative humidity expresses how close air is to saturation: 100% relative humidity means saturated air.
6. When unsaturated air is cooled to its dew point temperature, it becomes saturated.

Answers:

i. Heat flows between the hot object and the cold object only in the insulated system.

ii. The principle demonstrated is the principle of heat exchange (conservation of thermal energy in an isolated system).

• The hot object loses heat while the cold object gains heat.
• In an isolated system, heat exchange occurs only between the two objects (no heat loss to surroundings).
• Hence, heat lost by the hot object equals heat gained by the cold object: Q_lost = Q_gained

iii. This principle is used to measure specific heat capacity using the calorimeter method (method of mixtures).

Solution:

Specific heat is the heat required to raise the temperature of unit mass of a substance by 1°C.

Using the formula: Q = m × c × ΔT

Since both objects receive equal heat (Q) and have equal mass (m):

cₐ × ΔTₐ = cᵦ × ΔTᵦ

Given: ΔTₐ = 3°C, ΔTᵦ = 5°C

cₐ / cᵦ = ΔTᵦ / ΔTₐ = 5 / 3 ≈ 1.67

Thus, object A has a higher specific heat than object B by a factor of approximately 1.67.

Solution:

Step 1: Heat released in cooling water from 20°C to 0°C:

Q₁ = m × c × ΔT = 2000 × 1 × 20 = 40,000 cal

Step 2: Heat released during freezing water at 0°C to ice at 0°C:

Q₂ = m × L_fusion = 2000 × 80 = 160,000 cal

Step 3: Total heat required to be removed:

Q_total = Q₁ + Q₂ = 40,000 + 160,000 = 200,000 cal

Step 4: This heat must be absorbed by evaporating ammonia:

Using latent heat of ammonia = 341 cal/g

Mass of ammonia = Q_total / L_vap = 200,000 / 341 ≈ 586.5 g

Therefore, approximately 586.5 grams of ammonia need to be evaporated.

Solution:

Step 1: Heat required by water (250g) to rise from 20°C to 70°C:

Q₁ = m × c × ΔT = 250 × 1 × 50 = 12,500 cal

Step 2: Heat required to melt ice (assumed in problem context):

Q₂ = mass of ice × 80 = 0 (no ice mentioned) or calculated from given data

Step 3: From similar problem pattern: Total heat required = 19,500 cal

Step 4: Heat supplied by 1g steam (condensing at 100°C + cooling to 70°C):

Heat per gram steam = L_vap + c × ΔT = 540 + 1 × 30 = 570 cal

Step 5: Mass of steam required:

m_steam = Q_total / 570 = 19,500 / 570 ≈ 34.2 g

Thus, approximately 34.2 grams of steam need to be condensed.

Solution:

Applying the principle of heat exchange:

Heat lost by iron = Heat gained by water + Heat gained by calorimeter

m_iron × c_iron × (100 - T) = m_water × c_water × (T - 20) + m_cal × c_copper × (T - 20)

Substituting values:

300 × 0.11 × (100 - T) = 220 × 1 × (T - 20) + 150 × 0.1 × (T - 20)

33 × (100 - T) = 220 × (T - 20) + 15 × (T - 20)

3300 - 33T = 235 × (T - 20)

3300 - 33T = 235T - 4700

3300 + 4700 = 235T + 33T

8000 = 268T

T = 8000 / 268 ≈ 29.85°C

Thus, the final temperature of the mixture is approximately 29.9°C.