Chapter 5: Heat

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Grade 10 Q&A: Chapter 5: Heat

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Grade 10 Q&A: Chapter 5: Heat

Concept Questions

Q1: What is heat? How is it different from temperature?

Answer: Heat is a form of energy that flows from a body at a higher temperature to a body at a lower temperature. It is the total energy of molecular motion in a substance and is measured in joules (J) or calories (cal).

Temperature, on the other hand, is a measure of the average kinetic energy of the particles in a substance. It indicates how hot or cold a body is and is measured in degrees Celsius (°C), Kelvin (K), or Fahrenheit (°F).

The key differences between heat and temperature are: 1. Heat is the total energy of molecular motion, while temperature is the average kinetic energy of molecules 2. Heat is an extensive property (depends on the amount of substance), while temperature is an intensive property (independent of the amount) 3. Heat is measured in joules or calories, while temperature is measured in degrees 4. Heat flows from higher temperature to lower temperature, not necessarily from a body with more heat to one with less heat

An analogy to understand this difference: If we compare heat to the amount of water in a container, temperature would be like the water level. A large container with a low water level (large body at low temperature) can contain more water (heat) than a small container with a high water level (small body at high temperature).

Q2: Explain the three temperature scales: Celsius, Kelvin, and Fahrenheit. How are they related to each other?

Answer: The three common temperature scales are:

Celsius Scale (°C):
Developed by Anders Celsius
Water freezes at 0°C and boils at 100°C at standard atmospheric pressure
Commonly used in everyday life and scientific work in most countries
The unit is degree Celsius (°C)
Kelvin Scale (K):
Developed by Lord Kelvin
The SI unit of temperature
Absolute zero (the lowest possible temperature) is 0 K, which equals -273.15°C
No negative values on this scale
Water freezes at 273.15 K and boils at 373.15 K at standard atmospheric pressure
The unit is kelvin (K), not "degree kelvin"
Fahrenheit Scale (°F):
Developed by Daniel Gabriel Fahrenheit
Water freezes at 32°F and boils at 212°F at standard atmospheric pressure
Commonly used in the United States for everyday measurements
The unit is degree Fahrenheit (°F)

The relationships between these scales are given by the following conversion formulas:

Celsius to Kelvin: K = °C + 273.15
Kelvin to Celsius: °C = K - 273.15
Celsius to Fahrenheit: °F = (°C × 9/5) + 32
Fahrenheit to Celsius: °C = (°F - 32) × 5/9
Kelvin to Fahrenheit: °F = (K × 9/5) - 459.67
Fahrenheit to Kelvin: K = (°F + 459.67) × 5/9

These scales are used in different contexts: Kelvin is primarily used in scientific work, especially in physics and thermodynamics; Celsius is used in everyday life and scientific work in most countries; and Fahrenheit is mainly used in the United States for weather forecasts, body temperature, and other everyday measurements.

Q3: What is thermal expansion? Give examples of its applications and consequences in daily life.

Answer: Thermal expansion is the tendency of matter to increase in volume or length when heated and decrease when cooled. This occurs because as a substance is heated, its particles gain kinetic energy and vibrate more vigorously, taking up more space.

Types of thermal expansion: 1. Linear expansion: Increase in length of a solid when heated 2. Area expansion: Increase in surface area of a solid when heated 3. Volume expansion: Increase in volume of a substance (solid, liquid, or gas) when heated

Applications of thermal expansion in daily life: 1. Thermometers: Mercury or alcohol in glass thermometers expands when heated, allowing temperature measurement 2. Bimetallic strips: Used in thermostats, circuit breakers, and oven indicators; made of two different metals with different expansion rates bonded together, they bend when heated 3. Expansion joints: Gaps left in bridges, railway tracks, and concrete highways to allow for expansion during hot weather 4. Fitting metal rims on wooden wheels: The metal rim is heated, expanded, placed on the wheel, and then cooled to contract and fit tightly 5. Riveting: Rivets are heated before insertion and then cooled to create tight joints 6. Automatic fire alarms: Use the principle of differential expansion to trigger alarms when temperature rises

Consequences of thermal expansion: 1. Cracking of glass: When a hot liquid is poured into a glass container, uneven heating causes different parts to expand at different rates, creating stress that can crack the glass 2. Power lines sagging: Electrical transmission lines expand and sag in hot weather 3. Buckling of railway tracks: Without expansion joints, railway tracks can buckle on hot days 4. Bridges expanding: Large bridges can expand several centimeters on hot days 5. Gaps in concrete sidewalks: Concrete expands in hot weather, potentially causing buckling without expansion joints 6. Loosening of metal fillings in teeth: Different expansion rates between the filling and tooth can cause discomfort

Understanding thermal expansion is crucial for engineering, construction, and many everyday applications to prevent damage and ensure proper functioning of various systems.

Q4: What is the anomalous expansion of water? Why is it important for aquatic life?

Answer: The anomalous expansion of water refers to the unusual property of water to contract (decrease in volume) when cooled from room temperature down to 4°C, and then expand (increase in volume) when cooled below 4°C to its freezing point at 0°C. This behavior is contrary to most substances, which continuously contract as they cool.

Key points about anomalous expansion of water: 1. Water has maximum density at 4°C 2. Between 4°C and 0°C, water expands as it cools 3. When water freezes at 0°C, it expands by about 9% 4. This is why ice floats on water (ice is less dense than liquid water)

Importance for aquatic life: 1. Lakes freeze from top to bottom: Due to the anomalous expansion, water at 4°C sinks to the bottom of lakes in winter, while colder water (between 4°C and 0°C) and ice remain at the top.

Insulation effect: The layer of ice that forms on the surface acts as an insulator, preventing the water below from losing heat rapidly to the cold air above.
Survival of aquatic organisms: Because lakes and ponds freeze from the top down, the water at the bottom remains liquid (around 4°C) even in severe winter conditions. This allows aquatic organisms to survive in the unfrozen water beneath the ice.
Prevention of complete freezing: Without this property, lakes would freeze from the bottom up, potentially freezing solid in cold climates, which would make survival impossible for many aquatic species.
Spring turnover: When spring arrives and the ice melts, the surface water warms to 4°C, becomes denser, and sinks, causing a circulation that brings oxygen-rich water to the depths and nutrient-rich water to the surface.

This unique property of water is one of the reasons Earth can support such diverse aquatic life, even in regions with freezing winters. If water behaved like most other substances, many bodies of water would freeze solid in winter, making aquatic ecosystems as we know them impossible.

Q5: Define specific heat capacity. Why does water have a high specific heat capacity, and what are its implications?

Answer: Specific heat capacity is the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius (or one kelvin). It is measured in joules per kilogram per kelvin (J/kg·K) or joules per kilogram per degree Celsius (J/kg·°C).

The formula for calculating heat energy using specific heat capacity is: Q = m × c × ΔT

Where: - Q is the heat energy in joules (J) - m is the mass in kilograms (kg) - c is the specific heat capacity in J/kg·K - ΔT is the change in temperature in kelvin (K) or degrees Celsius (°C)

Why water has a high specific heat capacity: Water has an unusually high specific heat capacity (4,186 J/kg·°C) compared to most other substances due to:

Hydrogen bonding: Water molecules form hydrogen bonds with each other. These bonds absorb energy when broken and release energy when formed.
Molecular structure: Water's simple molecular structure (H₂O) allows it to store energy in various forms of molecular motion (vibration, rotation).
High polarity: Water molecules are highly polar, which affects how they interact and store energy.

Implications of water's high specific heat capacity:

Climate regulation: Oceans and large bodies of water absorb large amounts of heat with minimal temperature change, moderating Earth's climate and preventing extreme temperature fluctuations.
Coastal climates: Coastal areas experience milder temperatures compared to inland areas at the same latitude due to the moderating effect of nearby water bodies.
Thermal stability for aquatic life: Aquatic organisms benefit from the relatively stable temperatures in water environments.
Cooling systems: Water is an excellent coolant in car radiators, industrial processes, and power plants due to its ability to absorb large amounts of heat.
Cooking applications: Water requires significant energy to heat, making it useful for cooking methods like boiling and steaming that provide consistent, controlled heating.
Human body temperature regulation: The high water content in the human body helps maintain a stable internal temperature despite varying external conditions.
Thermal inertia: Water bodies warm up and cool down more slowly than land, creating seasonal lags in temperature changes.

This property of water is crucial for life on Earth and has significant implications for climate, ecosystems, and many practical applications in daily life and industry.

Q6: Explain the different types of phase changes in matter. What is latent heat?

Answer: Phase changes (or changes of state) refer to the transitions of matter between solid, liquid, and gas states. Each phase change involves energy transfer without a change in temperature.

Types of phase changes:

Melting (Fusion): Solid → Liquid
Occurs at the melting point
Requires heat energy (endothermic process)
Example: Ice melting into water at 0°C
Freezing (Solidification): Liquid → Solid
Occurs at the freezing point (same as melting point)
Releases heat energy (exothermic process)
Example: Water freezing into ice at 0°C
Vaporization: Liquid → Gas
Can occur as:
- Evaporation: Surface phenomenon that occurs at any temperature below the boiling point
- Boiling: Occurs throughout the liquid at the boiling point
Requires heat energy (endothermic process)
Example: Water boiling into steam at 100°C
Condensation: Gas → Liquid
Occurs when a gas is cooled below its boiling point
Releases heat energy (exothermic process)
Example: Water vapor condensing into water droplets on a cold surface
Sublimation: Solid → Gas (directly)
Occurs in substances like dry ice (solid CO₂), iodine, and naphthalene
Requires heat energy (endothermic process)
Example: Dry ice converting directly to carbon dioxide gas
Deposition: Gas → Solid (directly)
Occurs when water vapor forms frost or when iodine vapor forms crystals
Releases heat energy (exothermic process)
Example: Water vapor forming frost on a very cold window

Latent Heat:

Latent heat is the heat energy absorbed or released during a change of state without a change in temperature. The word "latent" means hidden, as this heat energy does not cause a temperature change but is used to change the phase of the substance.

There are two main types of latent heat:

Latent Heat of Fusion: The amount of heat energy required to convert 1 kg of a solid into liquid at its melting point without a change in temperature.
For water: 334,000 J/kg or 334 kJ/kg
Example: Melting 1 kg of ice at 0°C requires 334 kJ of heat energy
Latent Heat of Vaporization: The amount of heat energy required to convert 1 kg of a liquid into gas at its boiling point without a change in temperature.
For water: 2,260,000 J/kg or 2,260 kJ/kg
Example: Converting 1 kg of water at 100°C into steam requires 2,260 kJ of heat energy

The formula for calculating heat energy during a phase change is: Q = m × L

Where: - Q is the heat energy in joules (J) - m is the mass in kilograms (kg) - L is the latent heat in joules per kilogram (J/kg)

Understanding phase changes and latent heat is important in meteorology, refrigeration, air conditioning, cooking, and many industrial processes.

Q7: Compare and contrast the three methods of heat transfer: conduction, convection, and radiation.

Answer:

Aspect	Conduction	Convection	Radiation
Definition	Transfer of heat through a substance without the movement of the substance itself	Transfer of heat through the movement of a fluid (liquid or gas)	Transfer of heat through electromagnetic waves without requiring a medium
Medium Required	Requires a material medium	Requires a fluid medium (liquid or gas)	Can occur in vacuum; no medium required
Mechanism	Direct collision of particles; faster vibrating particles transfer energy to slower ones	Movement of heated fluid particles from one location to another	Emission of electromagnetic waves from all bodies above absolute zero
Particle Movement	No bulk movement of particles; only vibrations	Bulk movement of fluid particles	No particle movement involved
Speed	Slowest method	Intermediate speed	Fastest method (speed of light)
Direction	From higher temperature to lower temperature	Generally upward for natural convection (hot fluids rise)	In all directions from the source
Examples	- Heat traveling through a metal spoon in hot tea - Heat transfer through a cooking pan - Heat conduction through walls	- Hot air rising from a radiator - Ocean currents - Wind patterns - Boiling water	- Heat from the sun reaching Earth - Heat from a fire warming your hands - Heat from a light bulb
Factors Affecting Rate	- Temperature difference - Cross-sectional area - Length of conductor - Thermal conductivity of material	- Temperature difference - Fluid density - Fluid viscosity - Surface area	- Temperature of emitting body - Surface area - Nature of surface (color, texture) - Distance from source
Best Conductors	Metals (silver, copper, aluminum)	Gases and liquids with low viscosity	Black, rough surfaces are best absorbers and emitters
Mathematical Relationship	Q/t = (kA∆T)/L where k is thermal conductivity	Complex; depends on fluid properties and flow characteristics	P = eσAT⁴ where e is emissivity, σ is Stefan-Boltzmann constant
Applications	- Cooking utensils - Heat sinks in electronics - Thermal insulation in buildings	- Home heating systems - Refrigerators - Air conditioning - Weather patterns	- Solar heating - Infrared lamps - Microwave ovens - Thermal imaging

Key Differences: 1. Conduction requires direct contact between particles, convection involves fluid movement, and radiation requires no medium at all. 2. Conduction is most effective in solids (especially metals), convection in fluids, and radiation works in any environment including vacuum. 3. Radiation is the only method that can transfer heat through empty space.

Similarities: 1. All three methods transfer heat from higher temperature to lower temperature. 2. All three methods require a temperature difference to operate. 3. All three can occur simultaneously in many real-world situations.

In everyday situations, these methods often work together. For example, in a home heating system: the heater transfers heat to air by conduction, the warm air circulates by convection, and some heat is transferred to the walls and objects by radiation.

Q8: How does a thermos flask (vacuum flask) minimize heat transfer? Explain its construction and working.

Answer: A thermos flask (vacuum flask) is designed to minimize heat transfer through all three mechanisms: conduction, convection, and radiation. It keeps hot liquids hot and cold liquids cold for extended periods.

Construction of a thermos flask:

Double-walled container: The flask consists of two walls (usually made of glass or stainless steel) with a vacuum between them.
Vacuum layer: The space between the inner and outer walls is evacuated to create a vacuum.
Silvered surfaces: The walls facing the vacuum are silvered (coated with a reflective material like silver or aluminum).
Plastic or cork stopper: The mouth of the flask is sealed with an insulating stopper.
Protective outer casing: A durable outer casing (plastic or metal) protects the flask from damage.
Screw-on cup: Many thermos flasks have a cup that screws onto the top for drinking.

How it minimizes heat transfer:

Conduction:
The vacuum between the walls eliminates conduction through the sides since there are no particles to transfer heat
The small points of contact between the inner and outer containers (usually at the neck) are made of poor heat conductors like plastic
The stopper is made of insulating materials like plastic or cork to minimize conduction through the top
Convection:
The vacuum eliminates convection currents since there are no fluid particles to move and transfer heat
The narrow neck design minimizes air movement at the top of the flask
Radiation:
The silvered surfaces reflect thermal radiation back to its source
For hot contents: radiation trying to escape from the inner wall is reflected back inside
For cold contents: radiation from outside is reflected away from the inner container

Working principle:

When a hot liquid is placed in a thermos flask: - The vacuum prevents heat loss by conduction and convection through the sides - The silvered surfaces reflect radiant heat back into the liquid - The insulating stopper reduces heat loss through the top - The small contact points between inner and outer containers minimize conduction

When a cold liquid is placed in a thermos flask: - The vacuum prevents heat gain by conduction and convection from outside - The silvered surfaces reflect external radiant heat away from the cold liquid - The insulating stopper reduces heat gain through the top

Limitations: - Heat transfer still occurs through the stopper and the small contact points - Over time, the temperature will equalize with the surroundings - Physical damage can compromise the vacuum, reducing effectiveness - The effectiveness depends on how full the flask is (more air space means more convection inside)

The thermos flask is a practical application of heat transfer principles and demonstrates how understanding these principles can lead to useful everyday devices.

Q9: What is the greenhouse effect? How does it contribute to global warming?

Answer: The greenhouse effect is a natural process that warms Earth's surface. It occurs when certain gases in the atmosphere trap heat from the sun that would otherwise escape into space. This process is essential for life on Earth, as without it, the planet would be too cold for most organisms to survive.

How the natural greenhouse effect works:

Solar radiation reaches Earth: Short-wavelength radiation (visible light and ultraviolet) from the sun passes through the atmosphere and reaches Earth's surface.
Surface absorption and re-emission: Earth's surface absorbs this radiation and warms up. The warm surface then emits longer-wavelength infrared radiation (heat).
Atmospheric trapping: Greenhouse gases in the atmosphere (primarily water vapor, carbon dioxide, methane, and nitrous oxide) absorb some of this infrared radiation, preventing it from escaping directly to space.
Heat redistribution: The absorbed energy warms the atmosphere and is re-emitted in all directions, including back toward Earth's surface.
Temperature balance: This natural process maintains Earth's average surface temperature at about 15°C (59°F), rather than the -18°C (0°F) it would be without the greenhouse effect.

How the greenhouse effect contributes to global warming:

Global warming refers to the long-term increase in Earth's average temperature due to an enhanced greenhouse effect caused by human activities.

Increased greenhouse gas concentrations: Human activities have significantly increased the concentration of greenhouse gases in the atmosphere:
Carbon dioxide (CO₂): Released by burning fossil fuels (coal, oil, natural gas), deforestation, and industrial processes
Methane (CH₄): Released from livestock, rice paddies, landfills, and natural gas extraction
Nitrous oxide (N₂O): Released from fertilizers, industrial processes, and burning fossil fuels
Fluorinated gases: Synthetic gases used in various industrial applications
Enhanced heat trapping: Higher concentrations of these gases trap more infrared radiation, preventing more heat from escaping to space.
Positive feedback loops: Warming triggers processes that can release more greenhouse gases or reduce Earth's reflectivity:
Melting ice reduces Earth's reflectivity, causing more solar radiation to be absorbed
Warming oceans release dissolved CO₂
Thawing permafrost releases methane
Warmer conditions can lead to more water vapor in the atmosphere (a powerful greenhouse gas)
Climate system changes: The additional trapped heat leads to:
Rising global average temperatures
Changes in precipitation patterns
More frequent and intense extreme weather events
Rising sea levels due to thermal expansion of water and melting ice
Ocean acidification as more CO₂ dissolves in seawater

Distinction between natural and enhanced greenhouse effect: - The natural greenhouse effect is essential for life on Earth - The enhanced greenhouse effect, caused by human activities, is disrupting the climate system

Mitigation strategies: - Reducing fossil fuel consumption - Transitioning to renewable energy sources - Improving energy efficiency - Reforestation and preventing deforestation - Developing carbon capture and storage technologies - Changing agricultural practices to reduce methane and nitrous oxide emissions

Understanding the greenhouse effect and its role in global warming is crucial for developing effective strategies to address climate change, one of the most significant environmental challenges of our time.

Q10: Explain the working principle of a refrigerator. How does it transfer heat from a cooler region to a warmer region?

Answer: A refrigerator is a heat pump that transfers heat from its cooler interior to the warmer external environment, contrary to the natural direction of heat flow. This process requires external work, in accordance with the Second Law of Thermodynamics.

Working Principle of a Refrigerator:

The refrigerator operates on a refrigeration cycle, typically using a vapor-compression cycle with the following key components:

Compressor: Electrically powered, compresses the refrigerant gas, increasing its pressure and temperature
Condenser: Coils usually located at the back or bottom of the refrigerator where the hot, high-pressure refrigerant releases heat to the surroundings
Expansion valve: Restricts the flow of refrigerant, causing a pressure drop and cooling
Evaporator: Coils inside the refrigerator where the cold refrigerant absorbs heat from the interior
Refrigerant: A substance that circulates through the system, changing state between liquid and gas

The Refrigeration Cycle:

Compression Stage:
The compressor draws in low-pressure, low-temperature refrigerant gas from the evaporator
It compresses this gas, raising its pressure and temperature significantly
The refrigerant leaves the compressor as a hot, high-pressure gas
Condensation Stage:
The hot, high-pressure refrigerant gas flows through the condenser coils
As it passes through these coils, it releases heat to the surrounding air (which is cooler than the hot gas)
This heat release causes the refrigerant to condense into a high-pressure liquid
The heat released in this stage is what you feel as warm air behind or beneath the refrigerator
Expansion Stage:
The high-pressure liquid refrigerant passes through the expansion valve
This causes a sudden drop in pressure
The pressure drop leads to partial evaporation and significant cooling of the refrigerant
The refrigerant leaves the expansion valve as a cold mixture of liquid and vapor
Evaporation Stage:
The cold refrigerant mixture flows through the evaporator coils inside the refrigerator
It absorbs heat from the air inside the refrigerator
This heat absorption causes the refrigerant to completely evaporate into a gas
The air inside the refrigerator, having lost heat to the refrigerant, becomes cooler
A fan often circulates this cooled air throughout the refrigerator
Cycle Repetition:
The refrigerant, now a low-pressure gas, returns to the compressor
The cycle repeats continuously

How it transfers heat from cooler to warmer regions:

The refrigerator achieves the seemingly impossible task of moving heat from a cooler region (inside the refrigerator) to a warmer region (the kitchen) by:

Input of external work: The electrical energy supplied to the compressor provides the necessary work to drive the process, allowing heat to flow "uphill" against its natural direction.
Changing the state of the refrigerant: By manipulating the pressure and state of the refrigerant, the system creates conditions where:
In the evaporator, the refrigerant is colder than the refrigerator interior, allowing it to absorb heat
In the condenser, the refrigerant is hotter than the external environment, allowing it to release heat
Creating a temperature gradient: The compression process creates a temperature difference that drives heat transfer in the desired direction.

This process follows the laws of thermodynamics: - It doesn't violate the Second Law because external work is required - The total entropy of the system and surroundings increases - The coefficient of performance (efficiency) is limited by the temperature difference between the interior and exterior

The same principles apply to air conditioners, heat pumps, and freezers, with variations in design and operating temperatures.

Application-Based Questions

Q11: Why do we feel cooler when we step out of a swimming pool, even on a hot day?

Answer: When we step out of a swimming pool, even on a hot day, we feel cooler due to the process of evaporative cooling. This phenomenon involves several physical principles related to heat and phase changes:

Evaporation process: Water on our skin begins to evaporate when exposed to air. Evaporation is the phase change from liquid to gas that occurs at the surface of a liquid below its boiling point.
Latent heat of vaporization: During evaporation, the water molecules with the highest kinetic energy escape from the liquid surface. To change from liquid to vapor state, water requires significant energy (the latent heat of vaporization, which is 2,260 kJ/kg for water). This energy is absorbed from the surroundings—in this case, from our skin and the remaining water on our body.
Heat extraction from the body: As water molecules absorb heat energy from our skin to evaporate, our skin temperature decreases, creating the cooling sensation.
Factors affecting evaporation rate:
Air humidity: On days with lower humidity, evaporation occurs more rapidly because the air can accept more water vapor, increasing the cooling effect.
Air movement: Wind or breeze increases the evaporation rate by continuously replacing the humid air near our skin with drier air.
Surface area: The larger the wet surface area of our body, the more evaporation can occur simultaneously.
Temperature: Higher temperatures generally increase the evaporation rate, but the cooling effect is often more noticeable because of the greater temperature contrast.
Physiological response: Our nervous system is particularly sensitive to temperature changes rather than absolute temperatures. The rapid cooling from evaporation creates a strong sensation, especially when contrasted with the previous warmth of being in the pool.

This evaporative cooling is the same principle that our bodies use for thermoregulation through sweating. When we sweat, the evaporation of perspiration from our skin creates a cooling effect that helps maintain our body temperature.

The cooling sensation is often more pronounced after swimming than after sweating because: - A larger amount of water covers our body after swimming - Swimming pool water often evaporates more readily than sweat (which contains salts and oils) - The temperature contrast between being immersed in water and then exposed to air enhances the perceived cooling effect

This natural cooling mechanism has inspired various technologies, from traditional clay water coolers to modern evaporative cooling systems used in buildings in dry climates.

Q12: A metal ball just passes through a ring when both are at room temperature. When the ball is heated, it no longer passes through the ring. Explain this observation using the concept of thermal expansion.

Answer: This observation can be explained using the concept of thermal expansion, which is the tendency of matter to increase in volume when heated and decrease when cooled.

Initial condition: When both the metal ball and ring are at room temperature, the ball just passes through the ring. This means that the diameter of the ball is slightly smaller than or equal to the inner diameter of the ring, allowing it to pass through with minimal clearance.

After heating the ball: When the metal ball is heated: 1. The ball undergoes thermal expansion, increasing in volume 2. This volume increase results in an increase in the ball's diameter 3. The expanded diameter of the ball now exceeds the inner diameter of the ring 4. Consequently, the ball can no longer pass through the ring

Scientific explanation: The linear expansion of a solid object when heated is given by the formula: ΔL = L × α × ΔT

Where: - ΔL is the change in a linear dimension (in this case, the diameter of the ball) - L is the original dimension (the original diameter) - α is the coefficient of linear expansion (a property of the material) - ΔT is the change in temperature

For a spherical object like a ball, the expansion occurs in all three dimensions. The increase in diameter is directly proportional to: 1. The original diameter of the ball 2. The coefficient of linear expansion of the metal 3. The temperature increase

Quantitative example: If we consider a steel ball with a diameter of 50 mm at 20°C (room temperature), and the ring has an inner diameter of 50.05 mm (providing a clearance of 0.05 mm):

The coefficient of linear expansion for steel is approximately 12 × 10⁻⁶ per °C
If the ball is heated to 70°C (a 50°C increase):
ΔL = 50 mm × (12 × 10⁻⁶ per °C) × 50°C
ΔL = 0.03 mm
The new diameter of the ball would be 50 + 0.03 = 50.03 mm
This is still less than the ring's diameter of 50.05 mm, so the ball would still pass through
But if the ball is heated to 100°C (an 80°C increase):
ΔL = 50 mm × (12 × 10⁻⁶ per °C) × 80°C
ΔL = 0.048 mm
The new diameter would be 50.048 mm
This is very close to the ring's diameter, and considering manufacturing tolerances and non-uniform heating, the ball might not pass through
If heated to 120°C (a 100°C increase):
ΔL = 50 mm × (12 × 10⁻⁶ per °C) × 100°C
ΔL = 0.06 mm
The new diameter would be 50.06 mm
This exceeds the ring's diameter of 50.05 mm, so the ball definitely would not pass through

This phenomenon is used in various practical applications, such as: - Shrink-fitting parts in machinery - Installing metal rims on wooden wheels - Designing expansion joints in bridges and buildings - Creating tight seals in plumbing and other applications

If both the ball and the ring were heated to the same temperature, they would both expand, but since the ring has a larger diameter, its absolute expansion would be greater, and the ball would likely pass through again.

Q13: Why are cooking utensils often made with metal bodies but plastic or wooden handles? Explain using the concept of heat transfer.

Answer: Cooking utensils are often designed with metal bodies and plastic or wooden handles due to the different thermal properties of these materials and their roles in heat transfer. This design maximizes cooking efficiency while ensuring user safety and comfort.

Metal bodies for cooking surfaces:

High thermal conductivity: Metals like aluminum, copper, and stainless steel have high thermal conductivity, meaning they transfer heat rapidly and efficiently. This property allows:
Quick and even heating of food
Responsive temperature changes when adjusting the heat source
Uniform cooking with fewer hot spots
Specific metals and their properties:
Copper: Highest thermal conductivity among common cookware metals; provides excellent heat distribution but is expensive and reactive with some foods
Aluminum: Good thermal conductivity, lightweight, affordable, but can react with acidic foods
Stainless steel: Less conductive than copper or aluminum but durable, non-reactive, and resistant to corrosion
Cast iron: Moderate thermal conductivity but excellent heat retention; heats slowly but maintains temperature well
Durability at high temperatures: Metals can withstand the high temperatures required for cooking without melting, warping, or releasing harmful substances.

Plastic or wooden handles:

Low thermal conductivity: Plastics and wood are poor conductors of heat (good insulators), which means:
They transfer very little heat from the hot metal body to the handle
They remain relatively cool to touch even when the cooking surface is hot
They prevent burns when handling the utensil during cooking
Specific materials and their properties:
Wood: Natural insulator with a warm feel; can withstand moderate heat but may char at very high temperatures; has natural antibacterial properties
Bakelite (a type of plastic): Good heat resistance, doesn't conduct heat well, durable and lightweight
Silicone: Excellent heat resistance (typically up to 260°C/500°F), flexible, and comfortable to grip
Ergonomics and comfort: These materials are often more comfortable to hold than metal, providing better grip and reducing hand fatigue during cooking.

Heat transfer principles involved:

Conduction: The primary method of heat transfer in cooking utensils
Heat conducts rapidly through the metal body from the heat source to the food
The junction between the metal body and the handle is designed to minimize heat conduction
The poor thermal conductivity of the handle material prevents significant heat transfer to the user's hand
Thermal resistance at junctions: Many utensils incorporate design features at the junction between the metal body and handle:
Reduced cross-sectional area to decrease heat flow
Air gaps or insulating materials to create thermal barriers
Metal pins or rivets that are isolated from the main heat path
Specific heat capacity: Materials with high specific heat capacity require more energy to increase in temperature
Wood has a relatively high specific heat capacity, helping it stay cool longer

This design approach applies the principles of heat transfer to create cookware that efficiently transfers heat where needed (to the food) while preventing heat transfer where it's unwanted (to the user's hand). This balance of thermal properties makes cooking both effective and safe.

Q14: A student places a beaker of water at 20°C and a beaker of cooking oil at 20°C on identical hot plates. The oil reaches 80°C faster than the water. Explain why, using the concept of specific heat capacity.

Answer: The observation that cooking oil reaches 80°C faster than water when both start at 20°C and are heated on identical hot plates can be explained using the concept of specific heat capacity.

Specific heat capacity is the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius (or one kelvin). It is measured in joules per kilogram per kelvin (J/kg·K) or joules per kilogram per degree Celsius (J/kg·°C).

Comparison of specific heat capacities: - Water: approximately 4,186 J/kg·°C - Cooking oil: approximately 1,670-2,000 J/kg·°C (depending on the type of oil)

This means water requires more than twice the amount of energy to increase its temperature by the same amount compared to cooking oil.

Mathematical explanation: The heat energy required to change the temperature of a substance is given by the formula: Q = m × c × ΔT

Where: - Q is the heat energy in joules (J) - m is the mass in kilograms (kg) - c is the specific heat capacity in J/kg·°C - ΔT is the change in temperature in degrees Celsius (°C)

Assuming equal masses of water and oil (for simplicity, let's say 1 kg each) and the same temperature change (from 20°C to 80°C, so ΔT = 60°C):

For water: Q_water = 1 kg × 4,186 J/kg·°C × 60°C = 251,160 J

For cooking oil: Q_oil = 1 kg × 1,800 J/kg·°C × 60°C = 108,000 J

This calculation shows that water requires approximately 2.3 times more energy than oil to achieve the same temperature increase.

Factors contributing to the observation:

Molecular structure: Water molecules form hydrogen bonds, which require significant energy to overcome during heating. Cooking oil molecules (primarily triglycerides) have weaker intermolecular forces.
Heat transfer efficiency: While not directly related to specific heat capacity, oils might have different heat transfer characteristics than water, potentially affecting heating rates.
Density differences: If equal volumes rather than equal masses were used, the density difference would also affect the heating rate (water is generally denser than oil).
Evaporation effects: Water evaporates more readily than oil, which can remove heat energy from the system, though this effect is minimal until approaching boiling temperatures.

Practical implications:

This property explains why oil-based cooking methods (frying, sautéing) can achieve higher temperatures more quickly than water-based methods (boiling, steaming):

Oil can reach higher temperatures before boiling (water boils at 100°C at standard pressure, while most cooking oils begin to smoke at 175-230°C)
The lower specific heat capacity of oil means it heats up faster, allowing for quicker cooking times
When food is added to hot oil, the temperature drops less dramatically than when added to water, maintaining cooking efficiency

This experiment demonstrates an important principle in thermodynamics and has direct applications in cooking, industrial heating processes, and thermal engineering.

Q15: During winter in cold regions, water pipes sometimes burst. Explain this phenomenon using the concepts of freezing and anomalous expansion of water.

Answer: The bursting of water pipes during winter in cold regions is a direct consequence of the freezing process and the anomalous expansion of water. This phenomenon involves several physical principles:

1. Anomalous expansion of water: Unlike most substances that contract when cooled, water exhibits an unusual property: - Water contracts as it cools from room temperature down to 4°C (reaching its maximum density) - Below 4°C, water begins to expand as it continues to cool - When water freezes at 0°C, it expands by approximately 9% in volume

This behavior is called the anomalous expansion of water and is due to the formation of a crystalline structure with hexagonal symmetry as water molecules arrange themselves into ice, creating more space between molecules than in the liquid state.

2. The pipe-bursting process:

a) Initial cooling phase: - As the ambient temperature drops below freezing, the water in the pipes begins to cool - The water contracts slightly as it cools to 4°C - Below 4°C, it starts to expand slightly as it approaches 0°C

b) Freezing begins: - Ice formation typically starts along the pipe walls, where the water is coldest - As water freezes, it expands by about 9% - Initially, this expansion can be accommodated by pushing unfrozen water along the pipe

c) Pressure buildup: - As more water freezes, ice begins to block sections of the pipe - This creates isolated pockets of water trapped between ice plugs - When these trapped sections freeze, the expansion has nowhere to go - Tremendous pressure develops (up to 2,000 atmospheres or 30,000 psi)

d) Pipe failure: - When the pressure exceeds the tensile strength of the pipe material, the pipe ruptures - The burst typically occurs not at the ice blockage itself, but in unfrozen sections where the pipe wall is weakest - The actual rupture often happens before the entire pipe contents freeze solid

3. Factors affecting pipe bursting:

a) Pipe material: - Metal pipes (copper, steel) are strong but transfer heat quickly and have little flexibility - PVC and other plastic pipes have some flexibility to accommodate expansion but become brittle at very low temperatures - Pipe thickness and age affect resistance to bursting

b) Pipe location: - Exterior walls, unheated spaces, and areas with poor insulation increase freezing risk - Buried pipes are at risk if they're above the frost line - Wind exposure increases heat loss and freezing potential

c) Water flow: - Stagnant water freezes more readily than flowing water - Pipes serving rarely used fixtures are at higher risk

4. Prevention methods based on scientific principles:

a) Insulation: - Slows heat transfer from the water to the cold environment - Gives more time for residual heat in the water to prevent freezing

b) Heat tape or cables: - Provides heat energy to counteract the cooling process - Keeps water temperature above freezing

c) Allowing faucets to drip: - Creates water movement, reducing freezing risk - Relieves pressure in the system, even if some ice forms

d) Maintaining minimum indoor temperatures: - Keeps ambient temperature around pipes above freezing - Prevents the initial cooling process

e) Draining pipes: - Removes water from the system when freezing is likely - Eliminates the expansion problem by removing the medium

This phenomenon demonstrates the powerful forces that can develop due to the unique properties of water, particularly its anomalous expansion behavior during freezing. Understanding these principles is crucial for designing plumbing systems in cold climates and taking appropriate preventive measures.

Q16: Why does a desert cool down rapidly at night compared to a coastal area, even when both receive similar amounts of sunlight during the day?

Answer: Deserts cool down rapidly at night compared to coastal areas, even when both receive similar amounts of sunlight during the day, due to several interrelated factors involving heat capacity, humidity, and heat transfer mechanisms.

1. Specific heat capacity differences:

The primary factor is the difference in specific heat capacity between land and water: - Land (desert sand/soil): Has a relatively low specific heat capacity (around 800-1000 J/kg·°C) - Water (oceans/seas): Has a much higher specific heat capacity (4,186 J/kg·°C)

This means: - Land heats up more quickly during the day but also cools down more rapidly at night - Water requires more energy to heat up but also retains heat much longer - Coastal areas benefit from the thermal moderating effect of nearby water bodies

2. Heat storage and depth effects:

Desert: Solar radiation heats only the top few centimeters of soil
Coastal waters: Sunlight penetrates deeper into water, distributing heat through a greater volume
Heat distribution: Water circulates through convection, distributing heat throughout its depth
Thermal mass: Oceans represent an enormous thermal mass that changes temperature very slowly

3. Humidity and atmospheric moisture:

Desert air: Very dry with low humidity (often below 20%)
Coastal air: Higher humidity due to evaporation from water bodies (often 50-80%)

The effects of this difference: - Water vapor is a greenhouse gas that traps heat - Humid air retains heat better than dry air - Coastal areas benefit from this "blanket effect" at night - Desert air allows infrared radiation to escape more readily to space

4. Evaporative cooling:

Deserts: Minimal surface water means little evaporative cooling during the day, but also no water vapor to retain heat at night
Coastal areas: Evaporation from water bodies cools the area during the day but increases humidity, which helps retain heat at night

5. Surface properties and albedo:

Desert surfaces: Light-colored sand reflects more sunlight (higher albedo) but also radiates heat more efficiently at night
Water surfaces: Darker color absorbs more sunlight (lower albedo) but stores the energy more effectively

6. Radiative cooling:

Clear desert skies: Minimal cloud cover allows more direct solar radiation during the day but also permits more rapid radiative cooling at night
Coastal areas: Often have more cloud cover, which acts as an insulating layer, reducing nighttime cooling

7. Quantitative example:

If we consider equal masses of sand and water receiving the same amount of solar energy:

For 1 kg of material receiving 1,000,000 J of energy during the day: - Sand (c = 800 J/kg·°C): Temperature increase = Q/(m×c) = 1,000,000/(1×800) = 1,250°C (theoretical) - Water (c = 4,186 J/kg·°C): Temperature increase = 1,000,000/(1×4,186) = 239°C (theoretical)

These are simplified calculations, but they illustrate why desert temperatures fluctuate more dramatically.

8. Real-world temperature ranges:

Hot deserts: Daytime temperatures can reach 45-50°C, dropping to 0-10°C at night (a 35-45°C difference)
Coastal areas at similar latitudes: Might reach 30-35°C during the day, dropping to 20-25°C at night (a 10-15°C difference)

This dramatic temperature fluctuation in deserts has significant implications for desert ecosystems, human settlements, and infrastructure design in these regions. It also explains why coastal climates are generally considered more moderate, with less extreme temperature variations between day and night.

Q17: A student notices that when hot water is poured into a thick glass tumbler, the glass often cracks, but this doesn't happen with a thin glass. Explain this observation using the concept of thermal expansion.

Answer: When hot water is poured into a glass tumbler, the glass often cracks due to thermal stress caused by uneven expansion. The difference in behavior between thick and thin glass can be explained using the concept of thermal expansion and heat transfer.

Thermal expansion and stress in glass:

Basic principle: When heated, materials expand. The linear expansion of a material is given by: ΔL = L × α × ΔT Where:
ΔL is the change in length
L is the original length
α is the coefficient of linear expansion
ΔT is the temperature change
What happens when hot water is poured into a glass:
The inner surface of the glass in contact with hot water heats up rapidly
The heated inner surface expands according to the thermal expansion coefficient of glass
The outer surface remains relatively cooler and expands less or not at all
This differential expansion creates thermal stress (tension in the outer surface and compression in the inner surface)
If the stress exceeds the tensile strength of the glass, it cracks

Why thick glass is more prone to cracking than thin glass:

Temperature gradient effect:
Thick glass: Creates a steeper temperature gradient between the inner and outer surfaces
Thin glass: Has a smaller distance between inner and outer surfaces, allowing heat to transfer more quickly and creating a less severe temperature gradient
Heat transfer rate:
Glass is a poor conductor of heat (thermal conductivity ≈ 1 W/m·K)
Thick glass: Heat takes longer to conduct through the material, maintaining the temperature difference between inner and outer surfaces for longer
Thin glass: Heat conducts through the entire thickness more quickly, allowing the entire piece to expand more uniformly
Thermal mass:
Thick glass: Has greater thermal mass, meaning the outer portions stay cool longer
Thin glass: Has less thermal mass, allowing the entire piece to reach thermal equilibrium faster
Flexibility and stress distribution:
Thick glass: More rigid, concentrates stress and has less ability to flex to accommodate expansion
Thin glass: More flexible, can slightly deform to accommodate some thermal expansion, distributing stress more evenly

Mathematical perspective:

If we consider a temperature difference ΔT between the inner and outer surfaces:

The inner surface tries to expand by: ΔL_inner = L × α × ΔT
The outer surface remains at original length: ΔL_outer ≈ 0
The differential expansion ΔL_inner - ΔL_outer creates stress

For thick glass, this differential is maintained longer and across a greater distance, creating higher stress.

Practical applications of this principle:

Borosilicate glass (e.g., Pyrex) has a very low coefficient of thermal expansion (about one-third that of regular glass), making it more resistant to thermal shock
Pre-heating glass containers before adding hot liquids reduces the temperature gradient
Tempered glass is treated to create compressive stress in the outer layers, making it more resistant to thermal shock
Laboratory glassware is often made thin to reduce thermal shock risk

This phenomenon is an excellent example of how thermal properties and heat transfer principles affect everyday objects and experiences. Understanding these principles helps in designing better glassware and in handling glass safely in thermal environments.

Q18: In a solar water heater, water circulates through the system without using a pump. Explain how this circulation occurs using the concept of convection.

Answer: In a solar water heater, water circulates through the system without using a pump through a process called thermosiphoning, which is based on natural convection. This passive circulation system relies on fundamental principles of fluid dynamics and heat transfer.

The Thermosiphon Principle:

Thermosiphoning occurs due to density differences in water at different temperatures. When water is heated, it: 1. Becomes less dense (expands) 2. Rises due to buoyancy forces 3. Is replaced by cooler, denser water that flows in to take its place

This creates a natural circulation loop without requiring any mechanical pumps or external energy input beyond the solar heating itself.

Components of a Passive Solar Water Heater:

Solar collector: Usually flat panels or evacuated tubes positioned to receive maximum sunlight
Storage tank: Placed above the collector to facilitate thermosiphoning
Connecting pipes: Allow water to circulate between the collector and storage tank
Insulation: Minimizes heat loss from the system

How the Circulation Works:

Initial state: The system is filled with water, with the storage tank positioned above the solar collector
Heating phase:
Solar radiation heats the absorber plate in the collector
The absorber transfers heat to the water in the collector tubes
Water in the collector warms up and expands, becoming less dense
Convection circulation:
The warmer, less dense water naturally rises up through the outlet pipe to the top of the storage tank
Simultaneously, cooler, denser water from the bottom of the storage tank flows down through the inlet pipe to the collector
This creates a continuous circulation loop as long as the collector is receiving heat
Thermal stratification in the tank:
Hot water accumulates at the top of the tank (where it's drawn for use)
Cooler water settles at the bottom (where it flows back to the collector)
This stratification improves system efficiency

Physical Principles Involved:

Density variation with temperature:
Water at 20°C has a density of approximately 998 kg/m³
Water at 60°C has a density of approximately 983 kg/m³
This 1.5% difference in density creates sufficient buoyancy force to drive circulation
Buoyancy force:
The buoyancy force is proportional to the density difference and the height of the water column
Expressed as: F = g × V × (ρcold - ρhot)
Where g is gravitational acceleration, V is volume, and ρ is density
Flow rate factors:
Height difference between collector and tank (greater height creates stronger flow)
Pipe diameter (affects flow resistance)
Temperature difference (greater temperature difference creates stronger convection)

Design Requirements for Effective Thermosiphoning:

Vertical separation: The bottom of the storage tank must be at least 30 cm (preferably 60+ cm) above the top of the collector
Pipe sizing: Pipes must be adequately sized to minimize flow resistance while maintaining good heat transfer
Slope: All horizontal pipe runs should have a slight upward slope (minimum 1:20 or 5%) in the direction of flow to prevent air locks
No reverse flow: A one-way valve may be installed to prevent reverse thermosiphoning at night (when the collector becomes cooler than the tank)

Advantages of Thermosiphon Systems:

No electricity required: Functions during power outages and in off-grid locations
No moving parts: Higher reliability and lower maintenance
Long lifespan: Typically 15-30 years with minimal maintenance
Self-regulating: Flow rate naturally adjusts to solar intensity

Limitations:

Installation constraints: Requires the storage tank to be above the collector
Slower flow rates: Generally less efficient than active (pumped) systems
Freezing risk: In cold climates, special precautions are needed to prevent freezing

This natural convection process demonstrates how understanding basic physical principles can lead to simple, effective, and sustainable technologies that harness renewable energy without requiring external power sources.

Numerical Problems

Q19: A copper rod has a length of 1.5 m at 20°C. Calculate the increase in its length when heated to 120°C. (Coefficient of linear expansion of copper = 1.7 × 10⁻⁵ /°C)

Answer: To calculate the increase in length of the copper rod when heated, we'll use the formula for linear thermal expansion:

ΔL = L₀ × α × ΔT

Where: - ΔL is the change in length (what we're solving for) - L₀ is the initial length of the rod - α is the coefficient of linear expansion - ΔT is the change in temperature

Given: - Initial length (L₀) = 1.5 m - Initial temperature = 20°C - Final temperature = 120°C - Coefficient of linear expansion of copper (α) = 1.7 × 10⁻⁵ /°C

Step 1: Calculate the change in temperature (ΔT). ΔT = Final temperature - Initial temperature ΔT = 120°C - 20°C ΔT = 100°C

Step 2: Substitute the values into the linear expansion formula. ΔL = L₀ × α × ΔT ΔL = 1.5 m × (1.7 × 10⁻⁵ /°C) × 100°C ΔL = 1.5 m × 1.7 × 10⁻⁵ × 100 ΔL = 1.5 × 1.7 × 10⁻³ ΔL = 2.55 × 10⁻³ m

Step 3: Convert to millimeters for a more practical unit. ΔL = 2.55 × 10⁻³ m × 1000 mm/m ΔL = 2.55 mm

Therefore, the copper rod will increase in length by 2.55 millimeters when heated from 20°C to 120°C.

This calculation demonstrates how even a relatively small coefficient of expansion can lead to significant dimensional changes in larger objects when subjected to substantial temperature changes. This principle is important in engineering applications where thermal expansion must be accounted for, such as in the design of bridges, railroad tracks, and pipelines.

Q20: A metal sphere has a volume of 50 cm³ at 30°C. When heated to 80°C, its volume increases by 0.075 cm³. Calculate the coefficient of volume expansion of the metal.

Answer: To calculate the coefficient of volume expansion of the metal, we'll use the formula for volume thermal expansion:

ΔV = V₀ × β × ΔT

Where: - ΔV is the change in volume - V₀ is the initial volume - β is the coefficient of volume expansion (what we're solving for) - ΔT is the change in temperature

Given: - Initial volume (V₀) = 50 cm³ - Initial temperature = 30°C - Final temperature = 80°C - Change in volume (ΔV) = 0.075 cm³

Step 1: Calculate the change in temperature (ΔT). ΔT = Final temperature - Initial temperature ΔT = 80°C - 30°C ΔT = 50°C

Step 2: Rearrange the volume expansion formula to solve for β. ΔV = V₀ × β × ΔT β = ΔV / (V₀ × ΔT)

Step 3: Substitute the values into the rearranged formula. β = 0.075 cm³ / (50 cm³ × 50°C) β = 0.075 / (50 × 50) β = 0.075 / 2500 β = 3.0 × 10⁻⁵ /°C

Therefore, the coefficient of volume expansion of the metal is 3.0 × 10⁻⁵ per °C.

This value can help identify the metal, as different metals have characteristic coefficients of expansion. For comparison: - Aluminum: ~7.2 × 10⁻⁵ /°C - Iron: ~3.6 × 10⁻⁵ /°C - Steel: ~3.3 × 10⁻⁵ /°C - Brass: ~6.0 × 10⁻⁵ /°C

Based on this calculation, the metal could be steel or a similar alloy, as its coefficient of volume expansion is close to that of steel.

Note: For solids, the coefficient of volume expansion (β) is approximately three times the coefficient of linear expansion (α), i.e., β ≈ 3α. This relationship comes from the fact that volume expansion occurs in all three dimensions.

Q21: Calculate the heat energy required to raise the temperature of 2 kg of water from 20°C to 100°C. (Specific heat capacity of water = 4,186 J/kg·°C)

Answer: To calculate the heat energy required to raise the temperature of water, we'll use the formula:

Q = m × c × ΔT

Where: - Q is the heat energy required (what we're solving for) - m is the mass of the water - c is the specific heat capacity of water - ΔT is the change in temperature

Given: - Mass of water (m) = 2 kg - Initial temperature = 20°C - Final temperature = 100°C - Specific heat capacity of water (c) = 4,186 J/kg·°C

Step 1: Calculate the change in temperature (ΔT). ΔT = Final temperature - Initial temperature ΔT = 100°C - 20°C ΔT = 80°C

Step 2: Substitute the values into the heat energy formula. Q = m × c × ΔT Q = 2 kg × 4,186 J/kg·°C × 80°C Q = 2 × 4,186 × 80 J Q = 669,760 J

Step 3: Convert to kilojoules for a more manageable unit. Q = 669,760 J ÷ 1,000 Q = 669.76 kJ

Therefore, 669.76 kilojoules (669,760 joules) of heat energy are required to raise the temperature of 2 kg of water from 20°C to 100°C.

This calculation demonstrates why water is an excellent medium for storing and transferring heat energy. Its high specific heat capacity means it can absorb or release large amounts of heat with relatively small changes in temperature. This property makes water valuable for:

Cooling systems in engines and industrial processes
Home heating systems using water radiators
Moderating climate near large bodies of water
Cooking applications where consistent, controlled heating is needed

Note: This calculation assumes that all the heat goes into raising the temperature of the water and none is lost to the surroundings. In a real-world scenario, some heat would be lost to the environment, requiring additional energy input to achieve the same temperature change.

Q22: How much heat energy is required to convert 500 g of ice at 0°C to water at 0°C? (Latent heat of fusion of ice = 334 kJ/kg)

Answer: To calculate the heat energy required to convert ice at 0°C to water at 0°C (melting), we'll use the formula for latent heat:

Q = m × L_f

Where: - Q is the heat energy required (what we're solving for) - m is the mass of ice - L_f is the latent heat of fusion

Given: - Mass of ice (m) = 500 g = 0.5 kg (converting to kg for consistency with units) - Latent heat of fusion of ice (L_f) = 334 kJ/kg

Step 1: Substitute the values into the latent heat formula. Q = m × L_f Q = 0.5 kg × 334 kJ/kg Q = 167 kJ

Therefore, 167 kilojoules of heat energy are required to convert 500 g of ice at 0°C to water at 0°C.

Physical explanation: This process involves a phase change from solid (ice) to liquid (water) at the melting point. During this phase change:

The temperature remains constant at 0°C throughout the process
The heat energy is used to break the hydrogen bonds in the ice crystal structure
The molecules transition from a rigid, ordered arrangement to a more fluid state
The volume decreases by about 9% as ice melts to water

This calculation demonstrates why ice is effective for cooling: - It absorbs a large amount of heat energy during melting - The temperature remains at 0°C until all ice has melted - This provides a stable cooling temperature

Practical applications: 1. Ice packs for injuries or keeping food cold 2. Ice in drinks to maintain cold temperature 3. Ice baths for rapid cooling in emergency medical situations 4. Industrial cooling processes

Additional context: The latent heat of fusion of water (334 kJ/kg) is relatively high compared to many other substances, which is another unique property of water. For comparison: - Ethanol: 108 kJ/kg - Lead: 23 kJ/kg - Mercury: 11.8 kJ/kg

This high latent heat of fusion is beneficial for life on Earth as it helps stabilize temperatures during freezing and thawing cycles.

Q23: A hot piece of metal weighing 200 g at 95°C is dropped into 400 g of water at 20°C. The final temperature of the mixture is 30°C. Calculate the specific heat capacity of the metal. (Specific heat capacity of water = 4,186 J/kg·°C)

Answer: To solve this problem, we'll use the principle of conservation of energy. The heat lost by the metal equals the heat gained by the water.

Let's denote: - m_m = mass of metal = 200 g = 0.2 kg - T_m1 = initial temperature of metal = 95°C - T_m2 = final temperature of metal = 30°C - c_m = specific heat capacity of metal (what we're solving for) - m_w = mass of water = 400 g = 0.4 kg - T_w1 = initial temperature of water = 20°C - T_w2 = final temperature of water = 30°C - c_w = specific heat capacity of water = 4,186 J/kg·°C

Step 1: Set up the equation based on conservation of energy. Heat lost by metal = Heat gained by water m_m × c_m × (T_m1 - T_m2) = m_w × c_w × (T_w2 - T_w1)

Step 2: Substitute the known values. 0.2 kg × c_m × (95°C - 30°C) = 0.4 kg × 4,186 J/kg·°C × (30°C - 20°C) 0.2 kg × c_m × 65°C = 0.4 kg × 4,186 J/kg·°C × 10°C 13 × c_m = 0.4 × 4,186 × 10 13 × c_m = 16,744

Step 3: Solve for c_m. c_m = 16,744 ÷ 13 c_m = 1,288 J/kg·°C

Therefore, the specific heat capacity of the metal is 1,288 J/kg·°C.

Analysis of the result: This value (1,288 J/kg·°C) is within the range of specific heat capacities for common metals. For comparison: - Aluminum: ~900 J/kg·°C - Iron: ~450 J/kg·°C - Copper: ~385 J/kg·°C - Zinc: ~388 J/kg·°C - Brass: ~380 J/kg·°C - Lead: ~128 J/kg·°C

Based on this calculation, the metal could be a type of steel alloy or possibly a mixture of metals, as its specific heat capacity is higher than most pure metals but lower than that of aluminum.

Assumptions in this calculation: 1. No heat is lost to the surroundings (the system is perfectly insulated) 2. No phase changes occur during the process 3. The specific heat capacities remain constant over the temperature range 4. Thermal equilibrium is reached (both substances reach the same final temperature) 5. The container's heat capacity is negligible or accounted for in the metal's mass

This type of calorimetry problem demonstrates how we can determine material properties through careful measurement of temperature changes in controlled thermal interactions.

Q24: A room measures 5 m × 4 m × 3 m. Calculate the heat energy required to raise the temperature of the air in the room from 15°C to 25°C. (Density of air = 1.2 kg/m³, specific heat capacity of air = 1,005 J/kg·°C)

Answer: To calculate the heat energy required to raise the temperature of the air in the room, we'll use the formula:

Q = m × c × ΔT

Where: - Q is the heat energy required (what we're solving for) - m is the mass of air - c is the specific heat capacity of air - ΔT is the change in temperature

Given: - Room dimensions = 5 m × 4 m × 3 m - Initial temperature = 15°C - Final temperature = 25°C - Density of air (ρ) = 1.2 kg/m³ - Specific heat capacity of air (c) = 1,005 J/kg·°C

Step 1: Calculate the volume of the room. Volume = Length × Width × Height Volume = 5 m × 4 m × 3 m Volume = 60 m³

Step 2: Calculate the mass of air in the room using the density. Mass = Density × Volume Mass = 1.2 kg/m³ × 60 m³ Mass = 72 kg

Step 3: Calculate the change in temperature (ΔT). ΔT = Final temperature - Initial temperature ΔT = 25°C - 15°C ΔT = 10°C

Step 4: Substitute the values into the heat energy formula. Q = m × c × ΔT Q = 72 kg × 1,005 J/kg·°C × 10°C Q = 72 × 1,005 × 10 J Q = 723,600 J

Step 5: Convert to kilojoules for a more manageable unit. Q = 723,600 J ÷ 1,000 Q = 723.6 kJ

Therefore, approximately 723.6 kilojoules (723,600 joules) of heat energy are required to raise the temperature of the air in the room from 15°C to 25°C.

Practical context: This calculation is relevant for heating, ventilation, and air conditioning (HVAC) system design. However, in real-world applications, several additional factors would need to be considered:

Heat losses: Through walls, windows, doors, and other surfaces
Infiltration: Air exchange with the outside environment
Thermal mass: Heat absorbed by furniture, walls, and other objects in the room
Occupancy: Heat generated by people and equipment in the room
Humidity changes: Energy required to change the moisture content of the air

For a complete heating load calculation, these factors would need to be accounted for, making the actual energy requirement higher than the value calculated here, which represents only the theoretical minimum energy needed to heat the air itself.

Q25: A 2 kW electric heater is used to heat 5 liters of water from 20°C to 80°C. How long will it take? Assume 80% of the heat from the heater is transferred to the water. (Density of water = 1,000 kg/m³, specific heat capacity of water = 4,186 J/kg·°C)

Answer: To calculate the time required to heat the water, we'll use the relationship between power, energy, and time, along with the heat energy formula.

Given: - Power of heater = 2 kW = 2,000 W - Efficiency = 80% = 0.8 - Volume of water = 5 liters = 0.005 m³ - Initial temperature = 20°C - Final temperature = 80°C - Density of water = 1,000 kg/m³ - Specific heat capacity of water = 4,186 J/kg·°C

Step 1: Calculate the mass of water. Mass = Density × Volume Mass = 1,000 kg/m³ × 0.005 m³ Mass = 5 kg

Step 2: Calculate the heat energy required to raise the temperature of the water. Q = m × c × ΔT Q = 5 kg × 4,186 J/kg·°C × (80°C - 20°C) Q = 5 × 4,186 × 60 J Q = 1,255,800 J

Step 3: Calculate the effective power transferred to the water, considering the efficiency. Effective power = Power of heater × Efficiency Effective power = 2,000 W × 0.8 Effective power = 1,600 W

Step 4: Calculate the time required using the relationship between energy, power, and time. Time = Energy ÷ Power Time = 1,255,800 J ÷ 1,600 W Time = 784.875 seconds

Step 5: Convert to minutes for a more practical unit. Time = 784.875 seconds ÷ 60 Time = 13.08 minutes

Therefore, it will take approximately 13 minutes and 5 seconds to heat 5 liters of water from 20°C to 80°C using the 2 kW electric heater with 80% efficiency.

Practical considerations:

Heat losses: This calculation assumes that once the heat is transferred to the water, it stays there. In reality, some heat would be lost to the surroundings during the heating process, potentially increasing the required time.
Temperature-dependent properties: The specific heat capacity of water varies slightly with temperature, but this variation is small enough to ignore for this calculation.
Heating element coverage: The calculation assumes uniform heating. In practice, the heating rate might be affected by the design of the heating element and water container.
Power fluctuations: The calculation assumes constant power output from the heater.
Convection currents: Natural convection in the water helps distribute heat but might not be perfectly efficient, potentially leading to temperature stratification.

This calculation demonstrates the practical application of thermodynamics in everyday situations like heating water for household or industrial purposes. It also illustrates the importance of considering efficiency in energy transfer processes.

References

Maharashtra State Board 10th Standard Science Syllabus 2025-26
NCERT Science Textbook for Class 10
Thermal Physics: Concepts and Practice
Heat Transfer: Principles and Applications
Fundamentals of Thermodynamics and Heat Transfer
Applied Physics: Concepts and Applications

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