Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Heat Energy Measurements.
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Heat Energy Measurements.
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Subject: Physics
Class: Senior Secondary 2
Term: 1st Term
Week: 7
Theme: Conservation Principles
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Student should beable to:Explain the relationshipbetween the heat supplied toa substance and(a) its temperaturechange at constant mass.(b) It's mass at constanttemperaturechange Explain whythere is unequalrise in temperature for differentsubstances of the same masssupplied with the samequantity of heat Explain the terms, specificheat capacityand the rmalcapacity. calculateunknownquantities usingthe relation H =f\1C T when nochan~ of stateis in volved. determine the melting point ofa solid and boiling point of agiven iiquid. List the effectsof impurities and pressure on the melting point of solid and boilding point of liquid.
This section provides in-depth explanations of the core concepts related to heat energy measurements. 2.1 Relationship between Heat Supplied, Mass, and Temperature Change When heat is supplied to a substance, its temperature generally increases. The amount of heat (H) required to change the temperature of a substance is directly related to three factors: Mass (m) of the substance: For a constant temperature change, a larger mass requires more heat. (e.g., heating a large pot of water requires more energy than heating a small cup of water to the same temperature).
Mathematically: H ∝ m (when ΔT and type of substance are constant). Temperature change (ΔT): For a constant mass, a larger temperature change requires more heat. (e.g., boiling water from room temperature requires more heat than just warming it slightly).
Mathematically: H ∝ ΔT (when m and type of substance are constant).
Nature of the substance: Different substances respond differently to the same amount of heat. Some substances heat up quickly, while others heat up slowly. Combining these proportionalities, we get: H ∝ m ΔT To turn this into an equation, we introduce a constant of proportionality, which is called the specific heat capacity (c) of the substance. H = mcΔT Where: H = Quantity of heat supplied or absorbed (in Joules, J) m = Mass of the substance (in kilograms, kg) c = Specific heat capacity of the substance (in Joules per kilogram per Kelvin, J kg−1 K−1 or J kg−1 °C−1) ΔT = Change in temperature (in Kelvin, K or degrees Celsius, °C). Note that a change of 1 K is equal to a change of 1 °C. 2.2 Specific Heat Capacity (c)
Definition: The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of a unit mass (1 kg) of that substance by one degree Celsius (1 °C) or one Kelvin (1 K).
Units: J kg−1 K−1 or J kg−1 °C−
1. Explanation of Unequal Rise in Temperature (Performance Objective 2): Different substances have different specific heat capacities. For example, water has a high specific heat capacity (approx. 4200 J kg−1 K−1), while metals like iron have much lower specific heat capacities (e.g., iron approx. 450 J kg−1 K−1). If the same quantity of heat (H) is supplied to equal masses (m) of two different substances (e.g., water and cooking oil), the substance with the lower specific heat capacity (c) will experience a larger temperature change (ΔT). Rearranging H = mcΔT, we get ΔT = H / (mc). Since H and m are constant, ΔT is inversely proportional to c.
Therefore, a substance with a smaller 'c' will have a larger 'ΔT'. This is why a metal spoon heats up much faster than water when both are exposed to the same heat source. 2.3 Thermal Capacity (Heat Capacity - C)
Definition: The thermal capacity (or heat capacity) of a body is the amount of heat energy required to raise the temperature of the entire body by one degree Celsius (1 °C) or one Kelvin (1 K).
Units: J K−1 or J °C−
1. Relationship to Specific Heat Capacity: Thermal Capacity (C) = mass (m) × specific heat capacity (c) C = mc Therefore, the heat supplied (H) can also be expressed as: H = CΔT Example for Distinction: A 1 kg block of iron has a specific heat capacity of 450 J kg−1 K−
1. Its thermal capacity is 1 kg × 450 J kg−1 K−1 = 450 J K−
1. A 2 kg block of iron has a specific heat capacity of 450 J kg−1 K−
1. Its thermal capacity is 2 kg × 450 J kg−1 K−1 = 900 J K−
1. The specific heat capacity is an intensive property (independent of the amount of substance), while thermal capacity is an extensive property (depends on the amount of substance). 2.4 Calculations using H = mcΔT Worked Example 1: A student heats 500 g of water from 25 °C to 75 °C. Calculate the quantity of heat supplied to the water. (Specific heat capacity of water = 4200 J kg−1 K−1).
Solution: Given: Mass of water (m) higher temperatures, food cooks much faster in a pressure cooker. This is particularly useful in regions with lower atmospheric pressure (e.g., high-altitude areas like the Jos Plateau in Nigeria), where water boils at lower temperatures, making conventional cooking slower. 2.9 Effects of Humidity on Personal Comfort Humidity: The amount of water vapor present in the air. High humidity means there is a lot of moisture in the air.
Mechanism of Discomfort: The human body cools itself primarily through the evaporation of sweat from the skin. When the surrounding air has high humidity, it is already saturated with water vapor. This reduces the rate at which sweat can evaporate from the skin. * Effect: With reduced sweat evaporation, the body's natural cooling mechanism becomes less efficient. This leads to a sensation of being much hotter and more uncomfortable than the actual air temperature suggests, even making the air feel "sticky" or "heavy." This is a common experience in many coastal and southern parts of Nigeria, especially during the rainy season, where high temperatures combine with high humidity. --- Effects of Pressure: Melting Point: For most substances (like water), increasing pressure lowers the melting point slightly. This is an unusual property for water because ice is less dense than liquid water. For substances that contract on melting, increasing pressure raises the melting point. (e.g., the pressure from ice skates can slightly lower the melting point of ice, creating a thin layer of water for gliding).
Boiling Point: Increasing external pressure raises the boiling point of a liquid. Conversely, decreasing pressure lowers the boiling point. (e.g., water boils at a lower temperature at high altitudes like Jos Plateau due to lower atmospheric pressure, making cooking take longer. A pressure cooker, by increasing pressure, raises the boiling point of water, allowing food to cook faster). 2.7 Distinguishing Evaporation, Boiling, and Explaining Sublimation Evaporation: Definition: The process by which a liquid changes into a gas (vapor) at temperatures below its boiling point.
Mechanism: Occurs only at the surface of the liquid. Energetic molecules at the surface escape into the atmosphere.
Effect: Causes cooling (e.g., sweat evaporating from the skin cools the body).
Rate: Influenced by temperature, surface area, humidity, and air movement.
Boiling: Definition: The rapid vaporization of a liquid that occurs when the temperature of the liquid reaches its boiling point, forming bubbles of vapor throughout the liquid.
Mechanism: Occurs throughout the bulk of the liquid, not just at the surface.
Effect: Occurs at a specific constant temperature for a pure substance at a given pressure.
Rate: Requires continuous heat supply at the boiling point.
Sublimation: Definition: The process by which a substance changes directly from a solid to a gas (vapor) without passing through the liquid state.
Examples: Naphthalene balls (mothballs) used in Nigerian homes, solid carbon dioxide (dry ice) used for cooling, camphor. These substances gradually disappear over time without melting. 2.8 Working Principles of Common Devices Refrigerator/Air Conditioner: Principle: Both devices work on the principle of using the cooling effect of evaporation of a specialized fluid called a refrigerant (e.g., Freon, R-134a).
Cycle (Simplified):
1. Evaporator Coils (inside refrigerator/AC unit): The liquid refrigerant absorbs heat from the food compartment (refrigerator) or the room air (air conditioner) and evaporates into a low-pressure gas. This absorption of latent heat causes cooling.
2. Compressor: The gaseous refrigerant is compressed, increasing its pressure and temperature.
3. Condenser Coils (outside refrigerator/AC unit): The hot, high-pressure gaseous refrigerant releases heat to the surrounding environment (kitchen air or outdoor air) and condenses back into a high-pressure liquid. This is why the back of a fridge or the outdoor unit of an AC feels warm.
4. Expansion Valve: The high-pressure liquid refrigerant passes through an expansion valve, which causes its pressure and temperature to drop rapidly, preparing it to absorb heat again in the evaporator coils. The continuous cycle of evaporation and condensation, driven by the compressor, effectively transfers heat from inside to outside, cooling the desired space.
Pressure Cooker: Principle: Works by increasing the pressure inside the cooker, which in turn raises the boiling point of water.
Mechanism: The lid of the pressure cooker forms an airtight seal, trapping steam inside. As steam builds up, the pressure inside increases significantly (often to 2-3 times atmospheric pressure). According to the relationship between pressure and boiling point, this higher pressure causes the water to boil at a temperature much higher than 100 °C (e.g., 120 °C or more).
Advantage: Since cooking reactions (like the softening of tough meat or beans) speed up considerably at higher temperatures, food cooks much faster in a pressure cooker. This is particularly useful in regions with lower atmospheric pressure (e.g., high-altitude areas like the Jos Plateau in Nigeria), where water boils at lower temperatures, making conventional cooking slower. 2.9 Effects of Humidity on Personal Comfort Humidity: The amount of water vapor present in the air. High humidity means there is a lot of moisture in the air. * Mechanism of Discomfort: The human body cools itself primarily through the evaporation of sweat from the skin. When the surrounding air has J kg−1 K−
1. Its thermal capacity is 2 kg × 450 J kg−1 K−1 = 900 J K−
1. The specific heat capacity is an intensive property (independent of the amount of substance), while thermal capacity is an extensive property (depends on the amount of substance). 2.4 Calculations using H = mcΔT Worked Example 1: A student heats 500 g of water from 25 °C to 75 °C. Calculate the quantity of heat supplied to the water. (Specific heat capacity of water = 4200 J kg−1 K−1).
Solution: Given: Mass of water (m) = 500 g = 0.5 kg Initial temperature (T1) = 25 °C Final temperature (T2) = 75 °C Specific heat capacity of water (c) = 4200 J kg−1 K−1 Change in temperature (ΔT) = T2 - T1 = 75 °C - 25 °C = 50 °C (or 50 K) Using the formula H = mcΔT: H = 0.5 kg × 4200 J kg−1 K−1 × 50 K H = 2100 × 50 J H = 105,000 J or 105 kJ 2.5 Phase Changes and Latent Heat When a substance undergoes a change of state (e.g., from solid to liquid, or liquid to gas), heat is absorbed or released without a change in temperature. This heat is called latent heat.
Melting Point: The specific temperature at which a solid changes into a liquid at a given pressure. During melting, the temperature remains constant even though heat is being absorbed.
Boiling Point: The specific temperature at which a liquid changes into a gas (vaporizes rapidly) at a given pressure, forming bubbles throughout the liquid. During boiling, the temperature remains constant even though heat is being absorbed.
Latent Heat (L): Definition: The latent heat is the amount of heat energy absorbed or released by a substance during a phase change at a constant temperature.
Specific Latent Heat: The specific latent heat of a substance is the amount of heat energy required to change the state of a unit mass (1 kg) of the substance without a change in its temperature. Specific Latent Heat of Fusion (L_f): The heat absorbed per unit mass when a substance changes from solid to liquid (melts) at its melting point.
Formula: H = mL_f Units: J kg−1
Example: Specific latent heat of fusion of ice is approximately 334,000 J kg−
1. This means 334,000 J of heat is needed to melt 1 kg of ice at 0 °C into 1 kg of water at 0 °
C. Specific Latent Heat of Vaporization (L_v): The heat absorbed per unit mass when a substance changes from liquid to gas (vaporizes) at its boiling point.
Formula: H = mL_v Units: J kg−1
Example: Specific latent heat of vaporization of water is approximately 2,260,000 J kg−
1. This means 2,260,000 J of heat is needed to turn 1 kg of water at 100 °C into 1 kg of steam at 100 °C. 2.6 Effects of Impurities and Pressure on Melting and Boiling Points Effects of Impurities: Melting Point: Adding impurities to a solid generally lowers its melting point and makes the melting process occur over a range of temperatures, rather than at a single sharp temperature. (e.g., adding salt to ice to make it melt faster, often used to clear icy roads in colder climates, or to make local ice cream mixtures colder).
Boiling Point: Adding non-volatile impurities (like salt) to a liquid generally raises its boiling point. (e.g., water boils at a slightly higher temperature if salt is added, which can sometimes be useful in cooking certain foods).
Effects of Pressure: Melting Point: For most substances (like water), increasing pressure lowers the melting point slightly. This is an unusual property for water because ice is less dense than liquid water. For substances that contract on melting, increasing pressure raises the melting point. (e.g., the pressure from ice skates can slightly lower the melting point of ice, creating a thin layer of water for gliding).
Boiling Point: Increasing external pressure raises the boiling point of a liquid. Conversely, decreasing pressure lowers the boiling point. (e.g., water boils at a
Food Preservation and Storage in Homes and Markets: Application: Refrigerators are ubiquitous in Nigerian homes and businesses (e.g., market stalls selling cold drinks) for preserving food and beverages.
Integration: The lesson explicitly covers the working principle of a refrigerator, which uses the latent heat of vaporization of a refrigerant to remove heat from inside the unit and dissipate it into the surroundings. This prevents food spoilage by slowing down bacterial growth. Students can discuss how the design of a refrigerator, with its insulated walls, minimizes heat entry from the warm Nigerian climate.
Efficient Cooking and Meal Preparation: Application: Understanding specific heat capacity explains why some foods (like water-based soups) retain heat longer, and why different cooking methods are used. Pressure cookers are gaining popularity in Nigeria for faster cooking of tough staples like beans or meat.
Integration: The concept of specific heat capacity helps explain why a 'heavy' cast-iron pot retains heat longer than a 'light' aluminum pot. The explanation of the pressure cooker's principle (raising the boiling point of water by increasing pressure) directly relates to faster and more efficient cooking, particularly in high-altitude areas like Jos where atmospheric pressure is lower. This is crucial for energy conservation and time management in busy Nigerian households.
Personal Comfort and Climate Control: Application: Many Nigerians use fans or air conditioners to cope with heat. Understanding humidity is key to explaining why some hot days feel more oppressive than others.
Integration: The lesson explains how the body cools itself through sweat evaporation, and how high humidity (common in coastal cities like Lagos, Port Harcourt, Calabar) hinders this process, leading to discomfort. Air conditioners, whose working principle is discussed, provide comfort by not only cooling but also de-humidifying the air. Students can relate to experiences of 'sticky' heat and how a simple fan, by increasing air flow, aids evaporation and provides relief. ---