Содержимое курса
Additional and Extra materials
Here you can find different useful links, books and worksheets
0/6
General physics

    Specific Heat Capacity and Specific Latent Heat — Physics Lesson


    🎯 Learning Objectives

    By the end of this lesson, students will be able to:

    • Define and use specific heat capacity (14.3.1)
    • Define and use specific latent heat (14.3.2)
    • Distinguish between specific latent heat of fusion and specific latent heat of vaporisation
    • Apply thermal energy calculations to real-world scenarios
    • Analyze energy changes during phase transitions


    🗣️ Language Objectives

    Students will develop their ability to:

    • Use scientific terminology accurately when describing thermal properties
    • Explain energy transfer processes using appropriate vocabulary
    • Interpret and describe calculations involving thermal energy
    • Communicate thermal physics concepts clearly in written and oral form
    • Read and understand scientific texts about heat and energy


    📚 Key Terms
    English TermRussian TranslationKazakh Translation
    Specific heat capacityУдельная теплоёмкостьМеншікті жылу сыйымдылығы
    Specific latent heatУдельная теплота парообразованияМеншікті жасырын жылу
    Latent heat of fusionУдельная теплота плавленияБалқудың меншікті жасырын жылуы
    Latent heat of vaporisationУдельная теплота испаренияБуландырудың меншікті жасырын жылуы
    Phase transitionФазовый переходФазалық ауысу
    Thermal energyТепловая энергияЖылу энергиясы
    Temperature changeИзменение температурыТемператураның өзгеруі
    MassМассаМасса


    🎴 Study Flashcards

    Practice with these interactive flashcards to master thermal physics terminology:

    Click through each card to test your understanding of key thermal concepts!


    📖 Glossary

    Essential Thermal Physics Terminology

    Specific Heat Capacity (c): The amount of thermal energy required to raise the temperature of 1 kg of a substance by 1 K (or 1°C).

    Translation
    Russian: Удельная теплоёмкость — количество тепловой энергии, необходимое для повышения температуры 1 кг вещества на 1 K (или 1°C).
    Kazakh: Меншікті жылу сыйымдылығы — заттың 1 кг массасының температурасын 1 K (немесе 1°C) көтеру үшін қажетті жылу энергиясының мөлшері.

    Specific Latent Heat (L): The amount of thermal energy required to change the phase of 1 kg of a substance without changing its temperature.

    Translation
    Russian: Удельная скрытая теплота — количество тепловой энергии, необходимое для изменения фазового состояния 1 кг вещества без изменения его температуры.
    Kazakh: Меншікті жасырын жылу — заттың 1 кг массасының температурасын өзгертпей-ақ, фазалық күйін өзгерту үшін қажетті жылу энергиясының мөлшері.

    Specific Latent Heat of Fusion (Lf): The thermal energy required per unit mass to change a solid to a liquid at its melting point.

    Translation
    Russian: Удельная теплота плавления — тепловая энергия на единицу массы, необходимая для превращения твёрдого тела в жидкость при температуре плавления.
    Kazakh: Балқудың меншікті жасырын жылуы — қатты дененің балқу температурасында сұйыққа айналуы үшін бірлік массаға қажетті жылу энергиясы.

    Specific Latent Heat of Vaporisation (Lv): The thermal energy required per unit mass to change a liquid to a gas at its boiling point.

    Translation
    Russian: Удельная теплота испарения — тепловая энергия на единицу массы, необходимая для превращения жидкости в газ при температуре кипения.
    Kazakh: Буландырудың меншікті жасырын жылуы — сұйықтың қайнау температурасында газға айналуы үшін бірлік массаға қажетті жылу энергиясы.

    Phase Transition: The process of changing from one state of matter (solid, liquid, gas) to another.

    Translation
    Russian: Фазовый переход — процесс изменения одного агрегатного состояния вещества (твёрдое, жидкое, газообразное) в другое.
    Kazakh: Фазалық ауысу — заттың бір күйінен (қатты, сұйық, газ тәрізді) екінші күйіне ауысу процесі.


    🔬 Theory: Understanding Thermal Energy and Phase Changes

    Specific Heat Capacity

    Specific heat capacity is a property that describes how much thermal energy a substance can store per unit mass per degree of temperature change. Different materials have different specific heat capacities, which explains why some materials heat up faster than others.

    Kazakh Translation
    Меншікті жылу сыйымдылығы — бұл заттың бірлік массасының температурасы бір градусқа өзгергенде қанша жылу энергиясын сақтай алатынын сипаттайтын қасиет. Әртүрлі материалдардың меншікті жылу сыйымдылығы әртүрлі болады, сондықтан кейбір материалдар басқаларына қарағанда тезірек қызады.

    The formula for thermal energy transfer during temperature change is:

    Q = mcΔT

    Where:

    • Q = thermal energy transferred (J)
    • m = mass (kg)
    • c = specific heat capacity (J kg⁻¹ K⁻¹)
    • ΔT = temperature change (K or °C)

    Specific Latent Heat

    During phase transitions, energy is required to break or form intermolecular bonds without changing temperature. This energy is called latent heat because it remains hidden — the temperature doesn’t change even though energy is being added or removed.

    Kazakh Translation
    Фазалық ауысулар кезінде температураны өзгертпей-ақ молекулааралық байланыстарды үзу немесе түзу үшін энергия қажет. Бұл энергия жасырын жылу деп аталады, өйткені ол жасырын болып қалады — энергия қосылса да немесе алынып тасталса да температура өзгермейді.

    The formula for latent heat energy transfer is:

    Q = mL

    Where:

    • Q = thermal energy transferred (J)
    • m = mass (kg)
    • L = specific latent heat (J kg⁻¹)

    Types of Latent Heat

    Latent Heat of Fusion (Lf): Energy required for solid ↔ liquid transitions. Typical values: ice = 334,000 J kg⁻¹

    Latent Heat of Vaporisation (Lv): Energy required for liquid ↔ gas transitions. Typical values: water = 2,260,000 J kg⁻¹

    Kazakh Translation
    Балқудың жасырын жылуы: қатты ↔ сұйық ауысулары үшін қажетті энергия. Типтік мәндер: мұз = 334,000 Дж кг⁻¹
    Буландырудың жасырын жылуы: сұйық ↔ газ ауысулары үшін қажетті энергия. Типтік мәндер: су = 2,260,000 Дж кг⁻¹

    Practice Questions

    1. (Easy) What is specific heat capacity?
    2. Answer
      Specific heat capacity is the amount of thermal energy required to raise the temperature of 1 kg of a substance by 1 K (or 1°C).
    3. (Medium) Calculate the energy required to heat 2 kg of water from 20°C to 80°C. (cwater = 4,200 J kg⁻¹ K⁻¹)
    4. Answer
      Q = mcΔT = 2 kg × 4,200 J kg⁻¹ K⁻¹ × (80-20) K = 2 × 4,200 × 60 = 504,000 J = 504 kJ
    5. (Medium) Why is more energy needed to vaporise water than to melt ice of the same mass?
    6. Answer
      Vaporisation requires breaking all intermolecular bonds to separate molecules completely, while melting only requires weakening bonds enough for molecules to move past each other. Lv > Lf for the same substance.
    7. (Hard — Critical Thinking) A 0.5 kg ice cube at -10°C is heated until it becomes steam at 110°C. Calculate the total energy required. (cice = 2,100 J kg⁻¹ K⁻¹, Lf = 334,000 J kg⁻¹, cwater = 4,200 J kg⁻¹ K⁻¹, Lv = 2,260,000 J kg⁻¹, csteam = 2,000 J kg⁻¹ K⁻¹)
    8. Answer
      Step 1: Heat ice from -10°C to 0°C: Q₁ = 0.5 × 2,100 × 10 = 10,500 J
      Step 2: Melt ice at 0°C: Q₂ = 0.5 × 334,000 = 167,000 J
      Step 3: Heat water from 0°C to 100°C: Q₃ = 0.5 × 4,200 × 100 = 210,000 J
      Step 4: Vaporise water at 100°C: Q₄ = 0.5 × 2,260,000 = 1,130,000 J
      Step 5: Heat steam from 100°C to 110°C: Q₅ = 0.5 × 2,000 × 10 = 10,000 J
      Total: Q = 10,500 + 167,000 + 210,000 + 1,130,000 + 10,000 = 1,527,500 J = 1.53 MJ


    🧠 Exercises on Memorizing Terms

    Term Recognition Practice

    1. Define specific heat capacity and give its units.
    2. What is the difference between latent heat of fusion and latent heat of vaporisation?
    3. Write the formula for calculating thermal energy during temperature change.
    4. Write the formula for calculating energy during phase transitions.
    5. Why is latent heat called «latent» (hidden)?
    6. Name three examples of phase transitions.
    Answer
    1. Specific heat capacity is the thermal energy required to raise 1 kg of substance by 1 K. Units: J kg⁻¹ K⁻¹
    2. Fusion relates to solid-liquid transitions; vaporisation relates to liquid-gas transitions
    3. Q = mcΔT
    4. Q = mL
    5. Because temperature doesn’t change during phase transitions, so the energy appears «hidden»
    6. Melting, freezing, boiling, condensation, sublimation, deposition


    📹 Educational Video


    🔧 Worked Examples

    Problem Solving with Thermal Energy Calculations

    Example 1: Heating Water

    Calorimetry experiment setup

    Problem: How much energy is needed to heat 3 kg of water from 25°C to 95°C? (cwater = 4,200 J kg⁻¹ K⁻¹)

    Step-by-step Solution
    Given:
    Mass (m) = 3 kg
    Initial temperature = 25°C
    Final temperature = 95°C
    Specific heat capacity of water (c) = 4,200 J kg⁻¹ K⁻¹

    Step 1: Calculate temperature change
    ΔT = 95°C — 25°C = 70°C = 70 K

    Step 2: Apply the formula
    Q = mcΔT
    Q = 3 kg × 4,200 J kg⁻¹ K⁻¹ × 70 K
    Q = 882,000 J = 882 kJ

    Answer: 882 kJ of energy is required.

    Example 2: Phase Change Calculation

    Heating curve of water

    Problem: Calculate the energy required to convert 2 kg of ice at 0°C to water at 0°C. (Lf = 334,000 J kg⁻¹)

    Detailed Solution
    Given:
    Mass (m) = 2 kg
    Specific latent heat of fusion (Lf) = 334,000 J kg⁻¹
    Temperature remains constant at 0°C

    Step 1: Identify the process
    This is a phase change from solid to liquid at melting point

    Step 2: Apply the latent heat formula
    Q = mLf
    Q = 2 kg × 334,000 J kg⁻¹
    Q = 668,000 J = 668 kJ

    Answer: 668 kJ of energy is required to melt the ice.


    🎮 Interactive Investigation

    States of Matter and Phase Changes Simulator

    Use this simulation to explore how energy affects molecular motion and phase changes:

    Investigation Questions:

    1. What happens to molecular motion when you add energy to a solid?
    2. At what point does temperature stop increasing even though you’re adding energy?
    3. How does the energy required for melting compare to vaporisation?
    Brief Answers
    1. Molecular vibrations increase with energy addition, eventually breaking bonds for phase change.
    2. Temperature stops increasing during phase transitions when energy goes into breaking bonds.
    3. Vaporisation typically requires more energy than melting because all intermolecular bonds must be broken.


    👥 Collaborative Learning Activity

    Thermal Energy Challenge

    Work in pairs or small groups to complete this interactive activity:

    Group Discussion Points:

    • Compare specific heat capacities of different materials and explain practical implications
    • Discuss why water has such a high specific heat capacity and its environmental significance
    • Design experiments to measure specific heat capacity and latent heat values
    • Analyze energy efficiency in cooking and heating applications


    📝 Individual Assessment - Structured Questions

    Advanced Thermal Physics Analysis Problems

    Problem 1 — Analysis

    A kettle contains 1.5 kg of water at 20°C. The heating element provides 2.5 kW of power.

    a) Calculate the time required to heat the water to boiling point (100°C).

    b) If the kettle continues heating, how long would it take to boil away all the water?

    Use: cwater = 4,200 J kg⁻¹ K⁻¹, Lv = 2,260,000 J kg⁻¹

    Answer
    a) Energy needed: Q₁ = mcΔT = 1.5 × 4,200 × (100-20) = 504,000 J
    Time = Energy/Power = 504,000 J ÷ 2,500 W = 201.6 s = 3.36 minutes

    b) Energy to vaporise: Q₂ = mLv = 1.5 × 2,260,000 = 3,390,000 J
    Time = 3,390,000 J ÷ 2,500 W = 1,356 s = 22.6 minutes
    Total time = 3.36 + 22.6 = 26 minutes

    Problem 2 — Synthesis

    Design a calorimetry experiment to determine the specific heat capacity of an unknown metal. Include: equipment needed, procedure, calculations, and sources of error.

    Answer
    Equipment: Calorimeter, thermometer, balance, hot water, unknown metal sample
    Procedure: Heat metal to known temperature, place in calorimeter with cold water, measure final temperature
    Calculation: Heat lost by metal = Heat gained by water: mmetalcmetalΔTmetal = mwatercwaterΔTwater
    Sources of error: Heat loss to environment, calorimeter heat capacity, measurement uncertainties

    Problem 3 — Evaluation

    Coastal regions have more moderate temperatures than inland areas. Using thermal physics principles, evaluate this statement and explain the role of water’s high specific heat capacity.

    Answer
    Statement is correct. Water’s high specific heat capacity (4,200 J kg⁻¹ K⁻¹) means:
    — Large bodies of water resist temperature changes
    — Ocean/lake temperatures remain relatively stable
    — Coastal air temperatures moderated by water temperature
    — Inland areas lack this thermal buffer, experiencing greater temperature variations
    This creates maritime vs continental climate patterns.

    Problem 4 — Application

    A 60 W electric heater is used to melt ice in a well-insulated container. If 0.8 kg of ice at 0°C takes 45 minutes to completely melt, calculate the specific latent heat of fusion for ice and compare to the accepted value.

    Answer
    Given: Power = 60 W, mass = 0.8 kg, time = 45 min = 2,700 s
    Energy supplied: E = Pt = 60 × 2,700 = 162,000 J
    Calculated Lf: Lf = E/m = 162,000/0.8 = 202,500 J kg⁻¹
    Accepted value: 334,000 J kg⁻¹
    Percentage difference: (334,000-202,500)/334,000 × 100% = 39% lower
    Possible reasons: Heat losses, imperfect insulation, measurement errors

    Problem 5 — Critical Analysis

    Steam burns are often more severe than burns from boiling water at the same temperature. Using latent heat principles, critically analyze this observation and calculate the energy difference for 10 g of steam vs 10 g of boiling water cooling to skin temperature (37°C).

    Answer
    Steam condensation energy: Q₁ = mLv = 0.01 × 2,260,000 = 22,600 J
    Cooling condensed water: Q₂ = mcΔT = 0.01 × 4,200 × (100-37) = 2,646 J
    Total energy from steam: 22,600 + 2,646 = 25,246 J
    Energy from boiling water: Q₃ = 0.01 × 4,200 × (100-37) = 2,646 J
    Ratio: Steam releases 25,246/2,646 = 9.5 times more energy
    Conclusion: Steam burns are more severe because condensation releases enormous latent heat energy directly to skin.



    🤔 Lesson Reflection

    Self-Assessment and Reflection

    Take a moment to reflect on your learning by answering these questions:

    1. Understanding: Can you clearly explain the difference between specific heat capacity and specific latent heat?
    2. Application: What real-world examples can you now identify and explain using thermal physics principles?
    3. Calculations: How confident do you feel solving thermal energy problems involving both temperature changes and phase transitions?
    4. Connections: How does this lesson connect to other physics topics like conservation of energy?
    5. Questions: What aspects of thermal physics would you like to explore further?

    Learning Goals Check:

    Rate your confidence (1-5 scale) on each learning objective:

    • __ Defining and using specific heat capacity
    • __ Defining and using specific latent heat
    • __ Distinguishing between fusion and vaporisation
    • __ Applying thermal energy calculations
    • __ Analyzing phase change processes

    Areas where you rated yourself 3 or below should be revisited using the additional resources provided.

    Practical Applications:

    Consider how thermal physics principles apply to:

    • Cooking and food preparation
    • Weather and climate patterns
    • Industrial heating and cooling systems
    • Energy efficiency in buildings