Calorimetry
ICSE Grade 10 Physics - Chapter 11
📚 Smart Summary
1. Heat and Temperature
Heat: A form of energy that flows from hotter to colder body. SI unit: Joule (J).
Temperature: Measure of degree of hotness or coldness. SI unit: Kelvin (K).
Difference: Heat is energy transfer; temperature is a measure of thermal state.
Thermometer: Device to measure temperature using expansion/contraction of substances.
Scales: Celsius (°C), Fahrenheit (°F), Kelvin (K). Relation: K = °C + 273.
2. Specific Heat Capacity
Definition: Heat required to raise temperature of 1 kg of substance by 1°C (or 1 K).
Formula: Q = mcΔT, where Q = heat, m = mass, c = specific heat capacity, ΔT = temperature change.
SI Unit: J kg⁻¹ K⁻¹ or J kg⁻¹ °C⁻¹.
Water: Has high specific heat capacity (4200 J kg⁻¹ K⁻¹), used as coolant.
Practical Applications: Climate regulation near water bodies, car radiators, hot water bottles.
3. Heat Capacity
Definition: Heat required to raise temperature of entire body by 1°C (or 1 K).
Formula: Heat Capacity = mass × specific heat capacity = mc.
SI Unit: J K⁻¹ or J °C⁻¹.
Relation: Q = Heat Capacity × ΔT.
Difference: Specific heat capacity is per unit mass; heat capacity is for whole body.
4. Latent Heat
Definition: Heat required to change state of substance without changing temperature.
Latent Heat of Fusion (Lf): Heat required to convert 1 kg solid to liquid at melting point. For ice: 336 kJ/kg.
Latent Heat of Vaporization (Lv): Heat required to convert 1 kg liquid to gas at boiling point. For water: 2260 kJ/kg.
Formula: Q = mL, where L = latent heat (Lf or Lv).
Anomaly: During state change, temperature remains constant despite heat absorption.
5. Principle of Calorimetry
Principle of Heat Exchange: Heat lost by hot body = Heat gained by cold body (in isolated system).
Assumption: No heat exchange with surroundings (ideal calorimeter).
Formula: m₁c₁ΔT₁ = m₂c₂ΔT₂.
Applications: Determining specific heat capacity, heat capacity, and latent heat of substances.
Method of Mixtures: Used to find specific heat of solids using calorimeter.
6. Change of State
Melting: Solid to liquid (requires Lf). Melting point of ice: 0°C.
Freezing: Liquid to solid (releases Lf). Freezing point of water: 0°C.
Boiling: Liquid to gas (requires Lv). Boiling point of water: 100°C at 1 atm.
Condensation: Gas to liquid (releases Lv).
Evaporation: Liquid to gas at any temperature (surface phenomenon); causes cooling.
Sublimation: Direct conversion between solid and gas (e.g., camphor, dry ice).
📐 Formulas
Heat for temperature change
Heat for state change (L = Lf or Lv)
Total heat capacity of a body
Principle of calorimetry
Conversion between Celsius and Kelvin
🎯 Test Your Knowledge
Multiple Choice Questions
1. The SI unit of heat is:
Knowledge2. The SI unit of specific heat capacity is:
Knowledge3. The specific heat capacity of water is:
Knowledge4. Latent heat of fusion of ice is:
Knowledge5. During melting of ice at 0°C:
Comprehension6. The principle of calorimetry states that:
Knowledge7. Water is used as a coolant because:
Comprehension8. Evaporation causes cooling because:
Comprehension9. Assertion (A): Ice at 0°C is more effective for cooling than water at 0°C. Reason (R): Ice absorbs latent heat while melting.
Assertion-Reasoning10. The latent heat of vaporization of water is:
KnowledgePractice Numericals
Practice Problem 1: How much heat is required to raise the temperature of 2 kg of water from 20°C to 70°C? (Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹). Answer in kJ.
ApplicationPractice Problem 2: Calculate the heat required to melt 5 kg of ice at 0°C. (Latent heat of fusion = 336 kJ/kg). Answer in kJ.
ApplicationPractice Problem 3: How much heat is required to convert 0.5 kg of water at 100°C to steam at 100°C? (Lv = 2260 kJ/kg). Answer in kJ.
ApplicationPractice Problem 4: A metal piece of mass 0.5 kg at 100°C is dropped into 1 kg of water at 20°C. The final temperature is 25°C. Find the specific heat capacity of metal (in J kg⁻¹ K⁻¹). (Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹).
ApplicationPractice Problem 5: Calculate the total heat required to convert 2 kg of ice at 0°C to water at 50°C. (Lf = 336 kJ/kg, c of water = 4200 J kg⁻¹ K⁻¹). Answer in kJ.
Application