Lesson Note on Heat Energy Measurement: Specific Heat Capacity and Method of Mixtures for SSS 2

Class: Senior Secondary School 2 (SS2 / SSS2)
Term: 1st Term
Week: 2
Age: 16 years
Duration: 45 minutes
Subject: Physics
Curriculum Theme: Heat Energy
Previous Lesson: Heat Energy: Meaning, Thermometers and Temperature Scales.
Topic: HEAT ENERGY MEASUREMENT
Subject Matter: Meaning of specific heat capacity (heat needed to raise unit mass by 1°C or 1K), concept of specific heat capacity (its measurement, its significance), determination of specific heat capacity of a solid (method of mixtures), determination of specific heat capacity of a liquid (method of mixtures).

Specific Objectives

By the end of the lesson, pupils should be able to:

Cognitive Domain:

• Define specific heat capacity.
• Explain the significance of specific heat capacity in everyday life.
• State the principle of the method of mixtures.

Affective Domain:

• Appreciate the importance of specific heat capacity in various applications.
• Show interest in performing experiments to determine specific heat capacity.

Psychomotor Domain:

• Set up the apparatus for determining specific heat capacity of a solid and a liquid using the method of mixtures.
• Record accurate measurements of mass and temperature during experiments.
• Calculate the specific heat capacity of a solid and a liquid from experimental data.

Social Domain:

• Collaborate effectively in groups during practical activities.

Reference Materials

The following resources were used in planning this lesson:

• Senior Secondary Schools Education Curriculum
• State Unified Scheme of Work
• New School Physics by M.W. Anyakoha

Instructional Materials

The teacher will teach this lesson with the aid of:

• Physics textbook
• Calorimeter with stirrer
• Thermometers
• Electronic balance
• Stopwatch
• Known mass of solid (e.g., copper block)
• Known mass of liquid (e.g., water)
• Unknown liquid
• Steam heater/boiler
• Insulating materials

Rationale for the Lesson

This lesson enables pupils to understand how different materials store and release heat. It helps pupils to grasp the concept of specific heat capacity and its practical applications in engineering and daily life, such as in cooling systems and cooking.

Prerequisite/Previous Knowledge

Pupils should have prior knowledge of heat energy, temperature, mass, and basic algebraic calculations.

Lesson Content/Board Summary

Specific Heat Capacity

Meaning of Specific Heat Capacity (c)

Specific heat capacity of a substance is the quantity of heat energy required to raise the temperature of a unit mass (1 kg) of the substance by one degree Celsius (1°C) or one Kelvin (1K).

The formula for heat energy (Q) absorbed or released by a substance is:

Q = mcΔT

Where:

• Q = Quantity of heat energy (Joules, J)
• m = Mass of the substance (kilograms, kg)
• c = Specific heat capacity of the substance (Joules per kilogram per Kelvin, Jkg⁻¹K⁻¹ or Jkg⁻¹°C⁻¹)
• ΔT = Change in temperature (Kelvin, K or degrees Celsius, °C)

Significance of Specific Heat Capacity

The specific heat capacity of a material determines how quickly it heats up or cools down when heat is added or removed. Materials with high specific heat capacity absorb or release a large amount of heat for a small change in temperature, while materials with low specific heat capacity heat up or cool down quickly.

Examples of significance:

• Water as a coolant: Water has a high specific heat capacity, making it an excellent coolant in car engines and power stations, as it can absorb a large amount of heat without a drastic increase in its own temperature.
• Climate regulation: Large bodies of water (oceans, lakes) help to moderate the climate of coastal regions due to water’s high specific heat capacity, leading to smaller temperature fluctuations.
• Cooking utensils: Metals like aluminum and iron have relatively low specific heat capacities, allowing them to heat up quickly for cooking.

Determination of Specific Heat Capacity by Method of Mixtures

The method of mixtures is based on the principle of conservation of energy, which states that when hot and cold objects are mixed, the heat lost by the hot object(s) is equal to the heat gained by the cold object(s) and the calorimeter.

Principle: Heat Lost = Heat Gained

General equation: m₁c₁(T₁ – T₂) = m₂c₂(T₂ – T₃) + m₃c₃(T₂ – T₃)

Where subscripts 1, 2, 3 refer to different substances/calorimeter.

Determination of Specific Heat Capacity of a Solid

Apparatus: Calorimeter with stirrer, thermometer, balance, solid block (e.g., copper), steam heater, beaker, water.

Procedure:

1. Measure the mass of the calorimeter (m_c) and its specific heat capacity (c_c).
2. Pour a known mass of water (m_w) into the calorimeter and measure its initial temperature (T_1).
3. Heat the solid block (m_s) in a steam heater to a high temperature (T_s), which is usually the temperature of steam (100°C).
4. Quickly transfer the hot solid block into the water in the calorimeter.
5. Stir the mixture and record the final steady temperature (T_f).

Formula:

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

m_s * c_s * (T_s – T_f) = m_w * c_w * (T_f – T_1) + m_c * c_c * (T_f – T_1)

From this, c_s can be calculated.

Example 1:

A 0.2 kg copper block at 100°C is quickly transferred into a calorimeter containing 0.15 kg of water at 20°C. The mass of the calorimeter is 0.05 kg and its specific heat capacity is 400 Jkg⁻¹K⁻¹. If the final temperature of the mixture is 25°C, calculate the specific heat capacity of copper. (Specific heat capacity of water = 4200 Jkg⁻¹K⁻¹).

Solution:

Step 1: State the principle.

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

Step 2: List the given values.

• Mass of copper (m_s) = 0.2 kg
• Initial temperature of copper (T_s) = 100°C
• Mass of water (m_w) = 0.15 kg
• Initial temperature of water (T_1) = 20°C
• Mass of calorimeter (m_c) = 0.05 kg
• Specific heat capacity of calorimeter (c_c) = 400 Jkg⁻¹K⁻¹
• Final temperature (T_f) = 25°C
• Specific heat capacity of water (c_w) = 4200 Jkg⁻¹K⁻¹

Step 3: Apply the formula and substitute values.

m_s * c_s * (T_s – T_f) = m_w * c_w * (T_f – T_1) + m_c * c_c * (T_f – T_1)

0.2 * c_s * (100 – 25) = 0.15 * 4200 * (25 – 20) + 0.05 * 400 * (25 – 20)

0.2 * c_s * 75 = 0.15 * 4200 * 5 + 0.05 * 400 * 5

15 * c_s = 3150 + 100

15 * c_s = 3250

Step 4: Calculate c_s.

c_s = 3250 / 15

c_s = 216.67 Jkg⁻¹K⁻¹

Therefore, the specific heat capacity of copper is approximately 216.67 Jkg⁻¹K⁻¹.

Determination of Specific Heat Capacity of a Liquid

Apparatus: Calorimeter with stirrer, thermometer, balance, solid block (e.g., copper), steam heater, unknown liquid.

Procedure:

1. Measure the mass of the calorimeter (m_c) and its specific heat capacity (c_c).
2. Pour a known mass of the unknown liquid (m_L) into the calorimeter and measure its initial temperature (T_1).
3. Heat a solid block of known specific heat capacity (c_s) and mass (m_s) in a steam heater to a high temperature (T_s).
4. Quickly transfer the hot solid block into the unknown liquid in the calorimeter.
5. Stir the mixture and record the final steady temperature (T_f).

Formula:

Heat lost by solid = Heat gained by liquid + Heat gained by calorimeter

m_s * c_s * (T_s – T_f) = m_L * c_L * (T_f – T_1) + m_c * c_c * (T_f – T_1)

From this, c_L can be calculated.

Example 2:

A 0.1 kg brass block (specific heat capacity = 380 Jkg⁻¹K⁻¹) heated to 100°C is dropped into a calorimeter containing 0.2 kg of an unknown liquid at 20°C. The calorimeter has a mass of 0.08 kg and a specific heat capacity of 400 Jkg⁻¹K⁻¹. If the final temperature of the mixture is 28°C, calculate the specific heat capacity of the unknown liquid.

Solution:

Step 1: State the principle.

Heat lost by brass block = Heat gained by liquid + Heat gained by calorimeter

Step 2: List the given values.

• Mass of brass (m_s) = 0.1 kg
• Specific heat capacity of brass (c_s) = 380 Jkg⁻¹K⁻¹
• Initial temperature of brass (T_s) = 100°C
• Mass of liquid (m_L) = 0.2 kg
• Initial temperature of liquid (T_1) = 20°C
• Mass of calorimeter (m_c) = 0.08 kg
• Specific heat capacity of calorimeter (c_c) = 400 Jkg⁻¹K⁻¹
• Final temperature (T_f) = 28°C

Step 3: Apply the formula and substitute values.

m_s * c_s * (T_s – T_f) = m_L * c_L * (T_f – T_1) + m_c * c_c * (T_f – T_1)

0.1 * 380 * (100 – 28) = 0.2 * c_L * (28 – 20) + 0.08 * 400 * (28 – 20)

0.1 * 380 * 72 = 0.2 * c_L * 8 + 0.08 * 400 * 8

2736 = 1.6 * c_L + 256

Step 4: Solve for c_L.

1.6 * c_L = 2736 – 256

1.6 * c_L = 2480

c_L = 2480 / 1.6

c_L = 1550 Jkg⁻¹K⁻¹

Therefore, the specific heat capacity of the unknown liquid is 1550 Jkg⁻¹K⁻¹.

Teaching Methods/Instructional Techniques

Discussion, Lecture, Demonstration, Question and Answer, Visual Aids

Instructional Procedures

Step 1: Introduction

Time: 5 minutes

Teaching Skill: Set Induction

Teacher’s Activity: The teacher asks pupils what happens to a pot of water when placed on fire and how long it takes to boil compared to an empty pot. The teacher then introduces the topic: Heat Energy Measurement, focusing on Specific Heat Capacity.

Pupils’ Activity: Pupils respond based on their everyday experiences, noting that water takes longer to heat up.

Learning Point: Pupils recall prior knowledge of heat and temperature and are introduced to the lesson topic.

Step 2: Meaning of Specific Heat Capacity

Time: 7 minutes

Teaching Skill: Explanation/Definition

Teacher’s Activity: The teacher defines specific heat capacity, explains the formula Q = mcΔT, and identifies the units involved. The teacher gives examples of specific heat capacities of common materials.

Pupils’ Activity: Pupils listen attentively, take notes, and ask questions for clarification.

Learning Point: Pupils understand the definition, formula, and units of specific heat capacity.

Step 3: Significance of Specific Heat Capacity

Time: 5 minutes

Teaching Skill: Explanation/Illustrative

Teacher’s Activity: The teacher explains why different materials heat up or cool down at different rates, discussing practical applications like water as a coolant and its role in moderating climate.

Pupils’ Activity: Pupils contribute examples from their daily lives and discuss the importance of specific heat capacity.

Learning Point: Pupils grasp the real-world implications and significance of specific heat capacity.

Step 4: Introduction to Method of Mixtures

Time: 5 minutes

Teaching Skill: Principle Explanation

Teacher’s Activity: The teacher introduces the method of mixtures as a way to measure specific heat capacity, explaining the principle of heat lost equals heat gained and writing the general formula on the board.

Pupils’ Activity: Pupils listen, take notes, and understand the underlying principle of the method.

Learning Point: Pupils understand the theoretical basis for experimental determination of specific heat capacity.

Step 5: Determination of Specific Heat Capacity of a Solid

Time: 8 minutes

Teaching Skill: Demonstration/Calculation

Teacher’s Activity: The teacher demonstrates the setup for determining the specific heat capacity of a solid using the method of mixtures, guides pupils through the procedure, and solves a worked example on the board, emphasizing step-by-step calculations.

Pupils’ Activity: Pupils observe the demonstration, follow the calculation steps, and copy the example into their notes.

Learning Point: Pupils learn the practical steps and calculation method for determining the specific heat capacity of a solid.

Step 6: Determination of Specific Heat Capacity of a Liquid

Time: 8 minutes

Teaching Skill: Demonstration/Calculation

Teacher’s Activity: The teacher explains and demonstrates the procedure for determining the specific heat capacity of an unknown liquid, highlighting the differences from the solid determination, and solves another worked example on the board.

Pupils’ Activity: Pupils pay attention to the procedure and calculation, asking questions where necessary.

Learning Point: Pupils learn how to determine the specific heat capacity of a liquid through practical application and calculation.

Step 7: Evaluation/Review

Time: 5 minutes

Teaching Skill: Questioning/Assessment

Teacher’s Activity: The teacher evaluates the learning by asking the following questions:

1. Define specific heat capacity and state its S.I unit.
2. State the principle of the method of mixtures.
3. Explain two practical applications of specific heat capacity.
4. A 0.5 kg metal block at 90°C is placed in 0.3 kg of water at 25°C in a calorimeter of mass 0.1 kg and specific heat capacity 400 Jkg⁻¹K⁻¹. If the final temperature is 30°C, calculate the specific heat capacity of the metal. (Specific heat capacity of water = 4200 Jkg⁻¹K⁻¹).

Pupils’ Activity: Pupils answer orally and in writing.

Learning Point: Pupils demonstrate understanding of the lesson.

Step 8: Conclusion

Time: 2 minutes

Teaching Skill: Summarization

Teacher’s Activity: The teacher summarizes the key points of the lesson and assigns homework, which includes reviewing the examples and trying similar problems from their textbooks.

Pupils’ Activity: Pupils listen to the summary and copy down the homework assignment.

Learning Point: Pupils consolidate their learning and prepare for further practice.

Lesson Keywords

• Heat Energy – The total kinetic and potential energy of the particles in a substance.
• Specific Heat Capacity – The amount of heat required to raise the temperature of 1 kg of a substance by 1 K or 1 °C.
• Calorimeter – An apparatus used for measuring the amount of heat involved in a chemical reaction or physical change.
• Method of Mixtures – An experimental technique used to determine specific heat capacity by mixing substances at different temperatures.
• Temperature – A measure of the average kinetic energy of the particles in a substance.

Differentiation

For struggling learners, the teacher will provide simplified examples and extra guidance during practical setup. Advanced learners will be encouraged to explore more complex problems or research different methods of heat capacity determination.

Note for teachers using this lesson plan

Ensure all necessary laboratory equipment is available and in good working condition before the lesson. Emphasize safety precautions during practical demonstrations, especially when handling hot objects and steam. Encourage pupils to actively participate in calculations and discussions. Practical sessions should ideally be supervised closely, allowing pupils to gain hands-on experience.