Heat

In this section, we will look at heat capacity and specific heat capacity to understand how different materials take in and store heat. This will help explain why some substances or materials heat up or cool down faster than others. We will also explore phase or state changes, focusing firstly on the latent heat of fusion which will help us understand melting and freezing processes. Similarly, we will study the latent heat of vaporisation which tells us how much energy is needed to change a liquid into a gas. This will help us understand processes like boiling, evaporation, and condensation.

These thermal properties are important for understanding how materials behave when heated or cooled. They are also useful in real-world applications, like refrigerators and energy management systems, and help explain everyday phenomena, such as cooking and how weather patterns form.

KEY IDEAS

• Heat capacity is the quantity of heat that must be supplied or removed from a body to change its temperature by 1 °C or by 1 K without a change in state.

• Specific heat capacity is the quantity of heat that must be supplied or removed from a body to change the temperature of 1 kg of a substance of by 1 °C or by 1 K without a change in state.

• Latent heat of fusion is the quantity of heat required to change the solid state of a substance to the liquid state or from the liquid state of the substance to the solid state without a change in temperature.

• Specific latent heat of fusion is the quantity of heat required to change 1 kg of a substance from the solid state to the liquid state or from the liquid state to the solid state without a change in temperature.

• Latent heat of vaporisation is the quantity of heat required to change the liquid state of a substance to the gaseous state or from the gaseous state to the liquid state without a change in temperature.

• Specific latent heat vaporisation is the quantity of heat required to change 1 kg of the liquid state of a substance to the gaseous state of the substance or from the gaseous state to the liquid state without a change in temperature.

Figure 2.1: Two pots of water on fire Imagine you have two pots of water. One pot is small, and the other is large. If you put both pots on a stove and heat them for the same amount of time, which pot will reach a higher temperature? Usually, the smaller pot will heat up more than the larger one. This is because the smaller pot has a lower heat capacity.

Heat capacity is the quantity of heat that must be supplied or removed from a substance for the temperature to change by 1°C or 1 K. In simpler terms, it tells you how much heat it takes to warm something up. The standard unit for heat capacity is J/°C or J/K.

The concept of heat capacity helps classify materials as good or poor conductors of heat. Good conductors have low heat capacity, allowing them to transfer or absorb heat quickly with minimal energy needed to increase their temperature.

In contrast, poor conductors (insulators) have high heat capacity, meaning they absorb heat slowly and require more energy to change their temperature. This property is essential for applications in cooking, construction, and temperature control systems.

H_(c) = Q___ ∆θ Where H_(C)= heat capacity, ∆θ = temperature change, Q = quantity of heat (energy) Consider; if you have two pots of the same size, one made of aluminium and the other made of copper, both filled with the same amount of water and heated for the same amount of time, which pot will become hotter? Discuss your thoughts and potential explanations with your neighbour.

You might assume both pots would heat up equally, but that’s not necessarily true.

The material of the pot affects how quickly it heats. In this case, the aluminium pot will heat up faster than the copper pot because aluminium has a lower specific heat capacity than copper. Specific heat capacity is the heat capacity per unit mass.

Specific heat capacity is defined as the amount of heat required to change the temperature of 1 kg of a substance by 1°C or 1 K. The standard unit for specific heat capacity is J/kg°C or J/kgK.

c = Q/m ∆ θ Where c = specific heat capacity, m = mass, ∆θ = temperature change, Q = quantity of heat

Activity 2.1 Identify good and poor conductors of heat Order the materials in the table from the best to the worst conductor of heat based on it specific heat capacity:

Table 2.1: Good and poor conductors of heat Order from best (1) to worst (6) Material Specific heat capacity (J/ kg°C) Air 1003 Carbon dioxide 839 Iron 412 Steel 466 Water (room temperature) 4181 Water (boiling point) 2080

Activity 2.2 Investigating the effect of material on conduction Objective: To observe the temperature rise of two different materials of the same mass having absorbed the same amount of heat.

Materials

1. Heat source (oven or direct sunlight)

2. Two materials of the same size and dimensions (e.g. a plastic cup full of water and a plastic cup full of oil)

3. Thermometer Procedure

1. Prepare your two objects of identical mass and shape.

2. Measure the initial temperature of the objects using the thermometer.

3. Place the objects a) into an oven on a low temperature (50 degrees Celsius) or b) into direct sunlight. Leave them in place for several minutes until there is a notable change in temperature.

4. Measure the final temperature of the materials.

Conclusion: Which of the materials had the higher specific heat capacity?

How can you tell?

Activity 2.3 Comparing Methods for Determining Specific Heat Capacity Objective: To research and compare different methods for determining specific heat capacity, analysing their relative benefits and drawbacks.

Materials

1. Access to computers or devices with internet connectivity

2. Textbooks or scientific journals

3. Note-taking materials Procedure

1. Identify a friend or friends to form a small learner group.

2. Group Research

a. In groups, research the methods for determining specific heat capacity: Method of Mixtures, Electrical Method, Continuous Flow Method etc.

b. Use textbooks, online resources, or scientific journals to gather information about each method.

c. Search for videos demonstrating each method and watch.

3. Analysis Analyse the strengths and weaknesses of each method, considering factors such as accuracy, precision, user-friendliness, cost, time required, and applicability to different materials.

4. Comparison and Discussion

a. Compare and contrast the different methods in a class discussion.

b. Discuss the advantages and disadvantages of each method.

c. Identify the most suitable method for different applications based on the analysis.

5. Presentation

a. Prepare a presentation summarising your findings.

b. Present your findings to the class, explaining your preferred method and discussing its advantages and disadvantages.

Activity 2.4 Experimental determination of the specific heat capacity of a solid by method of mixtures Objective: To determine the specific heat capacity of a solid by observing the heat transfer between the solid and a liquid (usually water) of known specific heat capacity.

Materials Needed

1. Solid sample (e.g., metal)

2. Water (or any liquid with known specific heat capacity)

3. Calorimeter (can be a simple insulated container)

4. Thermometer

5. Balance (for measuring mass)

6. Stirrer

7. Thread

8. Bunsen burner/heating coil

9. Container for heating (e.g. glass beaker) What to do

1. Weigh and record the mass of the solid sample (mₛ).

2. Weigh and record the mass of the empty calorimeter (m_(c) ).

3. Pour water into the calorimeter to about ¾ full, weigh and record the mass of the calorimeter and the water (m_(cw)).

4. Obtain the mass of the water in the calorimeter m_(w) = m_(cw) − m_(c).

5. Measure the initial temperature of the water θᵢ.

Figure 2.2: Experiment to determine the specific heat capacity of solid by method of mixtures

6. Tie the metal sample with the thread and lower it gently into boiling water.

7. By leaving the metal in the water for approximately three minutes, allow the solid to reach thermal equilibrium with the heating source. Then, quickly measure the temperature of the water (and therefore the temperature of the metal) θₛ.

8. Transfer the hot solid into the calorimeter containing the liquid.

9. Stir gently to ensure uniform temperature distribution in the liquid.

10. Use the thermometer to monitor the temperature of the mixture.

11. Record the final equilibrium temperature once it stabilises (θ_(f)).

12. Determine the heat capacity from the theory below. Use c_(w) = 4181 J/ kg°C.

Assume there is no heat lost to the surroundings and no heat transferred to the calorimeter itself, [Heat lost by solid] = [Heat gained by cold water] mₛ cₛ(θₛ − θ_(f)) = m_(w) c_(w)(θ_(f) − θᵢ) Calculate the specific heat capacity of the solid using the formula cₛ = (m_(w) c_(w))( θ_(f) − θᵢ)____________ mₛ(θₛ − θ_(f)) Conclusion: Compare your experimental value of the specific heat capacity with one which you research from the internet or from a textbook. Why might yours be different? List any likely reasons for the discrepancy.

Activity 2.5 Calculating Specific Heat Capacity

Read the worked examples below before attempting the problems that follow.

Worked examples

1. A 2 kg block of copper is heated from 20°C to 80°C. Given that the specific heat capacity of copper is = 390 J/kg°C, calculate the amount of heat energy required to raise its temperature.

2. A 100 g block of an unknown metal at 100°C is dropped into 200 g of water at 25°C. The final temperature of the mixture is 30°C. Calculate the specific heat capacity of metal. (Assume no heat is lost to the surroundings and the specific heat capacity of water is 4.18 J/g°C.)

Step-by-Step Solution for Question 1

Step 1: Identify the given information m is the mass of the substance (in kg) = 2 kg c is the specific heat capacity of the substance (in J/kg°C) = 390 J/kg°C ΔT is the change in temperature (in °C) =80°C−20°C=60°C

Step 2: Introduce the heat energy (Q) Formula:

Q =m⋅c⋅ΔT Where:

• Q is the heat energy (in joules),

• m is the mass of the substance (in kg),

• c is the specific heat capacity of the substance (in J/kg°C),

• ΔT is the change in temperature (in °C).

Step 3: Substitute the known values:

Q =2 kg × 390 J/kg°C × 60°C

Step 4: Calculate and get your answer:

Q= 2 × 390 × 60=46,800 J So, the heat energy required is 46,800 J.

Step-by-Step Solution for Question 2

Step 1. Identify the known quantities:

Mass of metal, Mₘ= 100 g Initial temperature of metal, TIₘ= 100°C Mass of water, M_(W)= 200 g Initial temperature of water, T_(IW)= 25°C Final temperature, T_(F)= 30°C Specific heat capacity of water, C_(W)= 4.18 J/g°C

Step 2: Identify which of the substances loses or gains heat Metal loses heat Water gains heat

Step 3: Introduce the heat formula for heat lost by metal and substitute:

Qₘ= Mₘ× Cₘ× (TIₘ- T_(F)) Since the metal cools down, the temperature difference is TIₘ- T_(F)= 100

- 30 = 70°C.

Qₘ= 100 × Cₘ× 70

Step 4: Introduce the heat formula for heat gained by water:

Q_(W)= M_(W)× C_(W)× (T_(F)- T_(IW)) the temperature difference is T_(F)- T_(IW)= 30 - 25 = 5°C.

Q_(W)= 200 × 4.18 × 5 = 4180 J

Step 5: Apply the principle of conservation of energy Heat lost by metal = Heat gained by water Qₘ= Q_(W) Therefore: 100 × Cₘ× 70 = 4180

Step 6: Solve for Cₘ:

Cₘ = 180/100 × 70 = 0.597 J/g°C The specific heat capacity of the metal is 0.597 J/g°C or 597 J/kg°C.

Practice Problems

Now, using the worked example as a guide, solve the following problems individually or in groups.

1. A 1.5 kg block of aluminium is heated from 25 °C to 75 °C. Given that the specific heat capacity of aluminium is 900 J/kg°C, calculate the amount of heat energy required to raise its temperature.

2. A 500 g solid block at 150 °C is placed in 600 g of water at 25 °C. If the final temperature is 35 °C, calculate the specific heat capacity of the solid.

(Specific heat capacity of water = 4.18 J/g°C.)

3. A 200 g block of metal is heated to 200 °C and then dropped into 400 g of water at 25 °C. If the final temperature of the water and metal is 30 °C, and the specific heat capacity of water is 4.18 J/g°C, calculate the specific heat capacity of the metal.

Activity 2.6 Real world applications of specific heat capacity Objective: Research and present on the applications of specific heat capacity in various industries.

Materials Needed

1. Access to the internet for research (laptops/tablets)

2. Notebooks or digital devices for note-taking

3. Presentation software (e.g., PowerPoint, Google Slides)

4. Whiteboard or flip chart for brainstorming

5. Markers or pens What to do

1. Identify friends to form a group

2. Explore Applications of Specific Heat Capacity: Research on the various industries below using search engines.

a. Manufacturing

b. Construction

c. Energy efficiency

3. Gather Information

a. Take notes on key points, examples, and diagrams that illustrate the significance of specific heat capacity in each application.

b. Identify real-world examples or case studies that highlight its importance.

4. Presentation Preparation

a. Organise your findings into a clear presentation format.

b. Assign sections to each group member based on their research focus.

c. Include visuals such as charts, images, or videos to enhance your presentation.

d. Present your findings to the class, ensuring each member shares their section.

e. Engage with classmates by inviting questions and discussions about the applications of specific heat capacity.

The word “latent” means hidden or concealed. It refers to something present but not directly observable.

In the context of heat, “latent heat” refers to the heat energy absorbed or lost when there is a change of phase or state of a substance (e.g. from solid to liquid or liquid to gas), without any change in temperature (hence this process is ‘hidden’ as it is undetectable by a thermometer). The temperature does not change because the applied heat is being used to break the bonds between the molecules rather than increase the kinetic energy of the particles.

Figure 2.3: The temperature of a substance as heat is supplied.

From the diagram, Sections A, C and E show an increase in temperature while the substance is in its solid, liquid and gaseous states respectively. Sections B and D show no change in temperature whilst the substances changes state (melts and evaporates respectively) Latent heat of fusion When you put a bottle of water in the freezer, it turns into ice. When you take that frozen bottle out, the ice melts back into liquid water. This shows that water can switch between liquid and solid states.

When water freezes into ice, its temperature remains constant, even as it loses heat. This heat lost during the change from water to ice is called the latent heat of fusion. Similarly, ice absorbs this same amount of heat to melt back into water when taken out of the freezer.

Specific latent heat of fusion is the amount of heat required or removed to change 1 kg of a substance from solid to liquid (or vice versa) without changing its temperature. The standard unit for specific latent heat of fusion is J/kg.

Latent heat of vaporisation When you boil water, it changes into steam through boiling. If you cover the boiling water with a lid, you’ll see droplets forming on the underside of the lid as the steam cools and condenses back into liquid.

These two processes—evaporation and condensation—are opposites of each other.

When water boils, even though you keep adding heat, the temperature of the water stops rising once it reaches its boiling point. All the extra heat is used to turn the water into steam, and this energy is called the latent heat of vaporisation.

On the other hand, when steam cools and turns back into water (condensation), it releases this same amount of energy, still called latent heat of vaporisation because it’s the reverse of evaporation.

Specific latent heat of vaporisation is the quantity of heat required or removed when 1 kg of a substance changes from liquid to vapour or from vapour to liquid without a change in temperature. The standard unit for specific latent heat of vaporisation is J/kg.

Mathematical representation of specific latent heat l = Q/m Where;

l is specific latent heat [l_(f)for fusion, lᵥfor vaporisation] Q is the quantity of heat absorbed or lost m is the mass of the substance

Activity 2.7 Matching the process with the definition Match the letters with the numbers in the lists below to provide the correct definition for each word:

Table 2.3: Process matching Process Definition A Melting 1 The process of a solid changing directly into a gas without passing through the liquid state. This process requires the addition of heat energy to the solid.

B Freezing 2 the process of a liquid turning into vapour (gas) throughout the volume of the liquid, at its boiling point. It also involves the addition of heat.

C Boiling 3 The process of a solid turning into a liquid at its melting point. This involves the addition of heat energy.

D Condensation 4 The process by which a gas turns directly into a solid without passing through the liquid state. It occurs when a gas is cooling.

E Sublimation 5 The process of a gas turning into a liquid in the course of cooling. This occurs at the condensation point of the substance, which is the same as its boiling point.

F Deposition 6 The process of a liquid turning into a solid at its freezing point. This involves the removal of heat energy.

Activity 2.8 Modelling the behaviour of particles For this activity you should be in groups of approximately 20 learners. Split into two sub-groups of around 10 people. Name yourselves group 1 and group 2.

Group 1 are the advisors. They will watch group 2 and offer constructive feedback on how group 2 could improve their model.

Group 2 are the modellers. They should follow the instructions below, step by step:

1. Arrange yourselves as though you are each individual particle that are making up a solid.

2. Act as though you are increasing in temperature but remaining solid.

3. ‘Melt’ into a liquid.

4. Increase in temperature, remaining liquid.

5. ‘Evaporate’ into a gas.

Activity 2.9 Comparing Methods for Determining Specific Latent Heat Objective: To research and compare different methods for determining specific latent heat, analysing their relative benefits and drawbacks.

Materials needed

1. Access to computers or devices with internet connectivity

2. Textbooks or scientific journals

3. Note-taking materials What to do

1. Identify a friend or friends to form a small learner group.

2. Group Research

a. In groups, research the methods for determining specific latent heat of fusion and vaporisation: Method of Mixtures, Electrical Method, Continuous Flow Method etc

b. Use textbooks, online resources, or scientific journals to gather information about each method.

c. Search for videos demonstrating each method and watch.

3. Analysis Analyse the strengths and weaknesses of each method, considering factors such as accuracy, precision, ease of use, cost, time required, and applicability to different materials.

4. Comparison and Discussion

a. Compare and contrast the different methods in a class discussion.

b. Discuss the advantages and disadvantages of each method.

c. Identify the most suitable method for different applications based on the analysis.

5. Presentation

a. Prepare a presentation summarising your findings.

b. Present your findings to the class, explaining your method and discussing its advantages and disadvantages.

Activity 2.10 Determination of specific latent heat of fusion of ice Materials needed

1. Calorimeter

2. Stirrer

3. Thermometer

4. Ice (in small pieces)

5. Blotting paper (for drying ice)

6. Water

7. Balance/scale (for measuring mass)

8. measuring cylinder (optional, for measuring water volume),

9. Insulating material (to minimise heat loss) What to do

1. Weigh the empty container (calorimeter) and its stirrer. Write down its weight as m_(c).

2. Fill the container halfway with water, let it sit for five minutes, and record the water’s temperature as θ₁.

3. Weigh the container again to find out the mass of water m_(w)inside.

4. Dry some ice using a paper towel and add the ice to the water, a little at a time.

5. Stir the water until all the ice melts completely.

6. Check and record the new temperature of the water after all of the ice has melted, calling it θ₂.

7. Weigh the container again to figure out the mass of ice added mᵢ.

8. The heat lost by the water and container is equal to the heat needed to melt the ice and warm it up to the final temperature, θ₂. Calculate the SHC (l_(f)) of the ice as follows (assume no heat is transferred from surroundings or from the calorimeter itself):

[Heat gained by ice] = [Heat lost by water] mᵢl_(f)+mᵢc_(w)(θ₂-0) = m_(w)c_(w)(θ₁−θ₂) mᵢ(l_(f)+c_(w)θ₂) = m_(w)c_(w)(θ₁−θ₂) l_(f) = m_(w) c_(w)(θ₁ − θ₂)___________ mᵢ − c_(w) θ₂

Note: we have assumed the starting temperature of the ice to be zero degrees Celsius, but this could be verified using a thermometer.

If you do not have the equipment necessary to perform this practical, please watch the video below:

QR code for specific latent heat video

Activity 2.11 Determination of the specific latent heat of vaporisation of water Materials needed

1. Kettle

2. Power source

3. Stirrer

4. Stopwatch

5. Water

6. Balance/scale (for measuring mass) What to do

1. Note the power of the kettle by finding in on the label (usually 1200- 1500W). Note this down as Pₖ.

2. Half-fill the kettle with water.

3. Leave the kettle on top of the balance and set the balance to zero.

4. Turn the kettle on, with its lid off (or lifted). Start the stopwatch then the kettle begins to boil.

5. Let the kettle boil for t = 5 minutes. At the end of the time, note down the mass of water which has boiled away m_(w)(the value on the balance, which will be negative).

6. Calculate the energy supplied during the boiling by:

Energy Q = Power Pₖ× time t Calculate the specific latent heat of vaporisation by:

lᵥ = Q___ m_(w) Conclusion: Why is your value an approximation only? Is the value higher or lower than the true value, and why?

Activity 2.12 Calculating Specific Latent Heat

Read the worked example below before attempting the problems that follow.

Worked example

1. Calculate the amount of energy required to convert 250 g of water at 100°C into steam at 100°C? The latent heat of vaporisation of water is 2,260,000 J/kg Step-by-Step Solution for Question 1

Step 1: Identify the given information m = 250 g = 0.25 kg lᵥ = 2,260,000 J/kg

Step 2: Introduce the formula for the heat energy (Q) required for the phase change lᵥ= Q/m Q = m lᵥ

Step 3: Substitute the known values into the formula:

Q =0.25 kg × 2,260,000 J/kg

Step 4: Calculate and get your answer:

Q = 0.25⋅2,260,000=565,000 J So, the heat energy required is 565,000 J Practice Problems

1. How much heat energy is required to melt 500 g of ice at 0°C, given that the latent heat of fusion of ice is l_(f)=334,000 J/kg?

2. A 0.15 kg block of a solid substance at 0°C is added to 0.40 kg of water at 8.0°C in a well-insulated calorimeter. If the final temperature of the water after all the ice melts is 0°C, calculate the specific latent heat of fusion of the substance. Assume no heat is lost to the surroundings, and the specific heat capacity of water is 4200 J/kg°C.

Activity 2.13 Investigating Latent Heat and Its Real-World Applications Objective: Research and present on the applications of latent heat in various industries.

Materials Needed

1. Access to the internet for research (laptops/tablets)

2. Notebooks or digital devices for note-taking

3. Presentation software (e.g., PowerPoint, Google Slides)

4. Whiteboard or flip chart for brainstorming

5. Markers or pens What to do

1. Identify friends to form a group

2. Explore Applications of Latent Heat: Research on the various industries and applications below using search engines.

Applications of latent heat

i. Refrigeration and Air Conditioning

ii. Melting of Ice in Roads

iii. Phase Changes in Power Plants

iv. Cooking with Steam

v. Body Cooling through Sweating

3. Gather Information

a. Take notes on key points, examples, and diagrams that illustrate the significance of latent heat in each application.

b. Identify real-world examples or case studies that highlight its importance.

4. Presentation Preparation

a. Organise your findings into a clear presentation format.

b. Assign sections to each group member based on their research focus (e.g., one person covers HVAC systems, another covers food preservation).

c. Include visuals such as charts, images, or videos to enhance your presentation.

d. Present your findings to the class, ensuring each member shares their section.

e. Engage with classmates by inviting questions and discussions about the applications of latent heat.

Activity 2.14 Investigating the Effects of Salt on Ice Melting Time Objective: Conduct an experiment to compare the melting times of ice cubes made from pure water and those made with added salt.

Materials Needed

1. Ice cubes (made from pure water)

2. Ice cubes (made from saltwater)

3. Two transparent containers (e.g., bowls)

4. Room temperature water

5. Stopwatch or timer

6. Thermometer (optional)

7. Measuring cup (for water)

8. Notebook or digital device for recording observations What to do

1. Set Up the Experiment

a. Fill both transparent containers with the same amount of room temperature water.

b. Place one ice cube made from pure water in one container and one ice cube made from saltwater in the other.

2. Conduct the Experiment

a. Start the timer as soon as you place the ice cubes in the water.

b. Observe and record the time it takes for each ice cube to completely melt.

c. Note any differences in melting rates between the two types of ice cubes.

3. Record Observations

a. Write down your observations, including:

i. The time taken for each ice cube to melt.

ii. Any noticeable differences in how the ice melts (e.g., bubbling, temperature changes).

4. Write Your Explanation

a. In your notebook or digital document, write a detailed explanation of your observations:

i. Discuss how latent heat is involved in the melting process.

ii. Explain why the ice cube with added salt melted faster due to its lower freezing point.

5. Summarise Practical Applications

a. Research and summarise how this phenomenon applies to food preservation and refrigeration:

i. Explain how salt is used in food preservation methods (e.g., salting meats).

ii. Discuss how refrigeration systems utilize latent heat principles to maintain food quality.

Review Questions 2.1 (Specific Heat Capacity)

1. What is the formula used to calculate the heat energy required to raise the temperature of a substance?

2. a. Define specific heat capacity.

b. Explain how it affects the amount of heat needed to change a substance’s temperature.

3. A 500 g block of copper (specific heat capacity = 0.39 J/g°C) is heated from 20°C to 80°C. How much heat energy is absorbed by the block?

4. A 1 kg sample of a substance is heated, and its temperature rises by 10°C.

The specific heat capacity of the substance is 0.5 J/g°C. If a second sample with double the mass is heated under the same conditions, what will be the temperature rise?

5. Coastal areas often experience milder temperatures than inland areas due to the specific heat capacity of water. Explain how the specific heat capacity of water influences temperature regulation in coastal areas and contributes to the moderation of extreme temperatures.

Review Questions 2.2 (Latent Heat)

1. Define specific latent heat of vaporisation and state its SI units.

2. Explain why the temperature of a substance undergoing a phase change remains constant.

3. A block of ice of unknown mass at 0°C is added to 0.315 kg of water at 10°C in a well-insulated calorimeter. If the final temperature of the water after all the ice melts is 5°C, calculate the mass of the ice. Assume there is no heat loss to the surroundings and the specific heat capacity of water is 4200 J/kg°C, the specific latent heat of fusion of ice is 334000 J/kg.

4. After a rain shower on the skin, the drops of water begin to dry off and one begins to feel cold. Describe the process that leads to the drying of the water and explain why the sensation of coldness is felt.