Quantity of Heat, Calorimetry, Specific Heat, Latent Heat, Heat Transfer

TAKEAWAYS:

• Heat is energy transferred due to temperature differences, measured in joules (J) or calories (cal).
• Specific heat capacity tells how much heat a substance needs to raise its temperature by 1°C per gram.
• Latent heat is the hidden energy absorbed/released during phase changes (solid ↔ liquid ↔ gas).
• Calorimetry uses the principle of heat exchange to measure unknown quantities like specific heat or mass.
• Heat transfer happens via conduction (solids), convection (fluids), and radiation (all media).
• Applications include cooking, weather systems, and industrial processes like metalworking.

1. What is Heat?

Heat is a form of energy that flows from a hotter object to a colder object when they are in contact. It is measured in joules (J) in the SI system or calories (cal) in older systems.

• 1 calorie (cal) = amount of heat needed to raise 1 gram of water by 1°C.
• 1 kilocalorie (kcal) = 1000 calories (used in food labels).
• 1 joule (J) ≈ 0.239 calories.

Key Idea: Thermal Equilibrium

When two objects at different temperatures come into contact, heat flows until both reach the same temperature. This is called thermal equilibrium.

2. Specific Heat Capacity (c)

Different substances require different amounts of heat to raise their temperature by the same amount. This property is called specific heat capacity (c).

• Definition: Heat required to raise the temperature of 1 gram of a substance by 1°C.
• Formula: $$Q=m\cdot c\cdot \Delta T$$ where:
  • $Q$ = heat energy (J or cal)
  • $m$ = mass (g or kg)
  • $c$ = specific heat capacity (J/g°C or cal/g°C)
  • $\Delta T$ = change in temperature (°C)

Specific Heat Capacity of Common Substances

Worked Example 1: Calculating Heat Required

Problem: How much heat is needed to raise the temperature of 500 g of water from 20°C to 80°C? Solution: Given:

• $m=500$ g
• $c=4.18$ J/g°C (for water)
• $\Delta T=80°C-20°C=60°C$

Using $Q=m\cdot c\cdot \Delta T$: $$Q=500\times 4.18\times 60=125,400\text{ J}$$

Answer: 125,400 J of heat is required.

3. Latent Heat (L)

When a substance changes phase (solid ↔ liquid ↔ gas), it absorbs or releases heat without changing temperature. This heat is called latent heat (L).

• Latent Heat of Fusion ($L_f$): Heat required to change 1 g of a substance from solid to liquid (or vice versa) at its melting point.
• Latent Heat of Vaporization ($L_v$): Heat required to change 1 g of a substance from liquid to gas (or vice versa) at its boiling point.

Latent Heat of Common Substances

Formula for Latent Heat

$$Q=m\cdot L$$ where:

• $Q$ = heat energy (J)
• $m$ = mass (g)
• $L$ = latent heat (J/g)

Worked Example 2: Melting Ice

Problem: How much heat is required to melt 200 g of ice at 0°C? Solution: Given:

• $m=200$ g
• $L_f=334$ J/g (for ice)

Using $Q=m\cdot L_f$: $$Q=200\times 334=66,800\text{ J}$$

Answer: 66,800 J of heat is required to melt the ice.

4. Calorimetry: Mixing Liquids

Calorimetry is the science of measuring heat exchange. When two objects at different temperatures are mixed, the heat lost by the hotter object equals the heat gained by the colder object (assuming no heat is lost to the surroundings).

Principle of Calorimetry

$$\text{Heat lost}=\text{Heat gained}$$ $$m_1{c}_1\Delta T_1=m_2{c}_2\Delta T_2$$

Worked Example 3: Mixing Water and Alcohol

Problem: 100 g of water at 80°C is mixed with 200 g of alcohol at 20°C. The final temperature is 30°C. What is the specific heat capacity of alcohol? Solution: Given:

• $m_1=100$ g (water), $c_1=4.18$ J/g°C, $\Delta T_1=80°C-30°C=50°C$
• $m_2=200$ g (alcohol), $c_2=?$, $\Delta T_2=30°C-20°C=10°C$

Using the calorimetry principle: $$m_1{c}_1\Delta T_1=m_2{c}_2\Delta T_2$$ $$100\times 4.18\times 50=200\times c_2\times 10$$ $$20,900=2000\times c_2$$ $$c_2=\frac{20,900}{2000}=10.45\text{ J/g°C}$$

Answer: The specific heat capacity of alcohol is approximately 2.5 cal/g°C (since 1 cal = 4.18 J).

5. Heat Transfer Methods

Heat can be transferred in three ways:

1. Conduction

• Heat transfer through solids by direct contact.
• Example: A metal spoon gets hot when its handle is in hot soup.
• Good conductors: Metals (copper, aluminum, iron).
• Poor conductors (insulators): Wood, plastic, air.

2. Convection

• Heat transfer through fluids (liquids and gases) by movement of particles.
• Example: Boiling water (hot water rises, cold water sinks).
• Natural convection: Due to density differences (e.g., air currents).
• Forced convection: Due to external forces (e.g., fans, pumps).

flowchart TD
    A["Hot Water (Less Dense)"] -->|"Rises"| B["Top"]
    C["Cold Water (More Dense)"] -->|"Sinks"| D["Bottom"]
    B -->|"Cools"| E["Becomes Cold"]
    D -->|"Heats"| F["Becomes Hot"]
    E --> C
    F --> A

3. Radiation

• Heat transfer through electromagnetic waves (no medium required).
• Example: Heat from the Sun, fire, or electric heater.
• All objects emit radiation; hotter objects emit more.

6. Applications of Heat and Calorimetry

1. Cooking: Heat is used to cook food by conduction (pan), convection (oven), or radiation (microwave).
2. Refrigerators: Remove heat from inside to keep food cold.
3. Engines: Internal combustion engines use heat from fuel to produce work.
4. Weather Systems: Convection currents cause wind and rain.
5. Medical: Calorimetry is used in medical diagnostics (e.g., measuring metabolic rates).

Exam Tip: How to Score Full Marks

1. Understand the formulas:

   • $Q=m\cdot c\cdot \Delta T$ (Specific heat)
   • $Q=m\cdot L$ (Latent heat)
   • $\text{Heat lost}=\text{Heat gained}$ (Calorimetry)

2. Units matter!

   • Always check if the answer is required in Joules (J) or calories (cal).
   • Convert units if necessary (e.g., 1 cal = 4.18 J).

3. Draw diagrams for calorimetry problems:

   • Show the initial and final temperatures.
   • Label masses and specific heats.

4. Common mistakes to avoid:

   • Forgetting to convert grams to kilograms (if using J/kg°C).
   • Ignoring phase changes (latent heat) in problems involving melting/freezing or boiling/condensing.
   • Assuming all substances have the same specific heat (water is different!).

5. Practice NEB-style questions:

   • Problems often involve mixing two substances and finding final temperature or specific heat.
   • Always assume no heat is lost to the surroundings unless stated.

NEB Board-Style Questions

Short Answer Questions

1. Define specific heat capacity. Why does water have a high specific heat capacity?
2. What is the difference between latent heat of fusion and latent heat of vaporization?
3. Explain why a metal spoon gets hot faster than a wooden spoon when dipped in hot soup.

Numerical Problems

4. Calculate the heat required to raise the temperature of 5 kg of iron from 20°C to 220°C. (Given: $c=0.45$ J/g°C)
5. 300 g of water at 50°C is mixed with 200 g of water at 20°C. What is the final temperature of the mixture? (Assume no heat loss.)
6. How much heat is released when 100 g of steam at 100°C condenses into water at 100°C? (Given: $L_v=2260$ J/g)
7. A 200 g aluminum block ($c=0.9$ J/g°C) at 100°C is dropped into 500 g of water at 20°C. What is the final temperature of the mixture?

Descriptive Questions

8. Describe the three methods of heat transfer with one example each.
9. Explain the principle of calorimetry. How is it used in everyday life?
10. Why does sweating cool the body? Relate this to latent heat.

Good luck with your NEB exam preparation! 🚀 Keep practicing numerical problems and understanding concepts deeply.