Specific Heat Capacity and Latent Heat
CSEC Physics: Thermal Properties
Essential Understanding: Why does ice melt at 0°C while water temperature stays constant? Why does a metal spoon get hotter faster than a wooden spoon? These questions are answered by understanding specific heat capacity and latent heat – two fundamental concepts explaining how substances absorb heat energy during temperature changes and phase transitions.
Core Concepts
Specific Heat Capacity (c)
Definition: The amount of heat energy required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).
Formula: \[ Q = mc\Delta T \]
- \( Q \): Heat energy (Joules, J)
- \( m \): Mass (kg)
- \( c \): Specific heat capacity (J/kg°C or J/kgK)
- \( \Delta T \): Temperature change (°C or K)
High c: Substance heats slowly (e.g., water: 4200 J/kg°C)
Low c: Substance heats quickly (e.g., copper: 385 J/kg°C)
Latent Heat (L)
Definition: The heat energy required to change the phase of 1 kg of a substance without changing its temperature.
Two Types:
- Specific Latent Heat of Fusion (\( L_f \)): Solid ↔ Liquid
- Specific Latent Heat of Vaporization (\( L_v \)): Liquid ↔ Gas
Formula: \[ Q = mL \]
Unit: J/kg
Energy Conservation
Heat Gain = Heat Loss
In calorimetry problems, energy transferred from hot object equals energy gained by cold object (assuming no heat loss to surroundings).
For phase changes:
Heat to melt + heat to warm = Heat lost by hotter substance
Master Formulas for Thermal Physics
Temperature Change
\[ Q = mc\Delta T \]When substance temperature changes
Phase Change
\[ Q = mL \]When substance changes phase at constant temperature
Combined Process
\[ Q_{\text{total}} = mc\Delta T + mL \]Heating ice to steam involves both!
Specific Heat Capacity Values
| Substance | Specific Heat Capacity (J/kg°C) | Practical Significance |
|---|---|---|
| Water | 4200 | High value moderates coastal climates, used in cooling systems |
| Ethanol | 2400 | Lower than water, heats faster |
| Aluminum | 900 | Cooking pans heat evenly |
| Copper | 385 | Heats quickly, good for heat conductors |
| Iron | 450 | Moderate heating rate |
| Lead | 130 | Very low, heats very quickly |
Worked Examples
Example 1: Specific Heat Capacity
Problem: Calculate the energy required to heat 2 kg of water from 20°C to 100°C. (c = 4200 J/kg°C)
Solution:
\[ Q = mc\Delta T \]
\[ Q = (2)(4200)(100 – 20) \]
\[ Q = 2 \times 4200 \times 80 = 672,000 \, \text{J} \]
\[ Q = 672 \, \text{kJ} \]
Example 2: Latent Heat of Fusion
Problem: How much energy is needed to melt 0.5 kg of ice at 0°C? (Lf for ice = 334,000 J/kg)
Solution:
\[ Q = mL_f \]
\[ Q = 0.5 \times 334,000 = 167,000 \, \text{J} \]
\[ Q = 167 \, \text{kJ} \]
Note: Temperature remains at 0°C during melting!
Example 3: Combined Process (CSEC Style)
Problem: Calculate total energy to convert 1 kg of ice at -10°C to steam at 100°C.
Given: cice = 2100 J/kg°C, cwater = 4200 J/kg°C, Lf = 334,000 J/kg, Lv = 2,260,000 J/kg
Solution in 4 steps:
- Heat ice from -10°C to 0°C: \( Q_1 = 1 \times 2100 \times 10 = 21,000 \, \text{J} \)
- Melt ice at 0°C: \( Q_2 = 1 \times 334,000 = 334,000 \, \text{J} \)
- Heat water from 0°C to 100°C: \( Q_3 = 1 \times 4200 \times 100 = 420,000 \, \text{J} \)
- Boil water at 100°C: \( Q_4 = 1 \times 2,260,000 = 2,260,000 \, \text{J} \)
Total: \( Q = 21,000 + 334,000 + 420,000 + 2,260,000 = 3,035,000 \, \text{J} \)
\[ Q = 3.035 \, \text{MJ} \]
CSEC Past Paper Question (Adapted)
June 2019, Question 8: A 0.2 kg copper calorimeter contains 0.5 kg of water at 20°C. A 0.1 kg piece of iron at 80°C is placed in the water. The final temperature is 22°C.
Given: Specific heat capacity of water = 4200 J/kg°C, copper = 385 J/kg°C, iron = 450 J/kg°C.
(a) Calculate the energy gained by the water and calorimeter.
(b) Calculate the energy lost by the iron.
(c) Explain why the answers to (a) and (b) are slightly different in real experiments.
Energy gained by water: \( Q_w = 0.5 \times 4200 \times (22-20) = 4200 \, \text{J} \)
Energy gained by calorimeter: \( Q_c = 0.2 \times 385 \times (22-20) = 154 \, \text{J} \)
Total gained = \( 4200 + 154 = 4354 \, \text{J} \)
Energy lost by iron: \( Q_i = 0.1 \times 450 \times (80-22) = 0.1 \times 450 \times 58 = 2610 \, \text{J} \)
In real experiments, some heat is lost to the surroundings, and the calorimeter isn’t perfectly insulated. Also, some energy may be used to heat the thermometer and stirrer.
Key Examination Insights
Common Mistakes
- Using wrong specific heat capacity (e.g., using water’s c for ice)
- Forgetting that temperature doesn’t change during phase change
- Not converting units (grams to kg, kJ to J)
- In calorimetry: forgetting to include the calorimeter’s heat capacity
Success Strategies
- Draw heating/cooling curves for complex problems
- Label all known values on the diagram
- Use the principle: Heat lost = Heat gained (with sign for ΔT)
- Check if phase change occurs by comparing temperatures
CSEC Practice Arena
Test Your Understanding
CSEC Examination Mastery Tip
Interpreting Heating/Cooling Curves: In CSEC exams, you’ll often see temperature-time graphs for heating or cooling substances.
- Flat sections (plateaus): Phase change occurring (melting/freezing or boiling/condensing)
- Sloping sections: Temperature changing (specific heat capacity in action)
- Steeper slope: Lower specific heat capacity (heats/cools faster)
- Longer plateau: Higher latent heat value
Always label the phases on the graph and identify what’s happening at each stage!