Thermal Energy and Specific Heat
Specific heat capacity (often shortened to specific heat) is the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). It is a measure of a material's "thermal inertia."
Where:
- Q is the heat energy added or removed (Joules, J).
- m is the mass of the substance (kg).
- c is the specific heat capacity (J/kg·°C or J/kg·K).
- ΔT is the change in temperature (Tfinal - Tinitial).
Why is Water Special?
Water has an exceptionally high specific heat capacity (4186 J/kg·K). This means it can absorb a lot of heat without a large temperature rise. This property is crucial for regulating the Earth's climate and for biological life forms that are mostly water.
Phase Changes and Latent Heat
When a substance reaches its melting or boiling point, adding more heat does not increase its temperature. Instead, the energy is used to break intermolecular bonds. This is known as Latent Heat (Q = mL).
Common Specific Heat Values
| Material | c (J/kg·K) | Notes |
|---|---|---|
| Water | 4186 | Very high; excellent coolant |
| Aluminum | 900 | High for a metal |
| Copper | 385 | Heats up quickly |
| Iron | 450 | Common in cookware |
| Gold | 129 | Very low |
Deep Dive: Worked Examples
✅ Example 1: Heating a Cup of Tea
How much energy is needed to heat 0.25 kg of water (approx. one cup) from 20°C to 100°C?
Formula: Q = mcΔT
Solution:
Q = 0.25 × 4186 × 80
Q = 83,720 J = 83.7 kJ
✅ Example 2: Comparing Materials
If you supply 1000 J of heat to 1 kg of Water and 1 kg of Copper, which one will have the higher temperature increase?
Copper ΔT: ΔT = Q / (mc) = 1000 / (1 × 385) = 2.60°C
Result: Copper heats up more than 10 times faster than water for the same energy input.
✅ Example 3: Identifying an Unknown Metal
A 0.5 kg block of metal absorbs 5500 J of heat, causing its temperature to rise by 25°C. What is the metal most likely to be?
Formula: c = Q / (mΔT)
Solution:
c = 5500 / (0.5 × 25) = 5500 / 12.5 = 440 J/kg·K
Conclusion: Comparing to the table, this is most likely Iron (c ≈ 450).
✅ Example 4: Efficiency of a Kettle
A 2000W electric kettle takes 180 seconds to boil 1.0 kg of water from 20°C. Calculate its efficiency.
Electrical Energy Used: E = Power × time = 2000 × 180 = 360,000 J
Efficiency: (334,880 / 360,000) × 100% = 93%
✅ Example 5: Calorimetry (Mixing)
You mix 0.2 kg of hot water at 90°C with 0.5 kg of cold water at 20°C. What is the final temperature of the mixture?
m₁c(T₁ - Tf) = m₂c(Tf - T₂)
0.2 × (90 - Tf) = 0.5 × (Tf - 20)
18 - 0.2Tf = 0.5Tf - 10
28 = 0.7Tf → Tf = 40°C