Specific Heat Capacity and Thermal Insulation
Specific Heat Capacity and Thermal Insulation
When you heat a substance, its temperature rises. But different materials need different amounts of energy to warm up by the same amount. This is described by specific heat capacity.
Specific Heat Capacity
The specific heat capacity of a substance is the amount of energy needed to raise the temperature of 1 kg of the substance by 1 degree Celsius (or 1 K).
The equation is:
change in thermal energy = mass x specific heat capacity x temperature change
E = m x c x delta-theta
Where:
- E = change in thermal energy (joules, J)
- m = mass (kilograms, kg)
- c = specific heat capacity (J/kg degC)
- delta-theta = temperature change (degC or K)
Key values:
| Material | Specific heat capacity (J/kg degC) |
|---|---|
| Water | 4,200 |
| Copper | 390 |
| Aluminium | 900 |
| Concrete | 800 |
Water has a very high specific heat capacity. This means it takes a lot of energy to heat up, but it also releases a lot of energy as it cools. This is why water is used in central heating systems and why coastal areas have milder climates than inland areas.
Example calculation: How much energy is needed to heat 2 kg of water from 20 degC to 100 degC?
E = m x c x delta-theta
E = 2 x 4,200 x (100 - 20)
E = 2 x 4,200 x 80
E = 672,000 J (672 kJ)
Required Practical: Investigating Specific Heat Capacity
Aim: To measure the specific heat capacity of a material (e.g. an aluminium block or water).
Method:
1. Measure the mass of the material using a balance
2. Insert a thermometer and a heater into the block (or immerse in the liquid)
3. Record the starting temperature
4. Turn on the heater and use a joulemeter to measure the energy supplied (or use E = P x t with a voltmeter and ammeter: E = V x I x t)
5. Record the final temperature after a set time
6. Calculate specific heat capacity using c = E / (m x delta-theta)
Sources of error:
- Energy lost to the surroundings (the measured value will be higher than the true value because some energy heats the air, not the material)
- Use insulation around the block to reduce heat loss
- Ensure the thermometer is making good thermal contact with the material
Thermal Insulation
Reducing the rate of energy transfer from buildings saves energy and money. Heat is lost through:
| Part of building | Approximate heat loss | Insulation method |
|---|---|---|
| Roof | 25% | Loft insulation (fibreglass wool traps air) |
| Walls | 35% | Cavity wall insulation (foam injected between walls) |
| Windows | 10% | Double glazing (trapped air/gas between panes) |
| Floors | 15% | Underfloor insulation, carpets |
| Draughts | 15% | Draught excluders around doors and windows |
All of these methods work by reducing energy transfer through conduction and convection. Trapped air is a poor conductor and, when held in small pockets, cannot circulate (reducing convection).
Evaluating Insulation Methods
When deciding whether insulation is worthwhile, consider:
- Installation cost — How much does it cost to fit?
- Annual saving — How much money is saved on heating bills per year?
- Payback time — How many years before the savings cover the cost? (payback time = cost / annual saving)
Example: Cavity wall insulation costs 800 pounds and saves 200 pounds per year. Payback time = 800 / 200 = 4 years.
Thick walls, small windows, and fewer storeys reduce heat loss. The U-value of a material indicates how quickly heat is transferred through it — a lower U-value means a better insulator.
Exam Tips
- Always show the temperature CHANGE in calculations, not the final temperature
- In required practical questions, explain how to reduce heat loss and why the experimental value may differ from the accepted value
- Payback time is a common evaluation question — always give units in years