3.9–3.10Reducing unwanted energy transfers
You should be able to explain ways of reducing unwanted energy transfers — for example through lubrication and the use of thermal insulation.
Lubrication reduces friction between surfaces that rub together, so less energy is dissipated to their thermal stores. Thermal insulation reduces the rate at which energy is transferred by conduction out of a building or an object.
The higher the thermal conductivity of a material, the higher the rate of energy transfer by conduction across the material. You do not need to know the definition of thermal conductivity — only that a high value means a fast transfer.
You should be able to describe how the rate of cooling of a building is affected by the thickness and thermal conductivity of its walls.
Both work in the same direction: thicker walls cool the building more slowly, and walls made of a material with a lower thermal conductivity also cool it more slowly. A cavity wall does both at once — it is thicker, and the trapped air or foam in the cavity has a much lower thermal conductivity than brick.
Required practical 2
Investigate the effectiveness of different materials as thermal insulators
- Stand a small beaker inside a larger one and pack the chosen insulating material into the gap between them.
- Pour the same volume of hot water, at the same starting temperature, into the inner beaker for every run.
- Put a lid with a thermometer through it on the inner beaker, so that energy transfer from the open surface is the same each time.
- Record the temperature every minute for ten minutes.
- Repeat with each insulating material, changing nothing else.
- To investigate thickness rather than material, use the same material with different numbers of layers.
Common mistake
Compare rates of cooling, not final temperatures alone. Two materials can reach the same temperature after ten minutes if one started cooling faster and then levelled off — the graph shows this and a single reading does not.
3.11Efficiency
The energy efficiency for any energy transfer can be calculated using the equation:
efficiency = useful output energy transfer / total input energy transfer
Efficiency may also be calculated using the equation:
efficiency = useful power output / total power input
You should be able to recall and apply both equations. You may be required to calculate or use efficiency values as a decimal or as a percentage — multiply the decimal by 100 to get the percentage.
Worked example
A motor takes in 2000 J and transfers 1500 J usefully. Calculate the efficiency as a decimal and as a percentage.
- Identify which is which: useful output 1500 J, total input 2000 J.
- efficiency = useful output energy transfer ÷ total input energy transfer.
- 1500 ÷ 2000 = 0.75.
- As a percentage: 0.75 × 100 = 75%.
- The remaining 500 J is dissipated to the surroundings.
Answer: 0.75, or 75%
Worked example
A lamp has a power input of 60 W and is 8% efficient. Calculate the useful power output.
- Rearrange the power version: useful power output = efficiency × total power input.
- Convert the percentage to a decimal: 8% = 0.08.
- Substitute: 0.08 × 60.
- Give the unit: watts.
Answer: 4.8 W of useful power output — the other 55.2 W is dissipated, mostly heating the lamp.
Common mistake
Efficiency can never be greater than 1, or 100%. An answer above that means the useful output and the total input have been divided the wrong way round.
- Efficiency
- useful output energy transfer divided by total input energy transfer, or useful power output divided by total power input
3.12Increasing efficiencyHigher tier
Higher tier students should be able to describe ways to increase the efficiency of an intended energy transfer.
Every method comes down to the same idea: reduce the energy dissipated, so that a greater share of the input ends up in the useful store.
• Lubricate moving parts, so less energy is dissipated by friction to thermal stores. • Streamline a moving object, so less energy is dissipated in pushing air or water aside. • Insulate a heated system, so less energy is conducted away to the surroundings. • Reduce the mass of moving parts, so less energy goes into their kinetic stores rather than the intended one. • Use tighter-fitting or better-aligned components, so less energy is dissipated as sound and vibration.
No real machine can be made 100% efficient, because some energy is dissipated in all system changes.
Worked example
Suggest two ways of increasing the efficiency of an electric fan, and explain each.
- Ask where energy is being dissipated in a fan: at the bearings, and as sound from the blades and motor.
- Lubricating the bearings reduces friction, so less energy is dissipated to the thermal store of the bearings.
- Redesigning the blades so they cut through the air more smoothly reduces the energy dissipated as sound and turbulence.
- In both cases the input is unchanged, so a greater share reaches the useful output — the moving air.
Answer: Lubricate the bearings, and streamline the blades. Each reduces dissipation, so more of the same input becomes useful output and the efficiency rises.
Before you practise
- Friction makes mechanical processes wasteful: work done against it raises the temperature and that energy is dissipated.
- Lubrication reduces friction; thermal insulation reduces the rate of energy transfer out of a warm object.
- The thicker the walls and the lower their thermal conductivity, the slower a building cools.
- efficiency = useful energy transferred by the device ÷ total energy supplied to the device.
- The same equation works with powers instead of energies.
- Efficiency is between 0 and 1, or 0 and 100% — anything above that is an arithmetic error.
- Raise efficiency by reducing the wasteful transfer itself, not by trying to recover the energy afterwards.