P2: Conservation and Efficiency
Conservation of energy and efficiency calculations — using the efficiency equation with energy and power values, and understanding how to improve efficiency.
P2 Conservation And Efficiency
📋 Key Definitions
Conservation of energy: Energy can be transferred usefully, stored or dissipated, but it can never be created or destroyed.
Efficiency: A measure of how much of the total energy input to a device is transferred usefully. The greater the proportion of useful energy, the more efficient the device.
Useful energy: Energy transferred to the output store that is the intended purpose of the device.
Wasted energy: Energy that is transferred to stores other than the useful output — usually the thermal store of the surroundings.
📐 Efficiency Equations
Efficiency = useful energy out ÷ total energy in
Efficiency (%) = (useful energy out ÷ total energy in) × 100%
Efficiency (%) = (useful power out ÷ total power in) × 100%
No device is 100% efficient — some energy is always dissipated to the thermal store of the surroundings. The only exception is an electric heater, where all the wasted energy is thermal energy, which is the useful output anyway.
📊 Efficiency Comparison Table
| Device | Useful Output | Typical Efficiency | Main Waste |
|---|---|---|---|
| Electric heater | Thermal (heat) | 100% | None (wasted heat IS useful output) |
| LED lamp | Light | 80–90% | Heat |
| CFL lamp | Light | 60–70% | Heat |
| Filament lamp | Light | 5–10% | Heat (over 90% wasted as heat) |
| Electric motor | Kinetic (movement) | 70–80% | Heat (friction), sound |
| Petrol car engine | Kinetic (movement) | 20–30% | Heat, sound, exhaust gases |
| Diesel car engine | Kinetic (movement) | 30–40% | Heat, sound, exhaust gases |
🔧 Ways to Improve Efficiency
Reducing wasted energy improves efficiency. Any energy transferred to unwanted stores is wasted, so reducing these transfers makes a device more efficient.
| Method | How it works | Example |
|---|---|---|
| Reduce friction | Lubricate moving parts so less energy is dissipated as heat | Oiling gears in a motor |
| Reduce thermal energy loss | Use insulation to prevent heat escaping to the surroundings | Insulation around a hot water tank |
| Streamline shapes | Reduce air resistance so less energy is dissipated as heat | Aerodynamic design of cars |
| Use better conductors | Reduce electrical resistance so less heat is dissipated in wires | Thicker wires in cables |
📝 Worked Examples
Worked Example 1 — Calculating Efficiency from Energy
Question: A lamp transfers 500 J of electrical energy. 400 J is transferred as heat and 100 J as light. What is the efficiency of the lamp?
Solution:
Useful energy out (light) = 100 J
Total energy in = 500 J
Efficiency = (useful energy out ÷ total energy in) × 100% = (100 ÷ 500) × 100% = 0.2 × 100% = 20%
Worked Example 2 — Calculating Efficiency from Power
Question: A motor has a total input power of 800 W. The useful output power is 560 W. Calculate the efficiency.
Solution:
Efficiency = (useful power out ÷ total power in) × 100% = (560 ÷ 800) × 100% = 0.7 × 100% = 70%
Worked Example 3 — Finding Useful Energy from Efficiency
Question: A kettle has an efficiency of 80%. The total electrical energy input is 200 000 J. Calculate the useful energy transferred to the water.
Solution:
Useful energy out = (efficiency ÷ 100%) × total energy in = 0.80 × 200 000 = 160 000 J
Worked Example 4 — Finding Wasted Energy
Question: A car engine has an efficiency of 25%. The total energy input from fuel is 400 000 J. How much energy is wasted?
Solution:
Useful energy = 0.25 × 400 000 = 100 000 J
Wasted energy = total energy in − useful energy out = 400 000 − 100 000 = 300 000 J
This wasted energy is transferred to the thermal store of the surroundings (heat from engine, exhaust, sound).
Worked Example 5 — Comparing Efficiencies
Question: Lamp A is a filament lamp with efficiency 8%. Lamp B is an LED lamp with efficiency 85%. Both produce 850 J of useful light energy. Calculate the total energy input for each and state which uses less energy.
Solution:
Lamp A: Total energy in = useful energy ÷ efficiency = 850 ÷ 0.08 = 10 625 J
Lamp B: Total energy in = 850 ÷ 0.85 = 1000 J
Lamp B (LED) uses far less energy for the same useful output, making it much more efficient and cheaper to run.
Worked Example 6 — Efficiency and Power Together
Question: A machine has an input power of 1500 W and an efficiency of 60%. Calculate the useful output power and the power wasted.
Solution:
Useful power out = 0.60 × 1500 = 900 W
Wasted power = total power in − useful power out = 1500 − 900 = 600 W
❓ Practice Questions
Q1: Foundation A device has a total energy input of 1000 J. The useful energy output is 700 J. Calculate the efficiency as a percentage.
Q2: Foundation State why no machine can be 100% efficient (except an electric heater).
Q3: Foundation A motor has an input power of 400 W and an efficiency of 75%. Calculate the useful output power.
Q4: Higher An electric kettle is 90% efficient. It transfers 180 000 J of useful energy to the water. Calculate the total electrical energy input.
Q5: Higher Explain two ways to improve the efficiency of a machine that has moving parts. Refer to energy stores in your answer.
✅ Answers
- Efficiency = (700 ÷ 1000) × 100% = 70%.
- Some energy is always dissipated (wasted) to the thermal store of the surroundings due to friction, air resistance, or electrical resistance. An electric heater is the exception because all the "wasted" thermal energy IS the useful output.
- Useful power out = 0.75 × 400 = 300 W.
- Total energy in = useful energy out ÷ efficiency = 180 000 ÷ 0.90 = 200 000 J.
- 1) Lubricate the moving parts — this reduces friction, so less energy is dissipated to the thermal store of the surroundings. 2) Use more streamlined shapes — this reduces air resistance, so less energy is dissipated to the thermal store via drag. Both methods reduce wasted energy and increase the proportion of useful energy, improving efficiency.
🎯 Exam Tips
- Always show your working in efficiency calculations: write the equation, substitute values, then give the answer.
- Remember to multiply by 100% if the question asks for efficiency as a percentage.
- If asked "why can no device be 100% efficient?", the answer is always: some energy is always dissipated to the thermal store of the surroundings.
- The exception is an electric heater — be ready to explain why it can be considered 100% efficient.
- When asked to improve efficiency, think about reducing friction, reducing heat loss (insulation), or reducing air resistance (streamlining).
- If you know efficiency and useful energy, rearrange the equation: total energy = useful energy ÷ efficiency (as a decimal).
🔢 Maths Skills
Mathematical Skills
You need to calculate efficiency using both energy and power values, rearrange the efficiency equation to find any unknown, and compare efficiencies of different devices using decimal or percentage forms.
Maths Example
A motor has an efficiency of 65% and a useful output power of 780 W. Calculate the total input power.
Efficiency = useful power out ÷ total power in, so total power in = useful power out ÷ efficiency = 780 ÷ 0.65 = 1200 W.
⚠️ Common Misconceptions
Watch Out!
1. Wrong: Efficiency can be greater than 100% Correct: Efficiency can never exceed 100% — you cannot get more useful energy out than the total energy you put in (conservation of energy)
2. Wrong: Wasted energy disappears Correct: Wasted energy is transferred to the thermal store of the surroundings — it still exists but is no longer useful
3. Wrong: All devices have similar efficiencies Correct: Efficiencies vary hugely — an electric heater is 100% efficient, a filament lamp is only about 5–10% efficient
✍️ 6-Mark Question
Extended Answer
6 marks: Explain why no device (except an electric heater) can be 100% efficient, and describe two ways to improve the efficiency of a machine with moving parts.
No device can be 100% efficient because whenever energy is transferred, some energy is always dissipated to the thermal store of the surroundings. In machines with moving parts, friction between surfaces causes energy to be transferred to the thermal store rather than the useful output. Air resistance also dissipates energy to the thermal store. Electrical resistance in wires and components dissipates energy as heat. Sound energy from vibrations is also wasted. An electric heater is the only exception because its intended useful output IS thermal energy, so even the "wasted" heat counts as useful. Two ways to improve efficiency: (1) Lubricate the moving parts — this reduces friction, so less energy is dissipated to the thermal store of the surroundings and more energy goes to the useful output. (2) Streamline the shape — this reduces air resistance, so less energy is dissipated as heat due to drag.
Mark scheme: 1 mark — some energy always dissipated; 1 mark — friction causes thermal dissipation; 1 mark — electric heater exception explained; 1 mark — lubrication reduces friction; 1 mark — streamlining reduces air resistance; 1 mark — both methods reduce wasted energy and increase useful proportion
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student tests two lamps. Lamp X (LED) has a total input power of 10 W and useful light output power of 8.5 W. Lamp Y (filament) has a total input power of 60 W and useful light output power of 5.4 W. Both lamps produce a similar brightness of light.
(a) Calculate the efficiency of each lamp as a percentage.
(b) Calculate how much power is wasted by each lamp.
(c) A householder says "The filament lamp must be better because it uses more power." Evaluate this statement.
Answers: (a) Lamp X efficiency = (8.5 ÷ 10) × 100 = 85%. Lamp Y efficiency = (5.4 ÷ 60) × 100 = 9%. (b) Lamp X wasted power = 10 − 8.5 = 1.5 W. Lamp Y wasted power = 60 − 5.4 = 54.6 W. (c) The statement is incorrect. Using more power does not make a device better. Lamp Y uses 60 W of input power but only converts 5.4 W to useful light — most of the energy (54.6 W) is wasted as heat. Lamp X uses only 10 W of input power but converts 8.5 W to useful light, making it far more efficient (85% vs 9%). Lamp X is better because it produces similar light output with much less energy input, meaning lower electricity costs and less wasted energy. The filament lamp wastes 36 times more energy as heat.
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