Energy.
Notes
Energy stores and transfers
Stores: kinetic, gravitational potential, elastic potential, thermal, chemical, nuclear. Energy is transferred mechanically, electrically, by heating or by radiation (waves). Energy is conserved: it cannot be created or destroyed, only transferred or dissipated.
Key equations
| Quantity | Equation |
|---|---|
| Kinetic energy | Eₖ = ½mv² |
| Gravitational potential energy | Eₚ = mgh |
| Elastic potential energy | Eₑ = ½ke² |
| Specific heat capacity | ΔE = mcΔθ |
| Power | P = E ÷ t = W ÷ t |
| Efficiency | useful output energy ÷ total input energy |
Efficiency can be given as a decimal or a percentage. It can never exceed 100%.
Conservation in problems
If friction is ignored, loss in gravitational potential energy = gain in kinetic energy. So mgh = ½mv², and v = √(2gh). Mass cancels.
Specific heat capacity
The energy needed to raise the temperature of 1 kg of a substance by 1 °C. Water is 4200 J/kg °C.
Dissipation and efficiency
Wasted energy usually warms the surroundings. Ways to reduce it: lubrication (reducing friction), thermal insulation (reducing heat loss). Insulation with low thermal conductivity and greater thickness reduces the rate of heat transfer.
Energy resources
| Resource | Renewable? | Notes |
|---|---|---|
| Fossil fuels | no | reliable, release CO₂ |
| Nuclear | no | reliable, low CO₂, produces radioactive waste |
| Wind, solar | yes | no CO₂ in use, unreliable (depend on weather) |
| Hydroelectric, tidal | yes | reliable or predictable, may harm habitats |
| Biofuels | yes | carbon-neutral claim debated |
Required practical: specific heat capacity
Heat a metal block with an electric heater and measure the temperature rise. Use insulation to reduce heat loss and a joulemeter or measure current, p.d. and time.
Common mistakes
- Treating energy as "used up" instead of transferred to less useful stores.
- Forgetting to square the velocity or the extension.
- Using g = 10 when the question says 9.8.
- Quoting efficiency above 100%.
Exam tips
- Show the equation, substitution, answer and unit every time.
- For evaluation questions, give a reliability point and an environmental point.
Worksheet
Total: 28 marks. Use g = 9.8 N/kg. Specific heat capacity of water = 4200 J/kg °C.
- Calculate the kinetic energy of a 1500 kg car travelling at 20 m/s. (2)
- A 60 kg climber ascends 5.0 m. Calculate the gain in gravitational potential energy. (2)
- A roller coaster car of mass 450 kg is released from rest at a height of 30 m. Ignoring friction, calculate its speed at the bottom. (4)
- A spring with spring constant 200 N/m is extended by 0.15 m. Calculate the elastic potential energy stored. (2)
- A 2.0 kW kettle heats 0.50 kg of water from 20 °C to 100 °C.
(a) Calculate the energy needed. (2)
(b) Calculate the minimum time this takes. (2)
(c) Explain why the actual time is longer. (2) - An electric motor takes in 800 J of electrical energy and transfers 600 J usefully. Calculate its efficiency and state where the remaining energy goes. (3)
- Explain why wind turbines alone cannot supply a country's electricity needs reliably, and suggest one way to overcome this. (3)
- A student investigating specific heat capacity finds the value for aluminium is higher than the data book value. Suggest two reasons for the error. (2)
- A block of mass 2.0 kg slides down a ramp from a height of 1.5 m and reaches a speed of 4.0 m/s at the bottom. Calculate the energy dissipated by friction. (4)
Show mark scheme and answers (tutor copy)
Try the worksheet first, then check your work here.
- Eₖ = ½mv² = ½ × 1500 × 20² (M1) = 300 000 J = 300 kJ (A1).
- Eₚ = mgh = 60 × 9.8 × 5.0 (M1) = 2940 J (A1).
- mgh = ½mv² (M1), so v² = 2gh = 2 × 9.8 × 30 = 588 (M1, M1). v = 24.2 m/s (A1). The mass cancels.
- Eₑ = ½ke² = ½ × 200 × 0.15² (M1) = 2.25 J (A1).
- (a) ΔE = mcΔθ = 0.50 × 4200 × 80 (M1) = 168 000 J (A1).
(b) t = E ÷ P = 168 000 ÷ 2000 (M1) = 84 s (A1).
(c) Some energy heats the kettle itself (B1) and is lost to the surroundings, so more energy must be supplied. (B1) - Efficiency = 600 ÷ 800 (M1) = 0.75 or 75% (A1). The remaining 200 J is wasted, mainly as heat (and sound) dissipated to the surroundings. (B1)
- Wind output depends on the weather and varies (B1), so supply may not match demand (B1). Solutions: combine with other sources, or store energy, for example pumped-storage hydroelectricity or batteries. (B1)
- Any two: heat loss to surroundings means more energy was supplied than the block absorbed; the thermometer did not give the true average temperature; poor contact between heater and block; energy absorbed by the heater or insulation. (B1 each)
- Energy lost in gravitational potential energy = 2.0 × 9.8 × 1.5 = 29.4 J (M1). Kinetic energy at the bottom = ½ × 2.0 × 4.0² = 16 J (M1). Dissipated energy = 29.4 − 16 (M1) = 13.4 J (A1), transferred to thermal energy by friction.