Energy

:::info Board Coverage AQA Paper 1 | Edexcel Paper 1 | OCR A Gateway P1 | WJEC P1 :::

Energy Skate Park: Basics

Explore the simulation above to develop intuition for this topic.

1. Energy Stores and Systems

1.1 The Concept of Energy

Energy is the capacity of a system to do work. It is a scalar quantity measured in joules (J). Energy cannot be created or destroyed — this is the principle of conservation of energy, One of the most fundamental laws of physics.

A system is an object or group of objects. When a system changes, energy is transferred to Or from the system. We identify energy stores (where energy is held) and energy transfers (pathways by which energy moves).

Why this framing matters: you cannot “use up” energy. A car does not “use” energy — it transfers Chemical energy from fuel into kinetic energy, thermal energy, and sound. The total energy before And after any process is identical. What changes is the distribution of energy across stores. This Is not a cosmetic distinction; it underpins every energy calculation you will perform.

1.2 Energy Stores

Store	Description	Examples
Kinetic	Energy of a moving object	A rolling ball, a flowing river
Gravitational potential	Energy due to height in a gravitational field	A book on a shelf, water behind a dam
Elastic potential	Energy stored in a stretched or compressed object	A compressed spring, a drawn bow
Thermal	Energy due to the random motion of particles	A hot cup of tea, a warm room
Chemical	Energy stored in chemical bonds	Food, fuel, batteries
Magnetic	Energy stored in a magnetic field	Two repelling magnets
Electrostatic	Energy stored in an electric field	A charged capacitor
Nuclear	Energy stored in the nucleus of an atom	Uranium-235, the Sun

1.3 Energy Transfers

Energy is transferred by four main pathways:

Pathway	Mechanism
Mechanically	By a force acting over a distance (doing work)
Electrically	By charges moving through a circuit
By heating	By conduction, convection, or radiation
By radiation	By electromagnetic waves (light, infrared, etc.)

Worked Example. A person lifts a 5 kg box from the floor to a shelf 1.5 m high. Describe the Energy changes.

The system is the box and the Earth. Energy is transferred from the chemical store of the person to The gravitational potential store of the box.

$\Delta E_p = mgh = 5 \times 9.8 \times 1.5 = 73.5 \mathrm{ J$

1.4 Why “Stores and Transfers” Matters

This language exists because of a common conceptual error: students often say an object “has energy” Without specifying which kind, or that energy is “lost” when it is merely dissipated into thermal Stores. The stores-and-transfers model forces you to be explicit:

Store: Where energy resides right now (kinetic, gravitational potential, thermal, etc.).
Transfer: The mechanism by which energy moves between stores (mechanical working, electrical working, heating, radiation).

When you say “the light bulb transfers energy electrically from the mains to the thermal and light Stores of the surroundings,” you have given a complete, precise description. When you say “the light Bulb uses energy,” you have not.

2. Kinetic and Potential Energy

2.1 Kinetic Energy

$E_k = \frac{1}{2}mv^2$

Where $m$ is mass (kg) and $v$ is speed (m/s).

Derivation. Consider an object of mass $m$ accelerated from rest by a constant force $F$ over Distance $d$ :

$F = ma \implies a = \frac{F}{m}$ $v^2 = u^2 + 2as = 0 + 2 \times \frac{F}{m} \times d = \frac{2Fd}{m}$ $Fd = \frac{1}{2}mv^2$

Since work done $W = Fd$ The kinetic energy is $E_k = \frac{1}{2}mv^2$ .

2.2 Why the Factor of One-Half

The $\frac{1}{2}$ in $E_k = \frac{1}{2}mv^2$ is not arbitrary. It arises because the object Accelerates uniformly from rest: it covers distance $d$ in time $t$ And the average velocity during That interval is $\frac{0 + v}{2} = \frac{v}{2}$ . The work done is force times distance, which Equals $m \cdot \frac{v}{t} \cdot \frac{v}{2} \cdot t = \frac{1}{2}mv^2$ . If the factor were Different, the work-energy theorem would fail, and conservation of energy would be violated. The Factor of one-half is a direct, unavoidable consequence of Newton’s second law and the definition of Work.

2.3 Gravitational Potential Energy

$E_p = mgh$

Where $m$ is mass (kg), $g$ is gravitational field strength (N/kg), and $h$ is height (m).

:::info Near the Earth’s surface, $g \approx 9.8$ N/kg (often approximated as 10 N/kg in Calculations). :::

The height $h$ is measured from an arbitrarily chosen reference level (often the ground). Only changes in gravitational potential energy are physically meaningful, so the choice of reference Does not affect the answer to any real problem.

2.4 Conservation of Energy: Falling Objects

When an object falls freely (ignoring air resistance), gravitational potential energy is converted To kinetic energy.

$mgh = \frac{1}{2}mv^2$ $v = \sqrt{2gh}$

Worked Example. A 0.5 kg ball is dropped from a height of 20 m. Find its speed just before it Hits the ground (ignore air resistance).

$v = \sqrt{2 \times 9.8 \times 20} = \sqrt{392} = 19.8 \mathrm{ m/s$

2.5 Proof That the Final Speed Is Independent of Mass

In the equation $v = \sqrt{2gh}$ The mass $m$ cancels out. This means a bowling ball and a feather Dropped from the same height in a vacuum hit the ground at the same speed. The reason is that Although a heavier object has more gravitational potential energy, it also requires more force to Accelerate (it has more inertia), and these two effects cancel exactly. This is a deep result: it is A direct consequence of the equivalence of gravitational and inertial mass, which is the starting Point for Einstein’s general theory of relativity.

2.6 Elastic Potential Energy

$E_e = \frac{1}{2}ke^2$

Where $k$ is the spring constant (N/m) and $e$ is the extension (m).

This formula applies to objects that obey Hooke’s law: the extension is directly proportional to The force applied, up to the limit of proportionality.

Derivation. The force needed to extend a spring by $e$ is $F = ke$ . The average force during Extension from 0 to $e$ is $\frac{0 + ke}{2} = \frac{1}{2}ke$ . The work done (energy stored) is:

$E_e = \mathrm{average force \times \mathrm{distance = \frac{1}{2}ke \times e = \frac{1}{2}ke^2$

Alternatively, using calculus (for Higher tier students): the work done is the integral of force With respect to displacement:

$E_e = \int_0^e F\, dx = \int_0^e kx\, dx = \left[\frac{1}{2}kx^2\right]_0^e = \frac{1}{2}ke^2$

2.7 Energy Changes in a Vertical Spring-Mass System (Higher Tier)

When a mass is hung from a spring and released, the energy oscillates between elastic potential and Gravitational potential and kinetic energy. At the equilibrium position, the spring is extended but The mass is moving fastest (maximum kinetic energy). At the lowest point, the spring has maximum Extension (maximum elastic potential energy) and the mass is momentarily at rest (zero kinetic Energy).

2.8 Worked Example: Bungee Jumping

A bungee cord has spring constant $k = 50$ N/m and natural length $15$ m. A jumper of mass $60$ kg Falls from a bridge. Find the maximum extension of the cord.

At maximum extension, velocity is zero. All energy has been transferred from gravitational potential To elastic potential (measuring from the point where the cord first becomes taut):

$mgh = \frac{1}{2}ke^2$

Where $h = e$ (the distance fallen beyond the natural length equals the extension):

$60 \times 9.8 \times e = \frac{1}{2} \times 50 \times e^2$ $588e = 25e^2$ $e = \frac{588}{25} = 23.52 \mathrm{ m$

The total distance fallen is $15 + 23.52 = 38.52$ m.

3. Specific Heat Capacity

3.1 Definition

The specific heat capacity $c$ of a substance is the energy required to change the temperature Of 1 kg of the substance by $1^{\circ}\mathrm{C$ (or 1 K).

$\Delta E = mc\Delta T$

3.2 Microscopic Interpretation

Temperature is a measure of the average kinetic energy of the particles in a substance. When you Heat a material, you increase the kinetic energy of its particles. The specific heat capacity tells You how much energy is needed to produce a given temperature increase. Materials with high specific Heat capacities (like water) require more energy per degree of temperature rise because their Particles need more energy input to increase their average kinetic energy. This can be due to Intermolecular forces (hydrogen bonds in water, for instance, absorb energy without increasing Kinetic energy) or to the number of degrees of freedom available to the particles.

3.3 Specific Heat Capacities

Substance	Specific Heat Capacity (J/(kg $^{\circ}$ C))
Water	4200
Aluminium	900
Copper	390
Iron	450
Lead	130

Water has a very high specific heat capacity, which is why it is effective as a coolant and why Coastal climates are more moderate.

3.4 Why Water’s High Specific Heat Capacity Matters

Water’s specific heat capacity of 4200 J/(kg $^{\circ}$ C) is anomalously high compared to most Liquids. The consequence: a large body of water (ocean, lake) absorbs or releases enormous amounts Of energy with only a small temperature change. This is why coastal regions have milder climates Than inland areas at the same latitude. It is also why water is used as a coolant in car engines and Nuclear reactors: it can absorb a great deal of waste heat before its temperature rises to dangerous Levels.

Worked Example. A 2 kW kettle heats 1.5 kg of water from $15^{\circ}\mathrm{C$ to $100^{\circ}\mathrm{C$ . How long does this take?

$\Delta E = mc\Delta T = 1.5 \times 4200 \times 85 = 535,500 \mathrm{ J$

$t = \frac{\Delta E}{P} = \frac{535500}{2000} = 267.75 \mathrm{ s \approx 4.5 \mathrm{ minutes$

3.5 Required Practical: Specific Heat Capacity

Method:

Measure the mass of the metal block.
Place a heater in the block and connect it to a power supply.
Measure the initial temperature.
Switch on the heater and start a timer. Record the voltmeter and ammeter readings.
After a set time, switch off and record the final temperature.
Calculate: energy supplied $E = VIt$ Then $c = \frac{E}{m\Delta T}$ .

Sources of error:

Heat loss to the surroundings: the block loses thermal energy to the air, so the measured temperature rise is smaller than it should be. The calculated specific heat capacity will be higher than the true value.
Incomplete thermal contact between heater and block: some energy from the heater is not transferred to the block.
Reading the thermometer at an angle (parallax error).
The heater itself has a heat capacity and absorbs some of the energy.

Improvement: Plot a graph of temperature against energy supplied and take the gradient. This Reduces the impact of the initial temperature measurement error and allows you to account for a Constant rate of heat loss (the graph may not pass through the origin; the intercept represents Energy lost to the surroundings during the initial heating phase).

3.6 Specific Latent Heat (Higher Tier)

When a substance changes state (melts, boils, freezes, condenses), energy is transferred without any Change in temperature. The energy per unit mass for a change of state is the specific latent Heat.

$E = mL$

Where $L$ is the specific latent heat (J/kg).

Specific latent heat of fusion: energy per kg to change from solid to liquid (no temperature change).
Specific latent heat of vaporisation: energy per kg to change from liquid to gas (no temperature change).

Substance	Latent heat of fusion (J/kg)	Latent heat of vaporisation (J/kg)
Water	$3.34 \times 10^5$	$2.26 \times 10^6$
Ethanol	$1.05 \times 10^5$	$8.55 \times 10^5$

Why temperature stays constant during a change of state: The energy being supplied does not Increase the kinetic energy of the particles. Instead, it overcomes the intermolecular forces that Hold the particles in place (for melting) or that hold them close together in the liquid phase (for Boiling). Once all particles have been separated, further energy input raises the temperature again.

4. Power

4.1 Definition

Power is the rate of energy transfer.

$P = \frac{\Delta E}{t}$

Where $P$ is power (watts, W), $\Delta E$ is energy transferred (J), and $t$ is time (s).

One watt equals one joule per second: $1 \mathrm{ W = 1 \mathrm{ J/s$ .

4.2 Power and Force

$P = Fv$

Where $F$ is force (N) and $v$ is velocity (m/s).

Derivation. $P = \frac{W}{t} = \frac{Fd}{t} = F \times \frac{d}{t} = Fv$ .

This assumes the force is in the same direction as the velocity. If the force is at angle $\theta$ To the velocity, then $P = Fv\cos\theta$ .

4.3 Power, Force, and Constant Speed

When a vehicle travels at constant speed, the driving force equals the resistive force (friction Plus air resistance). The engine’s power output is:

$P = F_{\mathrm{drive} \times v = F_{\mathrm{resistive} \times v$

This means that for a given engine power, doubling the speed halves the available driving force (because $F = P/v$ ). This is why accelerating from 60 mph to 120 mph requires far more than double The power — the resistive force (especially air resistance) increases with speed.

Worked Example. A car engine provides a driving force of 3000 N at a speed of 25 m/s. Calculate The power output.

$P = Fv = 3000 \times 25 = 75000 \mathrm{ W = 75 \mathrm{ kW$

4.4 Human Power Output

A typical adult can sustain a power output of about 100 W for extended periods (walking, light Work). A trained cyclist might sustain 200—300 W. Sprinters can briefly reach 1000—2000 W. By Comparison, a car engine produces 50,000—100,000 W. This 500:1 ratio explains why human-powered Transport is limited.

5. Energy Resources

5.1 Non-Renewable Resources

Resource	Energy Source	Advantages	Disadvantages
Coal	Chemical (fossil fuel)	High energy density, established infrastructure	Releases CO $_2$ and pollutants, finite supply
Oil	Chemical (fossil fuel)	Versatile (transport, heating)	Releases CO $_2$ Finite, risk of spills
Natural gas	Chemical (fossil fuel)	Cleaner than coal, flexible power output	Releases CO $_2$ Finite
Nuclear	Nuclear (fission of U-235)	Very high energy density, no CO $_2$ during operation	Radioactive waste, risk of accidents, high decommissioning cost

5.2 Renewable Resources

Resource	Energy Source	Advantages	Disadvantages
Wind	Kinetic energy of moving air	Clean, unlimited	Intermittent, visual impact
Solar	Electromagnetic radiation from the Sun	Clean, low maintenance	Intermittent, low efficiency
Hydroelectric	Gravitational potential of water	Reliable, controllable	Habitat destruction, limited locations
Tidal	Gravitational energy (Moon)	Predictable	Limited locations, habitat impact
Wave	Kinetic energy of water waves	Abundant	Unreliable, technology immature
Geothermal	Thermal energy from radioactive decay in Earth	Reliable, low CO $_2$	Limited locations, high setup cost
Biofuel	Chemical (from recently living organisms)	Carbon-neutral in theory	Competes with food production, land use

5.3 Comparing Resources

Factors to consider:

Cost of setup and running
Reliability and predictability
Environmental impact (CO $_2$ emissions, habitat loss, visual impact)
Decommissioning costs
Energy density (energy per unit mass)
Efficiency of energy transfer

5.4 Energy Transfers in Power Stations

In a fossil fuel power station:

$\mathrm{Chemical \to \mathrm{Thermal \to \mathrm{Kinetic \to \mathrm{Electrical$

Fuel is burned to heat water, producing steam (chemical to thermal)
Steam drives a turbine (thermal to kinetic)
Turbine drives a generator (kinetic to electrical)

The overall efficiency is 35—40%, with the remaining energy transferred to the Surroundings as thermal energy.

In a nuclear power station, the initial energy store is nuclear rather than chemical, but the Subsequent transfers are the same.

5.5 Energy Density

Energy density is the energy per unit mass of a fuel. It determines how much fuel must be Transported and stored.

Fuel	Energy density (MJ/kg)
Uranium-235	$8.2 \times 10^7$
Petrol	46
Coal	30
Natural gas	54
Hydrogen	143
Lithium-ion battery	0.5 — 0.9

Uranium-235 has an energy density roughly a million times greater than fossil fuels. This is why a Nuclear power station needs only a few tonnes of fuel per year, whereas a coal station needs Millions of tonnes.

6. Efficiency

6.1 Definition

Efficiency is the ratio of useful energy output to total energy input.

$\mathrm{Efficiency = \frac{\mathrm{useful energy output}{\mathrm{total energy input} \times 100\%$

Equivalently, since power is energy per unit time:

$\mathrm{Efficiency = \frac{\mathrm{useful power output}{\mathrm{total power input} \times 100\%$

Worked Example. A light bulb transfers 60 J of energy per second. 6 J per second is transferred As light. Calculate the efficiency.

$\mathrm{Efficiency = \frac{6}{60} \times 100\% = 10\%$

:::caution No machine can be 100% efficient. Some energy is always dissipated (transferred to Thermal stores in the surroundings) due to friction, air resistance, or electrical resistance. :::

6.2 Why Efficiency Cannot Reach 100 Percent

The second law of thermodynamics sets a fundamental upper limit on the efficiency of any process That converts thermal energy into work. Even in principle, a perfect heat engine cannot convert all Its input thermal energy into useful work. In practice, all real machines lose energy to friction, Electrical resistance, sound, and thermal radiation. These losses are not a design flaw; they are a Consequence of the laws of physics.

6.3 Cascading Efficiency

When multiple energy transfers occur in sequence, the overall efficiency is the product of the Individual efficiencies. For example, if a power station is 40% efficient, transmission is 92% Efficient, and a light bulb is 5% efficient, the overall efficiency from fuel chemical energy to Light is:

$0.40 \times 0.92 \times 0.05 = 0.0184 = 1.84\%$

This means that only about 1.8% of the original chemical energy in the fuel ends up as light. The Rest is dissipated as thermal energy at various stages. This cascading effect is why improving Efficiency at every stage matters.

6.4 Improving Efficiency

Ways to improve efficiency:

Reduce friction (lubrication, streamlined shapes)
Reduce heat loss (insulation)
Use more efficient components (LED bulbs instead of filament bulbs)
Recover waste heat (combined heat and power systems)
Use cogeneration (using waste heat from electricity generation for district heating)

Example: An LED bulb has an efficiency of about 30—40%, compared to 5—10% for a compact Fluorescent and 2—5% for an incandescent bulb. Replacing all incandescent bulbs with LEDs in a Building can reduce lighting energy consumption by a factor of 8—10.

7. Thermal Energy Transfer

7.1 Conduction

Conduction is the transfer of thermal energy through a material without the material itself Moving. It occurs primarily in solids.

Mechanism: Particles at the hot end vibrate more vigorously. These vibrations are passed to Neighbouring particles through intermolecular bonds. Metals are good conductors because they have Free (delocalised) electrons that can transfer energy rapidly.

Metals vs. Non-metals: Metals conduct well (free electrons). Non-metals are poor conductors (insulators).

7.2 Why Metals Are Better Conductors Than Non-Metals

In a metal, the outermost electrons are delocalised: they are not bound to individual atoms but move Freely throughout the lattice. When one end of the metal is heated, these free electrons gain Kinetic energy and rapidly carry it to the cooler end by diffusion. This electron-based mechanism is Far faster than the vibration-based mechanism in non-metals, where energy transfer relies solely on Adjacent atoms passing vibrations through intermolecular bonds.

7.3 Convection

Convection is the transfer of thermal energy through a fluid (liquid or gas) by the movement of The fluid itself.

Mechanism: Fluid particles gain energy, expand (become less dense), and rise. Cooler, denser Fluid replaces them, creating a convection current.

7.4 Why Convection Cannot Occur in Solids

Convection requires bulk motion of the medium. In a solid, particles are locked in fixed positions By strong intermolecular forces and can only vibrate. They cannot move through the material, so no Convection current can form. This is why the handle of a metal spoon gets hot by conduction, not Convection.

7.5 Radiation

Thermal radiation is the transfer of energy by electromagnetic waves (infrared radiation). It Does not require a medium and can travel through a vacuum.

Key properties:

All objects emit and absorb thermal radiation
Hotter objects emit more radiation
Dark, matt surfaces are better emitters and absorbers
Light, shiny surfaces are better reflectors and poorer emitters

7.6 The Stefan-Boltzmann Law (Higher Tier)

The power radiated by a black body is proportional to the fourth power of its absolute temperature:

$P = \sigma A T^4$

Where $\sigma = 5.67 \times 10^{-8}$ W/(m $^2$ K $^4$ ) is the Stefan-Boltzmann constant, $A$ is the Surface area, and $T$ is the absolute temperature in kelvin.

This means that doubling the absolute temperature of an object increases its radiated power by a Factor of $2^4 = 16$ . A small increase in temperature produces a very large increase in radiated Power. This is why hot objects cool down rapidly at first (when they are hottest) and then cool more Slowly as their temperature drops.

7.7 Required Practical: Insulation

Method: Compare the rate of cooling of a hot object with and without insulation. Measure Temperature at regular time intervals and plot a temperature-time graph. The steeper the gradient, The faster the energy transfer.

Expected result: The insulated object cools more slowly (smaller gradient on the Temperature-time graph). The gradient is steepest at the start (when the temperature difference Between the object and its surroundings is greatest) and gradually decreases.

7.8 Vacuum Flasks: How They Work

A vacuum flask (Thermos) minimises all three mechanisms of heat transfer:

Vacuum between double walls: eliminates conduction and convection (no particles to transfer energy).
Silvered surfaces: reflect infrared radiation back into the flask, reducing radiative loss.
Stopper: reduces convection currents above the liquid surface.
Plastic or cork support: minimises conduction through the point where the inner and outer walls connect.

No flask is perfectly insulating. There will always be some conduction through the stopper and the Walls, and some radiation will escape. A good vacuum flask can keep a hot drink warm for several Hours.

8. Energy and Everyday Life

8.1 Energy in the Human Body

The average adult requires about 8000—10,000 kJ of energy per day from food. This energy is used For:

Basal metabolic rate (maintaining body temperature, organ function): about 60—70% of total
Physical activity: about 20—30%
Processing food (thermic effect of food): about 10%

The body converts chemical energy in food (carbohydrates, fats, proteins) into other forms: kinetic Energy (movement), thermal energy (maintaining body temperature at $37^{\circ}\mathrm{C$ ), and Potential energy (climbing stairs, lifting objects).

8.2 Energy Costs in the Home

A typical UK household uses about 3000—4000 kWh of electricity per year. Major energy users:

Appliance	Typical power (W)	Daily use (hours)	Daily energy (kWh)
Kettle	2000	0.2	0.4
Fridge	150	24	3.6
Washing machine	2000	1	2.0
TV	100	4	0.4
Electric shower	10000	0.2	2.0

The electric shower is one of the largest single energy consumers in a home because it heats water Very rapidly (high power = high rate of energy transfer).

9. Sankey Diagrams

A Sankey diagram is a visual representation of energy flows. The width of each arrow is Proportional to the amount of energy it represents.

Features:

Total energy input is shown on the left
Useful energy output goes to the right
Wasted energy branches off ( downwards)
The sum of useful output and wasted energy equals the input (conservation of energy)

Example: A power station receives 1000 J of chemical energy from fuel. It transfers 400 J to Electrical energy and 600 J to thermal energy (waste). The efficiency is 40%. On the Sankey diagram, The electrical energy arrow would be 40% as wide as the input arrow, and the waste arrow would be 60% as wide.

Common Pitfalls

Confusing energy stores and energy transfers. A store is where energy is held; a transfer is how it moves.
Forgetting that energy is conserved. In an inefficient machine, energy is not lost — it is transferred to unwanted stores ( thermal).
Using the wrong mass in $E_k = \frac{1}{2}mv^2$ . $v$ is speed, not velocity (though since it is squared, the direction does not matter).
Confusing specific heat capacity with specific latent heat. Specific heat capacity relates to temperature change; specific latent heat relates to change of state.
Writing efficiency as a decimal when a percentage is required (or vice versa).
Using $g = 10$ when the question specifies $g = 9.8$ . Always use the value given in the question.
Forgetting that gravitational potential energy depends on height above a reference level, not total height above the ground. If an object falls from 20 m to 5 m, the height change is 15 m, not 20 m.
Assuming the spring constant is the same in compression and extension. For many springs this is approximately true, but not all.
Stating that “energy is lost to the surroundings” without specifying that it is transferred to thermal stores. The energy still exists; it is just no longer in a useful form.

Practice Questions

A 1200 kg car is travelling at 20 m/s. Calculate its kinetic energy. If the brakes apply a force of 6000 N, calculate the stopping distance.
A 50 kg student climbs a flight of stairs 4 m high in 6 seconds. Calculate the gravitational potential energy gained and the power developed.
A 500 g block of aluminium is heated from $20^{\circ}\mathrm{C$ to $120^{\circ}\mathrm{C$ by a 200 W heater. How long does this take? (Specific heat capacity of aluminium = 900 J/(kg $^{\circ}$ C).)
A power station has an efficiency of 38%. It burns fuel at a rate of 500 kg per hour. If the fuel has an energy density of 45 MJ/kg, calculate the useful power output.
Explain why a vacuum flask keeps a hot drink warm. Refer to all three methods of thermal energy transfer.
A spring with spring constant 250 N/m is compressed by 0.08 m. Calculate the elastic potential energy stored.
A ball is thrown vertically upward with speed 14 m/s. Calculate the maximum height it reaches (ignore air resistance).
Compare the advantages and disadvantages of using wind power and nuclear power for electricity generation.
An electric motor lifts a 200 kg load through 5 m in 8 seconds. The motor is 80% efficient. Calculate the electrical power input.
Explain why the specific heat capacity of water being so high makes it a good coolant for car engines.
A pendulum bob of mass 0.2 kg is raised 0.5 m above its lowest point and released. Calculate its speed at the lowest point, and explain why it does not continue to swing forever.
A coal-fired power station burns coal with energy density 30 MJ/kg at a rate of 10 kg/s. The overall efficiency is 36%. Calculate the useful power output and the rate of energy wasted.
A student carries out the specific heat capacity practical and obtains a value of 510 J/(kg $^{\circ}$ C) for copper, compared to the accepted value of 390 J/(kg $^{\circ}$ C). Explain why the measured value is too high.
A car of mass 800 kg travelling at 20 m/s brakes to a stop. The brakes heat up. If 60% of the kinetic energy is transferred to the brakes, and the brakes have a total mass of 15 kg and specific heat capacity 460 J/(kg $^{\circ}$ C), calculate the temperature rise of the brakes.
Calculate the energy required to convert 0.5 kg of ice at $-10^{\circ}\mathrm{C$ to water at $50^{\circ}\mathrm{C$ . (Specific heat capacity of ice = 2100 J/(kg $^{\circ}$ C), specific latent heat of fusion of ice = $3.34 \times 10^5$ J/kg, specific heat capacity of water = 4200 J/(kg $^{\circ}$ C).)

10. Worked Example: Multi-Stage Heating

A common exam question involves calculating the energy needed to change both the temperature and the State of a substance. The key is to break the process into distinct stages and add the energies.

Question: Calculate the energy required to convert 0.5 kg of ice at $-10^{\circ}\mathrm{C$ to Water at $50^{\circ}\mathrm{C$ .

Stage 1: Heat ice from $-10^{\circ}\mathrm{C$ to $0^{\circ}\mathrm{C$ (no change of state).

$E_1 = mc_{\mathrm{ice}\Delta T = 0.5 \times 2100 \times 10 = 10500 \mathrm{ J$

Stage 2: Melt ice at $0^{\circ}\mathrm{C$ (change of state, temperature stays at $0^{\circ}\mathrm{C$ ).

$E_2 = mL_f = 0.5 \times 3.34 \times 10^5 = 167000 \mathrm{ J$

Stage 3: Heat water from $0^{\circ}\mathrm{C$ to $50^{\circ}\mathrm{C$ .

$E_3 = mc_{\mathrm{water}\Delta T = 0.5 \times 4200 \times 50 = 105000 \mathrm{ J$

$E_{\mathrm{total} = 10500 + 167000 + 105000 = 282500 \mathrm{ J \approx 283 \mathrm{ kJ$

Notice that the melting stage (Stage 2) requires by far the most energy — over half the total. This Is a common pattern: changing state requires more energy than changing temperature by the Same number of degrees.

11. Summary Table: Energy Stores and Their Formulas

Store	Formula	Variables	When to Use
Kinetic	$E_k = \frac{1}{2}mv^2$	$m$ = mass, $v$ = speed	Object moving
Gravitational potential	$E_p = mgh$	$m$ = mass, $g$ = field strength, $h$ = height	Object at height
Elastic potential	$E_e = \frac{1}{2}ke^2$	$k$ = spring constant, $e$ = extension	Spring stretched/compressed
Thermal	$\Delta E = mc\Delta T$	$m$ = mass, $c$ = specific heat capacity, $\Delta T$ = temp change	Temperature change
Latent heat	$E = mL$	$m$ = mass, $L$ = specific latent heat	Change of state

12. Derivation: Conservation of Energy from the Work-Energy Theorem

The conservation of mechanical energy (kinetic plus potential) is not an independent law — it Follows directly from the work-energy theorem when only conservative forces are present.

Starting from the work-energy theorem:

$W_{\mathrm{net} = \Delta E_k$

If only gravity does work (no friction, no applied forces), then $W_{\mathrm{net} = W_{\mathrm{gravity}$ . The work done by gravity is $W_{\mathrm{gravity} = -\Delta E_p$ (negative because when the object moves down, $\Delta E_p$ is Negative and gravity does positive work).

$-\Delta E_p = \Delta E_k$

$\Delta E_k + \Delta E_p = 0$

$E_{k,f} + E_{p,f} = E_{k,i} + E_{p,i}$

This is the statement of conservation of mechanical energy. It applies whenever only conservative Forces (gravity, spring force) do work. When non-conservative forces (friction, air resistance) are Present, the total mechanical energy decreases by an amount equal to the work done by those forces.

13. Worked Example: Power on an Incline

A cyclist of mass 70 kg (including the bicycle) travels up a hill inclined at $5^{\circ}$ to the Horizontal at a constant speed of $6 \mathrm{ m/s$ . The total resistive force (friction plus air Resistance) is 25 N. Calculate the power the cyclist must develop.

At constant speed, the driving force equals the component of weight down the slope plus the Resistive force:

$F_{\mathrm{drive} = mg\sin\theta + F_{\mathrm{resistive} = 70 \times 9.8 \times \sin 5^{\circ} + 25$

$= 70 \times 9.8 \times 0.0872 + 25 = 59.8 + 25 = 84.8 \mathrm{ N$

$P = Fv = 84.8 \times 6 = 508.8 \mathrm{ W$

This is roughly 500 W, which is elite cyclist-level output. It illustrates why cycling uphill is so Much harder than cycling on the flat: the additional force needed to overcome gravity at even a Modest gradient significantly increases the required power.

14. Worked Example: Efficiency of a Multi-Stage Process

A natural gas power station burns fuel with energy density 55 MJ/kg at a rate of 8 kg/s. The overall Efficiency is 42%. Calculate the useful power output and the rate of energy wasted per second.

$P_{\mathrm{input} = 55 \times 10^6 \times 8 = 4.4 \times 10^8 \mathrm{ W = 440 \mathrm{ MW$

$P_{\mathrm{useful} = 0.42 \times 440 = 184.8 \mathrm{ MW$

$P_{\mathrm{wasted} = 440 - 184.8 = 255.2 \mathrm{ MW$

Over one hour, the wasted energy is $255.2 \times 3600 = 918720 \mathrm{ MJ$ Enough to heat over 20000 homes. This highlights the enormous scale of energy waste in fossil fuel power stations and The importance of improving efficiency.

15. The Concept of Specific Heat Capacity: Microscopic View

At the microscopic level, temperature is a measure of the average kinetic energy of the particles in A substance. When you add energy to a substance, the particles gain kinetic energy (they move Faster) and the temperature rises. The specific heat capacity tells you how much energy is needed Per kilogram per degree of temperature rise.

Materials with high specific heat capacities have structural features that absorb energy without Increasing kinetic energy. In water, the hydrogen bonds between molecules absorb energy as the bonds Stretch and bend. This energy is stored as potential energy in the bonds rather than as kinetic Energy of the molecules. As a result, more total energy input is needed to produce a given Temperature rise.

Lead has a low specific heat capacity (130 J/(kg $^{\circ}$ C)) because its atoms are heavy and Tightly packed in a metallic lattice with weak intermolecular interactions. Nearly all the energy Input goes directly into kinetic energy, producing a rapid temperature rise.

16. Practice Questions (Additional)

A 200 g copper block at $150^{\circ}\mathrm{C$ is dropped into 300 g of water at $20^{\circ}\mathrm{C$ . Assuming no heat loss to the surroundings, find the final temperature. (Specific heat capacity of copper = 390 J/(kg $^{\circ}$ C), specific heat capacity of water = 4200 J/(kg $^{\circ}$ C).)
A crane lifts a 500 kg load through a height of 12 m in 15 s. If the motor is 75% efficient, calculate the electrical power input to the motor.
Explain, with reference to energy stores and transfers, what happens when a ball is thrown vertically upward and then falls back down. Account for why the ball does not return to its original height.
A student designs a solar water heater. The panel has an area of $2.5 \mathrm{ m^2$ and receives solar radiation at $800 \mathrm{ W/m^2$ . The efficiency of the panel is 35%. The water flows through the panel at a rate of 0.02 kg/s. Calculate the temperature rise of the water as it passes through the panel.
A bungee cord of natural length 20 m has spring constant 40 N/m. A jumper of mass 75 kg falls from a bridge. Find the maximum speed of the jumper during the fall. (Hint: the maximum speed occurs when the acceleration is zero, i.e., when the net force is zero.)

Practice Problems

Question 1: Gravitational potential energy

A $60 \mathrm{ kg$ climber ascends a $15 \mathrm{ m$ cliff. Calculate the change in gravitational potential energy. Take $g = 9.8 \mathrm{ N/kg$ .

Answer

$\Delta E_p = mgh = 60 \times 9.8 \times 15 = 8820 \mathrm{ J$ .

Question 2: Kinetic energy and speed

A car of mass $800 \mathrm{ kg$ is travelling at $20 \mathrm{ m/s$ . Calculate its kinetic energy. If the brakes apply a force of $4000 \mathrm{ N$ Calculate the stopping distance.

Answer

$E_k = \frac{1}{2}mv^2 = 0.5 \times 800 \times 400 = 160,000 \mathrm{ J$ .

Work done by brakes = $F \times d = E_k$ : $4000 \times d = 160,000$ , $d = 40 \mathrm{ m$ .

Question 3: Conservation of energy on a roller coaster

A roller coaster car starts from rest at the top of a hill $30 \mathrm{ m$ high. Calculate its speed at the bottom of the hill, assuming no energy is lost to friction.

Answer

$mgh = \frac{1}{2}mv^2$ So $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 30} = \sqrt{588} = 24.2 \mathrm{ m/s$ .

Question 4: Work done and power

A motor lifts a $200 \mathrm{ kg$ load through a height of $8 \mathrm{ m$ in $10 \mathrm{ s$ . Calculate (a) the work done and (b) the power of the motor.

Answer

(a) Work $= F \times d = mg \times h = 200 \times 9.8 \times 8 = 15,680 \mathrm{ J$ .

(b) Power $= W/t = 15,680/10 = 1568 \mathrm{ W$ .

Question 5: Specific heat capacity

$2.0 \mathrm{ kg$ of water is heated from $20^\circ\mathrm{C$ to $80^\circ\mathrm{C$ . The specific heat capacity of water is $4200 \mathrm{ J/(kg\cdot^\circ C)$ . Calculate the energy required.

Answer

$\Delta E = mc\Delta T = 2.0 \times 4200 \times (80 - 20) = 2.0 \times 4200 \times 60 = 504,000 \mathrm{ J = 504 \mathrm{ kJ$ .

Worked Examples

Example 1: Newton’s second law

A $2.0\,\text{kg}$ object is pulled across a rough horizontal surface by a horizontal force of $15\,\text{N}$ . The frictional force is $5.0\,\text{N}$ . Calculate the acceleration.

Solution:

$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 15 - 5.0 = 10\,\text{N}$

$a = \frac{F_{\text{net}}}{m} = \frac{10}{2.0} = 5.0\,\text{m\,s}^{-2}$

Summary

This topic covers the key concepts of Energy for GCSE Physics. Focus on understanding the fundamental principles, practising with exam-style questions, and applying your knowledge to unfamiliar contexts.

Mark this page as reviewed