Conservation and Dissipation of Energy

Conservation of energy is one of the great laws of physics:

Energy cannot be created or destroyed; it can only be transferred between stores.

But conservation does NOT mean energy stays useful. Real systems dissipate energy — it spreads out to the thermal stores of the surroundings, where it becomes too dilute to do useful work.

Where dissipation comes from:

• Friction between rubbing surfaces → thermal store of both surfaces.
• Air resistance on moving objects → thermal store of the air.
• Electrical resistance in wires → thermal store of the wire and surroundings.
• Sound radiated from vibrating parts → eventually thermal store of whatever absorbs the wave.

For a car engine: chemical store of petrol → useful kinetic store of car (about 25 %) + wasted thermal stores (about 75 %).

For a light bulb: electrical → useful radiation (light) + wasted thermal store of the bulb (large for incandescent; tiny for LED).

AQA spec lists three named techniques to reduce wasteful energy transfers.

1. Lubrication. Adding oil or grease between rubbing surfaces:

• Reduces direct contact between surfaces → reduces friction.
• Less work done against friction → less thermal dissipation.
• Examples: engine oil, bicycle chain lubricant, door hinge WD-40.

2. Streamlining. Shaping a body so air (or water) flows around it smoothly:

• Reduces the size of the air-resistance force.
• Less work done against air resistance → less thermal dissipation in the air.
• Examples: sports cars, racing cyclists' helmets, salmon body shape.

3. Thermal insulation. Surrounding a warm object with low-thermal-conductivity material:

• Slows the rate of heat transfer.
• Air pockets (in fibreglass, foam, double-glazing cavities) are key — air is a poor thermal conductor when trapped.
• Examples: loft insulation in UK homes, cavity wall insulation, vacuum flasks, double-glazed windows.

Thermal conductivity measures how easily a material lets energy transfer through it by conduction.

Material	Thermal conductivity (W/m·K)	Use
Copper	400	Saucepan base (high conductivity, fast heating)
Steel	50	Saucepan body
Glass	1	Windows
Brick	0.7	UK house walls
Wood	0.13	Window frames
Loft fibreglass	0.04	Roof insulation
Still air	0.025	Best 'insulator' when trapped

Two factors affect heat loss through a wall:

1. Thermal conductivity of the material (lower = better insulator).
2. Thickness — doubling thickness roughly halves the rate of heat transfer.

This is why UK building regulations specify minimum loft insulation thickness (270 mm is the current standard). Cavity wall insulation traps air bubbles to combine both low conductivity and adequate thickness.

UK cost comparison (typical 2024 retrofit, three-bed semi):

• Loft insulation upgrade: £400–£600.
• Cavity wall insulation: £600–£1500.
• Annual gas saving: £150–£400.

These payback calculations come up in evaluative exam questions about energy efficiency.

Apparatus: boiling kettle, small beaker (100 mL), thermometer (or thermocouple), stopwatch, lid, insulating materials to test (e.g. bubble wrap, newspaper, cotton wool, foam).

Method:

1. Pour the same volume (and same starting temperature) of hot water into the beaker for each trial.
2. Wrap with one layer of an insulating material; cover with lid.
3. Record temperature every minute for 10 minutes.
4. Repeat with different materials (and with different thicknesses of the best one).
5. Always start from approximately the same initial temperature for fair comparison.

Analysis:

• Plot temperature vs time graphs on the same axes.
• The shallowest curve indicates the best insulator (smallest temperature drop).
• Calculate the temperature drop after a fixed time (e.g. 10 minutes) for direct comparison.

Sources of error and mitigation:

• Lid removal during measurement → take care, replace immediately.
• Variable room temperature → repeat trials and average.
• Different starting temperatures → record exactly, or wait until water is at a consistent temperature.

Independent variable: material (or thickness for follow-up). Dependent variable: temperature (or temperature drop). Control variables: starting temperature, volume of water, lid used, ambient room temperature, time.

Efficiency is defined as:

$\text{efficiency} = \frac{\text{useful energy output}}{\text{total energy input}}$

or equivalently for power:

$\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}}$

The answer is between 0 and 1 (as a decimal) or 0 % and 100 % (as a percentage). Multiply the decimal by 100 to get the percentage.

Worked example. An electric heater receives 4000 J of electrical energy. 3600 J ends up as useful thermal energy in the room. The rest is wasted as sound radiation and unintended heating of the wall behind the heater.

$\text{efficiency} = \frac{3600}{4000} = 0.90 = 90\%$

Worked example (power version). A 60 W motor delivers 12 W of useful mechanical power.

$\text{efficiency} = \frac{12}{60} = 0.20 = 20\%$

The remaining 80 % is dissipated, mostly as thermal energy in the motor windings.

Why can't efficiency reach 100 %? Every device involves moving parts (friction), wires (resistance) or vibrating components (sound). All of these dissipate energy to thermal stores of the surroundings. The Second Law of Thermodynamics rules out 100 %-efficient devices — though some come close (electric kettles ≈ 95–98 %).

A Sankey diagram is a visual representation of energy flow through a device. The width of each arrow is proportional to the energy it represents.

For a filament bulb receiving 100 J of electrical input, with 5 J of useful light output and 95 J wasted as heat:

• The input arrow is 100 units wide.
• The useful (light) arrow branching off is 5 units wide.
• The wasted (heat) arrow branching off is 95 units wide.
• Total of branches = total input (conservation of energy).

For an LED bulb receiving 100 J: 80 J useful (light), 20 J wasted (heat).

Reading Sankey diagrams in exams. AQA might give you the diagram and ask:

• 'How much energy is wasted?' — measure the wasted arrow's width.
• 'Calculate efficiency.' — (useful arrow width) ÷ (total input width).
• 'Why is the filament bulb less efficient?' — most energy goes to the thermal store rather than light.

General strategies (from spec 4.1.2.1 and 4.1.2.2):

1. Lubrication of moving parts → reduces friction → less thermal dissipation.
2. Streamlining moving bodies → reduces air/water resistance.
3. Insulation of thermal devices → less heat lost to surroundings (so more remains in the useful store).
4. Low-resistance wires → less electrical energy dissipated in heat.
5. Modern materials — LED instead of filament, induction hob instead of resistance hob.
6. Regular maintenance — clean fans, replace worn bearings, descale kettles.

Real-world UK efficiency upgrades:

• Replacing an incandescent bulb (5 % efficient) with an LED (80 %+) reduces electricity use by ~90 %.
• Modern condensing gas boiler (90–95 %) vs old conventional boiler (70 %) — UK government grants available for replacements.
• Electric vehicles (75–90 % well-to-wheel efficient) vs petrol cars (25–30 %).
• Heat pumps (300–400 % "efficient" because they move thermal energy from outside; technically a coefficient of performance, not efficiency).

Calculating useful output from input + efficiency. If efficiency = 0.65 and input = 8000 J, useful output = 0.65 × 8000 = 5200 J. Wasted = 8000 − 5200 = 2800 J.