Power, conservation and dissipation of energy, efficiency (RP2) – GCSE Physics exam questions with mark scheme

1Grade 1–3Foundation3 marks

A lamp is supplied with 200 J of energy. It transfers 30 J of energy usefully as light.

(a) Calculate the efficiency of the lamp. [2]

(b) State what happens to the energy that is not usefully transferred. [1]

(a) efficiency $= \dfrac{30}{200}$(1)
(a) 0.15 allow 15%; do not accept 0.15% or 15 with no % sign(1)
(b) it is dissipated / transferred to the (thermal energy store of the) surroundings allow it heats the surroundings; allow transferred to the thermal store of the lamp; allow stored in less useful ways; allow (wasted as) thermal energy / heat (energy); 'wasted' or 'lost' alone does not score; reject 'destroyed' / 'used up'(1)
Answer: (a) 0.15 (15%) (b) dissipated to the surroundings
Examiner tip: Efficiency has no unit and can never be more than 1 (100%) — check you divided useful by total, not the other way.

2Grade 6–7Higher5 marks

An electric motor has an input power of 800 W. The efficiency of the motor is 0.65.

(a) Calculate the useful power output of the motor. [2]

(b) Calculate the energy wasted by the motor in 2.0 minutes. [3]

(a) $0.65 = \dfrac{\text{useful power output}}{800}$ or $0.65 \times 800$(1)
(a) 520 W(1)
(b) wasted power $= 800 - 520 = 280$ (W) or $0.35 \times 800 = 280$ (W) allow total input energy $= 800 \times 120 = 96\,000$ J (with or without useful output energy $= 520 \times 120 = 62\,400$ J); allow ecf from (a)(1)
(b) $E = 280 \times 120$ or $96\,000 - 62\,400$ or $0.35 \times 96\,000$ (time must be in seconds)(1)
(b) 33 600 J allow 34 000 J (2 s.f.); allow 3.36 × 10⁴ J; allow 33.6 kJ; allow ecf from (a); a correct answer with no working scores 3 marks(1)
Answer: (a) 520 W (b) 33 600 J
Examiner tip: Convert minutes to seconds before using $E = Pt$ — 2.0 min is 120 s.

3Grade 8–9Higher5 marks

An electric kettle has a power input of 2.4 kW. The kettle has an efficiency of 0.90.

The kettle heats 1.5 kg of water from 15 °C to 100 °C.

specific heat capacity of water = 4200 J/kg °C

Calculate the time taken to heat the water to 100 °C. Give your answer to 2 significant figures. [5]

$\Delta E = 1.5 \times 4200 \times 85$(1)
$\Delta E = 535\,500$ (J)(1)
useful power output $= 0.90 \times 2400 = 2160$ (W) allow total input energy = 535 500 / 0.90 = 595 000 J(1)
$t = \dfrac{535\,500}{2160}$ or $\dfrac{595\,000}{2400}$ (ecf)(1)
250 s (2 sf) an unrounded 248 s scores 4 marks only(1)
Answer: 250 s
Examiner tip: Power must be in watts (2.4 kW = 2400 W) and the efficiency applied once — either to the power or to the energy, not both.

4Grade 6–7Higher5 marks

A ball of mass 0.15 kg is dropped from a height of 20 m. It hits the ground at a speed of 18 m/s.

gravitational field strength = 9.8 N/kg

(a) Calculate the energy dissipated due to air resistance as the ball falls. [4]

(b) State what is meant by 'dissipated'. [1]

(a) $E_p = 0.15 \times 9.8 \times 20 = 29.4$ (J)(1)
(a) $E_k = 0.5 \times 0.15 \times 18^2$(1)
(a) $E_k = 24.3$ (J)(1)
(a) 5.1 J allow ecf: their $E_p$ − their $E_k$ correctly calculated(1)
(b) energy is spread out / transferred to the surroundings (in a way that is) not useful allow wasted energy transferred to thermal store of surroundings(1)
Answer: (a) 5.1 J (b) energy spread out to the surroundings in a non-useful way
Examiner tip: Air resistance energy = GPE lost − KE gained; you need both energies before you can subtract.

5Grade 4–55 marks

An electric motor lifts a load of mass 30 kg through a vertical height of 4.0 m in 6.0 s.

gravitational field strength = 9.8 N/kg

(a) Calculate the useful power output of the motor. [3]

(b) A second motor lifts the same load through the same height in 4.0 s.

Explain which motor has the greater power output. [2]

(a) $E_p = 30 \times 9.8 \times 4.0$ or work done $= 294 \times 4.0$(1)
(a) $E_p = 1176$ (J) allow 1180 J(1)
(a) $P = 1176 / 6.0 = 196$ W allow ecf; allow 197 W from 1180 J; allow 200 W; a correct answer with no working scores 3 marks(1)
(b) both motors transfer the same energy / do the same work (same load lifted through the same height)(1)
(b) the second motor does this in less time, so the second motor has the greater power (greater rate of energy transfer) allow for both marks a correct calculation showing the second motor's power is 294 W (greater than 196 W); naming the second motor with no reason scores 0(1)
Answer: (a) 196 W (b) the second motor: same energy transferred in a shorter time
Examiner tip: 'The second motor is faster' gets only 1 mark in (b): you must also say both motors transfer the same energy (do the same work).

6Grade 1–3Foundation4 marks

A cyclist stops pedalling and freewheels along a flat road. The bicycle gradually slows down and stops.

(a) Which energy store of the cyclist and bicycle decreases as the bicycle slows down?

Tick (✓) one box.

☐ chemical ☐ elastic potential ☐ gravitational potential ☐ kinetic [1]

(b) Energy cannot be destroyed. Describe what happens to the energy as the bicycle slows down. [2]

Give one way the cyclist could reduce this unwanted energy transfer. [1]

(a) kinetic (only one box ticked)(1)
(b) energy is transferred (by friction / air resistance) to the thermal energy store of the surroundings allow the bicycle / tyres / road / air get (slightly) hotter; allow 'heat' for thermal; do not accept 'energy is used up' or 'energy is lost' with no further detail(1)
(b) the energy is dissipated / spread out (into the surroundings) allow becomes wasted energy that cannot be used usefully; allow some transferred to sound(1)
(c) lubricate / oil / grease the chain allow lubricate the moving parts / axles / bearings; allow 'reduce friction in the chain' only if a method is given (e.g. oiling); ignore 'pump up the tyres'(1)
Answer: (a) kinetic (b) transferred to the thermal store of the surroundings (by friction / air resistance) and dissipated (c) lubricate (oil) the chain
Examiner tip: 'The energy is lost' scores nothing: say which store it goes to (thermal, surroundings) and that it is dissipated (spread out).

7Grade 6–7Higher6 marks

An electric drill is switched on for 60 s.

The graph shows how the total energy input to the drill and the useful energy output from the drill change with time.

(a) Use the graph to determine the total power input to the drill. [2]

(b) Calculate the efficiency of the drill. [2]

Suggest one way the manufacturer could increase the efficiency of the drill. Explain how your suggestion increases the efficiency. [2]

(a) $P = \dfrac{30\,000}{60}$ or any correct pair of values read from the 'total energy input' line, e.g. $\dfrac{15\,000}{30}$; energy must be in J (or answer given in kW)(1)
(a) 500 W allow 0.5 kW; a correct answer with no working scores 2 marks; 0.5 W or 500 kW scores 1 mark (power-of-ten error)(1)
(b) efficiency $= \dfrac{18}{30}$ or $\dfrac{18\,000}{30\,000}$ or $\dfrac{300}{500}$ (useful output ÷ total input, any consistent pair read from the graph at the same time)(1)
(b) 0.6 allow 60%; do not accept 60 with no % sign; allow ecf from (a) if power route used; 0.6 with no working scores 2 marks(1)
(c) lubricate the moving parts / gears / bearings (of the drill) allow use a motor with lower-resistance wires / thicker copper wires; allow use better-fitting / smoother moving parts(1)
(c) (this reduces friction / resistance) so less energy is dissipated / wasted (to the thermal store of the motor / surroundings), so a greater proportion of the input energy is transferred usefully the explanation must match the method given; ignore 'so less energy is used'(1)
Answer: (a) 500 W (b) 0.6 (60%) (c) e.g. lubricate the moving parts: reduces friction so less energy is dissipated as thermal energy
Examiner tip: Read the energy in kJ and convert to J before dividing by time — 30 kJ in 60 s is 500 W, not 0.5 W.

8Grade 6–7Triple only6 marks

A student wants to investigate how the number of layers of newspaper wrapped around a beaker affects how quickly hot water in the beaker cools down.

The student has a 250 cm³ beaker, sheets of newspaper, sticky tape, a kettle, a measuring cylinder, a thermometer and a stopwatch.

Describe a method the student could use for this investigation. [6]

wrap the beaker in one layer of newspaper (and tape it in place)(1)
use a measuring cylinder to pour a fixed volume of hot water (e.g. 200 cm³) from the kettle into the beaker(1)
put the thermometer in the water and record the starting temperature (e.g. 80 °C) / start the stopwatch when the water reaches a chosen starting temperature(1)
record the temperature of the water after a fixed time (e.g. 10 minutes) or every minute for 10 minutes(1)
calculate the temperature decrease (start temperature − final temperature)(1)
repeat with 2, 3, 4, 5 … layers of newspaper (and with no newspaper as a comparison)(1)
control variables: same volume / mass of water, same starting temperature, same beaker, same time interval, same type of newspaper, same room / surroundings(1)
put a lid (e.g. a piece of card) on the beaker or use the same arrangement each time, so energy transfer from the top is the same(1)
repeat each number of layers and calculate a mean temperature decrease(1)
plot a graph of temperature decrease (or final temperature) against number of layers(1)
safety: take care with hot water / stand the beaker on a heatproof mat / wipe up spills(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that could be followed by another student to get valid results. It clearly identifies the variable changed (number of layers), how the temperature change is measured over a fixed time, and how at least two control variables are kept the same.
Level 2	3–4	A method with the main steps in a sensible order that would allow some useful results to be collected, but some detail is missing, e.g. control variables only partly given or how the temperature change is found is unclear.
Level 1	1–2	Simple, separate statements about the investigation, e.g. 'wrap the beaker in newspaper and measure the temperature', with little or no sequence. Not enough to produce useful results.
No relevant content	0	Nothing worthy of credit.

Answer: Wrap beaker in n layers; add fixed volume of hot water at the same start temperature; record temperature after a fixed time; find temperature drop; repeat for different numbers of layers keeping volume, start temperature, time and beaker the same; repeat and average; plot graph.
Examiner tip: To reach Level 3 you must say what you keep the same (volume of water, starting temperature, time) — a method that only says 'measure the temperature' cannot give valid results.

9Grade 8–9Higher7 marks

A filament lamp and an LED lamp each have a useful power output of 3.0 W as light.

The efficiency of the filament lamp is 5.0%. The efficiency of the LED lamp is 0.40.

(a) A café replaces one filament lamp with one LED lamp. The lamp is switched on for 4.0 hours each day.

Calculate the energy saved in one year (365 days). Give your answer in megajoules (MJ) to 2 significant figures. [5]

(b) Explain, in terms of energy transfers, why the filament lamp has a much lower efficiency than the LED lamp. [2]

(a) total power input of filament lamp $= \dfrac{3.0}{0.050} = 60$ (W) must convert 5.0% to 0.050 (or use 3.0 × 100 / 5.0)(1)
(a) total power input of LED lamp $= \dfrac{3.0}{0.40} = 7.5$ (W) allow power saved $= 60 - 7.5 = 52.5$ W for this mark if both powers not shown separately(1)
(a) time $= 4.0 \times 3600 \times 365 = 5\,256\,000$ (s) allow $5.26 \times 10^{6}$ s(1)
(a) energy saved $= 52.5 \times 5\,256\,000$ or $315\,360\,000 - 39\,420\,000$ $= 275\,940\,000$ (J) allow ecf from incorrect power or time(1)
(a) 280 MJ (conversion to MJ and 2 s.f. both needed for this mark; allow ecf from an incorrect value in J correctly converted and rounded; 276 MJ or 2.8 × 10⁸ J scores 4 marks)(1)
(b) in the filament lamp most of the input energy is transferred to the thermal energy store of the lamp / surroundings (the filament gets very hot) allow 95% of the energy is wasted as heating; allow 'heat' for thermal(1)
(b) the LED transfers a greater proportion / fraction of the input energy to light (so less is dissipated to the surroundings) do not accept 'LED uses less energy' alone — it must refer to proportion / fraction of input energy(1)
Answer: (a) 280 MJ (2.8 × 10⁸ J) (b) the filament lamp dissipates most of its input energy by heating (thermal store of surroundings); the LED transfers a greater proportion to light
Examiner tip: Both lamps give the same light, so work out each lamp's input power (useful output ÷ efficiency) — and turn 5.0% into 0.050 before dividing.

10Grade 4–55 marks

The table gives the thermal conductivity of four building materials.

Material	Thermal conductivity in W/m K
brick	0.77
concrete	1.13
aerated block	0.15
mineral wool	0.04

(a) Which material in the table is the best thermal insulator? Give a reason for your answer. [2]

(b) Two houses, A and B, are identical except for their outside walls.

House A has walls made of brick 0.20 m thick. House B has walls made of concrete 0.20 m thick.

On a cold day both houses are at 20 °C when the heating is switched off.

Explain which house will cool down more slowly. [2]

State how this changes the rate of cooling of house B. [1]

(a) mineral wool(1)
(a) it has the lowest thermal conductivity allow lowest number / value in the table; allow it conducts energy most slowly; this mark is only awarded if mineral wool is chosen(1)
(b) house A because brick has a lower thermal conductivity (than concrete) allow 0.77 is less than 1.13; naming house A with no reason scores 0(1)
(b) so the rate of energy transfer (by conduction) through the walls (to the surroundings) is lower allow less energy is transferred through the walls each second; ignore 'heat is kept in' without reference to rate or energy transfer(1)
(c) the rate of cooling decreases allow house B cools more slowly; allow less energy transferred per second through the walls(1)
Answer: (a) mineral wool: lowest thermal conductivity (b) house A: brick has lower thermal conductivity so lower rate of energy transfer through the walls (c) rate of cooling decreases
Examiner tip: Always link low thermal conductivity to a lower rate of energy transfer — 'brick is a better insulator' alone is only half the explanation.

Still losing marks on this topic?

Book a free introductory session and we'll find exactly which step is costing you.

Book a free intro