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Quick subject briefing
Before this lesson: Consolidation after pupils have met stores, pathways, conservation and simple system boundaries. The supplied hoist gain avoids an omitted energy-equation calculation. Both tiers use the same route.
Energy is conserved in a complete account: a store decrease or input is matched by increases and outgoing transfers. Dissipation spreads energy into less useful stores, usually thermal stores of surroundings, without destroying it. For a stated closed boundary no energy crosses, although stores inside can exchange energy. A smaller system may gain or lose energy across its boundary. Useful versus unwanted depends on the intended task.
Words and symbols you will need
- Store / transfer pathway
- A store belongs to an object/system; electrically, mechanically, heating or radiation describes transfer.
- Conserved
- Not created or destroyed in a complete account.
- Dissipated
- Spread into less useful stores; still present.
- Boundary / closed
- Which objects are included; closed here means no energy enters or leaves that boundary.
- Useful / unwanted, J
- Useful serves the stated task; all account values use joules for the same interval.
Help pupils choose a method
- Name the task, time interval and objects inside the boundary.
- List distinct inputs/store decreases, internal increases and outgoing transfers. Balance the account without double counting.
- Check boundary crossings separately from conservation; decide usefulness from the intended task, and name receiving stores.
Use the supplied model — pupil page 2
Use page 2’s full 600 J battery account. The 420 J load increase and 180 J thermal increase sum to 600 J. Raising the load is useful; the thermal increase still exists. The transfer from battery to motor is electrical, not an electrical store.
During one lift a battery’s chemical store decreases by 600 J. The load’s gravitational potential store increases by 420 J and thermal stores increase by 180 J. There are no other changes. Include battery, motor, load and surroundings in the account. Does it conserve energy? Name the useful increase and the pathway from battery to motor.
Worked answer: 420+180=600 J, equal to the battery store decrease, so the full account conserves energy. The useful increase is the raised load’s gravitational potential store; 180 J is dissipated to thermal stores. The battery transfers energy electrically to the motor. No new energy is created.
If pupils are still stuck
Use the matching Say/Ask/Expect coaching and named new retry in its replacement time. If still insecure, help label the boundary and sort the supplied numbers into internal increases or crossing transfers. Record the exact remaining error for the usual teacher; next-lesson follow-up stays outside today’s 60 minutes.
KEEP THIS OPEN DURING THE LESSON
Timed teacher run sheet
- Print the complete 8-page pupil booklet, one per pupil, stapled in order. Double-sided printing is fine.
- Read teacher page 1, then use the timed run sheet on page 2. Keep the 10-page teacher guide separate from pupil copies.
- Pupils need pens, a ruler and a calculator. Reading, data and figures are supplied. No login, pupil devices or practical apparatus are required.
0-5 minPupil page / Q
1 / 1-3
Read aloud / check: Start from memory. Check the three responses at 5 minutes.
Next: If insecure A uses the first 2 reading minutes.
Answers: 1, 2, 3.
Open check 1 and its coaching →
5-15 minPupil page / Q
2 / reading
Read aloud / check: Read the account/boundary rules and complete lift model.
Next: Check A if used; then continue reading.
15-20 minPupil page / Q
3 / 4
Read aloud / check: Choose a letter and justify both store changes; stop.
Next: If insecure B replaces questions 5-6.
Answers: 4.
Open check 2 and its coaching →
20-30 minPupil page / Q
3-4 / 5-9
Read aloud / check: Apply the hoist givens, then explain heating/cooling in the dispenser.
Next: If B was used leave 5-6 blank. The 600 J hoist gain is supplied.
Answers: 5, 6, 7, 8, 9.
30-33 minPupil page / Q
5 / 10
33-45 minPupil page / Q
6-7 / 11-16
Read aloud / check: Use Figure 2, test the second record, then explain dissipation and counterweight.
Next: If C was used leave 11-12 blank; complete 13-16. Include every stated term once.
Answers: 11, 12, 13, 14, 15, 16.
45-55 minPupil page / Q
8 / 17
Read aloud / check: Correct practice in a different colour and explain a correction.
Next: Share only practice answers. Require named stores/boundaries.
Answers: 17.
55-60 minPupil page / Q
8 / 18
Read aloud / check: Cover earlier pages and answer question 18 independently.
Next: Collect all booklets; check afterwards. D is next lesson only.
Answers: 18.
Open check 4 and its coaching →
Estimated 60 minutes. Practice: 10 minutes, a 3-minute boundary check and 12 minutes. A uses reading time; B replaces 5-6; C replaces 11-12; D is next lesson only. Timings are not classroom-trialled.
Check entry, hinge and mid-practice at the marked pauses. At 45 minutes share only practice answers. Keep the independent exit and next-lesson answers hidden until submission.
Download teacher guide & answers ↓If several pupils are stuck
Use the matching Say/Ask/Expect coaching and named new retry in its replacement time. If still insecure, help label the boundary and sort the supplied numbers into internal increases or crossing transfers. Record the exact remaining error for the usual teacher; next-lesson follow-up stays outside today’s 60 minutes.
Estimated 60 minutes. Practice: 10 minutes, a 3-minute boundary check and 12 minutes. A uses reading time; B replaces 5-6; C replaces 11-12; D is next lesson only. Timings are not classroom-trialled.
ENTRY CHECK
Check 1: Question 1, Question 2, Question 3
Look for: Q1 60 J; Q2 kinetic decrease accounted for by thermal increases, not destruction; Q3 water warming is useful for its stated task.
If secure: Continue to page 2 reading and model.
If: The remaining transfer does not complete the 150 J account.Open coaching
Likely issue: Input and output bookkeeping is insecure.
- Say
- In this complete model, the useful and other transfers must add to the one input. Subtract the useful amount to find the remainder.
- Ask
- In question 1, what must be added to 90 J to make 150 J?
- Expect
- 60 J, so 90+60=150 J.
Next: Use page 2’s account/store/usefulness rules. Complete Try again A on page 1 in the first 2 reading minutes.
Pupil page 2, same-event and complete-account rules.
If: Calls brake heating destroyed energy or treats every thermal increase as unwanted.Open coaching
Likely issue: Conservation or task-dependent usefulness is insecure.
- Say
- Warmed objects have increased thermal stores. Energy is still present. Whether that warming is useful depends on the intended task.
- Ask
- For questions 2-3, which objects receive energy, and which warming is the water heater intended to produce?
- Expect
- Brakes/surroundings gain thermal energy without destruction. Warming the water is useful; warming the casing is unwanted for that task.
Next: Use page 2’s account/store/usefulness rules. Complete Try again A on page 1 in the first 2 reading minutes.
Pupil page 2, useful/unwanted and dissipated definitions.
HINGE CHECK
Check 2: Question 4
Look for: C: water -180 J and room +180 J give zero net change inside the stated closed boundary.
If secure: Start questions 5-9. Require complete accounts and named receiving stores.
If: A says 180 J is destroyed.Open coaching
Likely issue: A loss from one object has been mistaken for loss of total energy.
- Say
- Look for the receiving store. The room gains exactly what the water loses, inside the same stated boundary.
- Ask
- In question 4, where is the 180 J lost by the water accounted for?
- Expect
- In the room’s increased thermal store; it is not destroyed.
Next: Use page 2’s closed-boundary rule and model. Complete Try again B on page 3 instead of questions 5-6.
Pupil page 2, conservation and dissipation.
If: B counts only the water and says the combined total falls.Open coaching
Likely issue: The stated larger system boundary has been ignored.
- Say
- The system includes both water and room. Count both changes, not the water’s decrease alone.
- Ask
- For question 4, what is -180+180 for the complete water-and-room system?
- Expect
- 0 J net change: its total is unchanged.
Next: Use page 2’s closed-boundary rule and model. Complete Try again B on page 3 instead of questions 5-6.
Pupil page 2, closed-system rule.
If: D reverses the water’s change, or C is selected without an account.Open coaching
Likely issue: The direction of transfer is insecure or the correct letter may be a guess.
- Say
- The hot water is cooling, so its thermal store decreases. The cooler room receives that energy; equal and opposite changes cancel in the complete account.
- Ask
- Name both changes and their signs in question 4.
- Expect
- Water -180 J; room +180 J; total change zero.
Next: Use page 2’s closed-boundary rule and model. Complete Try again B on page 3 instead of questions 5-6.
Pupil page 2, boundary and named-store methods.
MID PRACTICE CHECK
Check 3: Question 10
Look for: Q10 energy leaving=200 J; toy-only boundary not closed because heating crosses it. The 100 J toy thermal increase is retained inside.
If secure: Continue 11-16; keep internal gains separate from outgoing transfers.
If: Gives 300 J or another remainder after omitting/double-counting an internal increase.Open coaching
Likely issue: Retained thermal energy and outgoing heating have been mixed.
- Say
- The spring decrease supplies two stated increases inside the toy and the energy leaving. Count the kinetic and retained thermal gains once each before finding the remainder.
- Ask
- In question 10, which increases stay inside the toy, and what is 500-200-100?
- Expect
- 200 J kinetic and 100 J toy thermal stay inside; 200 J leaves by heating.
Next: Use page 2, the boundary/account methods. Complete Try again C on page 5 instead of questions 11-12.
Pupil page 2, complete account and internal versus crossing transfers.
If: Calls the toy boundary closed because total energy is conserved.Open coaching
Likely issue: Universal conservation has been confused with no transfer across a chosen boundary.
- Say
- Energy conservation applies to a complete account, but the toy-only boundary loses energy to outside surroundings. That makes this chosen system non-closed for the interval.
- Ask
- In question 10, is the receiving outside room included inside the toy boundary? What crosses it?
- Expect
- No. Heating transfers 200 J out, so the toy-only system is not closed.
Next: Use page 2, the boundary/account methods. Complete Try again C on page 5 instead of questions 11-12.
Pupil page 2, closed-system definition.
EXIT CHECK
Check 4: Question 18
Look for: Q18 Q=100 J; boundary not closed because electrical input and sound output cross; 300 J remains in thermal stores and is unwanted for lifting.
If secure: Record independently secure account, boundary and usefulness reasoning.
If: Incorrect Q or describes missing/unwanted energy as destroyed.Open coaching
Likely issue: The full energy account or conservation still needs follow-up.
- Say
- Count the raised-load and thermal increases before finding the remaining outgoing transfer. Every term belongs in the same complete interval account.
- Ask
- For question 18, what are 500+300 and 900-800? Where is the stated thermal energy?
- Expect
- 800 J in stated store increases; 100 J leaves by sound. The 300 J still exists in thermal stores.
Next: Next lesson revisit the corresponding page 2 rules, then complete Try again D independently on page 8.
Pupil page 2, complete account and dissipation.
If: Calls the boundary closed or claims the thermal increase is useful simply because energy is conserved.Open coaching
Likely issue: Boundary crossings or usefulness relative to the intended task remains insecure.
- Say
- Check what crosses the actual boundary, then separately ask what the device is intended to do. Conservation does not make every output useful.
- Ask
- In question 18, which transfers cross the boundary and is the task to warm the room or raise a load?
- Expect
- Electrical energy enters and sound leaves, so it is not closed. The task is raising a load, so the thermal increase is unwanted for that task.
Next: Next lesson revisit the corresponding page 2 rules, then complete Try again D independently on page 8.
Pupil page 2, boundary and useful/unwanted rules.
Answers keyed to the pupil handout
- Question 1
Pupil page 1 - 150-90=60 J.
- Question 2
Pupil page 1 - No. The kinetic store decreases while thermal stores of brakes/surroundings increase through transfers associated with friction. Energy is conserved.
- Question 3
Pupil page 1 - The water’s thermal-store increase is useful; casing warming is unwanted for this task.
- Question 4
Pupil page 3 - C. Energy redistributes internally: water -180 J, room +180 J, net change 0 J. The water alone loses energy, but the stipulated larger system’s total stays constant.
- Question 5
Pupil page 3 - 1000-600=400 J. The load starts/finishes at rest and no other changes are in the stated model.
- Question 6
Pupil page 3 - Thermal stores of the hoist and surroundings increase through unwanted friction/heating transfers. Energy is conserved: it is dissipated into less useful stores, not destroyed. Name the receiving objects/stores.
- Question 7
Pupil page 4 - Increasing the water’s thermal store is useful for the intended water-heating task. Warming the casing is unwanted because it does not warm the required water. Thermal transfers are not automatically all unwanted.
- Question 8
Pupil page 4 - Water’s thermal store decreases; energy transferred by heating increases the cooler surroundings’ thermal store. Do not describe cold as an energy substance entering the water.
- Question 9
Pupil page 4 - Energy is redistributed internally from the water’s thermal store to the room’s thermal store. The loss from one equals the gain in the other, so total energy of the stipulated closed system is conserved.
- Question 10
Pupil page 5 - Energy leaving=500-200-100=200 J by heating to outside surroundings. The toy-only system is not closed because energy crosses its boundary. The retained 100 J thermal increase is inside the toy, not part of the 200 J leaving. A larger account including receiving surroundings still conserves energy.
- Question 11
Pupil page 6 - Q=2000-1200=800 J. Five equal input blocks each represent 400 J; the other branch contains two blocks.
- Question 12
Pupil page 6 - Energy is conserved with no other supply or store decrease: 3 water-heating blocks + 2 other blocks = 5 input blocks. The complete energy account must balance.
- Question 13
Pupil page 6 - 1200+950=2150 J, which is 150 J more than the input. Without another supply/store decrease or measurement error, the record cannot fit the stated model. It is not evidence that energy was created.
- Question 14
Pupil page 6 - No. The thin arrows organise the energy account; their direction on the page is not the real spatial direction of energy transfer. Equal blocks encode amounts; these thin arrows do not encode quantitative width.
- Question 15
Pupil page 7 - It spreads through the surroundings and is harder to gather and use for the intended task. It still exists; less useful does not mean destroyed or that no later transfer is possible.
- Question 16
Pupil page 7 - The counterweight store decreases by 900 J. Thermal increase=900-320=580 J. The raised load’s 320 J plus 580 J in thermal stores accounts for the 900 J decrease, so energy is redistributed, not created.
- Question 17
Pupil page 8 - Accept a corrected result with a physical reason, not a copied value. All-correct route: energy leaving the chosen system is received elsewhere; an enlarged complete account includes those gains, so total energy is conserved. The boundary determines the amount retained inside.
- Question 18
Pupil page 8 - (a) Q=900-500-300=100 J. (b) No: 900 J enters electrically and 100 J leaves by sound, so energy crosses the boundary. Stores inside increase by 800 J. (c) The 300 J remains in thermal stores, not destroyed; it is unwanted for the specified task of raising the load.
- Try again A
- (a) 240-150=90 J. (b) No, thermal stores of surroundings increase. (c) Useful for the water-heating task.
- Try again B
- Combined change=-300+300=0 J. Energy transfers internally to the room’s thermal store; it is not destroyed.
- Try again C
- Leaving energy=450-150-50=250 J. Not closed: 250 J crosses its boundary by heating. The 50 J thermal increase stays inside the toy.
- Try again D
- Sound energy=800-350-300=150 J. The boundary is not closed: electrical input and sound output cross it. Thermal energy remains in stores; it is not destroyed and is unwanted for the stated load-lifting task.
Worksheet selection and marking
The 11 selected source parts total 18 marks. Award complete accounts and named receiving stores; accept valid equivalent explanations. Do not require efficiency ratios or omitted energy equations.
Selection: Q2(b,e), Q4(b,c,d), Q5(a,b,d,e), Q6(d,e); pages 2,4,5,6.
Open the maintained worked solutions →Collect and follow up: Collect every named booklet with working and corrections. Use question 18 to distinguish arithmetic/account, boundary and usefulness/dissipation errors; report the exact difficulty and who needs Try again D.
