Why must gas laws use kelvin rather than Celsius?

The proportional relations hold for absolute temperature; Celsius can give negative volume.

Why may heat loss be ignored in the mixture method?

A well lagged calorimeter loses very little, so it is taken as zero.

Where does the energy go when efficiency is below 100%?

It escapes to the surroundings instead of raising the temperature.

S4 Physics: Heat and Gases Practice Paper 2

2026-11-01

Tutor Prep Guide · Secondary Physics

Paper two opens with E = mc dT and Boyle law, then latent heat, heating time, Charles law and mixtures, ending with specific heat capacity, efficiency and the missing-c mistake.

Quick answer　42 000 J, 21 000 J, 1.5 L, 113 000 J, 84 s, 8 m3, 34 C, 900 J/(kg K), 200 s, 63 000 J.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10 questions
Suggested time	25 minutes
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S4 Physics: Heat and Gases – Tutor Lesson Plan
Level	S4

02　Questions (10 items)

Foundation0.5 kg of water rises by 20 K. Find the energy absorbed. (c = 4 200 J/(kg K))
Foundation200 g of water rises by 25 K. Find the energy absorbed.
Foundation3 L of gas at 1 atm is compressed to 2 atm at constant temperature. Find the new volume.
Foundation0.05 kg of steam condenses to water. Find the energy released. (latent heat of vaporisation 2.26 x 10^6 J/kg)
IntermediateAn 800 W heater raises 0.4 kg of water by 40 K. Find the time needed.
IntermediateAt constant pressure 4 m³ of gas at 250 K is heated to 500 K. Find the new volume.
Intermediate0.1 kg of water at 90 C is added to 0.4 kg at 20 C. Find the final temperature, ignoring losses.
ChallengeA 0.5 kg metal block absorbs 22 500 J and rises by 50 K. Find its specific heat capacity.
ChallengeHeating 0.5 kg of water by 40 K needs 84 000 J. If the heater is only 70% efficient at 600 W, find the time.
Error redoHow much energy heats 0.25 kg of water by 60 K? A student answers 15 J; what was left out?

03　Answers and one-line notes

#	Answer	One-line note
1	42 000 J (0.5 x 4 200 x 20)	E = mc dT, with mass in kg and change in K.
2	21 000 J (0.2 x 4 200 x 25)	Convert 200 g to 0.2 kg first.
3	1.5 L (p1 V1 = p2 V2)	At constant temperature pressure and volume are inversely proportional.
4	113 000 J (0.05 x 2.26 x 10^6)	Condensing releases the same energy that vaporising needs.
5	67 200 J, so 84 s	First get 67 200 J, then divide by 800 W.
6	8 m³ (V / T is constant)	At constant pressure volume is proportional to absolute temperature.
7	34 C ((9 + 8) / 0.5)	Heat lost equals heat gained, with 0.5 kg in total.
8	900 J/(kg K) (22 500 / (0.5 x 50))	c = E / (m dT) with the denominator equal to 25.
9	Input of 120 000 J is needed, so 200 s	Divide useful energy by efficiency, then by power.
10	63 000 J; 15 is only m dT, the specific heat capacity 4 200 was left out	The product m dT must still be multiplied by c.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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