Why does melting absorb energy without a temperature rise?

The energy breaks the particle arrangement instead of raising temperature.

Why divide by the total mass when mixing water?

The energy is shared over all the mass, so the total mass is the divisor.

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

It leaks to the air and the container instead of heating the water.

S4 Physics: Heat and Gases Practice Paper 3

2026-11-01

Tutor Prep Guide · Secondary Physics

Paper three starts with heat capacity, unit conversion and Boyle law, moves to heating time, Charles law and mixing, then reverses for specific heat capacity and adds efficiency, ending with the missing c slip.

Quick answer　18 900 J, 16 800 J, 1 L, 6 680 J, 42 s, 3 m3, 56 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)

Foundation0.3 kg of water rises by 15 K. Find the energy absorbed. (c = 4 200 J/(kg K))
Foundation500 g of water rises by 8 K. Find the energy absorbed.
Foundation2 L of gas at 1.5 atm is compressed to 3 atm at constant temperature. Find the new volume.
Foundation0.02 kg of ice melts completely. Find the energy absorbed. (latent heat of fusion 3.34 x 10^5 J/kg)
IntermediateA 1 200 W heater raises 0.6 kg of water by 20 K. Find the time needed.
IntermediateA gas at 300 K and 2 m3 is heated to 450 K at constant pressure. Find the new volume.
Intermediate0.3 kg of water at 80 C is mixed with 0.2 kg at 20 C. Find the final temperature.
ChallengeA metal of mass 0.3 kg absorbs 18 900 J and rises 70 K. Find its specific heat capacity.
ChallengeA 700 W heater of 60% efficiency raises 0.5 kg of water by 40 K. Find the time needed.
Error redo0.3 kg of water rises by 50 K. Find the energy absorbed. A common mistake gives 15 J; why is that wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	18 900 J (0.3 x 4 200 x 15)	E = m c dT, a straight product of three numbers.
2	16 800 J (0.5 x 4 200 x 8)	500 g is 0.5 kg, so convert first.
3	1 L (p1 V1 = p2 V2)	At constant temperature doubling the pressure halves the volume.
4	6 680 J (0.02 x 3.34 x 10^5)	Melting happens at a fixed temperature, so use latent heat only.
5	50 400 J, so 42 s	Get the energy first, then divide by the power.
6	3 m3 (V is proportional to T, a factor of 1.5)	Use kelvin and scale the volume by the same factor.
7	56 C ((0.3 x 80 + 0.2 x 20) / 0.5)	Average the energy over the total mass.
8	900 J/(kg K) (18 900 / (0.3 x 70))	Rearrange E = m c dT to isolate c.
9	84 000 J useful, 140 000 J supplied, so 200 s	Divide the useful energy by the efficiency to get the supply.
10	63 000 J (0.3 x 4 200 x 50)	15 is only 0.3 x 50, which forgets the 4 200.

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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