How does specific heat capacity differ from latent heat?

Specific heat capacity is the energy per kg per kelvin; latent heat is the energy of a change of state.

Why does temperature stay constant during melting?

The energy breaks the regular arrangement instead of raising kinetic energy.

Why must gas calculations use absolute temperature?

Only in kelvin is volume simply proportional to temperature.

S4 Physics: Heat and Gases Practice Paper 1

2026-11-01

Tutor Prep Guide · Secondary Physics

Ten items from specific heat capacity and latent heat through Boyle law, gas volume at constant pressure and mixing, ending with efficiency and back-calculation.

Quick answer　84,000 J, 16,800 J, 0.5 L, 66,800 J, 168 s, about 2.67 m3, 44 degrees Celsius, 900 J per kg per K, 105 s, and 42,000 J rather than 4,200 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)

Foundation2 kg of water rises by 10 degrees Celsius. How much energy is absorbed? (c = 4,200 J per kg per K)
Foundation100 g of water is heated from 20 to 60 degrees Celsius. How much energy is absorbed?
FoundationAt constant temperature a gas occupies 2 L at 1 atm. It is compressed to 4 atm. Find the new volume.
Foundation0.2 kg of ice at 0 degrees Celsius melts completely. How much energy is absorbed? (specific latent heat of fusion 3.34 x 10^5 J per kg)
IntermediateA 500 W heater warms 1 kg of water by 10 degrees Celsius. Ignoring losses, how long does it take?
IntermediateA fixed mass of gas at constant pressure occupies 2 m3 at 300 K. It is heated to 400 K. Find the volume.
Intermediate0.2 kg of water at 80 degrees Celsius is mixed with 0.3 kg at 20 degrees Celsius. Ignoring losses, find the final temperature.
Challenge0.5 kg of metal cools from 100 to 30 degrees Celsius, releasing 31,500 J. Find its specific heat capacity.
ChallengeA 1,000 W heater of 80 per cent efficiency raises 0.5 kg of water by 40 degrees Celsius. How long does it take?
Error redoA student says 1 kg of water needs only 4,200 J to rise 10 degrees Celsius. What is wrong and what is the correct value?

03　Answers and one-line notes

#	Answer	One-line note
1	84,000 J (2 x 4,200 x 10)	Use E = mcDT.
2	16,800 J (0.1 x 4,200 x 40)	Convert the mass to kg first; the rise is 40 K.
3	0.5 L (P1V1 = P2V2)	Boyle law: at constant temperature pressure and volume are inversely proportional.
4	66,800 J (0.2 x 3.34 x 10^5)	Melting happens at constant temperature, so use E = mL.
5	168 s (84,000 J / 500 W)	Find the energy needed, then divide by the power.
6	About 2.67 m3 (2 x 400 / 300)	At constant pressure volume is proportional to absolute temperature.
7	44 degrees Celsius ((0.2×80 + 0.3×20) / 0.5)	Heat lost by the hot water equals heat gained by the cold water.
8	900 J per kg per K	Rearrange E = mcDT to c = E / (mDT).
9	105 s (84,000 J is needed, so 105,000 J must be supplied)	Only 80 per cent is useful, so divide by the efficiency.
10	The mass factor was left out; E = 1 x 4,200 x 10 = 42,000 J	4,200 J is the energy for 1 kg rising by just 1 K.

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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