Which four variables does the conduction formula use?

k the conductivity, A the area, delta T the temperature difference and L the thickness.

Why convert thickness to metres?

k is in W/(m K), so a length left in millimetres puts the rate out by a factor of a thousand.

S4 Physics heat transfer paper 1: ten items on the conduction formula

2026-11-10

Tutor Prep Guide · S4 Physics Heat Transfer

Ten items start from Q/t = kA delta T / L, converting thickness and area first.

Direct answers　Items 01 to 03 cover three materials, 04 to 06 reverse for difference, thickness and area, 07 to 09 ratios and vacuum. Allow 25 minutes.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S4 Physics Heat Transfer
Level	S4

02　Questions (10 items)

FoundationA copper plate has k = 400 W/(m K), area 0.5 m2, faces 20 K apart and thickness 0.01 m. Find the rate of heat flow.
FoundationGlass has k = 1.0 W/(m K). With area 1.5 m2, temperature difference 20 K and thickness 0.005 m, find the rate.
FoundationAir has k = 0.026 W/(m K). With area 1.0 m2, 10 K difference and thickness 0.1 m, find the rate.
IntermediateA board with k = 0.5 W/(m K), area 2.0 m2 and thickness 0.02 m carries 2000 W. Find the temperature difference.
IntermediateA board with k = 1.0 W/(m K), area 1.0 m2 and 20 K difference must carry only 1000 W. What thickness is needed?
IntermediateA window of 200 cm2 has k = 0.8 W/(m K), 50 K difference and thickness 0.01 m. Find the rate.
ChallengeIf the copper plate in the first item is 0.02 m thick instead, all else unchanged, find the rate.
ChallengeIf both the area and the temperature difference double, how many times larger is the rate?
ChallengeWhy can a vacuum gap block both conduction and convection?
Error redoA copper plate with k = 400 W/(m K), area 1.0 m2, 20 K difference and thickness 5 mm gives a student 1600 W. What is correct?

03　Answers and one-line notes

#	Answer	One-line note
1	400000 W	Q/t = kA(delta T) / L = 400 x 0.5 x 20 / 0.01 = 400000 W; thickness must be in metres.
2	6000 W	Q/t = 1.0 x 1.5 x 20 / 0.005 = 6000 W; the k of glass is four hundred times smaller than copper.
3	2.6 W	Q/t = 0.026 x 1.0 x 10 / 0.1 = 2.6 W; the tiny k of air makes it a natural insulator.
4	40 K	delta T = (Q/t) x L / (kA) = 2000 x 0.02 / 1.0 = 40 K.
5	0.02 m (2 cm)	L = kA(delta T) / (Q/t) = 20 / 1000 = 0.02 m, that is 2 cm.
6	80 W	200 cm2 = 0.02 m2, so Q/t = 0.8 x 0.02 x 50 / 0.01 = 80 W.
7	200000 W (halved)	The rate is inversely proportional to thickness, so doubling it halves the rate.
8	4 times	The rate is proportional to both A and delta T, so both doubling gives 2 x 2 = 4 times.
9	There are no particles, so neither collisions nor fluid flow can occur	Conduction needs particle collisions and convection needs fluid flow; a vacuum has neither, so only radiation crosses.
10	1600000 W	5 mm = 0.005 m, so Q/t = 400 x 1 x 20 / 0.005 = 1600000 W; the slip is the mm to m conversion.

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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