UAT-UK SPEC CODE: P4 (Thermal physics)ASSUMED MATHS 1 PREREQUISITE: M3.4FREQUENCY: HIGH

Thermal Conduction Rod Scaling (Q̇ ∝ AΔT/L), Convection Density Currents & Radiative Power (P ∝ R^2 T^4)

Deconstructed from 13 past UAT-UK questions (5.58% of all historical Physics papers). Students using standard A-Level classroom methods spend an average of 135s on this archetype—blowing past the 90s ESAT budget. Using the 2-line Sub-90 heuristic below compresses solve time to 35s (saving 100s per question).

Historical Frequency
13 Past Qs
5.58% of Physics
Classroom Trap Time
135s
+45s over 90s budget
Sub-90 Target Time
35s
Banks +100s on the clock
Dominant UI Format
5-Option A–E
Maths 1 Skill: M3.4

Card 1 & Card 2 Preview: The Classroom Time Trap vs. Sub-90 2-Line Rule

Excerpted from our interactive 6-Step Topic Lesson for ARCH-P-15.

Launch All 6 Active-Recall Cards
CARD 1 OF 6 • THE CLASSROOM TIME TRAP (135s)
Why Students Bleed Time:
By Fourier's conduction law, Q˙=kAΔTL=k(πd2/4)ΔTL\dot{Q} = \frac{k A \Delta T}{L} = \frac{k (\pi d^2 / 4) \Delta T}{L}. Taking ratios gives Q˙2Q˙1=(d2d1)2(ΔT2ΔT1)(L1L2)=22×1×12=2\frac{\dot{Q}_2}{\dot{Q}_1} = \left(\frac{d_2}{d_1}\right)^2 \left(\frac{\Delta T_2}{\Delta T_1}\right) \left(\frac{L_1}{L_2}\right) = 2^2 \times 1 \times \frac{1}{2} = 2. Thus Q˙2=20×2=40 W\dot{Q}_2 = 20 \times 2 = 40\text{ W} (~120s).
In 13 past UAT-UK questions (5.58% of Physics), candidates lose 100s by doubling cylinder diameter quadruples cross-sectional area a, multiplying thermal conduction rate by 4. Instead, apply P4 with Maths 1 skill M3.4.
CARD 2 OF 6 • THE 2-LINE SUB-90 HEURISTIC (<35s)
**Line 1 (Conduction Scaling Q˙∝d2ΔT/L\dot{Q} \propto d^2 \Delta T / L):** Square the diameter factor (22=42^2 = 4) and invert the length factor (1/21/2).
Line 2 (Multiply): Q˙2=20×4×12=40 W\dot{Q}_2 = 20 \times \frac{4 \times 1}{2} = 40\text{ W} (25s).
Sanity & Limit Check: Verify units/signs against factor_of_two_or_stoichiometry_trap (Scaling cylindrical rod conduction rate linearly with diameter d (×2) instead of cross-sectional area A = πd²/4 (×4)) and sign_or_direction_flip (Multiplying by rod length L instead of dividing by L in thermal conduction rate Q̇ ∝ A*(T₂ - T₁) / L).

Examiner Distractor Blueprint for ARCH-P-15

FOIL #1factor_of_two_or_stoichiometry_trap
Scaling cylindrical rod conduction rate linearly with diameter d (×2) instead of cross-sectional area A = πd²/4 (×4)

Doubling cylinder diameter quadruples cross-sectional area A, multiplying thermal conduction rate by 4.

Open Factor-of-2 / Stoichiometry Clinic
FOIL #2sign_or_direction_flip
Multiplying by rod length L instead of dividing by L in thermal conduction rate Q̇ ∝ A*(T₂ - T₁) / L

A longer bar has a smaller temperature gradient ΔT/L and therefore a lower rate of heat transfer.

Open Sign / Direction Flip Clinic
FOIL #3multi_statement_linchpin_foil
Claiming shiny/dull surface finish alters the rate of CONVECTION directly, or that individual air molecules expand when heated

Surface emissivity (dull black vs shiny) governs thermal RADIATION; convection depends on bulk fluid density decrease (molecules move further apart, molecules themselves do not expand).

Open Multi-Statement Linchpin Foil Clinic
FOIL #4fence_post_or_domain_trap
Raising temperature factor to power 2 instead of power 4 in Stefan-Boltzmann radiative power P = k * R² * T⁴

Halving surface temperature T reduces radiative power by 2⁴ = 16.

Open Fence-Post / Domain Trap Clinic

Try 1 Live Sample Question Under the 90-Second Stopwatch

Press W to open the 2-line dry-wipe scratchpad, select A–E, and lock in to reveal the Dual-View solution.

sub90 ESAT Stopwatch Emblemsub90 ESAT
ARCH-P-15: Thermal Conduction Rod Scaling (Q̇ ∝ AΔT/L), Convection Density Currents & Radiative Power (P ∝ R^2 T^4) (Sample)Physics
00:00.0
SUB-90 ZONE (<35S)
0.0s35s Sub-9090s ESAT Budget150s Classroom
Question 1 of 1ARCH-P-15 (6-Card Lesson)Format: Standard 5-Option
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Question 1 of 1 • 90s Target

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Physics•ARCH-P-15•Standard 5-Option
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