ARCH-P-15: Thermal Conduction Rod Scaling (Q̇ ∝ AΔT/L), Convection Density Currents & Radiative Power (P ∝ R^2 T^4)

Full 15-Q Timed Drill

Step 1 of 6 • The Core Idea

ARCH-P-15: Why the 135s Classroom Method Fails Under the 90s Clock

Standard Classroom Method (Too Slow)~135s (Blows 90s Clock)

By Fourier's conduction law,

\dot{Q} = \frac{k A \Delta T}{L} = \frac{k (\pi d^2 / 4) \Delta T}{L}

. Taking ratios gives

\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

\dot{Q}_2 = 20 \times 2 = 40\text{ W}

(~120s).

The Fast ShortcutTarget: 35s (Saves 100s)

**Line 1 (Conduction Scaling

\dot{Q} \propto d^2 \Delta T / L

):** Square the diameter factor (

2^2 = 4

) and invert the length factor (

1/2

).
Line 2 (Multiply):

\dot{Q}_2 = 20 \times \frac{4 \times 1}{2} = 40\text{ W}

(25s).

Where Students Lose Time: 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.

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