Fourier Conduction and Heat Flux
Fourier's law states that conductive heat flux is q_vector = -k grad T; for a uniform one-dimensional slab in steady state, heat current is H = kA(T_hot-T_cold)/L.
Why this shows up in the exam
Heat loss through insulation · Melting ice by conducted heat · Comparing rods of different dimensions
Learn the idea
Conduction rate is proportional to area, conductivity, and temperature gradient in the direction of heat flow. A better conductor, larger area, or steeper temperature drop carries more energy each second; length matters because it sets the gradient, not as an extra heat source.
🧠 Memory hook: Conductivity times area times temperature drop, divided by path length.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- H = dQ/dt = kA Delta T/L — steady one-dimensional heat current through a uniform slab
- q'' = H/A = -k dT/dx — local heat flux with the minus sign indicating flow down temperature
- Q = H t — energy conducted during a time interval at constant steady current
How to approach it
- 1Mark the actual heat-flow path
- 2Write its area, length, k, and endpoint temperatures
- 3Compute current and check watts or energy per time
Common slip-ups that cost marks
- •Using endpoint separation unrelated to the actual conduction path
- •Dropping cross-sectional area
- •Applying steady-state current before transients have settled
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.
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