In steady conduction through a plane wall, the heat transfer rate Q is given by which expression?

In steady conduction through a plane wall, the heat transfer rate Q is given by which expression?

• A. Q = ΔT / (k A L)
• B. Q = k ΔT / (A L)
• C. Q = k A ΔT / L
• D. Q = k A L / ΔT

Answer: (C)

Steady one‑dimensional conduction through a plane wall with uniform properties gives heat transfer rate proportional to the driving temperature difference, area, and conductivity, and inversely proportional to the wall thickness. Fourier’s law starts this reasoning: the heat flux q" = -k dT/dx. For a wall of thickness L with a linear temperature change from T1 to T2, the gradient is (ΔT)/L, so the total rate is Q = q" A = k A ΔT / L (taking magnitude since heat flows from hot to cold). This shows Q grows with higher k, larger cross‑sectional area A, and larger ΔT, and it shrinks as the wall gets thicker L. The other forms mix the factors incorrectly (for example, placing L in the numerator or inverting ΔT), which would give the wrong units or the wrong dependence.

Steady one‑dimensional conduction through a plane wall with uniform properties gives heat transfer rate proportional to the driving temperature difference, area, and conductivity, and inversely proportional to the wall thickness. Fourier’s law starts this reasoning: the heat flux q" = -k dT/dx. For a wall of thickness L with a linear temperature change from T1 to T2, the gradient is (ΔT)/L, so the total rate is Q = q" A = k A ΔT / L (taking magnitude since heat flows from hot to cold). This shows Q grows with higher k, larger cross‑sectional area A, and larger ΔT, and it shrinks as the wall gets thicker L. The other forms mix the factors incorrectly (for example, placing L in the numerator or inverting ΔT), which would give the wrong units or the wrong dependence.