Question

Derive the unit of k from Fourier's law of conduction equation: $q_{conduction} = -kA\frac{\Delta T}{\Delta x}$ q conduction = heat transfer rate (watts) k = thermal conductivity (unit=?) A = cross sectional area (m^2) delta T = change in temperature (Celsius or Kelvin) delta x = (meters)

          Derive the unit of k from Fourier's law of conduction equation:
$q_{conduction} = -kA\frac{\Delta T}{\Delta x}$
q conduction = heat transfer rate (watts)
k = thermal conductivity (unit=?)
A = cross sectional area (m^2)
delta T = change in temperature (Celsius or Kelvin)
delta x = (meters)

Derive the unit of k from Fourier's law of conduction equation:
qconduction = -kA(Δ T)/(Δ x)
q conduction = heat transfer rate (watts)
k = thermal conductivity (unit=?)
A = cross sectional area (m^2)
delta T = change in temperature (Celsius or Kelvin)
delta x = (meters)

Added by Kara W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Shahab Ullah

Step 1

- qconduction = heat transfer rate (watts or W) - k = thermal conductivity (unit to be determined) - A = cross sectional area (m²) - ΔT = change in temperature (Celsius or Kelvin) - Δx = distance (meters or m) Show more…

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Fourier's Law of heat conduction: Derive the unit of k from Fourier's law of conduction equation: Q = kA(delta T)/(delta x) where: Q = conduction heat transfer rate (watts) k = thermal conductivity (unit-?) A = cross-sectional area (m^2) delta T = change in temperature (Celsius or Kelvin) delta x = distance (meters)