Thermal diffusivity — correct engineering units Select the correct unit for thermal diffusivity when time is measured in hours (h).

Introduction / Context:
Thermal diffusivity (alpha) indicates how quickly a thermal disturbance spreads through a material. It appears in the transient heat equation and determines characteristic times for temperature equalization in solids and fluids.

Given Data / Assumptions:

• Definition: alpha = k / (rho * c_p).
• Units desired with time expressed in hours rather than seconds.

Concept / Approach:
Using SI base units, k has W/m·K which is equivalent to J/s·m·K. Density rho has kg/m^3 and c_p has J/kg·K. Substituting into alpha = k / (rho * c_p) yields units of m^2/s. If time is reported in hours, convert seconds to hours to get m^2/h. No temperature unit remains because K cancels in the ratio.

Step-by-Step Solution:
Start with alpha = k / (rho * c_p).Write units: k → J/(s·m·K); rho → kg/m^3; c_p → J/(kg·K).Compute denominator units: rho * c_p → (kg/m^3) * (J/(kg·K)) = J/(m^3·K).Divide: k / (rho * c_p) → [J/(s·m·K)] / [J/(m^3·K)] = m^2/s.Convert to hours: 1 s = 1/3600 h → units become m^2/h.

Verification / Alternative check:
Dimensional analysis of the transient conduction equation ∂T/∂t = alpha * ∇^2 T shows alpha must have length^2/time dimensions, confirming m^2/h when time is in hours.

Why Other Options Are Wrong:
(a) and (d) incorrectly retain Kelvin in the unit; K cancels. (b) has dimensions of velocity, not diffusivity. (e) places K in the numerator, which is dimensionally inconsistent for alpha.

Common Pitfalls:
Forgetting to cancel Kelvin and leaving an extra temperature unit; or confusing thermal diffusivity with conductivity (W/m·K).

Final Answer:
m2/h