Heat sterilization of lumber, timbers and pallets is used to kill insects to prevent their transfer between countries in international trade. Consider hot air heating that maintains the surface temperature at 70°C. The boards are stacked outside and in the winter time they can be considered to be at 0°C when they are brought in for heating. (a) Calculate the time for the slowest heating point to reach sterilization temperature of 56°C for a 2.54 cm thick board. (b) Calculate the heating time when four such boards are stacked together. Thermal diffusivity of the wood is 8.5×10-8 m2/s.

 Δt = 3.04 10³ s

Explanation:

a) This is a thermal conduction exercise that is described by the expression

         P = k A | dT/dx |     (1)

where P power, k thermal  conductivity, A the area, T the temperature and x the thickness

The power supplied to the wood is

          P = Q / Δt

where the heat is given by

         Q = m ΔT'

we substitute

         P = m c_{e} ΔT'/ Δt

We assume that the changes in temperature and time can be approximated by their differences,

we substitute in 1

          m c_{e} ΔT'/Δt = k A dT / dx

as the temperature changes are small, we can assume a linear variation, consequently the derivatives can be approximated to the variations

           m ce ΔT'/Δt = k A ΔT/Δx    (2)

let's write the temperature variations explicitly

        ΔT’= (T_f -T₀)

        ΔT’= 56 -0

        ΔT =

        ΔT = 70 -0    

the thermal diffusivity is

           α = k /ρ c_{e}

            k = α c_{e}

the definition of density

            ρ = m / V

            ρ = m / A L

            k = α c_{e} m / A L

we substitute   in 2

          m c_{e} Δx = (α c_{e} m / A L) A ΔT /ΔT'   dt

          Δx = α /L   ΔT / ΔT’  Δt

          Δt = L Dx /α ΔT'/ΔT

let's calculate

          Δt = 2.54 1.27 10⁻⁴ / 8.5 10⁻⁸    (56- 0) / (70-0)

          Δt = 3.04 10³ s

i believe the answer is a hope it and is correct