Answer:
Temperature at center of apples = 11.2⁰C
Temperature at surface of apples = 2.7⁰C
Amount of Heat transferred = 17.2kJ
Explanation:
The properties of apple are given as:
k = 0.418 W/m.°C
ρ = 840 kg/m³
Cр = 3.81 kJ/kg.°C
α = 1.3*10 ⁻⁷ m²/s
h = 8 W/m².°C
d = 0.09m
r = 0.045m
t = 1 hour = 3600s
Biot number is given as:
[tex]Bi = \frac{hr}{k}= \frac{8\cdot0.045}{0.418}=0.861[/tex]
The constants λ₁ and A₁ corresponding to Biot number (from the table) are:
λ₁ = 1.476
A₁ = 1.239
Fourier Number is:
[tex]T = \frac{a\cdot{t}}{r^2} = \frac{(1.3\cdot10^{-7})(3600)}{0.045^2}= 0.231> 0.2[/tex]
As Fourier Number > 0.2 , one term approximates solutions are applicable
The temperature at the center of apples, The temperature at surface of apples and Amount of heat transfer is found in the ATTACHMENT.
