Question

An orange grove in Florida is exposed to a cold front causing a sudden change in air
temperature from 25C to 0C. How long will it take for the oranges to reach 3C in
the center? Assume that the oranges are spherical with a radius of 4 cm, the thermal
conductivity of the orange is 0.6 W/mC, specific heat of the orange is 4.18 kJ/
kgC, the density of the orange is 1,000 kg/m3, the thermal conductivity of the air is
0.027 W/mC, the density of air is 1.17 kg/m3, the specific heat of air is 1 kJ/kgC,
and the heat transfer coefficient between orange and still air is 0.675 W/m2C.

Answer 1

This is a case of unsteady state heat conduction.

Let us now calculate Biots Number for this system.

h- heat transfer coefficent

L- charecteristic length (Volume/surface area= R/3 for sphere)

K - thermal conductivity of body

Given h = 0.675 w/m^{2 o}c , K =0.6w/m^oC, L = R/3 = 0.04/3

Bi < 0.1 Hence Lumped heat capacity*model can be assumed.

For lumped heat capacity model we have

T- temperature of solid after time t = 3 ^oC

T_i- initial temperature of solid= 25 ^oC

- Temperature od surrounding = 0 ^oC

- Density of solid = 1000 Kg/m³

C_p- Specific heat capacity of solid = 4.18 kJ/kg ⁰C = 4.18 X 10³ J/kg ⁰C

h- heat transfer coefficent = 0.675 w/m^{2 o}c

V/A = L  charecteristic length = R/3 = 0.04/3

Substitutuing all in the above equation

Simplfying and solving for t

take logaritham on both side

t = 175065.71 sec

t = 48.63 hr

NOTE: *In  lumped heat capacity assumption we assume that the temperature of the solid at any point is same or there is no temperature distribution in solid.