Roasting peanuts .One method of preparing fat-free roasted peanuts involves lowe

Question

Roasting peanuts .One method of preparing fat-free roasted peanuts involves lowering a wire basket of raw shelled peanuts into a vat of molten mannitol and sorbitol (nonsweet sugars) instead of into hot oil. When the peanuts are well roasted, they are removed, drained, lightly salted, and ready for packing. If peanuts, originally at 15 C, are lowered into the 165 C roasting medium:

(a) Find the time needed for their centers to reach 105 C.
(b) How hot does the surface temperature of the peanuts become?

Data and assumptions: Assume that the peanuts are close to spherical with diameters of 7.5 mm and have the following properties:

ks 14 0:5W=mK s 14 1,150 kg=m3 Cs 14 1,700 J=kg K

Between peanuts and molten sugar, take h 14 80 W/m2 K.

Explanation / Answer

Part 1: Heat transfer

Q/t = [T2-T1]/R
R = thermal resistance ; T2-T1 = temperature difference = 165 - 15 = 150 C

R = R1 + R1

R1 : resistance by sugars = 1/hA = 1/ 80 x (4 pi x r^2)

= 1/80 x 4 x 3.14 x 0.00375^2 = 70.77 K/W

R2 : resistance by peanut core = L/kA ... here L is equal to radius

= 0.00375 / 0.5 x 4 x 3.14 x 0.00375^2 = 42.46 K/W

R = 113.16 K/W

Q/t = 150/113.16 = 1.325 W

a)

To find time, we will use specific heat

Q = mcdT

m = density x volume = 1150 x 4/3 x 3.14 x 0.00375^3 = 2.54 x 10^-4 kg
dT = change in temp = 105-15 = 90

Q = 2.54 x 10^-4 x 1700 x 90 = 38.862 J

But Q/t = 1.325

so, t = Q/(Q/t) = 38.862 / 1.325 = 29.33 sec

b)

If surface temperature is required, we will use only outer resistance

[T2-Ts] / R1= Q/t

[165-Ts] / 70.77 = 1.325

Ts = 71.22 C

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