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Unformatted text preview: r rate of conductive heat transfer.
(c) Repeating the calculation above with the new value for k2 , we have
=
which leads to ∆ T2 = 13.8 °C. This is less than our part (a) result which implies that the
temperature gradients across layers 1 and 3 (the ones where the parameters did not
change) are greater than in part (a); those larger temperature gradients lead to larger
conductive heat currents (which is basically a statement of “Ohm’s law as applied to heat
conduction”).
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Problem 1865 ((My solution))
kice=0.0040 cal/s cm C,
ρice = 0.92 g/cm3
TC = 10°C, TH = 0°C. Lf = 3.33 x 105 J/kg = 79.50 cal/g ∆Q = L f ( Adx) ρice
℘= dQ
dT
∆T
∆T
= −kice A
= − kice A
= kice A
dt
dx
x
x From these equations, we have ℘ = Lf ( A dx
∆T
) ρice = kice A
,
dt
x where ∆T = 0 − (−10) = 10C
dx
k
∆T
= ice
dt ρice L f x
At x = 5 cm, we have
dx
= 1.09 × 10− 4 cm / s = 0.4cm / h
dt
((Note))
dx
k
∆T
= ice
,
dt...
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This note was uploaded on 05/14/2013 for the course MATH 346 taught by Professor Professormiguelarcones during the Spring '08 term at Binghamton University.
 Spring '08
 PROFESSORMIGUELARCONES

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