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Unformatted text preview: 4 kg, Cliq=3000 J/kg K (heat capacity of the liquid)
dQ
dT
= mCliq
dt
dt
Since T changes linearly with time t, dQ
= const .
dt dQ
270 − 300
= 0.4 × 3000 ×
= −900 J / min
dt
40
or
1. The latent heat LF,
mLF = 30min x 900 J/min, or
LF = 6.75 x 104 J/kg = 67.5 kJ/kg
2. The heat capacity Csol of the frozen phase
dQ
dT
= mCsol
dt
dt or
− 900 = 0.4Csol 250 − 270
= −0.4Csol
20 or
Csol = 2250 J/kg K
((WileyPlus))
34. While the sample is in its liquid phase, its temperature change (in absolute values) is
 ∆ T  = 30 °C. Thus, with m = 0.40 kg, the absolute value of Eq. 1814 leads to
°
Q = c m ∆ T  = ( 3000 J/ kg × C )(0.40 kg)(30 °C ) = 36000 J .
The rate (which is constant) is
P = Q / t = (36000 J)/(40 min) = 900 J/min,
which is equivalent to 15 Watts.
(a) During the next 30 minutes, a phase change occurs which is described by Eq. 1816:
Q = P t = (900 J/min)(30 min) = 27000 J = L m .
Thus, with m = 0.40 kg, we find L = 67500 J/kg ≈ 68 kJ/kg.
(b) During the final 20 minutes, the sample is solid and undergoes a temperature change
(in absolu...
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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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