Hello. Currently stuck on this problem. It\'s related to thermodynamics and it w

Question

Hello. Currently stuck on this problem. It's related to thermodynamics and it would be great if someone could solve it for me. Thanks.

A tub containing 42 kg of water is placed in a farmer's canning cellar, initially at 8 C. On a cold evening the cellar loses thermal energy through the walls at a rate of 1200J/s. Without the tub of water, the fruit would freeze in 4 h (the fruit freezes at 1 C because the sugar in the fruit lowers the freezing temperature).

By what time interval does the presence of the water delay the freezing of the fruit? The heat of fusion for water at 0 C is Lf=3.35×105J/kg. The specific heat of water is c = 4180 J/kg C.

Explanation / Answer

Q* = QLOSS* = ( QLdot ) ( t )

Q* = ( 1200 J/s ) ( 4 h ) ( 3600 s/h ) = 17.28 x 10^6 J = 17280 kJ

heat from 8 to -1 degre

QW = mW*C*dt + mW*Lf + mW*Cice*dt

here first water goes from 8 degree to 0 degree , then it will change the phase then ice will go to 0 to - 1 degree

m = 42 kg , dt = 8 , Lf = 3.35 x 10^5 J/kg , Cw = 4180 J/Kgoc , Cice = 2010 J/kg0c , dt = 1

put all the value

QW = 1404.48 kJ + 14070 kJ + 84.42 kJ = 15558.9 kJ

dt = QW/QLdot

dt = 15558.9kJ/1.2kJ/s = 12965.75 s = 3.6 hrs

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