A tub containing 48 kg of water is placed in a farmer\'s canning cellar, initial

Question

A tub containing 48 kg of water is placed in a farmer's canning cellar, initially at 12 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

energy needed in freezing the fruit = 1200 J /s   x ( 4 x 3600 s ) = 1728 x 10^4 J

fruit freezes at -1degree C so they will be able to freeze when temperature of water is also -1 deg C.

heat needed by water to reach -1 deg C = m Cwater deltaT + m Lfusion + m Cice deltaT

   = (48 x 4180 x (12 - 0 )) + (48 x 3.35 x 10^5) + ( 48 x 2030 x 1 )

= 1.86 x 10^7 J

now total energy = 1728 x 10^4   + 1.86 x10^7   = 3.587 x 10^7 J

now time required to freeze = 3.587 x 10^7 / 1200 =29887.6 sec    = 29887.6/3600 h = 8.30 hr

delay = 8.30 - 4 = 4.30 hr

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