Problem 12.35 A tub containing 40 kg of water is placed in a farmer\'s canning c
ID: 1794851 • Letter: P
Question
Problem 12.35 A tub containing 40 kg of water is placed in a farmer's canning cellar, initially at 15 °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 -1oC because the sugar in the fruit lowers the freezing temperature) Part A 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-335 x 105J/kg. The specific heat of water is c = 4180 J/kg·°C Express your answer to two significant figures and include the appropriate units. tdelay= 2844 Submit Mv Answers Give Up Incorrect; Try Again; 4 attempts remaining Provide Feedback ContinueExplanation / Answer
Get QLOSS in 4 h without the warm water, Q* :
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Q* = QLOSS* = ( QLdot ) ( t )
Q* = ( 1200 J/s ) ( 4 h ) ( 3600 s/h ) = 17.28 x 10^6 J = 17280 kJ
Now get heat loss to take the 40 kg of water from 15 C to ice at -1 C , QW*
QW* = ( mW ) ( CsnW ) ( 15.0 C - 0.0 C ) + ( mW ) ( h sub sfW ) + ( mW ) ( CshICE ) ( 0.0 - -1.0 )
QW* = ( 40 kg ) ( 4.180 kJ/kg ) ( 15.0 C ) + ( 40 kg ) ( 335 kJ/kg ) + ( 40 kg ) ( 2.10 kJ/kg-C ) ( 1.0 C )
QW* = 2508 kJ + 13400 kJ + 84 kJ = 15,992 kJ
Delta t = QW* / QLdot
Delta t = 15992 kJ / 1.2 kJ/s = 1.3 x 10^4 s (or 3.7 h )<-----------------