A 78.3-g aluminum ice tray in a home refrigerator holds 322 g of water. Calculat
ID: 1056471 • Letter: A
Question
A 78.3-g aluminum ice tray in a home refrigerator holds 322 g of water. Calculate the energy in kJ that must be removed from the tray and its contents to reduce the temperature from 19.7°C to 0.0°C, freeze the water, and drop the temperature of the tray and ice to –13.9°C. Assume the specific heat of aluminum remains constant at 0.900 J/g·°C over the temperature range involved. Data for water at 1 atm: Melting point = 0.0°C Specific heat liquid = 4.18 J/g·°C Specific heat solid = 2.06 J/g·°C Heat of fusion = 333 J/g ____ kJ
Explanation / Answer
From the data,
energy to be removed from the tray and contents to reduce temperature from 19.7 oC to 0 oC
= 322 x 4.184 x -19.7/1000 + 78.3 x 0.900 x -19.7
= -27.930 kJ
Energy to be removed to convert water to ice
= 322 x 333/1000
= 107.226 kJ
Energy released to cool ice and tray from 0 oC to -13.9 oC
= 322 x 2.06 x -13.9/1000 + 78.3 x 0.900 x -13.9
= -10.201 kJ
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