A 77.7 -g aluminum ice tray in a home refrigerator holds 334 g of water. Calcula

Question

A 77.7-g aluminum ice tray in a home refrigerator holds 334 g of water. Calculate the energy in kJ that must be removed from the tray and its contents to reduce the temperature from 15.6°C to 0.0°C, freeze the water, and drop the temperature of the tray and ice to –11.8°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
y ga ned by the water A calori cter contains 83.5 grams of water at 15. C. A 125 gram p ce ofan unknown metal s ted to 8 9 C and dropped into the water. The c tire system eventually reaches 178 ? Assum ng all of the cne comes from the cooling of the mctal no energy loss to the calorimeter or the surroundings calculate the specific heat of the metal The specific heat of water is 4.18 J/g·°C

Explanation / Answer

mass of water = 334 g

15.6   -----------   0 o C ------------- - 11.8 oC

Q1 = m Cp dT

     = 334 x 4.18 x 15.6

     = 21779.5 J

Q2 = m x delta Hfus

      = 334 x 333

      = 111222 J

Q3 = 334 x 2.06 x 11.8

     = 8118.87

total heat Q = Q1 + Q2 + Q3

                   = 141120.4

total heat removed = 141 kJ

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