A cube of ice is taken from the freezer at 89 C and placed i

A cube of ice is taken from the freezer at -8.9 C and placed in a 91 g aluminum calorimeter filled with 315 g of water at room temperature of 20.0 C. The final situation is observed to be all water at 15.0 C. What was the mass of the ice cube?

Solution

Specific heat of aluminum = 900 j/kg.K
Specific heat of liquid water = 4186 j/kg k
Specific heat of ice = 2050 j/kg.K
Heat of fusion of ice = 333,550 j/kg

As the situation moves towards new equilibrium, the following must happen:
The ice warms from -8.9c to 0c
The ice melts to liquid water at 0c
the ice water warms from 0c to 15c
the room-temperature water cools from 20c to 15c
the aluminum calorimeter cools from 20c to 15c

Energy given up by calorimeter and warm water: (DeltaE)
DeltaE = [(91g * 1 kg/1000 g * 900 j/kg.K) + (315 g * 1 kg/1000g * 4186 j/kg.K)] * (20c - 15c)
DeltaE = 7002.45 j

Energy required to raise temp of 1g of ice from -8.9c to 15c

K and C are the same scale with different zeros and we\'re using temperature changes, we don\'t need to convert C to K
DeltaE = 1g * 1 kg/1000g * [(2050 j/kg.K * 8.9K) + (333,550 j/kg) + (4186 j/kg.K * 15K)]
DeltaE = 414.585 j/g

Mass of ice:
m = 7002.45 j / 414.585 j/g

m = 16.89 g