A cube of ice is taken from the freezer at -5.5 C and placed in a 95-g aluminum

Question

A cube of ice is taken from the freezer at -5.5 C and placed in a 95-g aluminum calorimeter filled with 330 g of water at room temperature of 20.0 C. The final situation is observed to be all water at 16.0 C. The specific heat of ice is 2100 J/kgC, the specific heat of aluminum is 900 J/kgC, the specific heat of water is is 4186 J/kgC, the heat of fusion of water is 333 kJ/Kg. Part A What was the mass of the ice cube? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Heat Lost by Calorimeter = (95/1000)*900*(20-16) + (330/1000)*4186*(20-16) = 5867.5 J

Heat gained by ice = m*2100*5.5 + m*333000 + m*4186*17 = 411526*m J

Heat Gained = Heat Lost

411526*m = 5867.5

m = 14*10^-3 kg = 14.2 gms (ans)

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