A cube of ice is taken from the freezer at -6.5 C and placed in a 95-g aluminum

Question

A cube of ice is taken from the freezer at -6.5 C and placed in a 95-g aluminum calorimeter filled with 320 g of water at room temperature of 20.0 C. The final situation is observed to be all water at 17.0 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

specific heat of ice is 2100 J/kgC

specific heat of aluminum is 900 J/kgC

specific heat of water is is 4186 J/kgC

heat of fusion of water is 333 kJ/Kg

Let the mass of the ice cube was m kg

=>Q(gain) by ice cube = Q(lost) to by the calorimeter + Q(lost) by water

=>m x s(ice) x t + m x L(melting) + m x s(water) x t = m x s(aluminum) x t + m x s(water) x t

=>m x [2100 x 6.5 + 333000 + 4186 x 17] = 95 x 10^-3 x 900 x 3 + 320 x 10^-3 x 4186 x 3

=>417812m = 4275.06

=>m = 0.01023 kg

=>m = 10.23 gm

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