A cube of ice is taken from the freezer at -8.5 degree C and placed in a 85 g al

Question

A cube of ice is taken from the freezer at -8.5 degree C and placed in a 85 g aluminum calorimeter filled with 310 g of water at room temperature of 20.0 degree C. The final situation is observed to be all water at 15.0 degree C. The specific heat of ice is 2100 J/kg middot C degree, the specific heat of aluminum is 900 J/kg middot C degree, the specific heat of water is 4186 J/kg middot C degree, the heat of fusion of water is 333 kJ/Kg. What was the mass of the ice cube? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

here,

mass of water , m1 = 0.31 kg

mass of alumunium , m2 = 0.085 kg

let the mass of ice be m3

heat gained by ice = heat lost water + heat lost by alumunium

m3 * (Ci * ( 0 - (-8.5)) + Lf + Cw * ( 15 - 0)) = m1 * Cw * ( 20 - 15) + m2 * Cal * ( 20 - 15)

m3 * ( 2100 * 8.5 + 333000 + 4186 * 15) = 0.31 * 4186 * 5 + 0.085 * 900 * 5

solving for m3

m3 = 0.0166 kg

the mass of ice is 1.66 * 10^-2 kg

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