Question

A cube of ice is taken from the freezer at -8.50 oC and placed in an 85 g aluminum calorimeter filled with 310 g of water at room temperature of 20.0°C. The final situation is all water at 17.0°C. What was the mass of the ice cube in gram? (Lf = 3.3x105 J/kg, cwater = 4186 J/kg.C, cAl = 900 J/kg.C, and cice = 2100 J/kg.C)

Answer #1

Using energy conservation:

Heat gained by ice = heat released by water + calorimeter

Q1 = Q2

Q1 = heat absorbed by ice from -8.5 C to 0 C + heat gained by ice when phase changes + heat gained by ice when temperature increase from 0 C to 17 C

Q1 = Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2

Q2 = heat released by water + heat released by aluminum

Q2 = Mw*Cw*dT3 + Ma*Ca*dT3

Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2 = Mw*Cw*dT3 + Ma*Ca*dT3

Mi = mass of ice = ?

Mw = mass of water = 310 gm = 0.31 kg

Ma = mass of aluminum = 85 gm = 0.085 kg

Cw = 4186 J/kg-C, Ca = 900 J/kg-C, Ci = 2100 J/kg-C

Lf = 3.3*10^5 J/kg

dT1 = 0 - (-8.50) = 8.50 C

dT2 = 17 - 0 = 17 C

dT3 = 20 - 17 = 3 C

Using these values:

Mi*2100*8.50 + Mi*3.3*10^5 + Mi*4186*17 = 0.31*4186*3 + 0.085*900*3

Mi = (0.31*4186*3 + 0.085*900*3)/(2100*8.50 + 3.3*10^5 + 4186*17)

Mi = 0.0098 kg = 9.8 gm

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