In: Physics
A cube of ice is taken from the freezer at -9.5?Cand placed in a 95-g aluminum calorimeter filled with 310g of water at room temperature of20.0 ?C. The final situation is observed to be all water at 15.0?C. The specific heat of ice is2100 J/kg?C?, the specific heat of aluminum is900 J/kg?C?, the specific heat of water is is4186 J/kg?C?, the heat of fusion of water is333 kJ/Kg. What was the mass of the ice cube?
We have:
Specific heat of water = 4186
J/(kg.oC)
Specific heat of ice = 2100
J/(kg.oC)
Heat of fusion of ice = 3.33 x 105 J/kg
Mass of water = 326 g = 0.326 kg
The temperature of the water decreased 5o
For the water, Q = 0.326 * 4186 * 5 =
6823.18 joules
As the temperature of the water decreased 5o, 6823.18
joules of heat energy was released.
For the ice to melt, the temperature of the ice must increase to
9.5 oC
Q = m * 2100 * 9.5 = m *
19950
During the melting, Q = m * 3.33 *
105
Next, the temperature of the water increases from 0o to
15o
Q = m * 4186 * 15 = m *
62790
Total heat energy required by the ice cube = m * 19950 + m * 3.33 *
105 + m * 62790
Total heat energy required by the ice cube = Total heat energy
released by the water.
m * 19950 + m * 3.33 x 105 + m * 62790 = 6823.18
Mass of ice cube = 6823.18 / (19950 + 3.33 x 105 +
62790) =6823.18/415740
Mass of ice cube = 0.01641213258 kg = 16.41213258
grams.