A 0.220-kg piece of aluminum that has a temperature of -191 °C is added to 1.5...
A 0.220-kg piece of aluminum that has a temperature of -191 °C is added to 1.5 kg of water that has a temperature of 2.4 °C. At equilibrium the temperature is 0 °C. Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.
Homework Answers
The amount of heat added to the aluminum is m*c*deltaT =
0.22kg*900J/kgC*191deg C = 37818J
Therefore the water must give up 37818J.
This will be in two steps. The first step will be to cool all of the water to 0degC
This heat is m*c*deltaT = 1.5kg*4186J/kgC*2.4degC = 15069 J.
So in freezing the water must give up 37818-15069J = 22749
J
The latent heat of fusion for water is 3.34x10^5J/kg Use Q = m*Lf
so m = Q/Lf = 22749J/3.34x10^5J/kg = 0.068 kg . So, 0.068 kg or 68
g of the water will freeze.
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