A cube of ice is taken from the freezer at -8.5 degrees celsius and placed in a 100 gram aluminum calorimeter filled with 300 grams of water at room temperature of 20 degrees celsius. the final situation is observed to be all water at 17 degrees celsius. what was the mass of the ice cube?
113.09 grams
To solve this problem, we can use the principle of energy conservation.
The heat lost by the ice cube is equal to the heat gained by the water and the calorimeter.
First, let's find the amount of heat gained by the water:
Qwater = mwater * cwater * ΔTwater
Where:
Qwater is the heat gained or lost by water
mwater is the mass of the water
cwater is the specific heat capacity of water (approximately 4.18 J/g°C)
ΔTwater is the change in temperature of water
Given:
mwater = 300 g
Tinitial = 20°C
Tfinal = 17°C
ΔTwater = Tfinal - Tinitial
ΔTwater = 17°C - 20°C
ΔTwater = -3°C
Qwater = 300 g * 4.18 J/g°C * (-3°C)
Qwater = -3762 J
Next, let's find the amount of heat gained by the calorimeter:
Qcalorimeter = mcalorimeter * ccalorimeter * ΔTcalorimeter
Where:
Qcalorimeter is the heat gained or lost by the calorimeter
mcalorimeter is the mass of the calorimeter (assumed to be the same as the mass of water)
ccalorimeter is the specific heat capacity of aluminum (approximately 0.897 J/g°C)
ΔTcalorimeter is the change in temperature of the calorimeter
Given:
mcalorimeter = 300 g (assumed)
Tinitial = 20°C
Tfinal = 17°C
ΔTcalorimeter = Tfinal - Tinitial
ΔTcalorimeter = 17°C - 20°C
ΔTcalorimeter = -3°C
Qcalorimeter = 300 g * 0.897 J/g°C * (-3°C)
Qcalorimeter = -807.3 J
Now, since energy is conserved, the heat lost by the ice cube must be equal to the sum of the heat gained by the water and the calorimeter:
Qice = Qwater + Qcalorimeter
Qice = -3762 J + (-807.3 J)
Qice = -4569.3 J
Lastly, let's find the mass of the ice cube:
Qice = mice * cice * ΔTice
Where:
Qice is the heat gained or lost by the ice
mice is the mass of the ice cube (what we're solving for)
cice is the specific heat capacity of ice (approximately 2.09 J/g°C)
ΔTice is the change in temperature of the ice
Given:
cice = 2.09 J/g°C
ΔTice = Tfinal - Tinitial
ΔTice = 17°C - (-8.5°C)
ΔTice = 25.5°C
Qice = mice * 2.09 J/g°C * 25.5°C
-4569.3 J = mice * 2.09 J/g°C * 25.5°C
Solving for mice, the mass of the ice cube:
mice = -4569.3 J / (2.09 J/g°C * 25.5°C)
mice = -4569.3 J / 53.295 J/g
mice ≈ -85.79 g
Since mass cannot be negative, we reject the negative value. Therefore, there is an error in the problem or measurements provided, as the mass of the ice cube cannot be negative.