A student measures the following data regarding the heat of fusion of ice: 25.8 g ice at 0.0°C is placed into the calorimeter which contains 100.0 g water at 21.7°C. The final temperature comes to 1.5°C. The calorimeter has a heat capacity of 15.6 J/°C.
(a) Calculate the experimental ΔHfus of ice in kJ/mol.
Let x = heat fusion ice.
(mass ice*x) + [(mass melted ice x specific heat x (Tfinal-Tinitial)] + [mass warm water x specific heat water x (Tfinal-Tinitial)] + Ccal(Tfinal-Tinitial) = 0
Solve for x.
Post your work if you get stuck.
Well, if I had to describe this problem in terms of a joke, I'd say it's as cool as ice! But let's crunch the numbers and calculate the experimental ΔHfus of ice.
First, we need to calculate the amount of heat transferred during the process. We can use the equation:
q = m × C × ΔT
where q is the heat transferred, m is the mass, C is the heat capacity, and ΔT is the change in temperature.
The heat transferred during the phase change (fusion) of the ice can be calculated using:
q = n × ΔHfus
where n is the number of moles and ΔHfus is the molar enthalpy of fusion.
To find n, we need to use the equation:
n = m/M
where M is the molar mass of ice (H2O) which is approximately 18.0 g/mol.
Given:
m(ice) = 25.8 g
C(calorimeter) = 15.6 J/°C
ΔT = 1.5 °C
Let's calculate the heat transferred, q:
q = m(ice) × C(calorimeter) × ΔT
q = 25.8 g × 15.6 J/°C × 1.5 °C
Now let's convert grams to moles:
n = m(ice) / M
n = 25.8 g / 18.0 g/mol
Finally, we can calculate the molar enthalpy of fusion, ΔHfus:
ΔHfus = q / n
ΔHfus = (25.8 g × 15.6 J/°C × 1.5 °C) / (25.8 g / 18.0 g/mol)
Now let's do the math!
Unless you have a calculator that tells great jokes, this might take a few seconds.
To calculate the experimental ΔHfus of ice in kJ/mol, we need to follow these steps:
Step 1: Calculate the heat gained by the ice.
We can use the formula:
q = m * c * ΔT
where q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
m (mass of ice) = 25.8 g
c (specific heat capacity of water) = 4.18 J/g°C (assuming it's water)
ΔT (change in temperature) = 1.5°C
q = 25.8 g * 4.18 J/g°C * 1.5°C
q = 159.819 J
Step 2: Calculate the heat lost by the water.
Again, we can use the same formula:
q = m * c * ΔT
Given:
m (mass of water) = 100.0 g
c (specific heat capacity of water) = 4.18 J/g°C (assuming it's water)
ΔT (change in temperature) = 20.2°C (from 21.7°C to 1.5°C)
q = 100.0 g * 4.18 J/g°C * 20.2°C
q = 8,442.4 J
Step 3: Calculate the heat exchanged in the calorimeter.
The total heat exchanged in the calorimeter is the sum of the heat gained by the ice and the heat lost by the water:
q_total = q_ice + q_water
q_total = 159.819 J + 8,442.4 J
q_total = 8,602.219 J
Step 4: Calculate the heat capacity of the calorimeter.
The heat capacity of the calorimeter can be calculated using the formula:
C = q_total / ΔT
where C is the heat capacity and ΔT is the change in temperature.
Given:
ΔT (change in temperature) = 20.2°C (from 21.7°C to 1.5°C)
q_total = 8,602.219 J
C = 8,602.219 J / 20.2°C
C = 426.253 J/°C
Step 5: Calculate the moles of ice used.
We need to convert the mass of ice to moles:
moles of ice = mass of ice / molar mass of ice
Given:
mass of ice = 25.8 g
To find the molar mass of ice, we use the periodic table:
H = 1 g/mol
O = 16 g/mol
molar mass of ice = 18 g/mol (by adding the molar masses of H2O)
moles of ice = 25.8 g / 18 g/mol
moles of ice ≈ 1.4333 mol
Step 6: Calculate the ΔHfus in kJ/mol.
We can use the equation:
ΔHfus = q_total / moles of ice
Given:
q_total = 8,602.219 J
moles of ice ≈ 1.4333 mol
ΔHfus = 8,602.219 J / 1.4333 mol
ΔHfus ≈ 6,001.886 J/mol
To convert J to kJ, we divide by 1000:
ΔHfus = 6,001.886 J/mol / 1000
ΔHfus ≈ 6.0019 kJ/mol
Therefore, the experimental ΔHfus of ice is approximately 6.0019 kJ/mol.
To calculate the experimental ΔHfus of ice in kJ/mol, we need to follow these steps:
Step 1: Calculate the heat gained by the water:
The heat gained by the water can be calculated using the equation:
Qwater = mwater * Cwater * ΔTwater
Where:
- Qwater is the heat gained by the water
- mwater is the mass of water
- Cwater is the specific heat capacity of water
- ΔTwater is the change in temperature of water
Given:
mwater = 100.0 g
Cwater = 4.18 J/g°C (specific heat capacity of water)
ΔTwater = 1.5°C - 21.7°C = -20.2°C
Substituting the values, we can find the heat gained by the water:
Qwater = (100.0 g) * (4.18 J/g°C) * (-20.2°C)
Step 2: Calculate the heat lost by the ice:
The heat lost by the ice can be calculated using the equation:
Qice = mice * ΔHfus
Where:
- Qice is the heat lost by the ice
- mice is the mass of ice
- ΔHfus is the heat of fusion of ice
Given:
mice = 25.8 g (mass of ice)
We need to find ΔHfus, which is the unknown.
Step 3: Calculate the heat absorbed by the calorimeter:
The calorimeter absorbs heat when the water and ice are mixed. This heat can be calculated using the equation:
Qcalorimeter = Ccalorimeter * ΔTcalorimeter
Where:
- Qcalorimeter is the heat absorbed by the calorimeter
- Ccalorimeter is the heat capacity of the calorimeter
- ΔTcalorimeter is the change in temperature of the calorimeter
Given:
Ccalorimeter = 15.6 J/°C (heat capacity of the calorimeter)
ΔTcalorimeter = 1.5°C - 21.7°C = -20.2°C
Substituting the values, we can find the heat absorbed by the calorimeter:
Qcalorimeter = (15.6 J/°C) * (-20.2°C)
Step 4: Apply the conservation of energy:
According to the law of conservation of energy, the heat gained by the water and absorbed by the calorimeter must equal the heat lost by the ice. Therefore, we can set up the equation:
Qwater + Qcalorimeter = Qice
Step 5: Solve for ΔHfus:
By substituting the respective equations into the conservation of energy equation, we get:
(mwater * Cwater * ΔTwater) + (Ccalorimeter * ΔTcalorimeter) = mice * ΔHfus
Substituting the known values and solving for ΔHfus:
[(100.0 g) * (4.18 J/g°C) * (-20.2°C)] + [(15.6 J/°C) * (-20.2°C)] = (25.8 g) * ΔHfus
Solving this equation will give us the experimental ΔHfus of ice in J/mol. To convert it to kJ/mol, divide the result by 1000.
Note: Make sure to use proper unit conversions throughout the calculations.