How much heat is released when 75.0 g of steam at 100.0°C is cooled to ice at -15.0°C? The enthalpy of vaporization of water is 40.67 kJ/mol, the enthalpy of fusion for water is 6.01 kJ/mol, the molar heat capacity of liquid water is 75.4 J/(mol °C), and the molar heat capacity of ice is 36.4 J/(mol °C).
To calculate the amount of heat released during this process, we need to determine the different steps involved.
Step 1: Heating the steam from 100.0°C to its boiling point at 100.0°C.
Q1 = mass × specific heat capacity × temperature change
Given:
mass = 75.0 g
specific heat capacity of steam = 4.18 J/(g °C)
temperature change = 100.0°C - 100.0°C = 0°C
Q1 = 75.0 g × 4.18 J/(g °C) × 0°C
Q1 = 0 J
No heat is released during the heating of steam at its boiling point.
Step 2: Vaporizing the steam to water at its boiling point (100.0°C) and cooled to 0.0°C.
Q2 = moles of steam × enthalpy of vaporization
First, let's calculate the number of moles of steam:
moles = mass / molar mass of water
Given:
Mass = 75.0 g
Molar mass of water (H2O) = 18.015 g/mol
moles = 75.0 g / 18.015 g/mol
moles ≈ 4.16 mol (rounded to two decimal places)
Now, let's calculate the heat during vaporization:
Q2 = 4.16 mol × 40.67 kJ/mol = 169.17 kJ
Step 3: Cooling the liquid water from 0.0°C to the freezing point of water at 0.0°C.
Q3 = mass × specific heat capacity × temperature change
Given:
mass = 75.0 g
specific heat capacity of liquid water = 75.4 J/(mol °C)
temperature change = 0.0°C - 0.0°C = 0°C
Q3 = 75.0 g × 75.4 J/(mol °C) × 0°C
Q3 = 0 J
No heat is released during this step.
Step 4: Freezing the liquid water to ice at 0.0°C and cooled to -15.0°C.
Q4 = moles of water × enthalpy of fusion
First, calculate the number of moles of water:
moles = mass / molar mass of water
moles = 75.0 g / 18.015 g/mol
moles ≈ 4.16 mol (rounded to two decimal places)
Now, calculate the heat during freezing:
Q4 = 4.16 mol × -6.01 kJ/mol = -24.93 kJ
Step 5: Cooling the ice from 0.0°C to -15.0°C.
Q5 = mass × specific heat capacity × temperature change
Given:
mass = 75.0 g
specific heat capacity of ice = 36.4 J/(mol °C)
temperature change = -15.0°C - 0.0°C = -15°C
Q5 = 75.0 g × 36.4 J/(mol °C) × -15°C
Q5 = -24.93 kJ
Finally, let's calculate the total heat released during the entire process:
Total heat released = Q1 + Q2 + Q3 + Q4 + Q5
Total heat released = 0 J + 169.17 kJ + 0 J + -24.93 kJ + -24.93 kJ
Total heat released ≈ 119.41 kJ
Therefore, when 75.0 g of steam at 100.0°C is cooled to ice at -15.0°C, approximately 119.41 kJ of heat is released.
To calculate the amount of heat released when steam at 100.0°C is cooled to ice at -15.0°C, we can break down the process into several steps:
Step 1: Cooling steam from 100.0°C to 0°C
Step 2: Condensing steam to liquid water at 100.0°C
Step 3: Cooling liquid water from 100.0°C to 0°C
Step 4: Freezing water to ice at 0°C
Step 5: Cooling ice from 0°C to -15.0°C
Let's calculate the amount of heat released in each step:
Step 1: Cooling steam from 100.0°C to 0°C
To calculate the heat released during this step, we need to use the molar heat capacity of steam. The molar heat capacity of water is 75.4 J/(mol °C). First, we need to convert the mass of steam to moles using the molar mass of water (H2O), which is 18.02 g/mol.
moles of steam = mass of steam / molar mass of water
moles of steam = 75.0 g / 18.02 g/mol ≈ 4.16 mol
The heat released during this step can be calculated using the formula:
q = molar heat capacity * moles of steam * temperature change
q1 = 75.4 J/(mol °C) * 4.16 mol * (0 - 100) °C ≈ - 31,489 J
Step 2: Condensing steam to liquid water at 100.0°C
During the phase change of steam to liquid water, the enthalpy of vaporization is involved. The enthalpy of vaporization of water is 40.67 kJ/mol. We can calculate the heat released during this step by multiplying the moles of steam by the enthalpy of vaporization.
q2 = enthalpy of vaporization * moles of steam
q2 = 40.67 kJ/mol * 4.16 mol * 1000 J/1 kJ ≈ 169,371 J
Step 3: Cooling liquid water from 100.0°C to 0°C
Similar to step 1, we can calculate the heat released using the molar heat capacity of liquid water.
q3 = molar heat capacity * moles of liquid water * temperature change
We need to convert the moles of water from step 2 to moles of liquid water:
moles of liquid water = moles of steam - moles of steam condensed
moles of liquid water = 4.16 mol - 4.16 mol ≈ 0 mol (No change in moles)
q3 = 75.4 J/(mol °C) * 0 mol * (0 - 100) °C ≈ 0 J (No change in temperature)
Step 4: Freezing water to ice at 0°C
During the phase change of water to ice, the enthalpy of fusion is involved. The enthalpy of fusion for water is 6.01 kJ/mol.
q4 = enthalpy of fusion * moles of liquid water
q4 = 6.01 kJ/mol * 0 mol * 1000 J/1 kJ ≈ 0 J (No change in moles)
Step 5: Cooling ice from 0°C to -15.0°C
Using the molar heat capacity of ice, we can calculate the heat released during this step.
q5 = molar heat capacity * moles of ice * temperature change
Conversion from moles of liquid water to moles of ice:
moles of ice = moles of liquid water
moles of ice = 0 mol (No change in moles)
q5 = 36.4 J/(mol °C) * 0 mol * (0 - (-15.0)) °C ≈ 0 J (No change in temperature)
To calculate the total heat released, we sum up the values obtained in each step:
Total q = q1 + q2 + q3 + q4 + q5
Total q = -31,489 J + 169,371 J + 0 J + 0 J + 0 J
Total q ≈ 137,882 J
Therefore, approximately 137,882 J of heat is released when 75.0 g of steam at 100.0°C is cooled to ice at -15.0°C.
From the last problem you know how to handle the two different kinds of heat transfer; i.e., one kind at changing T when in just one state, the second when change state. Therefore, I will leave this to you except to point out the transitions that must be made.
q1 = heat released when steam condenses.
q2 = heat released moving from 100 C to 0 c.
q3 = heat released freezing liquid water at zero C to ice at zero C.
q4 = heat released moving ice from zero to -15 C.
Total heat released = q1 + q2 + q3 + q4.